Examples of common false beliefs in mathematics

Question

The first thing to say is that this is not the same as the question about interesting mathematical mistakes. I am interested about the type of false beliefs that many intelligent people have while they are learning mathematics, but quickly abandon when their mistake is pointed out -- and also in why they have these beliefs. So in a sense I am interested in commonplace mathematical mistakes.

Let me give a couple of examples to show the kind of thing I mean. When teaching complex analysis, I often come across people who do not realize that they have four incompatible beliefs in their heads simultaneously. These are

(i) a bounded entire function is constant;
(ii) $\sin z$ is a bounded function;
(iii) $\sin z$ is defined and analytic everywhere on $\mathbb{C}$;
(iv) $\sin z$ is not a constant function.

Obviously, it is (ii) that is false. I think probably many people visualize the extension of $\sin z$ to the complex plane as a doubly periodic function, until someone points out that that is complete nonsense.

A second example is the statement that an open dense subset $U$ of $\mathbb{R}$ must be the whole of $\mathbb{R}$. The "proof" of this statement is that every point $x$ is arbitrarily close to a point $u$ in $U$, so when you put a small neighbourhood about $u$ it must contain $x$.

Since I'm asking for a good list of examples, and since it's more like a psychological question than a mathematical one, I think I'd better make it community wiki. The properties I'd most like from examples are that they are from reasonably advanced mathematics (so I'm less interested in very elementary false statements like $(x+y)^2=x^2+y^2$, even if they are widely believed) and that the reasons they are found plausible are quite varied.

Answer 1

For vector spaces, $\dim (U + V) = \dim U + \dim V - \dim (U \cap V)$, so $$ \dim(U +V + W) = \dim U + \dim V + \dim W - \dim (U \cap V) - \dim (U \cap W) - \dim (V \cap W) + \dim(U \cap V \cap W), $$ right?

Answer 2

Everyone knows that for any two square matrices $A$ and $B$ (with coefficients in a commutative ring) that $$\operatorname{tr}(AB) = \operatorname{tr}(BA).$$

I once thought that this implied (via induction) that the trace of a product of any finite number of matrices was independent of the order they are multiplied.

Answer 3

The closure of the open ball of radius $r$ in a metric space, is the closed ball of radius $r$ in that metric space.

In a somewhat related spirit: the boundary of a subset of (say) Euclidean space has empty interior, and furthermore has Lebesgue measure zero. (This false belief is closely related to Gowers' example of the belief that there are no non-trivial open dense sets.)

More generally, point set topology and measure theory abound with all sorts of false beliefs that only tend to be expunged once one plays with the canonical counterexamples (Cantor sets, bullet-riddled squares, space-filling curves, the long line, $\sin\left(\dfrac{1}{x}\right)$ and its variants, etc.).

Answer 4

Many students believe that 1 plus the product of the first $n$ primes is always a prime number. They have misunderstood the contradiction in Euclid's proof that there are infinitely many primes. (By the way, $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 + 1$ is not prime and there are many other such examples.)

Much later edit: As pointed out elsewhere in this thread, Euclid's proof is not by contradiction; that is another widespread false belief.

Much much later edit: Euclid's proof is not not by contradiction. This is another very widespread false belief. It depends on personal opinion and interpretation what a proof by contradiction is and whether Euclid's proof belongs to this category. In fact, if the derivation of an absurdity or the contradiction of an assumption is a proof by contradiction, then Euclid's proof is a proof by contradiction. Euclid says (Elements Book 9 Proposition 20): The very thing (is) absurd. Thus, G is not the same as one of A, B, C. And it was assumed (to be) prime.

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Nb. The above edits were not added by the OP of this answer.

Edit on 24 July 2017: Euclid's proof was not by contradiction, but contains a small lemma in the middle of it that is proved by contradiction. The proof shows that if $S$ is any finite set of primes (not assumed to be the set of all primes) then the prime factors of $1+\prod S$ are not in $S$, so there is at least one more prime than those in $S.$ The proof that $\prod$ and $1+\prod$ have no common factors is the part that is by contradiction. All of this is shown in the following paper: M. Hardy and C. Woodgold, "Prime simplicity", Mathematical Intelligencer 31 (2009), 44–52.

Answer 5

Here's my list of false beliefs

• If $U$ is a subspace of a Banach space $V$, then $U$ is a direct summand of $V$.
• If $M/L$ and $L/K$ are normal field extensions, then the same is true for $M/K$.
• Submodules / subgroups / subalgebras of finitely generated modules / groups / algebras are finitely generated.
• For a subring $S \subseteq R$ of a commutative ring the Krull dimension satisfies $\dim(S) \leq \dim(R)$.
• The Krull dimension of a noetherian integral domain is finite.
• If $A \otimes B = 0$ for abelian groups $A,B$, then either $A=0$ or $B=0$.
• If $f$ is a smooth function with $df=0$, then $f$ is constant.
• If $X,Y$ are sets such that $P(X), P(Y)$ are equipotent, then $X,Y$ are equipotent.
• Every short exact sequence of the form $0 \to A \xrightarrow{f} A \oplus B \xrightarrow{g} B \to 0$ splits.
• $R[x]^{\times} = R^{\times}$ for any commutative ring $R$.
• Every presheaf on a site has an associated sheaf.
• (Co)limits may be computed in full subcategories. For example, $\mathrm{Spec}(\prod_i R_i) = \coprod_i \mathrm{Spec}(R_i)$ as schemes because $\mathrm{Spec}$ is an anti-equivalence between commutative rings and affine schemes.
• Every finite CW-complex is compact, thus every CW-complex is locally compact.
• The smash product of pointed spaces is associative, products of topological spaces commute with quotients, and so on.

Answer 6

I don't know if this is common or not, but I spent a very long time believing that a group $G$ with a normal subgroup $N$ is always a semidirect product of $N$ and $G/N$. I don't think I was ever shown an example in a class where this isn't true.

Answer 7

These are actually metamathematical (false) beliefs that many intelligent people have while they are learning mathematics, but usually abandon when their mistake is pointed out, and I am almost certain to draw fire for saying it from those who haven't, together with the reasons for them:

The results must be stated in complete and utter generality.

Easy examples are left as an exercise to the reader.

It is more important to be correct than to be understood.

(Applicable to talks as well as papers.)

Reasons: 1. Von Neumann is in the audience. 2. This is just a generalization of Lemma 1.2.3 in volume X of Bourbaki. 3. The results are impressive and speak for themselves.

Answer 8

a student, this afternoon: "this set is open, hence it is not closed: this is why [...]"

Answer 9

Some false beliefs in linear algebra:

• If two operators or matrices $A$, $B$ commute, then they are simultaneously diagonalisable. (Of course, this overlooks the obvious necessary condition that each of $A$, $B$ must first be individually diagonalisable. Part of the problem is that this is not an issue in the Hermitian case, which is usually the case one is most frequently exposed to.)

• The operator norm of a matrix is the same as the magnitude of the most extreme eigenvalue. (Again, true in the Hermitian or normal case, but in the general case one has to either replace "operator norm" with "spectral radius", or else replace "eigenvalue" with "singular value".)

• The singular values of a matrix are the absolute values of the eigenvalues of the matrix. (Closely related to the previous false belief.)

• If a matrix has distinct eigenvalues, then one can find an orthonormal eigenbasis. (The orthonormality is only possible when the matrix is, well, normal.)

• A matrix is diagonalisable if and only if it has distinct eigenvalues. (Only the "if" part is true. The identity matrix and zero matrix are blatant counterexamples, but this false belief is remarkably persistent nonetheless.)

• If $\mathcal L: X \to Y$ is a bounded linear transformation that is surjective (i.e. $\mathcal Lu=f$ is always solvable for any data $f$ in $Y$), and $X$ and $Y$ are Banach spaces then it has a bounded linear right inverse. (This is subtle. Zorn's lemma gives a linear right inverse; the open mapping theorem gives a bounded right inverse. But getting a right inverse that is simultaneously bounded and linear is not always possible!)

Answer 10

I once thought that if $A$, $B$, $C$, and $D$ were $n$-by-$n$ matrices, then the determinant of the block matrix $\pmatrix{A & B \\\ C & D}$ would be $\det(A) \det(D) - \det(B) \det(C)$.

Answer 11

Here are two things that I have mistakenly believed at various points in my "adult mathematical life":

For a field $k$, we have an equality of formal Laurent series fields $k((x,y)) = k((x))((y))$.

Note that the first one is the fraction field of the formal power series ring $k[[x,y]]$. For instance, for a sequence $\{a_n\}$ of elements of $k$, $\sum_{n=1}^{\infty} a_n x^{-n} y^n$ lies in the second field but not necessarily in the first. [Originally I had $a_n = 1$ for all $n$; quite a while after my original post, AS pointed out that that this actually does lie in the smaller field!]

I think this is a plausible mistaken belief, since e.g. the analogous statements for polynomial rings, fields of rational functions and rings of formal power series are true and very frequently used. No one ever warned me that formal Laurent series behave differently!

[Added later: I just found the following passage on p. 149 of Lam's Introduction to Quadratic Forms over Fields: "...bigger field $\mathbb{R}((x))((y))$. (This is an iterated Laurent series field, not to be confused with $\mathbb{R}((x,y))$, which is usually taken to mean the quotient field of the power series ring $\mathbb{R}[[x,y]]$.)" If only all math books were written by T.-Y. Lam...]

Note that, even more than KConrad's example of $\mathbb{Q}_p^{\operatorname{unr}}$ versus the fraction field of the Witt vector ring $W(\overline{\mathbb{F}_p})$, conflating these two fields is very likely to screw you up, since they are in fact very different (and, in particular, not elementarily equivalent). For instance, the field $\mathbb{C}((x))((y))$ has absolute Galois group isomorphic to $\hat{\mathbb{Z}}^2$ -- hence every finite extension is abelian -- whereas the field $\mathbb{C}((x,y))$ is Hilbertian so has e.g. finite Galois extensions with Galois group $S_n$ for all $n$ (and conjecturally provably every finite group arises as a Galois group!). In my early work on the period-index problem I actually reached a contradiction via this mistake and remained there for several days until Cathy O'Neil set me straight.

Every finite index subgroup of a profinite group is open.

This I believed as a postdoc, even while explicitly contemplating what is probably the easiest counterexample, the "Bernoulli group" $\mathbb{B} = \prod_{i=1}^{\infty} \mathbb{Z}/2\mathbb{Z}$. Indeed, note that there are uncountably many index $2$ subgroups -- because they correspond to elements of the dual space of $\mathbb{B}$ viewed as a $\mathbb{F}_2$-vector space, whereas an open subgroup has to project surjectively onto all but finitely many factors, so there are certainly only countably many such (of any and all indices). Thanks to Hugo Chapdelaine for setting me straight, patiently and persistently. It took me a while to get it.

Again, I blame the standard expositions for not being more explicit about this. If you are a serious student of profinite groups, you will know that the property that every finite index subgroup is open is a very important one, called strongly complete and that recently it was proven that each topologically finitely generated profinite group is strongly complete. (This also comes up as a distinction between the two different kinds of "profinite completion": in the category of groups, or in the category of topological groups.)

Moreover, this point is usually sloughed over in discussions of local class field theory, in which they make a point of the theorem that every finite index open subgroup of $K^{\times}$ is the image of the norm of a finite abelian extension, but the obvious question of whether this includes every finite index subgroup is typically not addressed. In fact the answer is "yes" in characteristic zero (indeed $p$-adic fields have topologically finitely generated absolute Galois groups) and "no" in positive characteristic (indeed Laurent series fields do not, not that they usually tell you that either). I want to single out J. Milne's class field theory notes for being very clear and informative on this point. It is certainly the exception here.

Answer 12

"Any subspace of a separable topological space is separable, too." Sounds natural.

Answer 13

Here are a few more: (Everything between quotation marks is a false belief.)

Basic logic: Among students: "If A implies B then B implies A" (or "if A implies B then not A implies not B").

Even among mature mathematicians a frequent false belief is to forget that the conclusion of a theorem need not hold once the conditions of the theorem fail. Another common frequent belief is to assume that once the conditions fail then the conclusion must fail too.

Calculus: "The derivative of a differentiable function is continuous."

"An infinite series whose general term tend to 0 is convergent."

Geometry: "The circle is the only figure which has the same width in all directions." (Feynman regarded this mistake as one reason for the space shuttle Challenger disaster).

Polytopes: Often people believe that "given a convex polytope P you can slightly move the vertices to rational positions keeping the structure of the polytope unchanged."

(From Udi de Shalit): Some people believe that "if you hold a cube along a main diagonal, the remaining vertices all lie on a plane." Some even say that their number is 4.

Algebra (Also from Udi) "I have encountered many misconceptions about solvability by radicals. Some people think that 'the solution of an irreducible equation of degree 5 and higher, say over $\mathbb Q$, is never expressible by radicals'. Some amateur mathematicians even say that 'equations of degree 5 and higher have no solutions'."

Probability: "If you play the casino patiently and carefully you will win in the long run" (and "you do not believe that?, this is my own experience on the matter!" and "Indeed when I am calm and patient I win, but when I lose my temper I lose big time".)

"an event which may occur has positive probability": (not true for infinite probability spaces)

Various places: "If you want to prove that a certain infinite structure exists it is enough to show that there is no upper bound to the sizes of such structures."

Combinatorics: "This is a finite problem, surely you can solve it with a computer."

"Hall marriage theorem is very nice and I am surprised no combinatorialist bothered to extend it to a matching built from triples instead of pairs." (It is unlikely that a general characterization when a hypergraph built from triples has a perfect matching (of triangles) will be found.)

Computer science: "It is known that quantum computers can solve NP complete problems in polynomial time."

Answer 14

From the Markov property of the random walk $(X_n)$ we have

$$P(X_4>0 \ |\ X_3>0, X_2>0) = P(X_4>0\ |\ X_3>0).$$

To paraphrase Kai Lai Chung in his book "Green, Brown, and Probability",

"The Markov property means that the past has no after-effect on the future when the present is known; but beware, big mistakes have been made through misunderstanding the exact meaning of the words "when the present is known"."

Answer 15

$$2^{\aleph_0} = \aleph_1$$

This is a pet peeve of mine, I'm always surprised at the number of people who think that $\aleph_1$ is defined as $2^{\aleph_0}$ or $|\mathbb{R}|$.

Answer 16

I think, there are different types of false beliefs. The first kind are statements which are quite natural to believe, but a moment of thought shows the contradiction. Of this type is the sin-example in the opening post or a favorite of mine (also occurred to me):

• The underlying additive group of the field with $p^n$ elements is $\mathbb{Z}/p^n\mathbb{Z}$.

The other type is also quite natural to believe, but one has really to think to construct a counter example:

• Every contractible manifold is homeomorphic to $\mathbb{R}^n$.
• Every manifold is homotopy equivalent to a compact one.
• Quotients commute with products in topological spaces.
• Every connected component of a topological space is open and closed. Or related to this:
• To give a continuous action of a topological group $G$ on a discrete space $X$ is the same as to give an action of the group of connected components of $G$ on $X$.

Answer 17

I used to believe that a continuous algebra homomorphism from $k[[x_1,\dots, x_m]]$ to $k[[y_1,\dots,y_n]]$, with $m > n$, could not be injective. Konstantin Ardakov set me straight on this.

Answer 18

Here are two group theory errors I've seen professionals make in public.

1) Believing that if $G_1 \subset G_2 \subset \cdots$ is an ascending union of groups such that $G_i$ is free, then $\bigcup_{i=1}^{\infty} G_i$ is free. Probably the vague idea they have is that any relation has to live in some $G_i$, so there are no nontrivial relations.

2) Consider a group $G$ acting on a vector space $V$ (over $\mathbb{C}$, say). Assume that $G$ acts as the identity on a subspace $W$ and that the induced action of $G$ on $V/W$ is trivial. Then I've seen people conclude that the action of $G$ on $V$ is trivial. Of course, this is true if $G$ is finite since then all short exact sequences of $G$-modules split, but it is trivial to construct counterexamples for infinite $G$.

Answer 19

The field of $p$-adic numbers has characteristic $p$.

Answer 20

"Either you can prove the statement, or you can find a counterexample."

This statement is usually applied to universal statements, those having the form $\forall x\ \varphi(x)$, where the concept of counterexample makes sense, but the general sentiment is the belief that every statement in mathematics is either provable or refutable.

The belief is false, because of the independence phenomenon.

Answer 21

I remember from my first analysis class thinking that if $\mathbb{Q}\subset E\subset\mathbb{R}$ with $E$ open, then $E$ would have to be all of $\mathbb{R}$ (at least more or less, maybe up to countably many points). And once we started measure theory I remember arguing with a friend over it for a good two hours.

Answer 22

If $f(x,y)$ is a polynomial with real coefficients, then the image of $f$ is a closed subset of $\mathbb{R}$. Note. Problem A1 on the 1969 Putnam exam asked to describe all possible images of $f$. I was told that the writers of this problem did not realize its subtlety.

Answer 23

In order to show that a polynomial $P \in F[x_1,\ldots,x_n]$ vanishes, it suffices to show that $P(x_1,\ldots,x_n) = 0$ for all $x_1,\ldots,x_n \in F$. True in infinite fields, but very false for small finite fields.

Closely related: if two polynomials $P$, $Q$ agree at all points, then their coefficients agree. Again, true in infinite fields, but false for finite fields.

(This is ultimately caused by a conflation of the concept of a polynomial as a formal algebraic expression, and the concept of a polynomial as a function. Once one learns enough algebraic geometry to be comfortable with concepts such as "the $F$-points $V(F)$ of a variety $V$" then this confusion goes away, though.)

Answer 24

"If any two of the $3$ random variables $X,Y,Z$ are independent, all three are mutually independent." In fact, they may be dependent; the simplest example is probably $(X, Y, Z)$ chosen uniformly from $\{(0, 0, 0), (1, 1, 0), (1, 0, 1), (0, 1, 1)\}$.

Answer 25

"Automorphisms of the symmetric group $S_n$ are inner (that is, each one is of the form $x \to axa^{-1}$ for some $a \in S_n$)" is a popular misconception, false for nontrivial reasons when $n=6$. That is both an easy mistake to make and important conceptually as an early hint of the complexities and special combinatorics that arise in finite group theory. Many people make it through a first class in group theory without understanding that something different happens for $S_6$ and in doing so have missed an important piece of the the big picture, as far as finite groups are concerned.

It is easy to implicitly or explicitly acquire this belief, because:

1. those really are all the automorphisms for $n$ other than 6, and

2. the inner automorphisms are used so often, for all values of $n$ (or $n>2$) without distinguishing any specific case as unusual.

3. $S_n$ behaves in many ways as a family of similar groups rather than a list of individual groups with their own diverse features. A typical proof might show some property of $S_n$ by induction on $n$, starting from a small value such as $n=1$ for basic properties, or $n=3$ to assure noncommutativity. Apart from the classification of symmetric group automorphisms itself (exposure to which would be an explicit articulation and correction of the false belief), these arguments never start as high as $n=7$ and I don't know of any that distinguish $n=6$ or some equivalent case as a lone nontrivial exception. So it is easy to get the idea of more uniformity in the $S_n$ than really exists.

In essence, there are no obvious clues in the environment that $n=6$ might be special, and a number of indicators that no special case should exist at all.

Answer 26

There are a couple of false beliefs regarding the $I$-adic completion functor, where $I$ is an ideal in a commutative ring $A$.

The first is that the completion of an $A$-module $M$ is complete, or in other words, that the completion functor is idempotent. This is true if $I$ is finitely generated (in particular when $A$ is Noetherian), but false in general. I find this quite unexpected - you take a module, "complete" it, and the result is not complete...

Another issue is the exactness of the completion functor. The completion functor is exact on the category of finitely generated modules over a Noetherian ring, but when you consider arbitrary modules, it is neither left-exact (this is easy to see) nor it is right-exact (this probably less known), even when $I$ is finitely generated and the modules in question are finitely presented.

Answer 27

Some people have trouble understanding that (and why is) 0.999... = 1

Answer 28

This is perhaps a misunderstood definition rather than a false belief, but:

"A subnet of a net $( x_\alpha )_{\alpha \in A}$ takes the form

$( x_\alpha )_{\alpha \in B}$ for some subset $B$ of $A$."

In truth, subnets are allowed to contain repetitions, and can be indexed by sets much larger than the original net. (In particular, there are subnets of sequences that are not subsequences.)

This false belief, incidentally, reinforces the false belief noted in a different answer, namely that compactness implies sequential compactness.

A precise Counterexample: The sequence $\sin(nx)$ is a sequence in the compact topological space $[-1, 1]^{\mathbb{R}}$ with product topology. So this net has a convergence subnet. But it is well known that the above sequence has no subsequence which is pointwise convergent (See the last page of the book of Walter Rudin's Principles of mathematical Analysis). So in this example the convergent subnet cannot be counted as a subsequence.

Answer 29

"It is impossible in principle to well-order the reals in a definable manner."

To be more precise, the belief I am talking about is the belief that well-orderings of the reals are provably chaotic in some sense and certainly not definable. For example, the belief would be that we can prove in ZFC that no well-ordering of the reals arises in the projective hierarchy (that is, definable in the real field, using a definition quantifying over reals and integers).

This belief is relatively common, but false, if the axioms of set theory are themselves consistent, since Goedel proved that in the constructible universe $L$, there is a definable well-ordering of the reals having complexity $\Delta^1_2$, which means it can be obtained from a Borel subset of $R^3$ by a few projections and complements. See this answer for a sketch of the definition of the well-order.

The idea nevertheless has a truth at its core, which is that although it is consistent that there is a definable well-ordering of the reals (or the universe), it is also consistent that there is no such definable well-ordering. Thus, there is no definable relation that we can prove is a well-ordering of the reals (although we also cannot prove that none is).

Answer 30

Occasionally seen on this site: if a polynomial $P:\mathbb{Q}\rightarrow\mathbb{Q}$ is injective, so must be its extension to $\mathbb{R}$.

Answer 31

It's easy when you're an amateur to topology to assume any continuous bijection has a continuous inverse. The inverse of an arbitrary continuous bijection in a topological space is open, but it's NOT necessarily continuous. Continuity turns out to be a stronger condition.

Answer 32

People are silly. Did you ever notice how at airports, say, people happily walk around but when they come to a moving walkway, they tend to stop and take a break? Walking on a moving walkway is not any harder than walking on an ordinary walkway, and resting for 10 seconds here or there will get you to your destination 10 seconds later. So why do people stop for a rest in one place but hardly ever in the other?

The embarrassing bit is that I believed this logic myself for some time, and thought that people were indeed silly, until my son corrected me. I'm not sure if this falls under "a common false belief in mathematics", but it's certainly an amusing and confusing mistake to make.

Answer 33

Some people believe there is no "formula" for the nth prime number. Of course there are many such formulas, even though not very useful: https://mathworld.wolfram.com/PrimeFormulas.html

The reason given for disbelieving the existence of a "prime number formula" is also curious: "because the primes are unpredictable". This belief is in contradiction with the simple fact that anyone can come up with an easy algorithm which gives the nth prime number. There is something mystical associated with this ill-defined term "formula".

Answer 34

"The image of a category under a functor is a category."

This is a small one, but it lasted for six months when I was starting in category theory.

A finite counterexample exists, with just 3 objects. Even under a connectivity requirement, a small finite counterexample still exists. In fact, it is dead wrong to think anything like this holds.

Answer 35

That the notion of "picking a random number" is well-defined without providing any further information.

Answer 36

It is a difficult open problem whether every finite group is isomorphic to the group of automorphisms of a finite extension $K$ of $\mathbb{Q}$. In fact, this result is true! See for instance M. Fried: A Note on Automorphism Groups of Algebraic Number Fields; https://www.jstor.org/stable/2043724, https://doi.org/10.2307/2043724. The actual "inverse Galois problem" also requires $K$ to be a Galois extension of $\mathbb{Q}$. The true theorem is equivalent to the statement that every finite group is isomorphic to a quotient group of a Galois group over $\mathbb{Q}$. I once observed a famous expert on algebraic number theory being confused on this issue.

Answer 37

Linear algebra: 1. If V is a vector space spanned by {e_i} and W is a subspace of V then W is spanned by e_k's contained in it. Actually, this is widely believed with bases in place of spanning sets. Or

2.   (U+V)∩W = U∩W + V∩W.

Both these "properties" are closely related to the current leader (by Tilman).

3. Every element of V⊗W is v⊗w with v∈ V, w∈ W.

All three are probably due to interpolating our intuition about sets to vector spaces.

4. Every symmetric matrix is diagonalizable.

Wait, didn't we prove this? ("True for the real matrices, so must be true in general").

Algebraic groups: if G is a linear algebraic group acting on a vector space V then the (Krull) dimension of the invariant ring satisfies the inequality

    dim k[V]^G ≥ dim V-dim G,

or even a more precise belief that dim k[V]^G=dim V-dim G_x for a generic x. This is true in the differentiable situation for the dimension of the quotient, when a compact Lie group acts smoothly on a manifold, and algebraic actions are "nicer", right?

Answer 38

False belief: "There are no known sub-exponential time algorithms for NP-complete problems."

This one is tricky for a couple of reasons. The first is that the term "sub-exponential" is sometimes defined in different ways. With a sufficiently strong definition of "sub-exponential" the above statement is true, in the sense that there is no known separation of the complexity classes NP and EXPTIME (EXPTIME being the class of languages decidable in time $2^{p(n)}$ where $p(n)$ is a polynomial). However, it is quite common to refer to an $O(2^{\sqrt{n}})$ algorithm as "sub-exponential." It is trivial to construct an NP-complete problem that can be solved in sub-exponential time in this sense, because the standard definition of a reduction allows you to expand the size of the input from $n$ to $n^2$ say (e.g., by padding with zeros). But less artificial examples also exist, such as the planar traveling salesman problem, which was shown by Smith to be solvable in $2^{O(\sqrt{n})}$ time without any artificial padding. What is true is that there are many NP-complete problems, such as 3SAT, for which no subexponential algorithms are known if you do not artificially pad the representation of the instances. (Reducing 3SAT to planar TSP does not work because the instance size blows up during the reduction.)

Often this false belief shows up in the following form: "Factoring cannot be NP-complete because there are subexponential algorithms for factoring." It is true that factoring is not known to be NP-complete but the reasoning is wrong. Showing that factoring is NP-complete would not automatically yield subexponential algorithms for all other NP-complete problems.

Answer 39

$\pi$ is equal to 22/7.

This was touched upon in the comments to a totally unrelated answer but I think this false belief is important enough to warrant its own answer (and as far as I could tell it does not have one yet, my apologies if I overlooked one.)

Of course, it's unlikely anyone on this site believes this, or ever believed it, which is why I think it's important to insist on this: it does not really resonate with us, we are unlikely to warn students against it, yet we probably see in front of us many students who have that false belief and then will move on to spread it around.

A Piece of Evidence

Let me offer as evidence this gem taken off the comments section of an unrelated (but quite thought-provoking) article on Psychology Today, of all places! When Less is More: The Case for Teaching Less Math in Schools (The title is a misnomer, it's a case for starting math later, but I think that with such a scheme you should be able to teach more math overall; anyway, read it for yourselves.)

Some years ago, my (now ex-) wife was involved in a "trivia night" fundraiser at her elementary school, and they wanted me on their "teacher team" to round out their knowledge. They had almost everything covered except some technology-related topics and I was an IT guy. In round four, my moment to shine arrived, as the category was "Math & Science" and one of the questions was, "give the first five digits of pi." I quickly said, "3.1415." The 9 teachers at the table ignored me and wrote down "22/7" on scrap paper and began to divide it out. I observed this quietly at first, assuming that 22/7ths gave the right answer for the first 5 digits, but it doesn't. It gives something like 3.1427. I said, "Whoops, that won't work." They ignored me and consulted among themselves, concluding that they had all done the division properly on 22/7ths out to five digits. I said, "That's not right, it's 3.1415." [...]

I'm cutting it off here because it's a long story: hilarity ensues when the non-teacher at the table stands up for the truth (when he finds out that the decimals of 22/7 were the expected answer!) The final decision of the judges:

"We've got a correction on the 'pi' question, apparently there's been confusion, but we will now be accepting 3.1415 as a correct answer as well" [as 3.1427].

The Moral of the Story

I used to dismiss out of hand this kind of confusion: who could be dumb enough to believe that $\pi$ is 22/7? (Many people apparently: in the portion of the story I cut was another gem - "I'm sorry, but I'm a civil engineer, and math is my job. Pi is 22/7ths.")

Now, I treat this very seriously, and depending on where you live, you should too. Damage wrought during the influential early years is very hard to undo, so that the contradictory facts "$\pi$ is irrational" and "$\pi$=22/7" can coexist in an undergraduate's mind. And when that person leaves school, guess which of the two beliefs will get discarded: the one implanted since childhood, or the one involving a notion (rational numbers) which is already getting fuzzy in the person's brain? I'm afraid it's no contest there, unless this confusion has been specifically addressed.

So if you have any future teachers in your classes (and even if you don't, cf. the civil engineer above), consider addressing this false belief at some point.

Answer 40

False belief: Saying that ZFC is consistent is the same as saying that if ZFC proves "there are infinitely many twin primes" (for example) then there really are infinitely many twin primes.

Everybody realizes that if ZFC is inconsistent then a formal ZFC proof of "there are infinitely many twin primes" tells us nothing about whether there really are infinitely many twin primes. A lot of people, without necessarily realizing it, turn this around and assume that the consistency of ZFC is the only condition needed to ensure that its theorems are "trustworthy." But this is not the case, even if we restrict our attention to first-order statements about the natural numbers. We say that ZFC is arithmetically sound if all first-order sentences about the natural numbers that are provable in ZFC are true. The arithmetical soundness of ZFC is a stronger condition than the consistency of ZFC. For example, Goedel's 2nd incompleteness theorem says that if ZFC is consistent, then ZFC doesn't prove "ZFC is consistent." So it's conceivable that ZFC is consistent but that "ZFC is inconsistent" is a theorem of ZFC. Then we would have an example of a theorem of ZFC that asserts something false, even though ZFC is consistent.

Answer 41

Real projective space ${\mathbb{RP}}^3 = (\mathbb R^4 - 0)/\mathbb R^*$ is non-orientable.

Answer 42

If $f$ is (Lebesgue) integrable on $\mathbb{R}$, then $f(x) \to 0$, as $x \to \infty$. False: there exists a continuous integrable function on $\mathbb{R}$ such that $\limsup_{\infty} f = \infty$ (an exercise in Stein and Shakarchi's Real Analysis).

Answer 43

The commutator [H,K] of two subgroups H,K is the set of commutators [h,k] with h in H and k in K. (Instead, it is the group generated by those commutators. Confusingly, the convention with products HK usually goes the other way.)

In a similar vein: a 2-vector $\omega \in \bigwedge^2 V$ is always the wedge product of two 1-vectors. (Instead, this is merely an important special case of a 2-vector.) Part of the difficulty here is that the statement is true in the important three-dimensional case. Once one is aware of the Plucker embedding, this confusion goes away, but that can take a while...

Answer 44

"A continuous image of a locally compact space is locally compact."

This is tempting because it is true without the "locally"s and it is often the case that topological properties and statements can be "localized". This came up in a problem session for my [number theory!] course this semester, and although the students were too experienced to accept it without proof, they had no alarm bells in their heads to prevent them from entertaining the possibility.

The way to quash this (as well as Andrew L.'s answer, which reminded me of this) is to realize that if it were true, every space $X$ would be locally compact, since the identity map from $X$ endowed with the discrete topology to $X$ is a continuous bijection.

Answer 45

Here's one that doesn't seem to be listed here: That ZFC isn't vulnerable to Russell's paradox because it has an axiom which implies that sets cannot be elements of themselves.

This one should be obviously false given even a moment's thought -- you cannot get rid of a paradox by adding axioms! But I've seen it repeated over and over.

The correct statement, of course, is that ZFC isn't vulnerable to Russell's paradox because it doesn't have the axiom of unrestricted comprehension; and that while ZFC does have an axiom which implies that all sets are well-founded, this is irrelevant to Russell's paradox. (It just means that the set of all sets which are not elements of themselves, if it existed, would happen to also be the set of all sets.)

Sometimes this is stated in an ambiguous way that allows for the charitable reading that modern set theory prevents Russell's paradox by preventing one from talking about a set being an element of itself, which is how Russell and Whitehead fixed the problem. But Principia Mathematica isn't exactly "modern set theory"; and I've seen this stated the blatantly wrong way often enough that I'm not too inclined to be charitable about it.

Answer 46

I just finished quadratic congruences in my number theory class. I am not any more surprised to see how strong is students' belief in the fact that $x^2\equiv a\pmod{m}$ has at most 2 solutions. Even after you discuss an example $x^2\equiv1\pmod{143}$ (with solutions $x\equiv\pm1,\pm12\pmod{143}$) in details.

And, of course, a lot of wrong beliefs in real analysis. Like an infinitely differentiable function, say in a neighbourhood of origin, must be analytic at the origin.

Answer 47

False belief:

• The category of commutative C*-algebras is opposite equivalent to the category of locally compact Hausdorff spaces.

It's actually not quite that simple! There is some discussion on math.SE.

Answer 48

$0^0$ is undefined.

EDIT: People write things like $\sum_{k=0}^\infty x^k$ all day, but somehow $x=k=0$ is still scary when written as $0^0$.

Answer 49

I do not think anybody mentioned this example:

If $M$ is a $C^k$-smooth manifold then the tangent space $T_pM$ is isomorphic to the space of derivations of germs (at $p$) of $C^k$-smooth functions on $M$.

This holds for $k=\infty$, $k=\omega$ (i.e. in the real-analytic category), but not otherwise.

For details and positive results (characterizing derivations which come from tangent vectors even when $k<\infty$) see

Newns, W. F.; Walker, A. G., Tangent planes to a differentiable manifold, J. Lond. Math. Soc. 31, 400-406 (1956). ZBL0071.15303.

Abraham, R.; Marsden, J. E.; Ratiu, T., Manifolds, tensor analysis, and applications., Applied Mathematical Sciences. 75. New York: Springer-Verlag. x, 654 p. (1988). ZBL0875.58002.

Answer 50

A false belief which I meet not infrequently this time of year while marking exams is the following:

The exponential map is surjective for a connected Lie group.

This is true for compact Lie groups, but certainly false in general. A (finite-dimensional) connected Lie group is generated by the image of the exponential map, but already $SL(2,\mathbb{R})$ shows that there are elements which are not in the image of the exponential map.

Interestingly, for a connected real Lie group, every element can be written as the product of at most two exponentials.

Answer 51

Let $A,B$ be Hermitian matrices. If $0_n\le A\le B$, then $A^2\le B^2$.

False, but subtle! Loewner's theory characterizes those numerical functions $f:[0,\infty)\rightarrow {\mathbb R}$ such that $0_n\le A\le B$ implies $f(A)\le f(B)$ (operator monotone functions). These are the traces over $[0,\infty)$ of holomorphic functions mapping the Poincaré half-plane ${\mathcal H}$ into itself, and of course real on $[0,\infty)$. Thus the square root is operator monotone: $$(0_n\le A\le B)\Longrightarrow(\sqrt A\le\sqrt B),$$ but the square map is not. Counter-example: $$A=\begin{pmatrix} 1 & 0 \\\\ 0 & 0 \end{pmatrix},\qquad B=\begin{pmatrix} 2 & 1 \\\\ 1 & 1 \end{pmatrix}$$

Answer 52

All higher homotopy groups of spheres are zero.

Proof: The higher homology groups of spheres are zero, and the higher homotopy groups are abelian, and since homology groups are abelianizations of homotopy groups the higher homotopy groups are also zero.

This misconception is also made more difficult by the fact that even the simple counterexamples can't be drawn easily.

Answer 53

This is a bit specialized, but a common misconception in low-dimensional topology (particularly in knot theory) is that any change of basis in homology is realized by a diffeomorphism, hence (for a surface) by an action of a mapping class. I think this is exactly the type of false belief being described (I falsely believed it for a long time myself).

Common misconception: Let F be a genus 2g surface, and let $b_1,\ldots,b_{2g}$ be a primitive basis for $H_1(F)$, represented as embedded curves in F. Any change of basis for $H_1(F)$ is realized by an action of the mapping class group of F on the embedded curves.

This is rubbish- the action of the mapping class group on homology is by $Sp_{2g}(\mathbb{Z}))$, which for $g>1$ is a proper subgroup of $GL_{2g}(\mathbb{Z})$, the group of base-changes of $H_1(F)$.
As an example of what you can't do with a diffeomorphism of a surface, consider a disc with 4 bands A,B,C,D attached, so the order of the end sements is $A^+B^-A^-B^+C^+D^-C^-D^+$, together forming a surface. A basis a,b,c,d for $H_1(F)$ is given by this picture as 4 loops going through the cores of the bands A,B,C,D correspondingly. You can add a to b, b to c, c to d, or d to a by diffeomorphism of F (sliding adjacent bands over one another). However, although you can add a to c algebraically, because bands A and C are "not adjacent in F", there is no corresponding diffeomorphism of $F$.
One place this mistake manifests itself (cranking up the level of terminology for a second) is in thinking that unimodular congruence of a Seifert matrix corresponds to ambient isotopy of a Seifert surface.
A related common mistake (closely related to this question):

Common misconception: Any homology class is represented as a submanifold. Maybe even as an embedded submanifold.

Answer 54

Suppose that two features $[x,y]$ from a population $P$ are positively correlated. We divide $P$ into two subclasses $P_1$, $P_2$. Then, it cannot happen that the respective features ( $[x_1,y_1]$ and $[x_2,y_2]$) are negatively correlated in both subclasses

Or more succinctly:

Mixing preserves the correlation sign.

This seems very plausible - almost obvious. But it's false - see Simpson's paradox

Answer 55

a function $f:\mathbb{R}\rightarrow \mathbb{R}$ is $n$ times differentiable if and only if for each real $x_0$ it may be approximated near $x_0$ by a polynomial of degree at most $n$ with remainder $o((x-x_0)^n)$

Answer 56

The quotient $G/Z(G)$ of a group by its center is centerless. I definitely thought this until it was pointed out to me in a Lie theory textbook that this wasn't true in general, but is true for (edit: connected) Lie groups with discrete center. (It is also true if $G$ is perfect by Grun's lemma.)

Answer 57

After learning that the Witt vectors of a finite field of size $p^n$ is the ring of integers of the unramified extension of ${\mathbf Q}_p$ of degree $n$, I think lots of people then think that the Witt vectors of $\overline{\mathbf F}_p$ (the algebraic closure of ${\mathbf F}_p$) is the ring of integers of the maximal unramified extension of ${\mathbf Q}_p$. It isn't: the integers of the maximal unramified extension is the union of the Witt vectors of the finite fields of $p$-power size whereas the Witt vectors of $\overline{\mathbf F}_p$ is the $p$-adic completion of the integers of the maximal unramified extension; the distinction turns on being able to write Witt vectors over $\overline{\mathbf F}_p$ as series with coefficients that are prime-to-$p$ roots of unity of increasingly large degree instead of having bounded degree.

I was at a conference last fall where a famous mathematician was confused by this point, although to be fair he really never worked seriously with Witt vectors before.

Answer 58

In group theory, if $G_1 \cong G_2$ and $H_1 \cong H_2$, then

$G_1 / H_1 \cong G_2 / H_2$.

For example, $\mathbb{Z} / 2\mathbb{Z} \not \cong \mathbb{Z} / \mathbb{Z}$. The point is that the inclusion of $H_j$ into $G_j$ is needed in order to define the quotient.

Answer 59

The product of two symmetric matrices is symmetric!

Answer 60

If a topological space has an open cover by Hausdorff spaces, it is Hausdorff.

Answer 61

The standard projection map in a first course in topology is open. How could it not be closed? I always forget the standard homework exercise in which people first try to use this non-fact.

Answer 62

Any subgroup of the direct product $G \times H$ of two groups is of the form $A \times B$, where $A$ is a subgroup of $G$ and $B$ is a subgroup of $H$.

Answer 63

As a teaching assistant in an elementary number theory course, I've seen the following quite often :

If $a$ divides $bc$ and $a$ does not divide $b$, then $a$ divides $c$.

That's of course true if $a$ is prime, but people seem to forget that hypothesis.

Answer 64

I'm not sure how common this is, but it confused me for years. Let $f : \mathbb{C} \to \mathbb{C}$ be an analytic function and $\gamma$ a path in $\mathbb{C}$. In your first class in complex analysis, you define the integral $\int_{\gamma} f(z) dz$.

Now let $a(x,y) dx + b(x,y) dy$ be a $1$-form on $\mathbb{R}^2$ and let $\gamma$ be a path in $\mathbb{R}^2$. In your first class on differential geometry, you define the integral $\int_{\gamma} a(x,y) dx + b(x,y) dy$.

It took me at least three years after I had taken both classes to realize that these notations are consistent. Until then, I thought there was a "path integral in the sense of complex analysis", and I wasn't sure if it obeyed the same rules as the path integral from differential geometry. (By way of analogy, although I wasn't thinking this clearly, the integral $\int \sqrt{dx^2 + dy^2}$, which computes arc length, is NOT the integral of a $1$-form, and I thought complex integrals were something like this.)

---

For the record, I'll spell out the relation between these notions. Let $f(x+iy) = u(x,y) + i v(x,y)$. Then $$\int_{\gamma} f(z) dz = \int_{\gamma} \left( u(x,y) dx - v(x,y) dy \right) + i \int_{\gamma} \left( u(x,y) dy + v(x,y) dx \right)$$ The right hand side should be thought of as multiplying out $\int_{\gamma} (u(x,y) + i v(x,y)) (dx + i dy)$, a notion which can be made rigorous.

Answer 65

Another false belief which I have been asked thrice so far in person is $$\lim_{x \rightarrow 0} \frac{\sin(x)}{x} = 1$$ even if $x$ is in degrees. I was asked by a student a year and half back when I was a TA and by couple of friends in the past 6 months.

Answer 66

A subgroup of a finitely generated group is again finitely generated.

Answer 67

I'm not sure that anyone holds this as a conscious belief but I have seen a number of students, asked to check that a linear map $\mathbb{R}^k \to \mathbb{R}^{\ell}$ is injective, just check that each of the $k$ basis elements has nonzero image.

Answer 68

Here's one that bugged me from point set topology: "A subnet of a sequence is a subsequence".

See here for the definitions. Using this one gives a great proof that compactness implies sequential compactness in any topological space:

Let $X$ be a compact space. Let $(x_n)$ be a sequence. Since a sequence is a net and it's a basic theorem of point set topology that in a compact topological space, every net has a convergent subnet (proof in the above link), there is a convergent subnet of the sequence $(x_n)$. Using the above belief, the sequence $(x_n)$ has a convergent subsequence and hence $X$ is sequentially compact.

For a counterexample to this "theorem", consider the compact space $X= \lbrace 0,1 \rbrace ^{[0,1]}$ with $f_n(x)$ the $n$th binary digit of $x$.

Answer 69

The ring $\mathbb{C}[x]$ has countable dimension over $\mathbb{C}$; therefore its field of fractions $\mathbb{C}(x)$ also has countable dimension over $\mathbb{C}$.

Answer 70

In measure-theoretic probability, I think there is sometimes an idea among beginners that independent random variables $X,Y$ should be thought of as having "disjoint support" as measurable functions on the underlying probability space $\Omega$. Of course this is the opposite of the truth.

I think this may come from thinking of measure theory as generalizing freshman calculus, so that one's favorite measure space is something like $[0,1]$ with Lebesgue measure. This is technically a probability space, but a really inconvenient one for actually doing probability (where you want to have lots of random variables with some amount of independence).

Answer 71

Yet another one:

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be differentiable. If $f'(x_0) > 0$, then there exists an interval $I$ containing $x_0$ such that $f$ is increasing in $I$.

Answer 72

A common misbelief for the exponential of matrices is $AB=BA \Leftrightarrow \exp(A)\exp(B) = \exp(A+B)$. While the one direction is of course correct: $AB=BA \Rightarrow \exp(A)\exp(B) = \exp(A+B)$, the other direction is not correct, as the following example shows: $A=\begin{pmatrix} 0 & 1 \\ 0 & 2\pi i\end{pmatrix}, B=\begin{pmatrix} 2 \pi i & 0 \\ 0 & -2\pi i\end{pmatrix} $ with $AB \neq BA \text{ and} \exp(A)=\exp(B) = \exp(A+B) = 1$.

Answer 73

"Euclid's proof of the infinitude of primes was by contradiction."

That is a very widespread false belief.

"Prime Simplicity", Mathematical Intelligencer, volume 31, number 4, pages 44--52, by me and Catherine Woodgold, debunks it. The proof that Euclid actually wrote is simpler and better than the proof by contradiction often attributed to him.

Answer 74

Perhaps the most prevalent false belief in math, starting with calculus class, is that the general antiderivative of f(x) = 1/x is F(x) = ln|x| + C. This can be found in innumerable calculus textbooks and is ubiquitous on the Web.

Answer 75

The gamma function is not the only meromorphic function satisfying $$f(z+1)=z f(z),\qquad f(1)=1,$$ with no zeroes and no poles other than the points $z=0,-1,-2\dots$.

In fact, there is a whole bunch of such functions, which, in general, have the form $$f(z)=\exp{(-g(z))}\frac{1}{z\prod\limits_{n=1}^{\infty} \left(1+\frac{z}{m}\right)e^{-z/m}},$$ where $g(z)$ is an entire function such that $$g(z+1)-g(z)=\gamma+2k\pi i,\qquad g(1)=\gamma+2l\pi i,\qquad k,l\in\mathbb Z, $$ ($\gamma$ is Euler's constant). The gamma function corresponds to the simplest choice $g(z)=\gamma z$.

Answer 76

False belief: Every commuting pair of diagonalizable elements of $PSL(2,\mathbb{C})$ are simultaneously diagonalizable. The truth: I suppose not many people have thought about it, but it surprised me. Look at $$\left(\matrix{i& 0 \cr 0 & -i\cr } \right), \left(\matrix{0& i \cr i & 0\cr } \right).$$

Answer 77

That Darboux functions are continuous is certainly a widely held belief among students, at least in France where it is induced by the way continuity is taught in high school.

I remember having gone through all the five "stages of grief" when shaken from this false belief with the $\sin(1/x)$ example : denial, anger ( "then the definition of continuity must be wrong ! Let's change it !), bargaining ("Ok, but a Darboux function must surely be continuous except at exceptional points. Let's prove that..."), depression (when shown a nowhere continuous Darboux function), acceptance ("Hey guys, you really think the intermediate value theorem has a converse? C'mon, you're smarter than that...")

Answer 78

"Some real numbers are not definable, by Cantor's diagonal argument."

There are subtleties involved in formalizing the statement "some real numbers are not definable", as explained in Joel's answer to this question. The statement can be seen to hold in some models and fail in other models. However, the claim that the statement follows from Cantor's diagonal argument is clearly false, yet seems to be fairly common.

The false reasoning typically proceeds in three steps:

1. There are only countably many definitions of real numbers: $\varphi_0(x),\varphi_1(x),\ldots$ (this part is ok.)

2. Consider the countably many real numbers so defined: $x_0,x_1,\ldots$ (this part is problematic for subtle reasons.)

3. Use Cantor's diagonal argument to obtain a real number $y$ that is not in the sequence from step 2, and is therefore not definable.

For the moment, let us assume that step 2 succeeds in the way that one might naively think it would. Then we have defined a sequence $x_0,x_1,\ldots$ containing all definable real numbers. Therefore Cantor's diagonal argument in step 3 defines, from this sequence, a real number $y$ that is not in the sequence. So $y$ is both definable and not definable, and we obtain a contradiction outright! Clearly, something is wrong (and it turns out to be in step 2.)

Answer 79

"Compact implies sequentially compact."

Answer 80

In descriptive set theory, we study properties of Polish spaces, typically not considered as topological spaces but rather we equip them with their "Borel structure", i.e., the collection of their Borel sets. Any two uncountable standard Borel Polish spaces are isomorphic, and the isomorphism map can be taken to be Borel. In practice, this means that for most properties we study it is irrelevant what specific Polish space we use as underlying "ambient space", it may be ${\mathbb R}$, or ${\mathbb N}^{\mathbb N}$, or ${\mathcal l}^2$, etc, and we tend to think of all of them as "the reals".

In Lebesgue Sur les fonctions representables analytiquement, J. de math. pures et appl. (1905), Lebesgue makes the mistake of thinking that projections of Borel subsets of the plane ${\mathbb R}^2$ are Borel. In a sense, this mistake created descriptive set theory.

Now we know, for example, that in ${\mathbb N}^{\mathbb N}$, projections of closed sets need not be Borel. Since we usually call reals the members of ${\mathbb N}^{\mathbb N}$,

it is not uncommon to think that projections of closed subsets of ${\mathbb R}^2$ are not necessarily Borel.

This is false. Note that closed sets are countable union of compact sets, so their projections are $F_\sigma$. The actual results in ${\mathbb R}$ are as follows: Recall that the analytic sets are (the empty set and) the sets that are images of Borel subsets of $\mathbb R$ by Borel measurable functions $f:\mathbb R\to\mathbb R$.

• A set is Borel iff it and its complement are analytic.

• A set is analytic iff it is the projection of the complement of the projection of a closed subset of ${\mathbb R}^3$.

• A set is analytic iff it is the projection of a $G_\delta$ subset of $\mathbb R^2$.

• There is a continuous $g:\mathbb R\to\mathbb R$ such that a set is analytic iff it is $g(A)$ for some $G_\delta$ set $A$.

• A set if analytic iff it is $f(\mathbb R\setminus\mathbb Q)$ for some continuous $f:\mathbb R\setminus\mathbb Q\to\mathbb R$. (Note that if $f$ is actually continuous on $\mathbb R$, then $f(\mathbb R\setminus\mathbb Q)$ is Borel.)

(See also here.)

Answer 81

Regard a reasonably nice surface in $\mathbb R^3$ that can locally be expressed by each of the functions $x(y,z)$, $y(x,z)$ and $z(x,y)$, then obviously

$\frac {dy} {dx} \cdot \frac {dz} {dy} \cdot \frac {dx} {dz} = 1$

(provided everything exists and is evaluated at the same point).

After all, this kind of reasoning works in $\mathbb R^2$ when calculating the derivative of the inverse function, it works for the chain rule and it works for separation of variables.

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Note that this product is in fact $-1$ which can either be seen by just thinking about what happens to the equation $ax+by+cz=d$ of a plane / tangent plane or by looking at the expression coming out of the implicit function theorem.

I recall someone claiming that this example proves that $dx$ should be regarded as linear function rather than infinitesimal, but I cannot reconstruct the argument at the moment as this discussion was 15 years ago.

In particular, it is true under appropriate conditions in $\mathbb R^4$ that $\frac {\partial y} {\partial x} \cdot \frac {\partial z} {\partial y} \cdot \frac {\partial w} {\partial z} \cdot \frac {\partial x} {\partial w} = 1$

Answer 82

Just today I came across a mathematician who was under the impression that $\aleph_1$ is defined to be $2^{\aleph_0}$, and therefore that the continuum hypothesis says there is no cardinal between $\aleph_0$ and $\aleph_1$.

In fact, Cantor proved there are no cardinals between $\aleph_0$ and $\aleph_1$. The continuum hypothesis says there are no cardinals between $\aleph_0$ and $2^{\aleph_0}$.

$2^{\aleph_0}$ is the cardinality of the set of all functions from a set of size $\aleph_0$ into a set of size $2$. Equivalently, it is the cardinality of the set of all subsets of a set of size $\aleph_0$, and that is also the cardinality of the set of all real numbers.

$\aleph_1$, on the other hand, is the cardinality of the set of all countable ordinals. (And $\aleph_2$ is the cardinality of the set of all ordinals of cardinality $\le \aleph_1$, and so on, and $\aleph_\omega$ is the next cardinal of well-ordered sets after all $\aleph_n$ for $n$ a finite ordinal, and $\aleph_{\omega+1}$ is the cardinality of the set of all ordinals of cardinality $\le \aleph_\omega$, etc. These definitions go back to Cantor.

Answer 83

My example is $G_{1}$ isomorphic to $G_{2}$'s subgroup and $G_{2}$ isomorphic to $G_{1}$'s subgroup implies $G_{1}$ and $G_{2}$ are isomorphic...

Answer 84

A common false belief is that all Gödel sentences are true because they say of themselves they are unprovable. See Peter Milne's "On Goedel Sentences and What They Say", Philosophia Mathematica (III) 15 (2007), 193–226. doi:10.1093/philmat/nkm015.

Answer 85

A projection of a measurable set is measurable. Not only students believe this. I was asked once (the quote is not precise): "Why do you need this assumption of a measurability of projection? It follows from ..."

A polynomial which takes integer values in all integer points has integer coefficients.

Another one seems to be more specific, I just recalled it reading this example. A sub-$\sigma$-algebra of a countably generated $\sigma$-algebra is countably generated.

Answer 86

In the past I have found myself making this mistake (probably fueled by the fact that you can indeed extend bounded linear operators), and I think it is common in students with a not-deep-enough topology background:

"Let $T$ be a compact topological space, and $X\subset T$ a dense subset. Take $f:X\to\mathbb{C}$ continuous and bounded. Then $f$ can be extended by continuity to all of $T$ ".

The classical counterexample is $T=[0,1]$, $X=(0,1]$, $f(t)=\sin\frac1t$ . It helps to understand how unimaginable the Stone-Cech compactification is.

Answer 87

The quaternions $\{x+yi+zj+wk\mid x,y,z,w\in \mathbb{R}$} is a complex Banach algebra (with usual operations). Hence it is apparently a counterexample to the Gelfand-Mazur theorem

So, what is the error?

The error is the following:

However the quaternions, being a skew field extention of the field of complex numbers, is a vector space over the field of complex number and it is also a ring, but there is no compatibility between scalar multiplication and quaternion multiplication). So it is not a complex algebra. This shows that in the definition of a complex algebra $A$, the commutative condition $\lambda (ab)=(a)(\lambda b),\;\;\lambda \in \mathbb{C},\;\;a,b\in A$, is very essential.

Answer 88

$\mathbb{R}^2$ has a unique complex manifold structure; it's just $\mathbb{C}$ right?

Answer 89

I got 2 well earned downvotes for a false belief I claimed proudly, it is time to balance that by exposing it here:

Let $(P,\le)$ be any poset, and let $\le^*$ be an order on $P$ extending $\le$. Any Endomorphism of $\le^*$ also is an endomorphism of $\le$

($f:P\to P$ endomorphism of $\le$ meaning $x\le y \implies f(x)\le f(y)$).

Of course this is a particular case of a very general fallacy: by extending $\le$ into $\le^*$ one weakens both the conclusion and the premise of the implication, so that there is no general relation between orders that extend one another.

Answer 90

I and several of my friends made our living off exploiting the fact that a space doesn't really have just one universal cover.

It has one at each basepoint. If the space is decent and connected, then these are all isomorphic, but the isomorphism requires a choice of a path connecting the points. You can clearly see the point if you make a bundle of universal covers over a moebius strip.

Answer 91

A false belief that I held until very recently, was that for a surface embedded in a three-manifold, orientability was the same as two-sidedness. However, it turns out that if the ambient manifold is not orientable, you can embed a torus so that it is one-sided, and a Möbius strip or a Klein bottle so that it is two-sided.

This picture, from Weeks' wonderful The shape of space, shows a two-sided Klein bottle embedded in a product of a Klein bottle and a circle, by identifying sides as indicated by the arrows.

This one shows a torus embedded single-sidedly in the same 3-manifold.

Answer 92

If $E$ is a contractible space on which the (Edit: topological) group $G$ acts freely, then $E/G$ is a classifying space for $G$.

A better, but still false, version:

If $E$ is a free, contractible $G$-space and the quotient map $E\to E/G$ admits local slices, then $E/G$ is a classifying space for $G$.

(Here "admits local slices" means that there's a covering of $E/G$ by open sets $U_i$ such that there exist continuous sections $U_i \to E$ of the quotient map.)

The simplest counterexample is: let $G^i$ denote $G$ with the indiscrete topology (Edit: and assume $G$ itself is not indiscrete). Then G acts on $G^i$ by translation and $G^i$ is contractible (for the same reason: any map into an indiscrete space is continuous). Since $G^i/G$ is a point, there's a (global) section, but it cannot be a classifying space for $G$ (unless $G=\{1\}$). The way to correct things is to require that the translation map $E\times_{E/G} E \to G$, sending a pair $(e_1, e_2)$ to the unique $g\in G$ satisfying $ge_1 = e_2$, is actually continuous.

Of course the heart of the matter here is the corresponding false belief(s) regarding when the quotient map by a group action is a principal bundle.

Answer 93

In his answer above, Martin Brandenburg cited the false belief that every short exact sequence of the form

$$0\rightarrow A\rightarrow A\oplus B\rightarrow B\rightarrow 0$$

must split.

I expect that a far more widespread false belief is that such a sequence can fail to split, when A, B and C are finitely generated modules over a commutative noetherian ring.

(Sketch of relevant proof: We need to show that the identity map in $Hom(A,A)$ lifts to $Hom(A\oplus B,A)$. Thus we need to show exactness on the right of the sequence

$$0\rightarrow Hom(B,A)\rightarrow Hom(A\oplus B,A)\rightarrow Hom(A,A)\rightarrow 0$$

For this, it suffices to localize and then complete at an arbitrary prime $P$. But completion at $P$ is a limit of tensorings with $R/P^n$, so to check exactness we can replace the right-hand $A$ in each Hom-group with $A/P^nA$. Now we are reduced to looking at modules of finite length, and the sequence is forced to be exact because the lengths of the left and right terms add up to the length in the middle. This is due, I think, to Miyata.)

Answer 94

It always confused me as an undergraduate that $\mathbb{Q}\subset\mathbb{R}$ has an open neighborhood $N_\epsilon \supset\mathbb{Q}$ of arbitrarily small measure $\epsilon$, because $\mathbb{Q}$ is dense in $\mathbb{R}$.

Answer 95

To my knowledge, noone has proven that the scheme of pairs of matrices (A,B) satisfying the equations AB=BA is reduced. But whenever I mention this to people someone says "Surely that's known to be reduced!"

(Similar-sounding problem: consider matrices M with $M^2=0$. They must be nilpotent, hence have all eigenvalues zero, hence $Tr(M)=0$. But that linear equation can't be derived from the original homogeneous quadratic equations. Hence this scheme is not reduced.)

Answer 96

"Every 1-dimensional knot in $R^n, n\ge 4,$ is trivial." This is true for tame knots and false for wild knots. See here.

Answer 97

By definition, an asymptote is a line that a curve keeps getting closer to but never touches. The teaching of this false belief at an elementary level is standard and nearly universal. Everybody "knows" that it is true. A tee-shirt has a clever joke about it. In the course of describing the function $f(x) = \dfrac{5x}{36 + x^2}$, I mentioned about an hour ago before a class of about 10 students that its value at 0 is 0 and that it has a horizontal asymptote at 0. One of them accused me of contradicting myself. What of $y = \dfrac{\sin x}{x}$? And even with simple rational functions there are exceptions, although there the curve can touch or cross the asymptote only finitely many times. And $3 - \dfrac{1}{x}$ gets closer to 5 as $x$ grows, and never reaches 5, so by the widespread false belief there would be a horizontal asymptote at 5.

Answer 98

The cost of multiplying two $n$-digit numbers is of order $n^2$ (because each digit of the first number has to be multiplied with each digit of the second number).

---

A lot of information is found on https://en.wikipedia.org/wiki/Multiplication_algorithm .

The first faster (and easily understandable) algorithm was https://en.wikipedia.org/wiki/Karatsuba_algorithm with complexity $n^{log_2 3} \sim n^{1.585}$.

Basic idea: To multiply $x_1x_2$ and $y_1y_2$ where all letters refer to $n/2$-digit parts of $n$-digit numbers, calculate $x_1 \cdot y_1$, $x_2\cdot y_2$ and $(x_1+x_2)\cdot(y_1+y_2)$ and note that this is sufficient to calculate the result with three such products instead of four.

Answer 99

This one has bit me and some very good mathematicians I know.

Let $X,Y$ be Banach spaces, and let $E \subset X$ be a dense subspace. Suppose $T : E \to Y$ is a bounded linear operator. Then $T$ has a unique bounded extension $\tilde{T} : X \to Y$. (True, this is the well-known and elementary "BLT theorem".)

If $T$ is injective then so is $\tilde{T}$. (False! See this answer for a counterexample.)

Answer 100

There are cases that people know that a certain naive mathematical thought is incorrect but largely overestimate the amount by which it is incorrect. I remember hearing on the radio somebody explaining: "We make five experiments where the probability for success in every experiment is 10%. Now, a naive person will think that the probability that at least one of the experiment succeed is five times ten, 50%. But this is incorrect! the probability for success is not much larger than the 10% we started with."

Of course, the truth is much closer to 50% than to 10%.

(Let me also mention that there are various common false beliefs about mathematical terms: NP stands for "not polynomial" [in fact it stands for "Nondeterministic Polynomial" time]; the word "Killing" in Killing form is an adjective [in fact it is based on the name of the mathematician "Wilhelm Killing"] etc.)

Answer 101

This is (I think) a fairly common misconception about maths that arises in connection with quantum mechanics. Given a Hermitian operator A acting on a finite dimensional Hilbert space H, the eigenvectors of A span H. It's easy to think that the infinite dimensional case is "basically the same", or that any "nice" operator that physicists might want to consider has a spanning eigenspace. However, neither the position nor the momentum operator acting on $L^2(\mathbb{R})$ have any eigenvectors at all, and these are certainly important physical operators! Based on an admittedly fairly small sample size, it seems that it's not uncommon to simultaneously believe that Heisenberg's uncertainty relation holds and that the position and momentum operators possess eigenvectors.

Answer 102

A stunning, ignorance-based false belief I have witnessed while observing a class of a math education colleague is that there is no general formula for the n-th Fibonacci number. I wonder if this false belief comes from conflating the (difficult) lack of formulas for prime numbers with something that is just over the horizon of someone whose interests never stretch beyond high-school math.

Behind a number of the elementary false beliefs listed here there is a widespread tendency among people to give up too easily (maybe when having to read at least to page 2 in a book), or to nourish an ego that allows to conclude that something is impossible if they cannot do it themselves.

Answer 103

False belief: ${\cal P}(\omega)$ has only countable chains with respect to $\subseteq$.

It seems mind-boggling to me that you can start with $\emptyset$, and "add stuff" uncountably many times until you reach $\mathbb{N}$ itself! I learnt this today in a comment by Andreas Blass that he wrote referring to this question.

Answer 104

I confess that I didn't carefully comb through all the answers, although I've read through this thread a few times in the past. So maybe these are repeats.

• "The category of compact Hausdorff spaces is complete but not cocomplete; for example, it doesn't have all coproducts."

• "The category of torsion abelian groups is cocomplete but not complete; for example, it doesn't have all products."

One of my professors in graduate school (quite a well-known and strong mathematician actually) insisted on the first, and quite a few people here at MO have mistakenly believed the second before the error was pointed out.

The moral of the story: sometimes categorical limits/colimits aren't computed the way you might first think of, e.g., colimits of compact Hausdorff spaces aren't always computed as colimits in $\mathrm{Top}$, and limits of torsion abelian groups aren't always computed as limits in $\mathrm{Ab}$.

Answer 105

Let $V$ be a vector space. Then the intersection of $n$ hyperplanes (i.e. subspaces of codimension 1) is a subspace of codimension at most $n$.

So, naturally, the intersection of countably many hyperplanes is a subspace of countable codimension. Hence if $V$ is of uncountable dimension, this intersection is non-trivial.

Except this is of course wrong. For example, consider the vector space $V:=\mathbb{K}^{\mathbb{Z}}:=\{ f:\mathbb{Z}\to\mathbb{K} \}$ of uncountable dimension. The kernels of the projections $\pi_i:V\to\mathbb{K},\ f\mapsto f(i)$ are hyperplanes. Their intersection is the trivial subspace of $V$, and thus has uncountable codimension.

Answer 106

By googling one sees that each of the following statements has a significant number of believers:

(1) the vector space {0} has no basis,

(2) the empty set is a basis of {0} by convention,

(3) the statements "{0} has no basis" and "the empty set is a basis of {0}" are equivalent,

(4) the statements "{0} has no basis" and "the empty set is a basis of {0}" are NOT equivalent,

(5) the statement "the empty set is a basis of {0}" is an immediate consequence of the definitions of the terms involved.

I think that we'll all agree that the 5 beliefs are not ALL true. My personal religion is to believe in (4) and (5). I don't think I'll ever understand the arguments in favor of (1), (2) or (3).

Answer 107

Here is a false belief I've been carrying in my head for 5 years and only woken up from recently. When one defines symmetric and anti-symmetric tensors on a vector space $V$, one usually starts with an action of the symmetric group on tensors, which is usually defined as:

$$ S_n \curvearrowright V^{\otimes n}, \quad \sigma (v_1 \otimes \ldots \otimes v_n) = v_{\sigma(1)} \otimes \ldots \otimes v_{\sigma(n)}. $$

Now, if $\tau \in S_n$ is another permutation, we have

$$ \tau(\sigma (v_1 \otimes \ldots \otimes v_n)) = \tau(v_{\sigma(1)} \otimes \ldots \otimes v_{\sigma(n)}) = v_{\tau(\sigma(1))} \otimes \ldots \otimes v_{\tau(\sigma(n))}, $$

right? Wrong! The action above is a right action, so

$$ \tau(v_{\sigma(1)} \otimes \ldots \otimes v_{\sigma(n)}) = v_{\sigma(\tau(1))} \otimes \ldots \otimes v_{\sigma(\tau(n))} = (\sigma \cdot \tau)(v_1 \otimes \ldots \otimes v_n). $$

The reason for that is that the defining formula for the action doesn't directly tell you where $v_i$ goes. Rather, it shows who comes to the $i$-th place.

Answer 108

I'm pretty sure I've heard both of the following multiple times:

1. Transfinite induction requires the axiom of choice. False, though many applications of transfinite induction require axiom of choice (either in the form of the well-ordering theorem, or directly (though using transfinite induction together with choice directly is essentially the same as just using Zorn's Lemma)).

2. Transfinite induction requires the axiom of foundation. I guess some people get transfinite induction mixed up with epsilon-induction?

Answer 109

There is a bijection between the set of [true: prime!] ideals of $S^{-1}R$ and the set of [true: prime!] ideals of $R$ which do not intersect $S$.

Answer 110

• Many students have the false belief that if a topological space is totally disconnected, then it must be discrete (related to examples already given). The rationals are a simple counter-example of course.

• It is common to imagine rotation in an n-dimensional space, as a rotation through an "axis". this is of course true only in 3D, In higher dimensions there is no "axis".

• In calculus, I had some troubles with the following wrong idea. A curve in a plane parametrized by a smooth function is "smooth" in the intuitive sense (having no corners). the curve that is defined by $(t^2,t^2)$ for $t\ge0$ and $(-t^2,t^2)$ for $t<0$ is the graph of the absolute value function with a "corner" at the origin, though the coordinate functions are smooth. the "non-regularity" of the parametrization resolves the conflict.

• When first encountering the concept of a spectrum of a ring, the belief that a continuous function between the spectra of two rings must come from a ring homomorphism between the rings.

Answer 111

In geometric combinatorics, there is a widespread belief that polytopes of equal volume are not scissor congruent (as in Hilbert's third problem) only because their dihedral angles are incomparable. The standard example is a cube and a regular tetrahedron, where dihedral angles are in $\Bbb Q\cdot \pi$ for the cube, and $\notin \Bbb Q\cdot \pi\ $ for the regular tetrahedron. In fact, things are rather more complicated, and having similar dihedral angles doesn't always help. For example, the regular tetrahedron is never scissor congruent to a union of several smaller regular tetrahedra (even though the dihedral angles are obviously identical). This is a very special case of a general result due to Sydler (1944).

Answer 112

Sequence $\{a_n\}$ has a limit $A$ in $\mathbb{R}$ and a limit $B$ in $\mathbb{Q}_p$. Then $A$ is rational iff $B$ is rational.

Answer 113

Here's a little factoid: (The Mean-value theorem for functions taking values in $\mathbb{R} ^n$.) If $\alpha : [a,b]\rightarrow \mathbb{R}^n$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists a $c\in (a,b)$ such that $\frac{\alpha (b)-\alpha (a)}{b-a}=\alpha '(c)$

A counterexample is the helix $(\cos (t),\sin (t), t)$ with $a=0$, $b=2\pi$.

Another common misunderstanding (although not mathematical) is about the meaning of the word factoid. In fact, the common mistaken definition of the word factoid is factoidal.

Answer 114

The following false belief enjoyed a certain success in the '70. (See R.S.Palais, Critical point theory and the minimax principle for an account.)

A second countable, Hausdorff, Banach manifold is paracompact.

Regular is necessary, otherwise there are counterexamples!

Answer 115

Piggybacking on one of Pierre's answers, I once had to teach beginning linear algebra from a textbook wherein the authors at one point stated words to the effect that the the trivial vector space {0} has no basis, or that the notion of basis for the trivial vector space makes no sense. It is bad enough as a student to generate one's own false beliefs without having textbooks presenting falsehoods as facts.

My personal belief is that the authors of this text actually know better, but they don't believe that their students can handle the truth, or perhaps that it is too much work or too time-consuming on the part of the instructor to explain such points. Whatever their motivation was, I cannot countenance such rationalizations. I told the students that the textbook was just plain wrong.

Answer 116

The distinction between convergence and uniform convergence. It even got Cauchy in its time.

Answer 117

It took me a bit too long to realize that these two beliefs are contradictory:

• Period 3 $\Rightarrow$ chaos: if a continuous self-map on the interval has a period-3 orbit, then it has orbits of all periods.
• The black dots on each horizontal slice of this picture above $x=a$ show the location of the periodic points of the logistic map $f_a(y) = ay(1-y)$:

You can clearly see a 3-cycle in the light area towards the right; yet we know that if there is a 3-cycle in that slice then there must be a cycle of any period in that slice... so where are they?

(The other cycles are there of course, but they are repelling and hence are not visible. You can see artifacts from these repelling cycles near the period-doubling bifurcations in this picture)

Answer 118

"The universal cover of $SL_2(R)$ is a universal central extension" (which I believed until recently...)

Answer 119

" Every open dense subset of $\mathbb{R}^n$ has full Lebesgue measure. "

Answer 120

Let $(X,\tau)$ be a topological space. The false belief is: "Every sequence $(x_n)$ in $X$ with an accumulation point $a\in X$ has a subsequence that converges to $a$". I subscribed to this intuitively until I stumbled over a counterexample, see https://dominiczypen.wordpress.com/2014/10/13/accumulation-without-converging-subsequence/

Answer 121

I had in mind that a $0$-sphere is only one point, but it is false, it is a collection of two points: $$\mathbb{S}^0 = \{ x \in \mathbb{R} \ \ | \ \ \|x\|=1 \} = \{-1,1\}$$

Answer 122

Here's one from basic set theory. Let k be a cardinal and consider the operation "adding k", meaning

$l \mapsto k+l$

on cardinals. We know that this operation "stabilizes" to the identity after $k$, that is, for any $l>k$, we have $l+k = l$. Similarly, the "multiplying by $k$" operation,

$l \mapsto l * k$

stabilizes to the identity after $k$.

Everyone also knows that if $l$ is an infinite cardinal then $l^2$ is equipotent to $l$, and more generally $l^n$ is equipotent to $l$ for every natural number $n$. I.e. all the finite power functions stabilize to the identity at $\omega$.

Well, obviously "exponentiation by $\omega$" also stabilizes at some point, right? Like, $l^\omega$ is equal to $l$ for sufficiently large $l$? Look, we probably already have the stabilization point at $2^\omega$.

Right?

Answer 123

Inversion is an automorphism of a group. ('Cause it, like, preserves the conjugacy classes and all that...)

Answer 124

The following seems not to be here yet.

Misconception.

$R[[x_1,x_2,x_3,\dotsc]]/(x_2,x_3,\dotsc)$ isomorphic to $R[[x_1]]$ ${}\hspace{118pt}$ (f)

Source of the misconception. A fallacy of type false generalization: for any $n\in\mathbb{N}$ it is true that

$R[[x_1,x_2,x_3,\dotsc,x_n]]/(x_2,x_3,\dotsc,x_n)\cong R[[x_1]]$ ${}\hspace{125pt}$ (t)

but to conclude from this that (f) was true by passing to the limit $n\to\infty$ is fallacious.

Reason for why the misconception is false. E.g. the formal power series $f:=x_2+x_3+\dotsm$ is an element of $R[[x_1,x_2,x_3,...]]$, but by the standard definition of $I:=(x_2,x_3,\dotsc)$, which after all means nothing more than the $R[[x_1,x_2,x_3,\dotsc]]$-module generated by the infinite set $\{x_i\colon i\in \omega,\ i\geq 2\}$, the ideal $I$ does not contain $f$. (Having coefficients from the huge power series ring $R[[x_1,x_2,x_3,\dotsc]]$ does not help.)

Reason for including the example. I saw this misconception in a dissertation. For obvious reasons, I won't give the source.

Further remarks. In the above, $R$ can be any commutative unital ring, and $R[[x_1,x_2,x_3,\dotsc]]$ as usual means the projective limit in the category of commutative unital rings of the diagram $\dotsm\twoheadrightarrow R[[x_1,x_2,x_3]]\twoheadrightarrow R[[x_1,x_2]]\twoheadrightarrow R[[x_1]]$ consisting of the canonical projections.

Answer 125

I would like to turn the attention of mathematical community to a false beliefs related to the direct limit topologies.

Many years ago in the theory of topological groups there was a false belief that for every space $X$ the free topological group carries the topology of direct limit of the sequence $F_n(X)$ of words of length $\le n$. This illusion was broken up by Fay, Ordman and Thomas who showed that even for the space of rational numbers the free topological group $F(\mathbb Q)$ is not a $k$-space.

The problems with direct limit topologies is that for the direct limit $X=lim X_n$ of an increasing sequence $(X_n)$ of topological spaces the topology on $X\times X$ does not coincide with the direct limit topology of the sequence $ (X_n\times X_n)$.

Now specialists in General Topology and Topological Algebra are conscious of pathological behaviour of direct limit topologies and are careful with this delicate topic.

On the other hand, I was quite surprised lerning that in Algebraic Geometry this misbelief still is alive. For example, in this paper posted to arxiv (maybe it is already published) in the very introduction (on page 3) it is written that for any topological space $X$ the Ran space (of all non-empty finite subsets of $X$, endowed with the topology of direct limit of the sequence $R_n(X)$ of sets of cardinality $\le n$ in $X$) is a topological semilattice. But this is not true in general, see Proposition 4 here.

So, some false beliefs that have died in some areas of mathematics can be still alive in others. By the way, this situation also explains why mathematicians should not neglect general topology.

Answer 126

A common belief of students in real analysis is that if $$ \lim_{x\to x_0}f(x,y_0),\qquad\lim_{y\to y_0}f(x_0,y) $$ exist and are both equal to $l$, then the function has limit $l$ in $(x_0,y_0)$. It is easly to show counter-examples. More difficult is to show that also the belief $$ \lim_{t\to 0}f(x_0+ht,y_0+kt)=l,\quad\forall\;(h,k)\neq(0,0)\quad\Rightarrow\quad\lim_{(x,y)\to(x_0,y_0)}f(x,y)=l $$ is false. For completeness's sake (presumably anybody who ever taught calculus has seen it, but it's easily forgotten) the standard counterexample is $$ f(x,y)=\frac{xy^2}{x^2+y^4} $$ at $(0,0$).

Answer 127

In a finite abelian $p$-group, every cyclic subgroup is contained in a cyclic direct summand.

Added for Gowers: Maybe one reason why people fall into this error goes something like this: First you learn linear algebra, so you know about vector spaces, bases for same, splittings of same. Then you run into elementary abelian $p$-groups and recognize this as a special case of vector spaces. Then you learn the pleasant fact that all finite abelian $p$-groups are direct sums of cyclic $p$-groups, and a corresponding uniqueness statement. You notice that all of the cyclic subgroups of order $p^2$ in $\mathbb Z/p^2\times \mathbb Z/p$ are summands, and if you have a certain sort of inquiring mind then you also notice that not every subgroup of order $p$ is a summand: one of them is contained in a copy of $\mathbb Z/p^2$, in fact in all of those copies of it. Having learned so much, both positive and negative, from the example of $\mathbb Z/p^2\times \mathbb Z/p$, you may think that it shows all the interesting basic features of the general case and overlook the fact that in $\mathbb Z/p^3\times \mathbb Z/p$ there is a $\mathbb Z/p^2$ not contained in any $\mathbb Z/p^3$.

In any case, reputable people sometimes make this blunder; it happened to somebody here at MO just the other day.

Answer 128

Before reading about it, I really thought that if $f \colon [0,1] \times [0,1] \to [0,1]$ is a function with the following properties:

1. for any $x \in [0,1]$ the function $f_x\colon [0,1] \to [0,1]$ defined by $f_x(y)=f(x,y)$ is Lebesgue measurable, and also the function $f^y \colon [0,1]\to[0,1]$ defined by $f^y(x)=f(x,y)$ is Lebesgue measurable, for all $y \in [0,1]$;
2. both $\varphi(x)=\int_0^1 f_x d\mu$ and $\psi(y)=\int_0^1 f_y d\mu$ are Lebesgue measurable.

Then the two iterated integrals $$ \int_0^1\varphi(x)dx \mbox{ and } \int_0^1\psi(y)dy $$ should be equal. This is false (see Rudin's "Real and Complex Analysis", pag. 167), at least if you assume the continuum hypothesis.

Answer 129

Let $X$ be a "nice" path-connected topological space, say a connected manifold or CW-complex.

False belief: "A universal covering $Y\to X$ of $X$ is unique up to unique isomorphism" and therefore can be called "the" universal covering.

The isomorphism far from unique in general (there are as many as elements in "the" fundamental group). However uniqueness (and the universal property) holds in the category of coverings of pointed topological spaces. (In particular, for topological groups there's a canonical choice.)

Browsing I found several textbooks teaching the above "false belief" (I saw several too that are careful with this issue).

Answer 130

If $X$ is uncountable, then $X^{\mathbb{N}}$ is in bijection with $X$.

König's theorem implies that $|X^{\mathbb{N}}|>|X|$ whenever the cardinality of $X$ has countable cofinality.

Answer 131

The following inexact belief can be spotted in many a textbook for undergraduates: the principle of mathematical induction and the well-ordering principle for $\mathbb{N}$ are equivalent.

Lars-Daniel Öhman wrote about this misbelief in his paper Are induction and well-ordering equivalent? (Math. Intelligencer, vol. 41 (2019), no. 3, pp. 33-40.). To make a long story short, one of the (main) points by Öhman in the said article is that if we define the natural numbers à la Peano, i.e. as a set $N$ endowed with a function $S \colon N \to N$ satisfying the following axioms:

1. $0\in N$,
2. $\forall n \in N \, (S(n) \neq 0)$,
3. $S$ is injective, and
4. (P.M.I.) if $M$ is a subset of $N$ such that $0 \in M$ and $S(m) \in M$ for every $m \in M$, then $M=N$.;

then, the P.M.I. is not equivalent to the well-ordering principle (every nonempty subset of $\mathbb{N}$ has a least element; heretofore, W.O.P.) relative to axioms 1 - 3.

One route that Öhman follows to evince that there are issues with the so-called equivalence of both principles is by illustrating that, in the popular proof of the implication W.O.P. $\Rightarrow$ P.M.I., the existence of an immediate predecessor for every $n \in \mathbb{N}$ is assumed: this is an assumption that can not be obtained as a consequence of 1, 2, 3, and W.B.O., "as evidenced by the existence of a model... in which this property [about immediate predecessors] does not hold [whereas 1, 2, 3, and W.B.O. do]"... In point of fact, the model he provides to exemplify the validity of the assertion between quotation marks is one in which the P.M.I. does not hold either.

Answer 132

False statement: If $A$ and $B$ are subsets of $\mathbb{R}^d$, then their Hausdorff dimension $\dim_H$ satisfies

$$\dim_H(A \times B) = \dim_H(A) + \dim_H(B). $$

EDIT: To answer Benoit's question, I do not know about a simple counterexample for $d = 1$, but here is the usual one (taken from Falconer's "The Geometry of Fractal Sets"):

Let $(m_i)$ be a sequence of rapidly increasing integers (say $m_{i+1} > m_i^i$). Let $A \subset [0,1]$ denote the numbers with a zero in the $r^{th}$ decimal place if $m_j + 1 \leq r \leq m_{j+1}$ and $j$ is odd. Let $B \subset [0,1]$ denote the numbers with a zero in the $r^{th}$ decimal place if $m_{j} + 1 \leq r \leq m_{j+1}$ and $j$ is even. Then $\dim_H(A) = \dim_B(A) = 0$. To see this, you can cover $A$, for example, by $10^k$ covers of length $10^{- m_{2j}}$, where $k = (m_1 - m_0) + (m_3 - m_2) + \dots + (m_{2j - 1} - m_{2j - 2})$.

Furthermore, if $\mathcal{H}^1$ denotes the Hausdorff $1$-dimensional (metric) outer measure of $E$, then the result follows by showing $\mathcal{H}^1(A \times B) > 0$. This is accomplished by considering $u \in [0,1]$ and writing $u = x + y$, where $x \in A$ and $y \in B$. Let $proj$ denote orthogonal projection from the plane to $L$, the line $y = x$. Then $proj(x,y)$ is the point of $L$ with distance $2^{-1/2}(x+y)$ from the origin. Thus, $proj( A \times B)$ is a subinterval of $L$ of length $2^{-1/2}$. Finally, it follows:

$$ \mathcal{H}^1(A \times B) \geq \mathcal{H}^1(proj(A \times B)) = 2^{-1/2} > 0. $$

Answer 133

(*) "Let $(I,\leq)$ be a directed ordered set, and $E=(f_{ij}:E_i\to E_j)_{i\geq j}$ be an inverse system of nonempty sets with surjective transition maps. Then the inverse limit $\varprojlim_I\,E$ is nonempty."

This is true if $I=\mathbb{N}$ ("dependent choices"), and hence more generally if $I$ has a countable cofinal subset. But surprisingly (to me), those are the only sets $I$ for which (*) holds for every system $E$. (This is proved somewhere in Bourbaki's exercises, for instance).

Of course, other useful cases where (*) holds are when the $E_i$'s are finite, or more generally compact spaces with continuous transition maps.

Answer 134

A random $k$-coloring of the vertices of a graph $G$ is more likely to be proper than a random $(k-1)$-coloring of the same graph.

(A vertex coloring is proper if no two adjacent vertices are colored identically. In this case, random means uniform among all colorings, or equivalently, that each vertex is i.i.d. colored uniformly from the space of colors.)

Answer 135

A common trap which sometimes I see people fall is that a Hermitian matrix $M$ is negative definite if and only if its leading principal minors are negative.

What is true is the Sylvester's criterion, which says that $M$ is positive definite if and only if its principal minors are positive. Thus, the true statement is that $M$ is negative definite if and only if the principal minors of $-M$ are positive.

Answer 136

I guess you don't want commonly held beliefs of students that for every real number there is a next real number, or that convergent sequences are eventually constant. A version I saw in a book asked whether points on a line "touch." Understanding the topology of a line is a challenge for many students, although presumably not for most mathematicians.

Here is a more esoteric belief that I have even seen in some books:

"The Banach-Tarski Paradox says that a ball the size of a pea can be cut into 5 pieces and reassembled to make a ball the size of the sun."

As a consequence of the Banach-Tarski paradox, a ball the size of a pea can be partitioned (not really "cut") into a finite number of pieces which can be reassembled into a ball the size of the sun, but a simple outer measure argument implies that the number of pieces must be very large (I roughly estimate at least $10^{30}$). The number 5 probably comes from the fact that the basic Banach-Tarski paradox is that a ball of radius 1 can be partitioned into 5 pieces which can be reassembled into two disjoint balls of radius 1. (It can almost, but not quite, be done with four pieces; one of the five pieces can be taken to be a single point.)

Answer 137

I don't know how common this mistake is, but I think it's worth mentioning. I used to think that existence of non-measurable sets is guaranteed by the axiom of choice only.

In the presence of AC, there cannot be a $\sigma$-additive measure on $\mathcal{P}(\mathbb{R})$ that extends the usual Lebesgue measure.

It is true that we cannot extend the Lebesgue measure in a translation-invariant way by various Vitali set constructions. On the other hand, if you do not insist that the extension is translation-invariant, it might be possible to do this relative to a real-valued measurable cardinal assumption.

Theorem (Ulam): If there exists a cardinal $\kappa$ such that there exists an atomless $\kappa$-additive probability measure on $\mathcal{P}(\kappa)$, then there exists a $\sigma$-additive measure on $\mathcal{P}(\mathbb{R})$ extending the Lebesgue measure.

Answer 138

A Banach space $X$ is reflexive if it is isomorphic to its double dual ${X^*}^*$.

(Couldn't find this is the list…)

Answer 139

"A 'random' number field has large class number"

I've heard this belief quite a few times. Usually random means taking a not-too-small degree (7?) and then somehow taking integer coefficients (around 10,000?).

But in fact class number tend to be much smaller than one expects. Usually they are logarithmic in the size of the discriminant.

The main reasons for the belief are the common examples of fields given in undergraduate and early graduate courses - imaginary quadratic fields and cyclotomic fields. In more advanced courses students see abelian extensions and CM-fields, which also have special arithmetic properties that make their class groups somewhat larger. In the courses I have taken the actual size of 'random' number fields was not addressed, and, say, the Cohen-Lenstra heuristics were not mentioned.

Answer 140

False belief. If a family $(x_n)_{n\geq 1}$ is commutatively convergent (i.e. summable) in a normed space $(V,\|\ \|)$ then $$ \sum_{n\geq 1} \|x_n\|<+\infty\ . $$

This is true in finite dimensions and has counterexamples in infinite dimensions. Details and counterexamples can be found there.

Recall that a family $(x_i)_{i\in I}$ is called summable with sum $S$ iff $$ (\forall \epsilon>0)(\exists F\subset_{finite} I)(\forall F_1\subset_{finite} I)(F\subset F_1\Longrightarrow\\ \|S-\sum_{i\in F_1}x_i\|<\epsilon) $$ This is equivalent with commutative convergence in case $I\subset \mathbb{N}$ is infinite.

Answer 141

True: The solution operator of the linear one-dimensional time-dependent ordinary differential equation (ODE) $x' = a(t) x$ is $$ \exp\Bigl(\int\limits_{t_0}^{t} a(s) \, ds\Bigr). $$ True: The solution operator of the linear multi-dimensional time-independent ODE $x' = A x$ is $$ \exp\,(A(t - t_0)). $$ A quite popular misconception, even among research mathematicians:

The solution operator $\Phi(t;t_0)$ of the linear multi-dimensional time-dependent ODE $x' = A(t) x$ is $$ \exp\Bigl(\int\limits_{t_0}^{t} A(s) \, ds \Bigr), $$

perhaps strengthened by Liouville's formula:

$$ \det{\Phi(t;t_0)} = \exp\Bigl(\int\limits_{t_0}^{t} \operatorname{tr}{A(s)} \, ds\Bigr). $$

Answer 142

Here's a mistake I've seen from students taking a first course in linear analysis. For a vector $g$ in a Hilbert space $H$, it is true that $\langle f,g\rangle=0$ for every $f\in H$ implies $g=0$. This leads us to the mistaken:

“Let $(g_n)$ be a sequence in $H$. If, for every $f\in H$, $\langle f,g_n\rangle\to0$, then $g_n\to 0$.”

Answer 143

I don't know how common this is, but I've noticed it half an hour ago in some notes I had written: If $J$ is a finitely generated right ideal of a not necessarily commutative ring $R$, and $n$ is natural, then $J^n$ is finitely generated, isn't it?

No, it isn't. For an example, try $R=\mathbb Z\left\langle X_1,X_2,X_3,...\right\rangle $ (ring of noncommutative polynomials) and $J=X_1R$.

Answer 144

Here are mistakes I find surprisingly sharp people make about the weak$^{*}$ topology on the dual of $X,$ where $X$ is a Banach space.

-It is metrizable if $X$ is separable.

-It is locally compact by Banach-Alaoglu.

-The statement $X$ is weak$^{*}$ dense in the double dual of $X$ proves that the unit ball of $X$ is weak$^{*}$ dense in the unit ball of the double dual of $X.$

The first two are in fact never true if $X$ is infinite dimensional. While both statements in the third claim are true, the second one is significantly stronger, but a lot of people believe you can get it from the first by just "rescaling the elements" to have norm $\leq 1.$ (Although the proof of the statements in the third claim is not hard). The difficulty is that if $X$ is infinite dimensional then for any $\phi$ in the dual of $X,$ there exists a net $\phi_{i}$ in the dual of $X$ with $\|\phi_{i}\|\to \infty$ and $\phi_{i}\to \phi$ weak$^{*},$ so this rescaling trick cannot be uniformly applied. Really these all boil down to the following false belief:

-The dual of $X$ has a non-empty norm bounded weak$^{*}$ open set.

Again when $X$ is infinite dimensional this always fails.

Answer 145

Common false belief: a space that is locally homeomorphic to $\mathbb{R}^n$ must be Hausdorff. More generally, many people forget that the usual definition of a manifold contains the Hausdorff and paracompact conditions.

There are of course examples that show that forgetting this assumption leads to unexpected result, and they are in fact much wilder than I knew a few weeks ago. Notably, among examples of (Hausdorff) non-paracompact "manifolds" are the well-known long line, but also the Prüfer manifold constructed from a closed half-plane by attaching to it a half plane at each boundary point.

Added: Let me give a particular case of this false belief to illustrate what kind of weird things can happen that most people would not realize when they are sloppy with the paracompact hypothesis: there exists a path-connected, locally contractible, simply-connected space that admits non-trivial locally trivial bundles with fiber $[0,1]$. Indeed, the first octant in the product of two long line is not homeomorphic to a product a long ray with an interval, but has a natural bundle structure over a long ray.

Answer 146

Teaching introduction to analysis, I had students using the "fact" that if $f: [a,b] \rightarrow \mathbb{R}$ is continuous, then $[a,b]$ can be divided to subintervals $[a,c_1],[c_1,c_2],...,[c_n,b]$ such that $f$ is monotone on every subinterval. For instance you can use this "fact" to "prove" the (true) fact that $f$ must be bounded on $[a,b]$. Also, some students used the same "fact", but with countably many subintervals. I found this mistake hard to explain to students, because constructing a counterexample (such as the Weierstrass function) is impossible at the knowledge level of an introduction course.

Answer 147

Many students believe that every abelian subgroup is a normal subgroup.

Answer 148

If $H$ and $K$ are subgroups of $G$, then $HK$ is a subgroup of $G$.

Answer 149

False belief: Any orthonormal basis of a subvectorspace $W\subset V$ of an inner product space $V$ can always be extended to an ONB of $V$.

Counterexample: Let $V$ be $\bigoplus_{i\ge 1} \mathbb{R}$ with the inner product given by $\langle a_*,b_*\rangle =\sum_{i\ge 1} a_ib_i$ and let $W$ be the subvectorspace of $V$ spanned by $e_1+e_i$ for $i\ge 2$. The given set is basis and we can apply Gram-Schmidt to obtain an ONB.

However $W^\perp = 0$ so there is no way to complete it. Related false belief: $(W^\perp)^\perp=W$. These beliefs are all true in finite dimensions, but false in general.

Answer 150

An incredibly common false belief is:

For a (say smooth, projective) algebraic variety $X$ the $K_X$-negative part of the cone $NE(X)$ is locally polyhedral.

A right statement of the theorem of the cone is

$\overline{NE(X)} = \overline{NE(X)}_{K_X \geq 0} + \sum_{i} \mathbb{R}[C_i]$ for a denumerable set $\{ C_i \}$ of rational curves, which accumulate at most on the hyperplane $K_X = 0$.

At a first glance this seems to imply that $\overline{NE(X)}_{K_X < 0}$ is locally poyhedral, but this is not true. It depends on the shape of the intersection $\overline{NE(X)} \cap \{ K_X = 0 \}$.

For instance if this latter intersection is round, and there is only one curve $C_i$, the half-cone $\overline{NE(X)}_{K_X < 0}$ is actually a circular cone! Definitely not polyhedral in any sense. I believe this behaviour can happen even with varieties birational to abelian varieties.

The strange thing about this false belief is that it is held true by many competent mathematicians (and indeed I don't believe that many undergraduates meet the theorem of the cone!).

Answer 151

Some things from pseudo-Riemannian geometry are a bit hard to swallow for students who have had previous exposure to Riemannian geometry. Aside from the usual ones arising from sign issues (like, in a two dimensional Lorentzian manifold with positive scalar curvature, time-like geodesics will not have conjugate points), an example is that in Riemannian manifolds, connectedness + geodesic completeness implies geodesic connectedness (every two points is connected by a geodesic). This is not true for Lorentzian manifolds, and the usual example is the pseudo-sphere.

Answer 152

I just realized yesterday that, given $A \to C, B \to C$ in an abelian category, the kernel of $A \oplus B \to C$ is not the direct sum of the kernels of $A \to C, B \to C$.

Answer 153

"the quadratic variation of a Brownian motion between $0$ and $T$ is equal to $T$"

this is only true that if $\mathcal{D}^N$ is a nested sequence of partitions of $[0,T]$ (with mesh size going to $0$) then the quadratic variation of a Brownian motion along these partitions converges towards $T$, almost surely. If we define the quadratic variation of a continuous function $f$ as we would like to, $$Q(f,[0,T]) = \sup_{0=t_0<\ldots, t_n=T } \sum |f(t_k)-f(t_{k+1})|^2,$$ then the Brownian paths have almost surely infinite quadratic variation.

This was something I had never noticed until I read the wonderful book "Brownian motion" by Peter Morters and Yuval Peres.

Answer 154

Complex variables: "An entire function that is onto and locally one-to-one is globally one-to-one."

Counterexample: $f(z) := \int_0^z \exp(\zeta^2)\,d\zeta$

I'll leave the proof that this is indeed a counterexample as a pleasant exercise.

(I believe this example is due to Lawrence Zalcman.)

Answer 155

The fundamental group of the Klein bottle is $D_\infty$, the infinite dihedral group (which is $\mathbb Z \rtimes \mathbb Z_2$).

I believed this for some time, and I seem to recall some others having the same confusion.

The group that has been mistaken for $D_\infty$ is in fact $\mathbb Z \rtimes\mathbb Z$, which can also be written with the presentation $x^2y^2=1$. The former abelianizes to $\mathbb Z_2\oplus \mathbb Z_2$, the latter to $\mathbb Z\oplus \mathbb Z_2$.

A 2-dimensional Lie group is a product of circles and lines, in particular it is abelian.

I don't know if anyone else suffered this one. The mistake is (a) in forgetting that the classification of surfaces doesn't apply since homeomorphic Lie groups are not necessarily isomorphic (e.g., the (bijective, orientation preserving) affine transformations $x\mapsto ax+b$, where $a>0, b\in \mathbb R$ are homeomorphic to $\mathbb R^2$, though not isomorphic) and (b) that Lie groups aren't necessarily connected, in particular $\mathbb R^2$ cross any finite non-abelian group is non-abelian.

Answer 156

As is well known, if $V$ is a vector space and $S, T \subset V$ are subspaces, then $S \cup T$ is a subspace iff $S \subset T$ or viceversa. However, $S \cup T \cup U$ can be a subspace even if no two spaces are contained in each other (think finite fields...)

Answer 157

Duality reverses inclusions of vector spaces.

Answer 158

A degree $k$ map $S^n\to S^n$ induces multiplication by $k$ on all the homotopy groups $\pi_m(S^n)$.

(Not sure if this is a common error, but I believed it implicitly for a while and it confused me about some things. If you unravel what degree $k$ means and what multiplication by $k$ in $\pi_m$ means, there's no reason at all to expect this to be true, and indeed it is false in general. It is true in the stable range, since $S^n$ looks like $\Omega S^{n+1}$ in the stable range, "degree k" can be defined in terms of the H-space structure on $\Omega S^{n+1}$, and an Eckmann-Hilton argument applies.)

Answer 159

"A real symmetric matrix is positive-definite iff all the leading principal minors are positive, and positive-semidefinite iff all the leading principal minors are nonnegative."

This paper collects some evidence that this belief is "common", and presents a counterexample (of size $3\times 3$. Exercise: find an example of size $2\times 2$).

(Related to, but not the same as this answer.)

Answer 160

Some undergraduate common false beliefs that I found

(1) If $H$ is a subgroup of $\mathbb{Z}$ and $H$ and $\mathbb{Z}$ are isomorphic, then $H = \mathbb{Z}$;

(2) In a metric space every two open balls are homeomorphic;

(3) For $ \ p \in [1, \infty] \, $, $ \, L^p(X, \mathfrak{M}, \mu) = \big\{ f \in \mathbb{C}^X : \int_X |f|^p \, d \mu < \infty \big\} \, $ is a $\mathbb{C}$-normed vector space, with norm $ \ \lVert f \rVert_p = \big( \int_X |f|^p \, d \mu \big)^{1/p} \, $.

Belief (1) is very naive, for every nontrivial subgroup of $\mathbb{Z}$ is of the form $n \mathbb{Z}$, all of them isomorphic to $\mathbb{Z}$. For (2) people tend to think of normed vector spaces and forget the discrete metric spaces. For (3) some people just forget that one have to consider the quotient space, where the classes $[f]=[g] \ $ iff $ \ f=g \ $ $\mu$-almost everywhere.

Answer 161

When I was studying Banach spaces, I was confused with the following: We know that, in any Banach Space $V$, the closed unit ball is compact in the topology generated by the norm if, and only if, the dimension of $V$ is finite. But thinking about $\mathbb R$ as a vector space over $\mathbb Q$, we have an infinite-dimensional vector space which is complete in the norm (given by the modulus) but the closed unit ball is, of course, compact in topology generated by the norm.

I took some time to discover that my mistake was that I thought about $\mathbb R$ over $\mathbb Q$ as a Banach space. In fact, this vector space is a complete metric space (in the sense of Cauchy sequences), but I realized later that the word Banach space is reserved only for vector spaces defined over the fields $\mathbb R$ or $\mathbb C$.

Answer 162

As a student, I thought (for quite a while) that our textbook had stated that tensoring commutes with taking homology groups. It wasn't until calculating the homology groups of the real projective plane over rings Z and Z/2Z that I realized my mistake.

Answer 163

While this example was already mentioned by Skopenkov in a comment, here it is, explicitly:

If $G\times X\to X$ is a free continuous action of a compact metrizable group on a compact metrizable space $X$, then $X\to X/G$ is a principal $G$-bundle.

This is true if $G$ is a Lie group (Gleason) and, more generally, for proper Lie group actions on locally compact spaces (Palais), but false in general (the example is essentially due to Kolmogorov), see:

R. F. Williams, A useful functor and three famous examples in topology. Trans. Amer. Math. Soc. 106 (1963) 319–329.

In this example, $\dim(X/G)=2, \dim(X)=1$.

Answer 164

False belief: A function being continuous in some open interval implies that it is also differentiable on some point in that interval:

Counterexample:

The Weierstrass function is an example of a function that is continuous everywhere but differentiable nowhere:

$f(x) = \sum_{n=0}^\infty a^n \cos(b^n \pi x)$

Where $a \in (0, 1)$, $b$ is a positive odd integer, and $ab > 1 + \frac{3\pi}{2}$. The function has fractal-like behavior, which leads to it not being differentiable. This notion is rather disheartening to most calculus students, though.

Answer 165

If $A$ and $D$ are $n \times n$ matrices and $D$ is diagonal, then $A \cdot D=D \cdot A$.

Many of my Linear Algebra students believe this, also because it's written in the textbook in some form, but it's only true if $D$ is a multiple of the identity matrix or if both $A$ and $D$ are diagonal.

Answer 166

Draw the graph of a continuous function $f$ (from $\mathbb{R}$ to $\mathbb{R}$). Now draw two dashed curves: one which everywhere a distance $\epsilon$ above the graph of $f$ and one which is everywhere a distance $\epsilon$ below the graph of $f$. Then the open $\epsilon$-ball around $f$ (with respect to the uniform norm) is all functions which fit strictly between the two dashed curves.

Answer 167

1- A very common mistake that 1st year students (but not even a single mathematician) think that it is true is "a transitive and symmetric relation on a set is reflexive". But as the empty set is a transitive and symmetric relation but not reflexive on any non-empty set. Of course there lots of non-trivial examples also.

2- Another common mistake is that the expression "countable union of countable sets is again countable" is independent of axiom of choice (AC). Many people make the proof of this statement without mentioning axiom of choice. Indeed, in his holly book Algebra, Lang proves this statement just by taking an ordering from each countable set and continues without the mentioning AC.

Answer 168

Let $B(r_1) \subset B(r_2)$ be two open balls of radius $r_1$ and $r_2$ respectively. Then $r_1 \leq r_2$.

Bounded metric spaces give trivial counterexamples. Also, $B \left( \frac{1}{6}, \frac{2}{3} \right) \subsetneq B \left( \frac{1}{2}, \frac{1}{2} \right)$ in $(0,+ \infty)$.

Answer 169

"Let $E$ be a complete locally convex topological vector space (or a complete topological vector space or a complete topological group) and let $F$ be closed vector subspace (or a closed subgroup). Then the quotient $E/F$ is complete."

This just has to be true. One can almost see the proof. And in fact it is true for Banach spaces. So it has to be true for locally convex spaces as well.

Another one with completions:

"Every topological group is a dense subgroup of a complete topological group." True for abelian groups but false in general (take the homeomorphism group of $[0,1]$ with the compact open topology)

Answer 170

Two very common errors I see in (bad) statistics textbooks are

(i) zero 3rd moment implies symmetry (though generally stated in terms of "skewness", where skewness has just been defined as a scaled third moment)

(ii) the median lies between the mean and the mode

(I have seen a bunch of related errors as well.)

Another one I often see is some form of claim that the t-statistic goes to the t-distribution (with the usual degrees of freedom) in large samples from non-normal distributions.

Even if we take as given that the samples are drawn under conditions where the central limit theorem holds, this is not the case. I have even seen (flawed) informal arguments given for it.

What does happen is (given some form of the CLT applies) Slutzky's theorem implies that the t-statistic goes to a standard normal as the sample size goes to infinity, and of course the t-distribution also goes to the same thing in the limit - but so, for example, would a t-distribution with only half the degrees of freedom - and countless other things would as well.

The first two errors are readily demonstrated to be false by simple counterexample, and to convince people that they don't have the third usually only requires pointing out that the numerator and denominator of the t-statistic won't be independent if the distribution is non-normal, or any of several other issues, and they usually realize quite quickly that you can't just hand-wave this folk-theorem into existence.

Answer 171

Yet another common false belief is :

"For non constant periodic functions $f: \mathbb{R} \to \mathbb{R}$ and $g: \mathbb{R} \to \mathbb{R}$ with smallest positive periods $p_1 , p_2$ respectively, the sum $f+g$ is periodic if and only if $\frac{p_1}{p_2}$ is rational."

One side of the statement above is true (the latter implies the former, but the former does not necessarily imply the latter. (But it's true for continuous functions.)

As an exercise one may like to prove the following : (Source : Miklos Schweitzer competition)

Given any two positive real numbers $p_1,p_2$, there exists functions $f_1 : \mathbb{R} \to \mathbb{R}$ with smallest positive period $=p_1$, and $f_2 : \mathbb{R} \to \mathbb{R}$ with smallest positive period $=p_2$, such that $f_1+f_2$ is also periodic.

One more interesting false belief is :

"If $f: \mathbb{R} \to \mathbb{R}$ is a continuous function taking unequal values at points $x_1,x_2 \in \mathbb{R}$ with $x_1 < x_2$ then there is some sub-interval of $(x_1,x_2)$ on which $f$ is either strictly increasing or strictly decreasing."

This is wrong. Among the most familiar counterexamples is the Devil's staircase.

I believe this false belief is often due to the habit of seeing continuous functions wearing the glasses provided by the Intermediate Value Property these functions have.

Addendum : Some nice examples from Wikipedia, here : https://en.wikipedia.org/wiki/Normally_distributed_and_uncorrelated_does_not_imply_independent

Answer 172

Here's one I was reminded recently during lunch in the common room.

A maximal abelian subalgebra of a semisimple Lie algebra is a Cartan subalgebra.

This is true for compact real forms of semisimple Lie algebras, but fails in general. The missing condition is that the subalgebra should equal its normaliser.

Answer 173

An elementary false belief in elementary number theory: for $a, b, c\hspace{.1cm}\varepsilon\hspace{.1cm} \mathbb{N}$

$LCM\left(a,b\right)\times GCF\left(a,b\right) = ab$ .

Thus, $LCM\left(a,b,c\right)\times GCF\left(a,b,c\right) = abc$.

In general, $\left(a_1,a_2,\ldots,a_n\right)[a_1,a_2,\ldots,a_n] = a_1a_2\ldots a_n$.

Answer 174

This is more of a false philosophy than a clear mistake, but nevertheless it is very common:

A compact topological space must be "small" in some sense: it should be second countable or separable or have cardinality $ \le 2^{\aleph_0}$, etc.

This is all true for compact metric spaces, but in the general case, Tychonoff's theorem gives plenty of examples of compact spaces which are "huge" in the above sense.

Answer 175

A possible false belief is that "a maximal Abelian subgroup of a compact connected Lie group is a maximal torus". Think of the $\mathbf Z_2\times\mathbf Z_2$-subgroup of $SO(3)$ given by diagonal matrices with $\pm1$ entries.

Answer 176

Multiplication of differential forms is inherently anti-commutative. Thus, if $x$ and $y$ are coordinates on a surface, then $dx \wedge dy$ makes sense but $(dx)^2+(dy)^2$ is either nonsense or, if it means anything, is $0$.

I'm not sure why I believed this, but I did for several years. I tried my best to avoid creating this impression in my students, but I think it still happened in some of them, simply because the curriculum spends a lot of time on integration and Stokes theorem and very little time on metrics, curvature, etc.

Answer 177

Here is a short list of some false beliefs I had when I was studying mathematics, I suppose they may be common but I have never checked:

• I was in the last year of high school and studying university-level math in advance. I remember trying for a week to prove that a continuous injective map from an open subset of $\mathbb{R}^2$ to $\mathbb{R}^2$ that preserves "being aligned" (I mean that maps aligned triples to aligned triples) must be the restriction of an affine map (over $\mathbb{R}$). That is disproved by restrictions of projective transformations... Which I knew of but I was not able to see they contradicted my belief. When my teacher told me "What about projective transformations?"... I felt dumb.
• I was in the 1st year of PhD studies. My advisor, Adrien Douady, had an idea to build polynomial Julia sets with positive Lebesgue measure. Julia sets are fractals, often with complicated topological structures at every scale. Surely that must be the source of measure? So as an exercise, I tried for a week to prove that Jordan curves are necessarily of Lebesgue measure 0. I told Adrien about my attempts. He gave me a counter-example. I felt dumb.
• Learning that there are closed subsets of the interval with positive Lebesgue measure but no interior did not surprise me as much, as the construction is very simple, but still that's a bit counterintuitive.
• When you zoom on the Mandelbrot set, you see all that round components with smooth boundaries. They look so round. Surely they must be circles, for otherwise the difference would be visible. Well... they are not (except one). Guess how I felt when I learned.
• Frankly, when learning the first time about complex numbers, did anybody here expect that, adding the square root of -1 to the reals would add the roots of all other polynomials?
• I was giving a lecture to math teachers about sensitivity to initial condition (call it chaos) and showing strange attractors on the computer, one told me that by the very presence of chaos, what we see may be quite far from the actual behaviour of the equation, save reality. It turns out hyperbolic systems are stable, so I believe this is still representative (it does not prove it but it is an encouraging hint).
• ... Chaos in deterministic systems. I won't develop on that.
• Surely before hearing of set theory and Cantor's argument, you will believe that all sets are countable. Then after learning that this is not the case, you will think that $\mathbb{R}^2$ must be bigger than $\mathbb{R}$, right?
• You have a $C^\infty$ function on the right half plane, all of whose derivatives have a continuous extension to the boundary line. Surely, it must be easy to extend it to a $C^\infty$ function of the whole plane, isn't it? Well... You can but I would not call it easy.
• Short statements have short proofs. Disproved by Fermat's last theorem (among others).
• I was quite disappointed to learn that there cannot be a finite non-commutative field (division algebra).

I have a few other examples, that I would not term "common false beliefs" but rather "fun and surprising math facts". Is there already a MO question about that?

Answer 178

Probably my fault for not paying enough attention in analysis, but:

Any continuous function on the interval that has derivative equal to zero almost everywhere is constant.

Answer 179

A polynomial $p(x)$ of degree $n$, with coefficients in a commutative ring $R$, has at most $n$ roots, counting multiplicity. This is true if $R$ is an integral domain, but it can fail in the presence of zero divisors.

For instance, $p(x) = x^2+5x$ has four solutions when $R = \mathbb{Z}/6\mathbb{Z}$. I realized this mistake when a colleague asked me about factorization over a non-commutative ring, and I realized that I did not even know what would happen in the presence of zero-divisors.

This does motivate a question I have not found an answer to: is the number of solutions of $p(x)$ bounded by a function of the degree and the characteristic of the ring?

Answer 180

For some reason, I believed until recently that every compact operator on a Hilbert space must have eigenvalues. It's obviously true for finite-rank operators and Hermitian operators, so I must have subconsciously generalized.

This was despite being well aware of (say) the Volterra operator, say; I simply had two contradictory ideas in my head.

Answer 181

Consider the following well-known result: Let $(E,\leq)$ be an ordered set. Then the following are equivalent: (i) Every nonempty subset of $E$ has a maximal element. (ii) Every increasing sequence in $E$ is stationary.

It is immediate that (i) implies (ii). To prove the converse, one assumes that (i) is false and then "constructs step by step" a strictly increasing sequence.

The common mistake (which I have seen in textbooks) is to describe the latter construction as a proof by induction. In fact, the construction uses the axiom of choice (or at least the dependent choice axiom).

(As a special case, I don't think ZF can prove that every PID is a UFD.)

Answer 182

This is a common error made by mature mathematicians in many books and papers in analysis, especially in differential equations: If $X$ is a closed subspace of a Banach space $Y$, then the $Y^*$ (the dual of $Y$) is isomorphic to a subspace of $X^*$ (the dual of $X$). It is false (of course) since Euclidian space $\mathbb R$ is a subspace of $\mathbb R^2$, yet the dual of $\mathbb R^2=\mathbb R^2$ is not isomorphic to a subspace of the dual of $\mathbb R=\mathbb R$. I guess, sometimes they really, really want it to be true. Cheers Boris

Answer 183

"If a field $K$ has characteristic 0 and $G$ is a group, then all $KG$-modules are completely reducible."

True for finite groups but very false in general.

Answer 184

Here's another howler some people commit: If $m$, $n$ are integers such that $m$ divides $n^2$ then $m$ divides $n$.

It's true sometimes, for example if $m$ is prime (or more generally squarefree, i.e. a product of distinct primes). But in general all one can conclude is that there exists integers $p$, $q$, $r$ with $p$ squarefree such that $ m = p q^2 $ and $ n = p q r $

The usual counterexample is that $8$ divides $4^2$ but not $4$ ;-)

Answer 185

I have heard the following a few times :

"If $f$ is holomorphic on a region $\Omega$ and not one-to-one, then $f'$ must vanish somewhere in $\Omega$."

$f(z)=e^z$ of course is a counterexample.

Answer 186

If $\alpha>0$ is not an integer, the set of functions $f:[a,b]\rightarrow\mathbb R$ such that $$\sup_{y\ne x}\frac{|f(y)-f(x)|}{|y-x|^\alpha}<+\infty$$ is ${\mathcal C}^\alpha([a,b])$.

False for $\alpha>1$, because this set contains only constant functions.

Answer 187

"The set A = {a, b} has two elements..."

It's quite simple to notice that a can be the same as b, but after 5 years of university there were people still believing it...

Answer 188

Taylor's Formula and displacement operator: I (too often) see in papers (mathematical physics but a recent paper (a) by mathematicians also) the statement

False belief 1 : a) Let $D=\frac{d}{dx}$ be the derivation operator. Then, for all $f\in C^\infty(\mathbb{R})$, $$ e^{tD}[f](x)=f(x+t) $$

which is false (take any $\phi\in C^\infty(\mathbb{R})$ with compact support, for instance).

False belief 2 : b) In the same vein, for formal power series (``our object is formal then we do not have to ensure convergences''). Let $S(x)\in \mathbb{R}[[x]]$ ($x$ is a formal variable) then for $t\in \mathbb{R}$, one has $$ e^{tD}[S](x)=S(x+t) $$

which is false as we must have $t$ in the domain of convergence of $S$.

Remarks (i) The function $f\in C^\infty(\mathbb{R})$ is analytic over $\mathbb{R}$ iff $$ (\forall x\in \mathbb{R})(\exists R>0)(\forall t\in ]-R,R[) (\sum_{n\geq 0}\frac{t^n}{n!}D^n[f](x)=f(x+t))\qquad (1) $$ (ii) Even if $f\in C^\omega(\mathbb{R})$, it can happen that the left hand side of eq. (1) do not converge otherwise $f$ would be the restriction of an entire function (which e.g. $\frac{1}{1+x^2}$ is not, for example).

(iii) Even if the LHS of (1) converges for all $x,t\in \mathbb{R}$, $f$ need not be analytic. Consider the following function (classic in theory of distributions)
$$ f(x)=0\mbox{ if } x\notin ]-1,1[\mbox{ and } f(x)=e^{\frac{1}{1-x^2}} \mbox{ if } x\in ]-1,1[ $$ (iv) In the (b) case $S=\sum_{n\geq 0}n!\, x^n$ for example cannot be displaced.

Answer 189

Heisenberg-Weyl and enveloping algebras

I have heard (and have read) the following belief even among distinguished mathematical physicists.

Let $HW_\mathbb{C}$ be the (associative with unit) algebra generated by two elements $\{a,a^\dagger\}$ subjected to the relation $[a,a^\dagger]=1$. It is easy to check that $$ \mathfrak{g}=span_\mathbb{C}\{a,a^\dagger,1_{HW_\mathbb{C}}\} $$ is a Lie subalgebra.

False belief The algebra $HW_\mathbb{C}$ is the universal enveloping algebra of $\mathfrak{g}$ i.e. $$ HW_\mathbb{C}=U(\mathfrak{g})\ . $$

One way to see that this is false at once is to observe that any enveloping algebra $U(\mathfrak{g})$ possesses (at least) a character $\varepsilon$, but there is none on $HW_\mathbb{C}$ (one would have indeed $1=\varepsilon([a,a^\dagger])=0$).

Late edit (After Darij's post) Indeed, the enveloping algebra of $$ \mathfrak{g}=span_\mathbb{C}\{a,a^\dagger,1_{HW_\mathbb{C}}\} $$ is the algebra (associative with unit) generated by $\{a,a^\dagger,e\}$ ($e$ is a clone of $1_{HW_\mathbb{C}}$) subjected to the relations $$ [a,a^\dagger]=e ; [a,e]=[a^\dagger,e]=0\qquad \mbox{[Rel 1]} $$ and the canonical arrow $U(\mathfrak{g})\to HW_\mathbb{C}$ transforms an expression in $a,a^\dagger,e$, already reduced by [Rel 1] into its image in $HW_\mathbb{C}$. For relations [Rel 1] one can take, for instance, the normal form: all $a^\dagger$ (creations) on the left, all $a$ (annihilations) on the right and $e$ anywhere (as they are central).

Combinatorially, in the reduced expressions of $U(\mathfrak{g})$, the power of e counts the number of times Wick commutations have been applied. I'd include this in my post if it were not out of place.

Answer 190

"This algebraic variety is a $C^\infty$-smooth manifold, therefore it must be non-singular". This sounds obvious (and in fact it is true over $\mathbb{C}$) however it is false in general (for instance over $\mathbb{R}$). See the discussion here for many details.

Answer 191

A somewhat common belief among students starting out in cryptography:

Breaking RSA requires factoring the modulus.

Although it is not quite known to be a "false" belief, there is no known reduction showing that breaking RSA implies finding the prime factors of the modulus. This in contrast to e.g. Rabin's cryptosystem, and various cryptographic schemes built on other hard problems, whose security provably relies on the underlying hard problems.

Answer 192

"Frac" and "Const" commute: A differential ring is simply a pair $(R,\partial)$ where $R$ is a ring and $\partial$ is a derivation in $R$ (i.e. $\partial\in End_{\mathbb{Z}}(R)$ follows identically Liebniz rule $\partial(ab)=\partial(a)b+a\partial(b)$). The constants of $R$, $\ker(\partial)$ form a subring $Const(R)\subset R$ called subring of constants.

If $R$ is a commutative domain (i.e. without zero divisors), it is standard to consider the field of fractions $Frac(R)$ and to extend $\partial$ by the formula of calculus $\partial(1/g)=-\partial(g)/g^2$.

False belief: $Const(Frac(R))=Frac(Const(R))$, in other words, every constant in $Frac(R)$ is of the form $\alpha/\beta$, where $\alpha,\beta\in Const(R)$.

Until yesterday morning, I postponed to prove this (false) lemma (thinking it was an easy exercise). Then, preparing a talk, I could not prove this and searched for a counterexample. Finally I found a simple one in MSE (see below).

See there for a discussion and counterexamples.

Answer 193

A common misconception in analysis is that if two $C^1$-functions are linearly independent then their Wronskian is non-zero at some point.

This is the result of carelessly reversing a true implication. The mistake was first made in 1882 by Thomas Muir (who is responsible for the name "Wronskian") in his "Treatise on the Theory of Determinants" and Giuseppe Peano pointed out a counterexample as early as 1889 (it suffices to take $y=x^2$ and $y=x|x|$).

For more information on the history of this mistake see "Peano on Wronskians: A Translation". by Susannah M. Engdahl and Adam E. Parker.

Answer 194

I saw many students using the "fact" that for a subset $S$ of a group one has $SS^{-1}=\{e\}$

Answer 195

Every matrix is the sum of a symmetric and an antisymmetric matrix. Hence:

If $V$ is a vector space and $k$ is a number, then the $k$-th tensor product of $V$ with itself decomposes as a direct sum into symmetric and antisymmetric tensors: $$ \underbrace{V \otimes ... \otimes V}_{k\text{ times}} = \Lambda^kV \oplus \mathrm{Sym}^kV $$

Recall (in the finite-dimensional case) the dimensions: $$ \dim \Lambda^k V = \binom{n}{k} \quad\text{ and }\quad \dim\mathrm{Sym}^kV = \binom{n+k-1}{k} $$

Looking at $k=1$ shows that we have non-trivial intersection.

Looking at $n=k=3$ shows that the sum is not exhausting.

Answer 196

Maybe I am a bit late to the party, but here is something that I falsely believed in for a while:

False Belief: If $f$ is a continuously differentiable function with a horizontal asymptote, i.e. $\lim_{x\to +\infty}f(x)=L$, then $f'(x)\to 0$ as $x$ goes to infinity.

Of course, this is incorrect even for smooth functions; take $g$ to be a smooth $L^1$ function with oscillatory behavior so that $\lim_{x\to +\infty}g(x)$ does not exist and then set $f(x)=\int_{0}^xg(s)ds$.

I came to this misconception during an ODE course I took in my undergrad. In the study of the asymptotic behavior of solutions of an autonomous equation $$ \frac{dy}{dx}=F(y) $$ one argues that if a solution $y=y(x)$ tends to a limit $L$ as $x\to +\infty$, that limit must be a zero of the vector field. We then proceeds with determining which zeros of $F$ are asymptotically stable to find $L$. The point is that here not only $\lim_{x\to +\infty}y$, but also $\lim_{x\to +\infty}\frac{dy}{dx}$ exists due to the ODE. Hence:

Correct Statement: If $f$ is a continuously differentiable function with a horizontal asymptote, and if $f'$ admits a limit as $x$ goes to infinity, that limit must be zero.

This readily follows from the mean value theorem.

Answer 197

It’s tempting to believe that a knot and its mirror-image, if tied in succession on a rope, can cancel. View https://youtu.be/lwWeRMmXIoU to see John Conway’s lucid explanation of why this is false.

I would be interested in knowing whether Lord Kelvin was under this misapprehension when he proposed his vortex theory of the atom. (Knots don’t cancel but vortices can.)

I once went to a conference on recreational mathematics in which a speaker claimed that a knot and its mirror-image can be cancelled; I was kind enough not to ask for a demonstration. (There is a way to pretend to make them cancel, but I don’t know the details of the trick.)

Answer 198

Maybe that's a cute one:

If $f$ is a continuous real function, then $f^{-1}(x)$ is at most countable unless $f$ is constant on some interval.

Answer 199

The "curse of dimensionality" means that in a hypercube the volume is increasingly concentrated in the corners as the number of dimensions increase.

---

In fact half the volume of a hypercube is closer to the centre than to the nearest vertex, with any number of dimensions.

The real curse is that the vast majority of the points of a unit hypercube of dimension $n$ are a distance less than $\frac{5}{n}$ from the outside of the hypercube, distances $\sqrt{\frac{n}{12}} \pm \frac12$ both from the centre and from the nearest vertex, and a distance $\sqrt{\frac{n}{6}} \pm 1$ from the vast majority of other points, which for large $n$ are narrow bands.

Answer 200

Surprisingly, I didn't find this answer in the ten pages above, so...

If two measures agree on a class generating their sigma-algebras, they are equal, right?

(And there actually are two ways in which this is false. Even knowing the total measure and assuming that it is finite is not enough to make this true)

Answer 201

I just learned incidentally in the comments of another question here that it is not true that every proper subgroup is contained in a maximal (proper) subgroup. A counterexample is easy to find: the additive group $\mathbb{Z}[\frac{1}{2}]$ has no maximal proper subgroup. But the confusion with proper ideals which are contained in a maximal (proper) ideal, by Zorn's lemma, is certainly something that will surprise others. (The reason for the difference, as I see it, is that in the case of ideals we can detect properness by $1 \not\in I$, whereas for subgroups there is no such criterion.)

Answer 202

As a sequel of this famous answer on $\dim(U+V+W)$, the following inequality is not true $\forall n \ge 4$:
$$ \dim(\sum_{i = 1}^{n} U_i) \le \sum_{r=1}^{n} (-1)^{r+1} \sum_{i_1 < i_2 < \dots < i_r} \dim(\bigcap_{s=1}^{r}U_{i_s}) = \alpha$$
Darij Grinberg has found a counter-example (see this post).

Same flavor: for $n \le 5$, it is true that $\alpha \ge 0$ (see this proof), but it's false for $n>5$ (see this comment).

Answer 203

Many people believe that Cantor proved the uncountability of the real line using a diagonal argument. This paper does not that provide that proof; Cantor's stated purpose was to prove the existence of `uncountable infinities' without using the theory of irrational numbers.

Answer 204

This may not count as a false belief, but it is an amusing misconception I had. In college I had a numerical analysis professor who had both a strong accent and messy handwriting, so it was hard to know exactly what he was talking about sometimes.

I was not yet familiar with the Greek letter $\xi$, and that was the variable he always used to represent the error of a computation, but with his handwriting it just looked like a purposeful scribble. So he would say,"Here we have the calculated value and then of course with some error" (scribble).

I thought he was just being dismissive about the error and trying to represent it in a pejorative way.

To be fair, the letter $\xi$ is not one of the easier ones to draw by hand.

Answer 205

(Cauchy and/or ordinary) product of two summable families. Until recently, I thought that, in a topological ring (i.e. a ring $R$ with topology $\tau$ such that, the maps $x\mapsto -x;\ (x,y)\mapsto x+y;\ (x,y)\mapsto x.y$ are continuous), products of two summable families were summable. In the following contexts, were my (false) beliefs

• $(a_i)_{i\in I},\ (b_j)_{j\in J}$ supposed summable and then $(a_ib_j)_{(i,j)\in I\times J}$ is summable
• Same situation with $I=J=\mathbb{N}$ and $c_n=\sum_{p+q=n}a_pb_q$ (Cauchy product).

But, I found this question and discussion (which proved me that this belief was false in general), returned to Bourbaki General Topology Chapter III, § 6, and there were Exercises 4-5 which proved me that this question was very delicate. Then I could debunk it.

Late addition: See also discussions and the beautiful answer by Robert Furber here.

Answer 206

Uncorrelatedness implies independence

This statement is indeed false. Suppose $X \sim U[-1, 1]$. Then $Cov(X, X^2) = EX^3 - EXEX^2 = 0$, but $X$ and $X^2$ are clearly not independent. However, that mistake is quite popular...

Answer 207

"A closed unit-ball (in a Banach space) is compact!"

It only true in finite-dimensional spaces.

Answer 208

True: Given a graded algebra $A$, there is a notion of a "homogeneous" ideal of $A$. It is a property that connects an ideal of $I$ with the grading and is often necessary to require. For example, if $I$ is a homogeneous ideal of $A$, then the algebra $A / I$ is graded again. If $I$ is not homogeneous, then it is not graded in general (since the projections of different graded components of $A$ onto $A / I$ might have nonzero intersection).

False: Given a filtered algebra $A$, there is a notion of a "filtered" ideal of $A$.

There is no such notion. We can require $I$ to be generated by $I\cap A_n$ for some $n$, or actually to lie inside $A_n$ for some $n$, but in most cases none of these is actually needed. (Correct me if I am wrong.) Formulations like "Let $I$ be an ideal compatible with (or respecting) the filtration" are cargo cult.

But: Given a filtered algebra $A$ and a generating set $G$ of an ideal $I$ of $A$, it is an important question whether $I\cap A_n$ is generated by $G\cap A_n$ for every $n\in \mathbb N$. This is not always satisfied, often nontrivial (in many cases it can be proved by using the diamond lemma to show that every element of $A_n$ has a unique "remainder" modulo $I$ in a certain sense, and this remainder can be obtained by repeated subtraction multiples of elements of $G\cap A_n$) and used tacitly in various texts.

Answer 209

For a bounded subset of a metric space the diameter is two times the radius!

Let $S\subset X$ be bounded. The definitions are:

$\mathrm{diameter}(S):=\sup\{d(x,y)\,|\,x,y\in S\}$

$\mathrm{radius}(S):=\inf\{r>0\,|\,\exists x\in X:\,S\subset B(x,r)\}$

where $B(x,r)$ denotes the open ball of radius $r$ around $x$.

Answer 210

Here are some various examples (I hope that some of them weren't already mentioned):
1. If a space $X$ have two different norms $\| \cdot \|_i, i=1,2$ such that $\| \cdot \|_1 \leq \| \cdot \|_2$ then the completion with respect to $\| \cdot \|_1$ is contained in the completion with respect to $\| \cdot \|_2$.
2. If $M_1,M_2$ are isomorphic modules and $N_1,N_2$ are isomorphic submodules then $M_1/N_1$ and $M_2/N_2$ are isomorphic.
3. If $A,B$ are subsets of topological spaces $X,Y$ (resp.) and $A,B$ are homeomorphic then the closures $\overline{A}$ and $\overline{B}$ are also homeomorphic.
4. The standard construction of adjoining unit to the Banach algebra $A$ yields nothing new if $A$ already was unital.
5. The phrase "a function is almost everywhere continuous" means the same as: "the function is almost everywhere equal to the continuous function".
6. Suppose you are trying to prove that some function space $F$ is complete (say that functions are defined on $X$ and real valued): you take a Cauchy sequence $\{f_n\}_n$ and prove that for each point $x \in X$ the sequence $\{f_n(x)\}_n$ is Cauchy. Then form the completeness of $\mathbb{R}$ you obtain a function $f$. The false belief is that it is now enough to show that $f$ belong to $F$.
7. If you have an ascending family $\{A_i\}_i$ then to obtain it's union $\bigcup_{i}A_i$ it is enough to take some countable subfamily
8. A convergent net $\{x_i\}_i$ in a metric space is bounded and the set $\{x_i\}_i \cup \{x\}$ is compact (where $x$ is the limit).
9. If $D$ is an open dense subset of a topological space $X$ then $card \; D= card \; X$

Answer 211

I have checked the existing answers, but I think this one is not given yet. Sorry, if I missed it.

Although the incompleteness theory of Gödel is generally correctly understood, the consequence of it has multiple false beliefs:

• Due to the incompleteness theory it is not possible to make an AI. Humans will always be be superior to the AI. This assumes that human thinking is complete and will eventually find the answer on any question.

• Due to the incompleteness theory, it is not possible to formalize mathematics. This is refuted by many proof systems, which can formalize almost all mathematics.

As side note, I think this is partly fueled how logic is taught. It puts more emphasis on impossibilities (incompleteness theory), than possibilities (a proof system).

Answer 212

For awhile, I used to think:
If $depth\ M\ge depth\ N$ then $depth\ M_p\ge depth\ N_p$; for any prime ideal $p$ and finite R-modules $M$ and $N$ (Which is not true).

Answer 213

Here are two beliefs. I think everybody will agree that one of them, at least, is false. I adhere to the second one.

Belief 1. There is no simple generalization of the Hodge Theorem to noncompact manifolds.

Belief 2. The most naive statement which would, if true, generalize the Hodge Theorem to noncompact manifolds is this.

The inclusion of the complex of coclosed harmonic forms into the de Rham complex of a riemannian manifold is a quasi-isomorphism.

This statement happens to be true.

Here is a reference:

http://www.iecl.univ-lorraine.fr/~Pierre-Yves.Gaillard/DIVERS/Hodgegaillard/ (Wayback Machine)

The simplest example is that of the real line with its standard metric. In degree zero the complex of coclosed harmonic forms is $\mathbb C\oplus\mathbb Cx$, and in degree one it is $\mathbb Cdx$, which gives the right cohomology.

Here is the (trivial) algebra background.

Let $A$ be a module over some unnamed ring, and let $d,\delta$ be two endomorphisms of $A$ satisfying $d^2=0=\delta^2$. Put $\Delta:=d\delta+\delta d$. Assume $A=\Delta A+A_{d,\delta}$ where $A_{d,\delta}$ stands for $\ker d\cap\ker\delta$. Write $A_{\delta,\Delta}$ for $\ker\Delta\cap\ker\delta$.

We claim that the natural map $$H(A_{\delta,\Delta},d)\to H(A,d)$$ between homology modules is bijective.

Injectivity. Assume $\delta da=0$ form some $a$ in $A$. We must find an $x$ in $A_{\delta,\Delta}$ such that $dx=da$. We have $a=\Delta b+c$ for some $b\in A$ and some $c\in A_{d,\delta}$. One easily checks that $x:=\delta db+c$ does the trick.

Surjectivity. Let $a$ be in $\ker d$. We must find $x\in A$, $y\in A_{d,\delta}$ such that $a=dx+y$. We have $a=\Delta b+c$ for some $b\in A$ and some $c\in A_{d,\delta}$. One easily checks that $x:=\delta b$, $y:=\delta db+c$ works.

Answer 214

This example is similar to this earlier answer.

If $k$ is a field, then $k[x] \otimes_k k[y] \cong k[x,y]$. Therefore also $k[[x]] \otimes_k k[[y]] \cong k[[x,y]]$, right?

Answer 215

I didn't notice this in the long list. A student beginning to learn group theory may believe that the converse of Lagrange's Theorem is true, because it is true for subgroups of prime power. They may also believe that a Sylow subgroup is normal because it has a special name. A counter example to both is $A_{4}$ of order $12$ which has no subgroup of order $6$ and whose four different Sylow $3$-subgroups are all conjugates of one another.

Answer 216

I always thought that the GCD of two elements existed if and only if the LCM of those two elements existed, because of my experience with GCDs and LCMs in the ring $\mathbb{Z}$. Only recently did I discover that this isn't true in other commutative rings.

Answer 217

This is more sort of a convention issue than an outright false belief (connected to the usual $\emptyset$ vs $\{\emptyset\}$ stuff), but I find it funny. I guess a fair share of mathematicians believe that: \begin{equation} \bigcap\emptyset=\emptyset\label{eq}\end{equation} while retaining the standard definition for intersection: $$\bigcap S:=\{x\ \text{such that}\ \forall Y(Y\in S\implies x\in Y)\}$$ according to which in fact: $$\bigcap\emptyset=V$$ where $V$ is the universal class. The condition in round brackets is of course vacuously true. So in a way - this is what I find funny - the former is the worst possible tentative solution of an equation ever.

Answer 218

Let $A$ be an $n\times n$-matrix over a noncommutative (but associative and unital) ring. Then, $A$ is invertible if and only if its transpose $A^T$ is invertible, right?

Wrong. A counterexample (which I just learnt from Ulrich Krähmer and Blessing Bisola Oni, Pointed Hopf algebra (co)actions on rational functions, arXiv:2209.06516v3, Example 2.1.1) is the matrix $\begin{pmatrix} a&1 \\ 0&d \end{pmatrix}$ over the ring with two generators $a$ and $d$ and one relation $da=1$ (but not $ad=1$). Its inverse is $\begin{pmatrix} d&-1 \\ 1-ad&a \end{pmatrix}$, and the fact that the two matrices are mutually inverse is straightforward to verify. But its transpose $\begin{pmatrix} a&0 \\ 1&d \end{pmatrix}$ is not invertible, as its inverse would have to contain a right inverse to $a$, but then $a$ would be (two-sidedly) invertible.

Of course, this all stems from the fact that $\left(AB\right)^T$ is not $B^T A^T$ in general when the matrices $A$ and $B$ are over a noncommutative ring.

Answer 219

This might not be common, but it gave me a headache once. I'll delete if it gets heavily downvoted.

I once had to think really hard about a contradiction in the great scheme of things that followed from my unwitting assumption that if $f$ was a function from a semigroup to a semigroup, then if its kernel was a congruence, $f$ had to be a homomorphism. I encountered a function whose kernel clearly was a congruence but which clearly wasn't a homomorphism, and it took about an hour's walk in a park for my vague notions and incoherent thought to produce the necessary realization.

Answer 220

A few mistakes I remember:

• The quotient groups $\frac{G}{N}$ and $\frac{H}{K}$ are isomorphic if $G \thicksim H$ and $N\thicksim K$.
• A closed interval of a complete lattice is a complete sublattice.
• Two homeomorphic topologies on a set are the same.
• The set of all compatible uniformities of a topological group forms a complete lattice.
• The trace of the identity matrix is 1.

Answer 221

I'm seven years late to the game, but here is mine:

False belief: The irrational numbers, in their usual topology as a subset of $\mathbb{R}$, are not a complete metric space.

Answer 222

Given a bundle $ E \to X$, let $\mathcal{E}$ denote its sheaf of sections.

False belief: Given a map $f: Y \to X$, the inverse image sheaf $f^{-1}\mathcal{E}$ is the sheaf of sections of the pullback bundle $f^* E \to X$.

This is true if $E \to X$ is a local homeomorphism (e.g. a covering space), or if $f: U \hookrightarrow X$ is the inclusion of an open subset, but not for general maps and bundles.

For instance, taking $x^{-1}\mathcal{E}= \mathcal{E}_x$ for $x: 1 \to X$ the inclusion of a point and $\mathcal{E}$ the sheaf of smooth functions on a manifold will demonstrate that it is false.

For vector bundles (or sheaves of modules over the structure sheaf of a ringed space in general), the correct statement is obtained by using the pullback functor $$f^*\mathcal{V} = \mathcal{O}_Y \otimes_{f^{-1}\mathcal{O}_X} f^{-1} \mathcal{V}$$ which is the inverse image followed by extension of scalars.

One issue which leads to this false belief is that texts on sheaves often use $f^*$ in place of $f^{-1}$ for the inverse image functor, rather than reserving the former for sheaves of modules over ringed spaces.

Answer 223

False belief: It is obvious how to prove that $\sin'=\cos$.

Not so much... if $\cos$ and $\sin$ are defined geometrically. You need to prove geometrically that $$\lim\limits_{x\to 0}\frac{\sin x}x=1$$ and a (non-circular) proof of that is not obvious (see here).

Personally I'm aware of that just today! (thanks to a remedial course given to my niece).

Answer 224

False belief: << Let $M$ be the von Neumann algebra generated by a $\rm{C}^{\star}$-algebra $\mathcal{A}$. >>

The false belief is to think that the above sentence makes sense. In fact, a von Neumann algebras and a $\rm{C}^{\star}$-algebra don't have the same status. A von Neumann algebra is an operator algebra by definition, i.e. it is defined inside $B(H)$ for some separable Hilbert space $H$. Now, some subalgebras of $B(H)$ are (separable) $\rm{C}^{\star}$-algebras, but a $\rm{C}^{\star}$-algebra can also be defined abstractly. It can next be represented and a given representation $H$ (defined for example by GNS construction for a given state), if it is faithful, induces an embedding in $B(H)$.
So to make sense, the sentence above should be modified as:

<< Let $M$ be the von Neumann algebra generated by $(\mathcal{A},\rho)$, a couple of $\rm{C}^{\star}$-algebra and state. >>
or
<< Let $M$ be the von Neumann algebra generated by a $\rm{C}^{\star}$-algebra $\mathcal{A}$ represented on $H$. >>

Then, $M = \pi_H(\mathcal{A})''$. We can use $M$ to characterize the representation $H$, for example, we can talk about a representation of type ${\rm I}$, ${\rm II}$ or ${\rm III}$ if $M$ is a von Neumann algebra of type ${\rm I}$, ${\rm II}$ or ${\rm III}$. There is a $\rm{C}^{\star}$-algebra with representations of every type, for example the Cuntz algebra.

Finally, there exists a universal representation for every $\rm{C}^{\star}$-algebra (i.e. the direct sum of the corresponding GNS representations of all states; it is faithful). The associated von Neumann algebra is called the enveloping von Neumann algebra (it can also be defined as the double dual); it contains all the operator-algebraic information about the given $\rm{C}^{\star}$-algebra.

Answer 225

I already thought that the following two sets of matrices are one. $$M(\color{blue}{\Bbb R},2n)\qquad \text{and}\qquad M(\color{red}{\Bbb C},n).$$

Answer 226

It is a common mistake to believe that epimorphisms are either identical to surjections or that they are a better concept. Unfortunately this is rarely the case; epimorphisms can be very mysterious and have unexpected behavior

Answer 227

The following analogue of this result is clearly false:

Falsehood. If $M$ is a module over a commutative ring $R$, then $M^\vee = \operatorname{Hom}_R(M,R)$ is at least as big as $M$ (e.g. in terms of cardinality or rank).

For example, if $R$ is a domain and $M$ is torsion, then $M^\vee = 0$. But what's much more surprising is that the following is still false:

False belief: If $M$ is a torsion-free module over a principal ideal domain $R$ (even $R = \mathbf Z$), then $|M^\vee| \geq |M|$ and/or $\operatorname{rk}(M^\vee) \geq \operatorname{rk}(M)$.

(Even assuming $M$ has no divisible elements doesn't help.)

Answer 228

"It is less risky to place 100 one-dollar bets on 100 independent coin flips than to place a single one-dollar bet on a single coin flip".

This is quite evidently false for any reasonable definition of "risky"; for example, in the former case the probability of losing more than a dollar is about 38%; in the latter case the probability is zero.

Nevertheless I am prepared to attest that at least two excellent mathematicians have told me that they believed this all their lives until it was pointed out to them that nothing like it can be true.

What is true is that it is less risky to place 100 one-dollar bets on a 100 independent coin flips than to place a single hundred dollar bet on a single coin flip.

A related fallacy (or the same fallacy in different language), which I have heard many times from both students and professional colleagues, is that "insurance works because the insurance company takes on many independent risks, thereby reducing their overall risk to nearly zero". This is of course false. The more independent risks the company takes on, the more risk it faces.

The correct statement is that insurance works because each risk is apportioned among a great many shareholders, each of whom is now effectively betting a very small amount on each of a great many independent coin flips. If there were only one shareholder, there would be no point in insurance unless either a) the single shareholder was pathologically risk-preferring or b) the premiums were sufficiently actuarially unfair for the company to earn supernormal profits, which for some reason did not get competed away.

Answer 229

I searched through the answers and hope that I haven't missed this one:

The Minkowski sum of two closed and convex sets is closed and convex.

Answer 230

A set is compact iff it is closed and bounded.

Answer 231

The "conditional Vitali convergence theorem": Let $X_n$ be a uniformly integrable sequence of random variables with $X_n \to X$ almost surely, and $\mathcal{G}$ a sub-$\sigma$-field. Then $\mathbb{E}[X_n \mid \mathcal{G}] \to \mathbb{E}[X \mid \mathcal{G}]$ almost surely (FALSE).

I believed this one until I read Uniformly integrable sequence such that a.s. limit and conditional expectation do not commute. It is particularly startling because the conditional versions of the monotone convergence theorem, the dominated convergence theorem, and Fatou's lemma are all true!

What is true is that $\mathbb{E}[X_n \mid \mathcal{G}] \to \mathbb{E}[X \mid \mathcal{G}]$ in $L^1$, so you do have a subsequence converging almost surely.

Answer 232

In ${\mathbb F}_p^\times$, the non-squares are the opposite of the squares. In other words, $a$ is square iff $-a$ is not a square.

This is a confusion with the facts that the kernel of $x\mapsto x^2$ is $\{1,-1\}$ and the subgroup of squares has index $2$.

Answer 233

Most people that study Riemannian geometry for their first time make the following assumption at some point: "Let $(e_1,\dots,e_n)$ be a local orthonormal frame of $TM$ such that all Lie brackets $[e_i,e_j]$ vanish..."

This one is not so common (maybe special to me), but here we go: "$\mathbb{RP}^\infty$ and $\mathbb{CP}^\infty$ are Eilenberg-Mac Lane spaces, so $\mathbb{HP}^\infty$ is one, too."

Answer 234

(1) All Lebesgue-null sets are countable, or are strongly measure zero. (2) The following,verbatim, was a Q in American Mathematical Monthly : " A student asserted that any uncountable real set has a closed uncountable subset. Is this true ?" .

Answer 235

Here are two beliefs. I think everybody will agree that one of them, at least, is false. I adhere to the second one.

Belief 1. The simplest way to compute the exponential $e^A$ of a complex square matrix $A$ is to use the Jordan decomposition.

Belief 2. It's simpler and more efficient to use the following fact.

Let $f(z)$ be the minimal polynomial of $A$, let $g(z)$ be $f(z)$ times the singular part of $e^z/f(z)$, and observe $e^A=g(A)$.

(By abuse of notation $z$ is at the same time an indeterminate and a complex variable.) (The problems of computing the exponential of $A$ and that of computing the Jordan decomposition of $A$ have the same difficulty level. But, to solve one of them, there is no need to refer to the other.) Here are two references

https://en.wikipedia.org/wiki/Matrix_exponential#Evaluation_by_Laurent_series (current revision)

http://www.iecl.univ-lorraine.fr/~Pierre-Yves.Gaillard/DIVERS/Constant_coefficients/

Jordan decomposition is often mentioned in relation with matrix exponentials. I'm convinced (rightly or wrongly) that the association of these notions in this context is purely irrational. I think somebody once made this association by accident, and then many people repeated it mechanically.

Here is another attempt to describe the situation.

Put $B:=\mathbb C[A]$. This is a Banach algebra, and also a $\mathbb C[X]$-algebra ($X$ being an indeterminate). Let $$\mu=\prod_{s\in S}\ (X-s)^{m(s)}$$ be the minimal polynomial of $A$, and identify $B$ to $\mathbb C[X]/(\mu)$. The Chinese Remainder Theorem says that the canonical $\mathbb C[X]$-algebra morphism $$\Phi:B\to C:=\prod_{s\in S}\ \mathbb C[X]/(X-s)^{m(s)}$$ is bijective. Computing exponentials in $C$ is trivial, so the only missing piece in our puzzle is the explicit inversion of $\Phi$. Fix $s$ in $S$ and let $e_s$ be the element of $C$ which has a one at the $s$ place and zeros elsewhere. It suffices to compute $\Phi^{-1}(e_s)$. This element will be of the form $$f=g\ \frac{\mu}{(X-s)^{m(s)}}\mbox{ mod }\mu$$ with $f,g\in\mathbb C[X]$, the only requirement being $$g\equiv\frac{(X-s)^{m(s)}}{\mu}\mbox{ mod }(X-s)^{m(s)}$$ (the congruence taking place in the ring of rational fractions defined at $s$). So $g$ is given by Taylor's Formula.

This can be summarized as follows:

---

There is a unique polynomial $E$ such that $\deg E<\deg\mu$ and $e^A=E(A)$. Moreover $E$ can be uniquely written as $$E=\sum_{s\in S}\ E_s\ \frac{\mu}{(X-s)^{m(s)}}$$ with (for all $s$) $\deg E_s < m(s)$ and $$E_s\equiv e^s\ e^{X-s}\ \frac{(X-s)^{m(s)}}{\mu}\mbox{ mod }(X-s)^{m(s)},$$ the congruence taking place in $\mathbb C[[X-s]]$.

---

Answer 236

I once misunderstood the definition of monads, and thought that for a monad $(T,\eta,\mu)$, we have $T\eta_X = \eta_{TX}$ (or fmap return == return in Haskell). Of course this is not the case (in case of $T=$[], fmap return [1,2] is [[1],[2]], whereas return [1,2] is [[1,2]]).

Answer 237

People seem to believe that conventional computation (for example, running a chaotic irreversible cellular automaton) can be as efficient as one wants simply with good engineering, but this is not the case. Landauer's principle states that erasing a bit of information always takes $\ln(2)\cdot k\cdot T$ energy where $k$ is Boltzmann's constant ($k=1.38065\cdot 10^{-23}$ Joules/Kelvin) and $T$ is the temperature. Landauer's principle is a consequence of the second law of thermodynamics since if Landauer's principle were violated, then entropy would decrease. Landauer's principle means that conventional irreversible computation always must take $\ln(2)\cdot k\cdot T$ energy per bit erased (and one can erase data just by running it through AND and OR gates, so every irreversible gate must take a minimum amount of energy by Landauer's principle). However, Landauer's principle does not apply to reversible computation since reversible computers are not allowed to erase data.

Answer 238

False belief: relativization is well-defined and the corresponding notation $C^A$ is unambiguous. Which is not quite true because $P=NP$ would not imply $P^A=NP^A$.

Answer 239

I used to think that the subset of even norm vectors in an integral lattice is a sub-lattice. This is true for the "classically integral" lattice defined by $<u,v> \in \mathbb{Z}$ for $u,v$ in the lattice because the even vectors is the kernel of the group homomorphism $v \rightarrow <v,v>$ mod 2. However this fails for the more general notion of "integer norm" lattice where we only require the quadratic form is integer valued (ie. the coefficients are all integral or that the off diagonal entries in the Gram matrix may be half integral eg $x^2+xy+2y^2$). For the hexagonal lattice $x^2+xy+y^2$ which is not classically integral, the even vectors is a sub-lattice but for a different reason that it is the lattice scaled by 2.

If $t = 3$ mod 4, the lattice $L_t$ with quadratic form $x^2+ty^2$ is classically integral. Its even sublattice $L_{t0}$ has quadratic form $4(x^2+xy+(t+1)y^2/4)$ which clearly equals 2$W_t$ where $W_t$ is the lattice with quadratic form $(x^2+xy+(t+1)y^2/4)$. If $t=7$ mod 8, the coefficients of the form is [odd,odd,even], the even vectors is not a subgroup since for example $[0,1]$ and $[1,1]$ has even norm but $[0,1]+[1,1]=[1,2]$ has odd norm. If $t$ is 3 mod 8, the form $(x^2+xy+(t+1)y^2/4)$ can only be even only if both $x,y$ are even since all coefficients are odd. So the even vectors in $W_t$ turn out to be $2W_t$. It is a sub-lattice and it is the subset of even vectors but it is index 4 in $W_t$. It is the 2-scaled sub-lattice.

If $v \in L_t$, $2v \in 2L_t \subset L_{t0}=2W_t$, so $L_t \subset W_t$. So the picture is $2W_t=L_{t0} \subset L_t \subset W_t$ with each containment is index 2.

Answer 240

Just today, I realised that I had been mis-interpreting the FTFGAG. One can speak of the $2^\infty$-torsion subgroup or the $2'$-torsion subgroup, the $3^\infty$-torsion or the $3'$-torsion subgroup, or even just the torsion subgroup of a FGAG or the maximal torsion-free quotient … so surely one can speak of the torsion-free subgroup, right? In fact, when I was corrected on this, my first thought was to reply: "just take the subgroup consisting of all infinite-order elements", and it only occurred to me as I was saying it to wonder how the identity element would squeeze its way into this so-called 'subgroup'.

Answer 241

The GNS construction

Let $\phi$ be a state on a $C^*$-algebra $\mathcal{A}$, and put $N_{\phi}:=\{a\in\mathcal{A}\;|\;\phi(a^*a)=0\}$. Then $N_{\phi}$ is a norm-closed left ideal in $\mathcal{A}$. The sesquilinear form $\left<\cdot,\cdot\right>:\mathcal{A}/N_{\phi}\times\mathcal{A}/N_{\phi}\to\mathbb{C}$ defined by $\left<a+N_{\phi},b+N_{\phi}\right>:=\phi(b^*a)$ is a well-defined inner product on $\mathcal{A}/N_{\phi}$. The completion of $\mathcal{A}/N_{\phi}$ establishes a Hilbert space.

False belief: The completion is in the quotient norm.

Surprisingly, Wikipedia (as of April 27, 2018) presents a false statement "The quotient space of the A by the vector subspace I is an inner product space. The Cauchy completion of A/I in the quotient norm is a Hilbert space, which we label H."(https://en.wikipedia.org/wiki/Gelfand%E2%80%93Naimark%E2%80%93Segal_construction#The_GNS_construction) First of all, the quotient of a Banach space by its closed subspace is again a Banach space in the quotient norm, which is a very elementary fact in functional analysis. Thus A/I is already complete in the quotient norm, and hence there is no need to complete it in the quotient norm!

The correct completion is, of course, in the norm induced by the inner product, and this norm is not equivalent to the quotient norm in general. In fact, let $\mathcal{H}$ be a separable infinite-dimensional Hilbert space and $\{\xi_n\}_{n=1}^{\infty}$ be an orthonormal basis for $\mathcal{H}$. The linear functional $\phi:\mathbb{B}(\mathcal{H})\to\mathbb{C}$ defined by $\phi(a):=\sum_{n=1}^{\infty}\frac{1}{2^n}\left<a\xi_n,\xi_n\right>$ is a state on $\mathbb{B}(\mathcal{H})$, and $N_{\phi}=\{0\}$. Let $\xi_k\otimes\xi_k$ be the canonical rank-one operator, and put $p_n:=\sum_{k=1}^n\xi_k\otimes\xi_k$. Then $(p_n)_{n=1}^{\infty}$ is a Cauchy sequence in $\mathbb{B}(\mathcal{H})/N_{\phi}$ in the norm induced by the inner product defined at the beginning, but it is NOT a Cauchy sequence in the quotient norm.

I have never seen a remark which clearly states the distinction between the norm induced by the inner product and the quotient norm in the literature on $C^*$-algebras. Since a quotient space is involved, students are easily tempted to think that the completion is in the quotient norm. (Even the Wikipedia editor was confused!) Or, they may thoughtlessly assume that these two norms are the same. So it will be instructive to clearly state the distinction between these two norms when one teaches this subject to undergraduate students.

Answer 242

These are 2 instances which i have seen to happen with my friends. If $A$ and $B$ are 2 matrices, then they believe that $(A+B)^{2}=A^{2}+ 2 \cdot A \cdot B +B^{2}$.

Another mistake is if one i asked to solve this equation, $ \displaystyle\frac{\sqrt{x}}{2}=-1$, people generally square both the sides and do get $x$ as $4$.

Answer 243

Given a finite dimensional vector space $V$ over $\mathbb{R}$ it cannot be written as a countable union of proper subspaces. (This can be proved by algebraic arguments or by the Baire category theorem.)

This may lead one to believe that the same is true if $V$ is infinite dimensional. However, that is false!

The vector space $P$ of polynomials with real coefficients is the union of the subspaces $P_n$ of polynomials of degree $n$.

Answer 244

From Kleiman's article "Misconceptions about $\mathcal{K}_X$", the sheaf of meromorphic functions:

Denote by $A_\mathrm{tot}$ the total fraction ring of a ring $A$.

(1) $\mathcal{K}_X$ can be defined as the sheaf associated to the presheaf of total fraction rings $$U \mapsto \Gamma(U, \mathcal{O}_X)_\mathrm{tot}$$

(2) The stalks of the meromorphic functions are the total fraction rings of the stalks: $$\mathcal{K}_{X,x} = (\mathcal{O}_{X,x})_\mathrm{tot}$$

(3) If $U = \mathrm{Spec}(A) \subset X$ is an affine open, then $$\Gamma(U,\mathcal{K}_X) = A_\mathrm{tot}$$

The first two misconceptions apply to any ringed space $X$, and the third applies to a scheme. Please see his nice, three-page article for discussion and examples.

Answer 245

If $R$ is a commutative ring with $char(R)=p$ ($p$ is prime), then

$$\varphi:R \to R$$ $$x\mapsto x^p$$

is an automorphism.

Which is false of course.

(If $R$ is a field then see comments).

Answer 246

"It cannot be shown without some form of AC that the union (or disjoint union) of countably many countable sets is countable. I have a countably infinite set X of countably infinite sets. Therefore, the union of X cannot be shown to be countable without Choice."

The fallacy is that in many cases of interest, it is possible to exhibit an explicit counting of every element of X. In such a case a counting of X by antidiagonals is easily constructed. The usual counting of the rationals is an example of this.

I think this may even be an example of a more general phenomenon of "people think AC is necessary for a certain construction, but in fact it turns out not to be necessary for the example they have in mind". For example, AC is necessary to find a maximal ideal in an arbitrary ring ... but it isn't if you're prepared to assume the ring is Noetherian.

Answer 247

Cambridge mathematicians (with the notable exception of the exceptional Dirac) have often misunderstood the Euler-Heaviside fractional integro-derivative calculus, disadvantaging their students.

False belief: A erroneous claim that has been often repeated is that the fractional calculus (FC) as envisioned by Euler and Heaviside (Hadamard, Pincherle, and others) doesn't obey the law of exponents

$D^{\alpha}D^{\beta} =D^{\alpha+\beta},$

and, specifically, differentiation $D$ and integration $D^{-1}$ do not commute, and consequently, neither do they obey the law of exponents.

One example of this apparent lack of commutativity is given on the webpage Fractional Calculus III (circa 2008) by Beardon:

$$D^{-1}D \; e^x = \int_0^x e^t dt = e^x-1 \neq DD^{-1} e^x = D(e^x-1) = e^x = D^0e^x = D^{-1+1}e^x.$$

In fact, however, commutativity applies once the Heaviside step function $H(x)$ is correctly introduced since

$$D^{-1}D \; H(x) e^x =H(x) \int_0^x (\delta(t) +e^t) dt =H(x)( 1+ e^x-1) =H(x) e^x = D^0H(x) e^x .$$

The mistaken claim of lack of commutativity is repeated in another example by Beardon illustrated below (and similar examples by others).

More generally, the Euler-Heaviside FC can be framed a number of ways to ensure the fundamental operator action is interpreted as

$$D^{\alpha}D^{\beta} H(x) \frac{x^{\gamma}}{\gamma!} =D^{\beta} D^{\alpha} H(x) \frac{x^{\gamma}}{\gamma!} =D^{\alpha+\beta}H(x) \frac{x^{\gamma}}{\gamma!} = H(x) \frac{x^{\gamma-\alpha-\beta}}{(\gamma-\alpha-\beta)!}$$

for $\alpha,\beta,$ and $\gamma$ any real numbers.

In one interpretation (e.g., see Gelfand and Shilov's Generalized Functions Vol. I, p. 57),

$$H(x) \frac{x^{-n-1}}{(-n-1)!}= D^n \delta(x)= \delta^{(n)}(x)$$

such that, under a finite part construction or other analytic continuation,

$$D^{n} H(x)\frac{x^{\alpha}}{\alpha!} =H(x) \int_0^x \frac{(x-t)^{-n-1}}{(-n-1)!} \frac{t^{\alpha}}{\alpha!}dt = H(x)\oint_{|z-x|=x} \frac{n!}{(z-x)^{n+1}}\frac{z^{\alpha}}{\alpha!}dz =H(x) \frac{x^{\alpha-n}}{(\alpha-n)!}$$

and, more generally,

$$D^{\beta} H(x)\frac{x^{\alpha}}{\alpha!} = H(x)\int_{-\infty}^\infty H(x-t) \frac{(x-t)^{-\beta-1}}{(-\beta-1)!} H(t)\frac{t^{\alpha}}{\alpha!}dt$$

$$= H(x)\int_0^x \frac{(x-t)^{-\beta-1}}{(-\beta-1)!} \frac{t^{\alpha}}{\alpha!}dt = H(x)\oint_{|z-x|=x} \frac{\beta!}{(z-x)^{\beta+1}}\frac{z^{\alpha}}{\alpha!}dz =H(x) \frac{x^{\alpha-\beta}}{(\alpha-\beta)!}.$$

Then the Euler-Heaviside FC gives

$$D^{\frac{1}{2}} H(x)\frac{x^{\frac{-1}{2}}}{(\frac{-1}{2})!} = H(x)\frac{x^{-1}}{(-1)!} = \delta(x)$$

and

$$D^{\frac{1}{2}}D^{\frac{1}{2}} H(x)\frac{x^{\frac{-1}{2}}}{(\frac{-1}{2})!}$$$$ = D^{\frac{1}{2}} H(x)\frac{x^{-1}}{(-1)!} = H(x)\frac{x^{\frac{-3}{2}}}{(\frac{-3}{2})!} = D H(x)\frac{x^{\frac{-1}{2}}}{(\frac{-1}{2})!},$$

so, in this case,

$$D^{\frac{1}{2}}D^{\frac{1}{2}} = D$$

whereas Beardon concludes that

$$D^{\frac{1}{2}}x^{\frac{-1}{2}} = 0,$$

implying that the law of exponents is violated since then

$$D^{\frac{1}{2}} D^{\frac{1}{2}}x^{\frac{-1}{2}} = D^{\frac{1}{2}}0 = 0 \neq D x^{\frac{-1}{2}} =\frac{-1}{2}x^{\frac{-3}{2}} .$$

H. and B. Jeffreys on p. 229 of their book Methods of Mathematical Physics (Second Ed., 1950), often referenced in discussions of operational calculus, assert unqualifiedly that integration and differentiation do not commute. And, Lavoie, Osler, and Trembley in "Fractional derivatives and special functions" (1976) repeated the false argument of the unqualified violation of the law of exponents. A lack of appreciation of the roles of the Heaviside step and delta functions in the Euler-Heaviside FC seems to be at the core of this oversight with a concomitant tendency (issuing from some blend of laziness, carelessness, and quasi-authoritarianism) to reprint the claims of previous researchers who have omitted discussions of the core constructs--the step and delta functions--in their analyses.

If the FC is constructed using an infinitesimal generator, the analytic continuations can be dodged, or a Pochhammer contour integral can be invoked for generalizing the beta function integral. These constructions are consistent with Laplace- and Mellin-transform approaches over the domains of common convergence and analytic continuation of the reps, with Pincherle's axiomatic treatment of a canonical FC, and with the calculus of Appell Sheffer polynomial sequences.

Mikusinski provided an analogous algebraic convolutional approach, and Sato, a hyperfunction approach, reflecting earlier presentations by Niels Nielsen and then much later than Nielsen by Feynman and others.

(An equally erroneous claim is made to the opposite extreme in an answer to this MO-Q that differentiation and integration operators commute unqualifiedly.)

Answer 248

If every collection of disjoint open sets in a topological space is at most countable, then the space is separable

Answer 249

Here is a false belief I had. Let $f:X \to Y$ be a map of topological spaces having the property that for every finite CW complex $K$, the induced map $f_{\ast}:[K,X] \to [K,Y]$, on unpointed homotopy classes of maps, is a bijection. Then $f$ is a weak homotopy equivalence (that is, it induces isomorphisms on all homotopy groups relative to all basepoints). A counterexample is given by the stabilization map $B \Sigma_{\infty}\xrightarrow{+1} B \Sigma_{\infty}$, which is not an isomorphism on $\pi_1$.

Answer 250

Anytime I wanted to write an answer to this question, I doubted maybe it is not as common as worthy of mentioning here. In fact, I am also not sure how common is the false belief that I observed today in a PDE class. I didn't observe that in many years of teaching calculus, but today four or five students in a small PDE class when calculating a definite integral by parts only applied the limits of the integral to the "second" integral, that is:

$$\int_{a}^b{f(x) g'(x) dx}=f(x) g(x) - \int_{a}^b{f'(x) g(x) dx}$$

Haven't I observed well enough in my calculus classes?

Answer 251

Here's one that I think will surprise some number theorists:

False belief. Let $E$ be an elliptic curve over an algebraically closed field $k$ of characteristic $p > 0$. Then $\operatorname{End}^\circ(E)$ is strictly larger than $\mathbb Q$.

While this is true for all elliptic curves defined over finite fields, most elliptic curves whose field of definition is transcendental over $\mathbb F_p$ have $\operatorname{End}^\circ(E) \cong \mathbb Q$. The extra automorphism on elliptic curves over a finite field comes from the geometric Frobenius. For varieties over larger fields, this is not a thing.

Answer 252

I don't think I've seen it in here:

Every vector space has a non-trivial dual space ($L^p$ for $0 < p < 1$ was a counter-example only mentioned during one of the classes in measure theory)

And of course there's the common false belief of people outside of mathematics that "mathematicians work with numbers and formulae all day long" :)

Answer 253

The derived subgroup of a finite group equals to the set of all its commutators

or equivalently

A product of two commutators in a finite group is always a commutator

This mistake is very widespread, probably because counterexamples to it tend to be quite large. The smallest group, for which it is not true has order $96$.

Answer 254

Let $M_1$ be a finitely generated module over a PID and let $M_2$ be a submodule.

We may pick $L_i$ and $T_i$ submodules of $M_i$ such that $L_i$ is free, $T_i$ is torsion, $M_i = L_i \oplus T_i$, $L_2\subseteq L_1$ and $T_2\subseteq T_1$.

Answer 255

If we regard a ring $R$ (with identity) as a right module ($R_{R}$), then there is a ring isomorphism $\text{End}(R_{R}) \simeq R$, however the same does not happen if we regard $R$ as a left module!

The correct is $\text{End}(_{R}R) \simeq R^{\text{op}}$.

Answer 256

I don’t know how common is the following false belief, but I had it for several years, so maybe some other people also have it. I apologize to those to whom I shared this false belief. I hope this post will help.

Kaplansky’s 6th conjecture (here, 1975) states that if $H$ is a finite dimensional semisimple Hopf algebra and $V$ an irreducible representation of $H$, then $\dim (V)$ divides $\dim (H)$. This conjecture is open over the complex field $\mathbb{C}$, but false in positive characteristic. So we assume to be over $\mathbb{C}$.

For the group case, this property was proved by Frobenius, that is why a finite dimensional semisimple Hopf algebra (over $\mathbb{C}$) with this property is called of Frobenius type.

A finite dimensional Hopf algebra (over $\mathbb{C}$) is called a finite quantum group (or Kac algebra) if it has a $*$-structure. And then it is also semisimple. It is an open problem whether such a $*$-structure always exists.

False belief: George Kac proved Kaplansky’s 6th conjecture for the finite quantum groups.

This false belief was pointed out to me by Pavel Etingof after this talk I gave for Harvard University, and where I mentioned it. Fortunately, that does not affect the content of the talk.

What I had in mind is Theorem 2 in the following paper:
G. I. Kac, Certain arithmetic properties of ring groups., Funct. Anal. Appl., 6 (1972), pp. 158–160.

In modern language, Theorem 2 proves the following: let $H$ be a finite quantum group, and let $\mathcal{C} = Corep(H)$ be the fusion category of complex corepresentations of $H$. For every simple object $X$ of the Drinfeld center $Z(\mathcal{C})$ which contains the trivial object of $\mathcal{C}$ under the forgetful functor, $FPdim(X)$ divides $FPdim(\mathcal{C}) = \dim(H)$ (the quotients are called the formal codegrees).

Note that these $X$ correspond to the irreducible representations of the Grothendieck ring $K(\mathcal{C})$ of $\mathcal{C}$ (see Theorem 2.13 here). In particular, for $G$ a finite group, $\mathcal{C} = Corep(G) = Vec(G)$, and $Irr(K(\mathcal{C})) = Irr(G)$. That is why Theorem 2 implies Kaplansky’s 6th conjecture in the group case (i.e. covers Frobenius theorem). But it is not clear for a finite quantum group in general. It could be relevant to search in this direction, in particular to check whether for every object $Y$ of $Irr(H)$ there exists an $X$ as above such that $\dim(Y)$ divides $FPdim(X)$, because this would prove that $H$ is a Frobenius type.

Note that Theorem 2 (as stated above) holds more generally for every (complex) fusion category $\mathcal{C}$. The case $\mathcal{C} = Rep(G)$, with $G$ a finite group, recovers the fact that the size of each conjugacy class of $G$ divides $|G|$. Finally, according to Pavel, the theorem holds more generally without the assumption ‘which contains the trivial object’ (I don’t have the exact reference for that, so if you know it, please put it in comment).

Answer 257

False Belief: "The suspension spectrum map from spaces to (edit: symmetric) spectra preserves smash-products"

The facts that one denotes the smash product of spectra and the smash product of a space with a spectrum (levelwise) with the same $\wedge$ and tends to leave away the $\Sigma^\infty$ when one embeds a space into spectra are also not helpful in getting used to the harsh reality that the above is wrong.

Answer 258

“A reversible computer can factor integers efficiently in polynomial-time”: Since a reversible computer can multiply two integers efficiently in polynomial-time, it can also factor an integer efficiently in polynomial-time by just putting the computer in reverse.

I wish this were true.

Answer 259

I’m not sure if this counts as “reasonably advanced”, but given the wide-ranging nature of the answers here, I feel like this is worth a mention, given that it plagued my mind for so long when I was in high school.

$\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$ for all real numbers $a, b$.

Of course, what exactly does $\sqrt{a}$ even mean when $a < 0$ is not completely clear. I’ll just take $\sqrt{a} = i\sqrt{|a|}$ for negative $a$, which seems perfectly reasonable given that we usually teach students $i = \sqrt{-1}$. Under this definition, the above statement is false. Indeed, $\frac{\sqrt{1}}{\sqrt{-1}} = \frac{1}{i} = -i$ but $\sqrt{\frac{1}{-1}} = \sqrt{-1} = i$. When I first encountered complex numbers, I obtained a “proof” that $1 = -1$ by multiplying both sides of $i = \sqrt{-1} = \sqrt{\frac{1}{-1}} = \frac{\sqrt{1}}{\sqrt{-1}} = \frac{1}{i}$ by $i$ and I had been wondering all throughout my high school years where the error was in the proof until I worked out why exactly I thought $\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$ and figured out it doesn’t work for negative $a$ or $b$.

Answer 260

The assumption that a cubic surface expressed as a foliation of Weierstrass curves cannot be rational, because a general Weierstrass curve is not rational.

I've seen this false assumption more than once on sci.math over the years. But there are simple counterexamples, such as:

$ (x + y) (x^2 + y^2) = z^2 $

On defining $ u = x/y $ and $ v = z/y $ one obtains $ y (u + 1) (u^2 + 1) = v^2 $, and hence x, y, z as rational functions of u, v.

I'd love to have a reference to a procedure for calculating the geometric genus and algebraic genus of surfaces like this, because they are rational if and only if both these quantities are zero, and for other cubic surfaces that interest me it would save a lot of fruitless hacking around trying to find a rational solution that probably doesn't exist! Are there any symbolic algebra packages that can do this?

I mean for example is $ x y (x y + 1) (x + y) = z^2 $ rational? I'm almost sure it isn't; but how can one be sure?

Answer 261

If a matrix $A$ is self-adjoint/skew-self-adjoint with respect to a symmetric bilinear form, then it is diagonalizable.

True for matrices over $\mathbb{R}$, with respect to a positive definite inner product.

False over other fields. For example, over $\mathbb{C}$, $\left( \begin{smallmatrix} 1 & i \\ i & -1 \end{smallmatrix} \right)$ and $\left( \begin{smallmatrix} 0 & 1 & i \\ -1 & 0 & 0 \\ -i & 0 & 0 \end{smallmatrix} \right)$ are nilpotent, but self-adjoint and skew self-adjoint respectively with respect to the standard inner product.

False for other nondegenerate symmetric bilinear forms: $\left( \begin{smallmatrix} 1 & 1 \\ -1 & -1 \end{smallmatrix} \right)$ and $\left( \begin{smallmatrix} 0 & -1 & -1 \\ 1 & 0 & 0 \\ -1 & 0 & 0 \end{smallmatrix} \right)$ are nilpotent, but self-adjoint and skew self-adjoint respectively with respect to $\left( \begin{smallmatrix} 1 & 0 \\ 0 & -1 \end{smallmatrix} \right)$ and $\left( \begin{smallmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{smallmatrix} \right)$.

You can exponentiate the skew-self-adjoint matrices to get examples of matrices preserving a nondegenerate symmetric bilinear form, with Jordan blocks of the form $\left( \begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix} \right)$.

Answer 262

If $a$ is a real zero of a cubic polynomial with rational coefficients then $a$ can be written as a combination of cube roots of rational numbers.

More generally if $a$ is a real zero of an irreducible polynomial with rational coefficients that is solvable by radicals then students expect the following:

1. Any expression inside a radical evaluates to a real number
2. Any sub-expression of the expression for $a$ evaluates to an algebraic number of order less than or equal to the order of $a$

Of course the problem is that from Cardan's solution to the cubic we can have negative rational numbers inside a square root. Let $c$ = $4*(-1 + \sqrt{-3})$.

$a$ = $\frac{\sqrt[3]{c}}{4} + \frac{1}{\sqrt[3]{c}}$

$f(x) = 4x^3 - 3x + \frac{1}{2}$.

So while $a$ is an algebraic number of degree three, it can not be written as combination of cube roots of rational numbers. Indeed, it is counter-intuitive that $\sqrt[3]{c}$ has degree 6 over the rational numbers yet we can use this number and simple arithmetic to produce an algebraic number of degree 3.

Also $a$ = $\sin(50^{\circ})$. For many values of $\theta$, $\sin \theta$ is a radical number. See also radical values for sine and cosine

Answer 263

This might not be common, but I once believed the following.

Let $ A, B $ be integers, and define a sequence by the linear recurrence $ s_n = A s_{n-1} + B s_{n-2} $ with the base case $ s_0 = 0 $, $ s_1 = 1 $. Two important special cases are the Fibonacci sequence ($ A = B = 1 $) and the sequence $ s_n = 2^n - 1 $ (where $ A = 3 $, $ B = -2 $). Then, for any integers $ n $ and $ k $, $ \gcd(s_n, s_k) = s_{\gcd(n,k)} $.

This is true in the two mentioned special cases, so it's tempting to believe it's true in general. But there's a counterexample: $ A = B = k = 2 $, $ n = 3 $.

Update: corrected the powers of two minus one example from $B = 2$ to $B = -2$. Thanks to Harry Altman.

Answer 264

Something I was sure about until earlier today:

Suppose $\kappa$ is an $\aleph$ number, then $AC_\kappa$ is equivalent to $W_\kappa$, namely the universe holds that the product of $\kappa$ many sets is non-empty if and only if every cardinality is either of size less than $\kappa$ or has a subset of cardinality $\kappa$.

In fact this is only true if you assume full $AC$, and $(\forall \kappa) AC_\kappa$ doesn't even imply $W_{\aleph_1}$, I was truly shocked.

Furthermore, $W_\kappa$ doesn't even imply $AC_\kappa$ in most cases.

The strongest psychological implication is that most people actually think of the well-ordering principle as a the "correct form" of choice, when it is actually Dependent Choice (limited to $\kappa$, or unbounded) which is the "proper" form, that is $DC_\kappa$ implies both $AC_\kappa$ and $W_\kappa$.

Answer 265

Hopefully this isn't a repeat answer. False belief: a matrix is positive definite if its determinant is positive.

Answer 266

Coordinates on a manifold do not have an immediate metric meaning. Until becoming familiar with differential geometry one tends to think they do. (Einstein wrote that he took seven years to free himself from this idea.)

For example, linear control theory is for the most part metric with variables in $R^n$. When moving away from linear control theory, variables are represented as coordinates on a manifold. Nevertheless, much of the literature tends to either abandon metric notions altogether, or to keep using an Euclidean metric though it is no longer very useful.

Answer 267

A common false assumption is that that two non-orthogonal pure states of a quantum mechanical system may never be unambiguously distinguished by a measurement. (See https://arxiv.org/pdf/quant-ph/9807022.pdf)

Another false belief is that a quantum computer is similar to an analogue computer, in that large computations will necessarily fail because of accumulated error. (See, for example, https://arxiv.org/abs/quant-ph/9712048)

For that matter, another common false believe is that Bell Inequalities aren't violated, although it is mostly held by people who have never heard of Bell Inequalities.

Answer 268

Let $R$ be a ring with identity $e$, $A, B\in R$, $A\neq 0$, $B$ is invertible element. If $A\cdot B = A$ then $B = e$.

Answer 269

Assume that $a,b\in \mathbb{R}\setminus \{0\}$ which satisfy $a^{3}= 2b^{3}$.

Then $a-2b$ is a non zero nilpotent element of group ring $\mathbb{Z}_{3} \mathbb{R}$, that is $(a-2b)^{3}=0$.

This would be a counterexample to the zero divisor Kaplansky conjecture

The false lies in an obvious abuse in the definition of the group ring multiplication.

Answer 270

Let $M \subset B(H)$ be a von Neumann algebra, $p \in B(H)$ a projection and $q=I-p$.

False belief: If $pM=Mp$ then $M=pMp \oplus qMq$.
(I think it is a quite common careless mistake)

Counter-example: diagonal embedding of $\mathbb{C}$ into $M_2(\mathbb{C})$.

Answer 271

Fans: (related to the one of polytopes written above) all convex cones are rational, i.e. one would expect that a line would eventually hit a point in the lattice. It is obviously not true, just take the one-dimensional cone generated by $(1,\sqrt{2})$. A similar one was thinking that if I rotate the cone a bit, I can always make it rational.

Answer 272

A dense subspace of a Hilbert space $H$ must contain an ONB for $H$.

This is, of course, true if $H$ is separable, but false in general. See, for example, this answer in MSE: https://math.stackexchange.com/a/201149/465145.

Answer 273

I don't know how common this is, but it occurs as a corollary of a theorem in the fine, and widely used, text by Shafarevich on algebraic geometry: namely, if $f \colon X \longrightarrow Y$ is a surjective algebraic map of varieties, then 1) for all $y \in Y$, the fiber over $y$ has dimension $≥ \dim(X)-\dim(Y)$; 2) on some non empty open set in $Y$ the dimension of the fibers equals $\dim(X)-\dim(Y)$; 3) for all $r$, the set of $y \in Y$ such that the fiber over $y$ has dimension $≥ r$, is closed in $Y$.

The first two are true, but the third is false. Upper semicontinuity of fiber dimension is true on the source, not the target. For the conclusion as stated to hold, one can add properness to the hypothesis on the map. I think this is not at all widely believed by experts, but for some reason it persists in the text, hence may be believed by students.

Since I have myself written notes in which blatantly false statements occur, I do not think for a moment that Shafarevich himself believed this false statement. But such things do slip by, and may mislead beginners. In fact I believed it for some time until enlightened by a friend.

In keeping with the OP's desire to know the psychological reason for the error, it seems for some reason common in my experience for people to assume unconsciously that maps are proper.

Answer 274

False belief: a subgroup isomorphic to a quotient is a retract.

Formally: Let $H,N$ be subgroups of $G$ with $N$ normal and $H \simeq G/N$, then $H$ is a retract of $G$.

It is false, because otherwise $C_2$ would be a retract of $C_4$, but it is not.

In fact, $H$ is a retract of $G$ if and only if $G$ is isomorphic to $H \ltimes N$ (semidirect product).

This false belief caused this post.

Answer 275

If an Abelian category $\mathcal{A}$ is a full subcategory of an Abelian category $\mathcal{B}$, then for all objects $M,N$ of $\mathcal{A}$, we have an injection $$\operatorname{Ext}^i_{\mathcal{A}}(M,N) \hookrightarrow \operatorname{Ext}^i_{\mathcal{B}}(M,N).$$

As an example, let $G$ be the free group on $2$ letters, $A$ its abelianization, $\mathcal{B} = G-mod$, $\mathcal{A}=A-mod$, and $M=N=\mathbb{Z}$ with the trivial action. Then $\operatorname{Ext}^i_{\mathcal{A}}(M,N) \cong \mathbb{Z}$, while $\operatorname{Ext}^i_{\mathcal{B}}(M,N) \cong 0$.

(This example comes from the topological fact that a torus has nontrivial $H^2$, while a punctured torus has trivial $H^2$. In algebra, it's related to the idea that group homology $H_1$ is space of generators for a group while $H_2$ is a space of relations.)

This false belief came up a context where $\mathcal{B}$ was the category of all Galois representations while $\mathcal{A}$ was a certain subcategory. See the comments to Status of the conjectured vanishing of Bloch-Kato H^2.

Answer 276

Initially when I started studying sequences I believed that:

Consider $(x_n)$ and $(y_n)$ are two convergent real sequences having limits $x$ and $y$ as limits respectively. If $ x_n < y_n \quad \forall n\in \mathbb{N}$ then $x < y$

which turned out to be false.

Answer 277

Common false belief:

Consider a non-linear system $\dot{x} = f(x)$ such that $f(0) = 0$. Suppose that $V : D \rightarrow \mathbb{R}$ is a Lyapunov function, i.e. $V$ is positive-definite in $D$ and $\dot{V}$ is negative-definite in $D$. Then $D$ is contained in the basin attraction of the equilibrium point $0$.

Here is a example of a system with a Lyapunov function that is positive definite everywhere, its time-derivative is negative definite everywhere and there are trajectories that does not converge to $0$.

This misconception is common among students in some engineering courses where the proof of the Lyapunov stability theorem is sometimes omitted. Many students rely on their intuition from simple examples and skip or forget the proof, leading to this false belief.

Answer 278

“There are no trigonometric proofs, because all the fundamental formulae of trigonometry are themselves based upon the truth of the Pythagorean Theorem.”

that quote is from Elisha Scott Loomis' book, "The Pythagorean Proposition: Its Demonstrations Analyzed and Classified, and Bibliography of Sources for Data of the Four Kinds of Proofs" and is cited in the abstract of Ne'Kiya D Jackson and Calcea Rujean Johnson's latest American Mathematical Society paper, "An Impossible Proof of Pythagoras":

In the 2000 years since trigonometry was discovered it's always been assumed that any alleged proof of Pythagoras’s Theorem based on trigonometry must be circular. In fact, in the book containing the largest known collection of proofs (The Pythagorean Proposition by Elisha Loomis) the author flatly states that “There are no trigonometric proofs, because all the fundamental formulae of trigonometry are themselves based upon the truth of the Pythagorean Theorem.” But that isn’t quite true: in our lecture we present a new proof of Pythagoras’s Theorem which is based on a fundamental result in trigonometry—the Law of Sines—and we show that the proof is independent of the Pythagorean trig identity $\sin^2x + \cos^2x = 1$.

Answer 279

Root(s) of a $3^{rd}$ degree polynomial over $ \Bbb Q$ are expressible using radicals with the imaginary $i$. If a root $r$ is real, by taking only the real part, $r$ is expressible using radicals over the rational numbers.

Why not? See here on Wikipedia.

Answer 280

Another common mistake. If $W = _P(e_1,\ldots, e_{n})$ is a vector space and $V$ is a subspace of $W$ of dimension $k$, then $V = _P(e_{i_1},\ldots, e_{i_k})$.

Answer 281

I cannot believe this example was not yet given (but, if my belief is false, I will happily delete this answer):

It is very common among "lay people" (who do not understand what it means for lines to be parallel) to believe that "in some kind of geometry" (frequently described as non-euclidean) parallel lines can intersect.

One finds many instances (my guess, the count is in hundreds of thousands) of this false believe just by searching the internet. Here is a random example, the article "How Looking At A Basketball Disproves Something Everybody Learns In High School Geometry" from "Business Insider", 2014. The article concludes with

And voila! We’ve successfully disproven “parallel lines never intersect” using just a basketball.

Answer 282

Characters of a Hopf algebra form a group under convolution .-

It is, of course, the set of non-commutative (or vector valued) characters as it is well-known that, a Hopf algebra $(\mathcal{H},\mu_\mathcal{H},1_\mathcal{H},\Delta,\epsilon,S)$ [1] and a commutative algebra $(\mathcal{A},\mu_\mathcal{A},1_\mathcal{A})$ being given (all over the same commutative ring $k$), the set $Hom_{k_AAU}(\mathcal{H},\mathcal{A})$ is a group under convolution (the inverse being performed through precomposition with $S$). The story is the following.

Until yesterday and based on hasty readings (or rather non-revisited readings of hasty writers :), I believed that the above was true even if the algebra $\mathcal{A}$ was non-commutative. Which is false as shows the counterexample below.

Take any commutative ring $k$ and two letters (or non-commuting variables), $\{a,b\}$, then, with the Hopf algebra $\mathcal{H}=(k<a,b>,conc,1_{\{a,b\}^*},\Delta,\epsilon,S)$ and the (non-commutative) algebra $\mathcal{A}=(k<a,b>,conc,1_{\{a,b\}^*})$, we get a $\mathcal{A}$-valued character $Id:\ \mathcal{H}\to \mathcal{H}$ such that its convolutional square $Id^{*\,2}$ is NOT a character.

If one prefers a more standard world and matrix-valued characters, one can consider the following (and essentially the same counterexample)

1. $k$, a field
2. a two-dimensional vector space $V=k.a\oplus k.b$
3. The Hopf algebra $T(V)$
4. The algebra of 2x2 matrices over $k$
5. The character on $T(V)$, given by $a\to \Big(\begin{matrix}0 & 1\\ 0 & 0\end{matrix}\Big)$ and $b\to \Big(\begin{matrix}0 & 0\\ 1 & 0\end{matrix}\Big)$.

[1] The order is always (space, product,unit,coproduct,counit,antipode).

Answer 283

Way late to the party...

"$ \mathrm{polymod}\ p$ and $\mathrm{mod}\ p$ are the same thing."

And it's cousin: "$\forall{x}, f(x) \cong g(x) \pmod{q} \implies f(x) = g(x)$"

Answer 284

From Keith Devlin

"Multiplication is not the same as repeated addition", as put forward in Devlin's MAA column.

I'm not really sure how I feel about this one; I might be one of the unfortunate souls who are still prey to that delusion.

Caution

In case you missed it, the column ended up spilling a lot of electronic ink (as evidenced in this follow-up column), so I don't believe it would be wise to start yet a new one on MO. Thanks in advance!

Answer 285

For $p$ prime and the chain of embeddings $\mathbb{Z}/p\mathbb{Z} \hookrightarrow \mathbb{Z}/p^2\mathbb{Z} \hookrightarrow \cdots$ given by multiplication by $p$, then $\bigcup_n \mathbb{Z}/p^n\mathbb{Z}$ is not the group of $p$-adic integers $\mathbb{Z}_p$, but its Pontryagin dual, the Prüfer $p$-group $\mathbb{Z}(p^{\infty})$.

Answer 286

In algebraic topology, I thought for a while:

• "For $k \geq 2$, $H_k$ is the abelianization of $\pi_k$." False. True for $k = 1$. Also for all $k$ up to $n-1$ if the space is $(n-1)$-connected for $n \geq 2$ (vacuously, since this says the first $n-1$ homotopy groups are trivial and for these, the Hurewicz homomorphism is the isomorphism, $\pi_k \cong H_k$). See the Hurewicz theorem for more.
• "Generically, all the $\pi_k$ are nonabelian." False. For $k \geq 2$, $\pi_k$ is abelian.

---

Edit: I had a third error in thinking when I first posted this, mangling the above into something further from true. Which I suppose makes the first version of this post meta-appropriate for this thread (but I've fixed it anyway). Thankfully, user Michael gently pointed out my mangling.

Answer 287

I once very briefly thought that:

Given a vector space $V$ and a sub-space $U \subset V$ that $V-U$ is also a subspace.

I've heard this several times as a TA also.

Answer 288

One common misconception is that Category Theory is abstract. It is in fact, no more abstract than set theory or calculus. This common complaint is the usual complaint that people have when they first meet set theory or category theory. Since people interested in category theory are usually already interested in mathematics, it is strange - but perhaps not that strange - that they voice this complaint. It's underlying reason is that it's notions are unfamiliar, rather than abstract. And it doesn't help that there isn't a bestiary of easy examples that we would have for algebra and calculus.

This needs to be developed.

Answer 289

The sigma function

$$\sigma_{1}({p_i}^{\alpha_i}) = \displaystyle\sum_{j=0}^{\alpha_i}{{p_i}^j}$$

satisfies the inequalities

$$\sigma_{1}({p_i}^{\alpha_i}) \gt (\alpha_i + 1)(\sqrt{p_i})^{\alpha_i}$$

$$\sigma_{1}({p_i}^{\alpha_i}) \gt 1 + \alpha_i(\sqrt{p_i})^{1 + \alpha_i}$$

for prime $p_i$ and $\alpha_i \ge 1$.

The "proof" uses the Arithmetic Mean-Geometric Mean Inequality.

As a particular application of this result, Sorli's Conjecture implies the OPN Conjecture.

Answer 290

I'm not sure how common it is but I've certainly been able to trick a few people into answering the following question wrong:

Given $n$ identical and independently distributed random variables, $X_k$, what is the limiting distribution of their sum, $S_n = \sum_{k=0}^{n-1} X_k $, as $n \to \infty$?

Most (?) people's answer is the Normal distribution when in actuality the sum is drawn from a Levy-stable distribution. I've cheated a little by making some extra assumptions on the random variables but I think the question is still valid.

Answer 291

When I was a kid (8th grade), I solved a bunch of math problems in an exam using the ``well-known identity'' that $(x+y)^2=x^2+y^2$, which I was sure I had been taught the year before. It was of course way before I heard about characteristic two and I didn't get a good grade that day!

Answer 292

I don't know if this is what you are looking for, but I keep hearing that "a differentiable function is one that is locally linear", not one whose local variation can be approximated linearly. No one stops to think about e.g, $x^2$, and the fact that its graph does not look like a line at any value of $x$.