Interesting examples of vacuous / void entities

Question

I included this footnote in a paper in which I mentioned that the number of partitions of the empty set is 1 (every member of any partition is a non-empty set, and of course every member of the empty set is a non-empty set):

"Perhaps as a result of studying set theory, I was surprised when I learned that some respectable combinatorialists consider such things as this to be mere convention. One of them even said a case could be made for setting the number of partitions to 0 when $n=0$. By stark contrast, Gian-Carlo Rota wrote in \cite{Rota2}, p.~15, that 'the kind of mathematical reasoning that physicists find unbearably pedantic' leads not only to the conclusion that the elementary symmetric function in no variables is 1, but straight from there to the theory of the Euler characteristic, so that 'such reasoning does pay off.' The only other really sexy example I know is from applied statistics: the non-central chi-square distribution with zero degrees of freedom, unlike its 'central' counterpart, is non-trivial."

The cited paper was: G-C.~Rota, Geometric Probability, Mathematical Intelligencer, 20 (4), 1998, pp. 11--16. The paper in which my footnote appears is the first one you see here, doi: 10.37236/1027.

Question: What other really gaudy examples are there?

Some remarks:

• From one point of view, the whole concept of vacuous truth is silly. It is a counterintuitive but true proposition that Minneapolis is at a higher latitude than Toronto. "Ex falso quodlibet" (or whatever the Latin phrase is) and so if you believe Toronto is a more northerly locale than Minneapolis, it will lead you into all sorts of mistakes like $2 + 2 = 5,$ etc. But that is nonsense.

• From another point of view, in its proper mathematical context, it makes perfect sense.

• People use examples like propositions about all volcanoes made of pure gold, etc. That's bad pedagogy and bad in other ways. What if I ask whether all cell phones in the classroom have been shut off? If there are no cell phones in the room (that is more realistic than volcanoes made of gold, isn't it??) then the correct answer is "yes". That's a good example, showing, if only in a small way, the utility of the concept when used properly.

• I don't think it's mere convention that the number of partitions of the empty set is 1; it follows logically from some basic things in logic. Those don't make sense in some contexts (see "Minneapolis", "Toronto", etc., above) but in fact the only truth value that can be assigned to $\text{“}F\Longrightarrow F\text{''}$ or $\text{“}F\Longrightarrow T\text{''}$ that makes it possible to fill in the truth table without knowing the content of the false proposition (and satisfies the other desiderata?) is $T.$ That's a fact whose truth doesn't depend on conventions.

Answer 1

How many open covers does the empty topological space have? Not one, not none, but two: the empty cover $\varnothing$, since its union is $\bigcup\varnothing=\varnothing$, and the cover $\{\varnothing\}$, since its union is also $\bigcup\{\varnothing\} =\varnothing$.

This comes up when using the Grothendieck plus-construction to sheafify a presheaf. Apply the construction to the (nonseparated) presheaf $P:\mathcal{O}(X)^\mathrm{op}\to \mathrm{Set}$ sending every open set to the set $A$, with $|A|\geq 2$. Then the presheaf $P^+:\mathcal{O}(X)^\mathrm{op}\to\mathrm{Set}$ agrees with $P$ on every open set except $\varnothing\subseteq X$, where $P^+(\varnothing)$ is now a one-element set $\{*\}.$ This is because the matching families for the cover $\{\varnothing\}$ of $\varnothing$ (of which there is one for each $a\in A$) are all set equal to the unique matching family for the refining cover $\varnothing\subseteq\{\varnothing\}$ of $\varnothing$.

This elementary example comes from III.5 of "Sheaves in Geometry and Logic", by Moerdijk and MacLane.

Answer 2

There is a big difference between statements such as, one the one hand "the empty sum is zero" or "0!=1" and on the other hand "1 is not a prime number". In my opinion, the latter does involve a convention (i.e., a choice) but the former does not.

The first definition of a prime that comes to mind (and came historically, I guess) is "a natural number with no divisors except 1 and itself". This is a perfectly reasonable notion, but it leads to unpleasant contortions when one tries to state the prime decomposition theorem, including uniqueness. A similar phenomenon explains why an irreducible space is nonempty by definition. In these cases, the definition has been tailored to the need of getting cleaner statements. The question "is the empty space connected?" falls into the same category; I find it strange that the more common convention (which is yes) does not match the other two.

In the case of the empty sum, 0 is the only conceivable value, the other choice being "undefined": a mathematician hostile to the empty set might define finite (nonempty) sums by induction, starting from the one-term case and leaving the empty case meaningless. This would not lead to contradictions, only to lots of traps in proofs because whenever you take the sum of some finite set of numbers you first have to check that it is not empty, or treat the empty case separately.

And of course, if you run the inductive definition "backwards" from 1 term to 0 term you immediately find the right value for the empty sum. This is an efficient way to convince students.

Answer 3

Over the reals, $\sup \emptyset = -\infty$ and $\inf \emptyset = \infty$.

Answer 4

An elementary example, but pedagogically nice: a standard early induction proof example is that you can tile any $2^n \times 2^n$ square with one unit square removed, using L-shaped tiles of three unit squares each.

Surprisingly (to me), many textbooks take the base case as $n=2$. The better ones use $n=1$. But the version in The Book, though, surely starts at $n = 0$!

(Of course, I understand the pedagogy of not starting at 0: it’s usually best to make one point at a time. Trying to use this single example to teach about both induction and vacuity simultaneously would end up confusing most students. But when it’s not needed for the former, it does work nicely for the latter, I think!)

Answer 5

Counting is a special case I think: the number of ways of doing nothing is always 1, because you do exactly that, nothing. The number of ways of doing something impossible is 0, because you can't do it. That's why we have: $$ \binom{n}{0}=1 \quad \text{but} \quad \binom{n}{n+1}=0.$$ So I don't think your partition example or the cell phone example are really about vacuous truth the same way the Minneapolis example is. Though if pressed I'm not sure how I would formulate precisely how to make the distinction.

Answer 6

1. The value of any sheaf on the empty set is the terminal object. (Consider the gluing condition for the empty open cover of the empty set.)

2. If $A→B$ is a morphism of sets, then we can define the factor set $B/A$. We have $B/∅=B⊔*$, where $*$ is a one-element set. (Consider the left adjoint of the forgetful functor from the category of pointed sets to the category of morphisms of sets. Alternatively, observe that $B/A≅(B∖A)⊔*$.)

3. Sometimes the norm of a morphism of normed spaces $f: X→Y$ is defined as $$\sup_{x∈X: x≠0} ‖f(x)‖/‖x‖$$ or as $$\sup_{x∈X: ‖x‖=1} ‖f(x)‖.$$ This does not work for $X=0$. The correct definition is $$‖f‖=\sup_{x∈X: ‖x‖≤1} ‖f(x)‖.$$ It also works for seminorms.

4. The zero ring is the terminal object in the category of unital rings. It is not an integral domain, nor a local ring or a field.

5. The empty manifold is not connected. Its number of connected components is 0. (Think of the following theorem: Every manifold is the coproduct of a unique family of connected manifolds. The cardinality of the family equals the number of connected components.)

6. Examples in elementary mathematics abound. The zero vector space has an empty basis and a unique endomorphism $A$. The matrix of $A$ in the unique basis is empty and the determinant of $A$ is 1. There is exactly one function from the empty set to any other set (the empty function). Zero is a natural number, $0^0=1$, the sum of the empty family of numbers is 0, the product of the empty family of numbers is 1, the product or the coproduct of an empty family of objects in a category is the terminal or the initial object of this category, the monoidal product of the empty family of objects in a monoidal category is the monoidal unit.

Answer 7

If you've ever written code to convert an integer into a string of decimal digits, you may have come to the conclusion that the integer 0 should map not to the string 0, but to the empty string instead. Most algorithms I've seen need to introduce a kludge to make 0 come out right. After all, when we write 0 we are violating the usual rule of "no leading zeros".

A nice, natural recursive expression of the conversion process is

def itoa(n):
  if n==0: return ""
  return itoa(n/10) + chr(ord('0') + n%10)

which can be thought of as
The string representation of an integer consists of its leading digits (n/10) followed by its last digit (n%10).

Trying to fix this by returning "0" instead of "" would result in everything getting a superfluous leading zero.

On the other hand writing 0 as the empty string would be rather annoying.

Answer 8

The usual axiomatizations of set theory (without urelements) mean that every set in the entire set-theoretic universe is ultimately built from copies of the emptyset, in complex empty-box-in-a-box-in-a-box constructions.

Answer 9

What about the two orientations of a point? $$*$$

(Pro trivialogia). I'd like to add some general remarks about the question you raised in the comment below, which seems to me a question of general interest. I see at least three general good reasons why it is worth dealing with trivial cases of mathematical notions.

• Sometimes we simply do not know whether $x$ is a trivial object. Even if our main interest is in non-trivial cases, in the course of a proof or a computation we deal with unknown objects $x,\, y\dots,$ that may possibly degenerate. Therefore, we would like theorems, methods, rules, to hold with the minimum of assumptions, avoiding special separate treatments for degenerate cases (think to some classifications into unnecessary special cases, for algebraic equations, used at the beginning of algebra).

• Abstraction. It is a great feature of modern mathematics the ability of translating a complicated notion belonging to a simple setting, into a simple notion belonging to a possibly more complicated setting (in many abstract contexts the cost of this operation is zero). Example: a limit or a colimit in a category, and in fact any universal construction, is just a zero-object in a suitable category (as an application, Freyd's theorem about existence of adjoint functors, &c.)

• Constructions and proofs by induction. As soon as the induction step from $n$ to $n+1$ is suitably clarified, the validity of the general fact is reduced to that of the trivial case, that becomes the heart of the whole story. So for instance, the very reason of some facts about spheres is some (trivial, but important) fact about $\mathbb{S}^0$. This is of course also the case of constructions and operations with orientations.

Answer 10

$\bigcap \emptyset = V$

Unfortunately, I have read more than one philosophical comment on the "set theoretic depth" of this logical triviality.

Answer 11

The terminal object of a category is the product over the empty set of objects.

Answer 12

I regard "negative thinking" in category theory as an example of cool vacuity: see e.g. https://ncatlab.org/nlab/show/negative+thinking. As category theory is not set theory, such vacuity does not necessarily involve the empty set directly, but the same principle of backwards generalization is used.

The fact that a set is uniquely determined by its elements (i.e., has no additional structure beyond the equality relation between its elements) is summarized by saying that a (-1)-category is a truth value: a morphism between two elements in a set is either true (the elements are the same) or false (they are not). So the morphisms in a 0-category (a set) are (-1)-categories (either true or false), just as the morphisms in a (1-)category are 0-categories (sets). This admits generalizations to situations where "truth" is a more subtle concept (e.g. parameter dependent).

Answer 13

The determinant of the $0\times 0$ matrix is $1$. First of all, it's the only way to make determinant of a direct sum of matrices be a product of determinants. Second, it is the sum of $0!$ terms each of which is a product of $0$ factors.

Answer 14

When defining a topology.

Do not say

the intersection of any two open sets is open

but instead say

any finite intersection of open sets is open

That way, you assert in particular that the empty intersection of open sets (i.e. the whole space) is open.

Of course we also postulate

an arbitrary union of open sets is open

telling us that the empty union (i.e. $\varnothing$) is open.

similar
Caratheodory's "Method I" for constructing an outer measure $m^*$ in a set $X$ starting from a set-function $E : \mathcal E \to [0,+\infty]$ is done like this:

$m^*(A) = \inf\sum_{i=1}^\infty E(A_i)$ where the inf is over all sequences $(A_i)_{i=1}^\infty \subseteq \mathcal E$ such that $\bigcup_{i=1}^\infty A_i \supseteq A$.

But then we have to include artificial hypotheses like "$\mathcal E$ covers the whole space" and "$\varnothing \in \mathcal E$ and $E(\varnothing) = 0$".
But instead we should do it like this

$m^*(A) = \inf\sum_{F \in \mathcal F}E(F)$ where the inf is over all countable collections $\mathcal F \subseteq \mathcal E$ such that $\bigcup_{F \in \mathcal F} F \supseteq A$

Then: we need not postulate that $\mathcal E$ covers $X$, since if there is no countable $\mathcal F$ with $\bigcup_{F \in \mathcal F} F \supseteq A$, then we simply get $m^*(A) = \inf\varnothing = +\infty$. And for the empty set, we have (possibly among others) the empty cover for $\varnothing$, so that $m^*(\varnothing) \le 0$, the empty sum. So we get $m^*(\varnothing) = 0$ even if (for example) $E$ is identically $+\infty$.

Answer 15

'Silly' but I like them anyways:

$$\prod_{y\in\emptyset}\left(\sum_{x\in \emptyset}x\right)=1,$$

and

$$\sum_{y\in\emptyset}\left(\prod_{x\in \emptyset}x\right)=0.$$

But, careful:

$$\sum_{S\subseteq \emptyset}\prod_{x\in\emptyset}x=1,$$

and

$$\prod_{S\subseteq \emptyset}\sum_{x\in\emptyset}x=0.$$

Answer 16

The ‘divides’ relation $\mid$ should properly be called a ‘has-a-multiple-of’ relation and defined without any reference to division as

$$ a \mid b \iff \exists c: ac = b $$

This definition implies $\forall a: a \mid 0$ (including $0 \mid 0$) and $\forall a: 0 \mid a \Longrightarrow a = 0$. And over the non-negative integers in particular, the relation becomes a complete lattice, where the infimum is the GCD and the supremum is the LCM.

This comes up as the inductive base case when computing the GCD/LCM of arbitrary sets of numbers. Sometimes people say that the GCD of no numbers doesn’t exist, but when the GCD is defined as the infimum of the $\mid$ relation, it becomes perfectly natural to set it to 0 at the empty set.

This can be taken advantage of in other mathematical definitions. The characteristic of a ring $R$ is commonly defined as ‘the smallest non-zero natural number $k$ such that $\forall x \in R: \sum^k x = 0_R$, or 0 if there is no such number’. Knowing the above, however, we can reformulate the definition thus:

$$ \operatorname{char}(R) := \operatorname{gcd}\, \left\{ k \in \mathbb N : \forall x \in R : \sum^k x = 0_R \right\}$$

which does not require separately considering zero and non-zero cases.

---

Somewhat related, though perhaps too silly to mention: division by zero is undefined. However, remainder of division by zero is not: it is simply the identity function.

One way to see this is to observe that no matter how Euclidean division by zero $(a \mapsto a \div 0)$ is defined, there is only one way to define the remainder $a \bmod 0$ such that the identity

$$ (a \div b) \cdot b + (a \bmod b) = a $$

is maintained. (Of course, if $a \div 0$ is defined, it will necessarily break the usual property that $|a - (a \div b) \cdot b| < |b|$ for every $a$ and $b$ where it is defined.)

Another is to notice that $(a \mapsto a \bmod b) : \mathbb{Z} \to \mathbb{Z}/b\mathbb {Z}$ is a unity-preserving homomorphism of rings. For $b = 0$, the codomain becomes $\mathbb{Z}/0\mathbb {Z} = \mathbb{Z}$, and the requirement that $1 \bmod 0 = 1$ forces the function to be the identity function. Setting $a \bmod 0 = a$ also agrees with the property that $b \mid a$ if and only if $a \bmod b = 0$.

Answer 17

The Generalized Continuum Hypothesis is the assertion that $2^\kappa=\kappa^+$ for all infinite cardinals $\kappa$, or in other words that the power set of a set of size $\kappa$ has the next larger cardinal size above $\kappa$.

If we consider all cardinals, rather than only the infinite cardinals, then the two provable instances of this equation occur in the following vacuous and near-vacuous facts:

• The power set of a set with $0$ members has $1$ member.

• The power set of a set with $1$ member has $2$ members.

All other instances of $2^\kappa=\kappa^+$, finite or infinite, are either false or independent of ZFC.

Answer 18

Zero is a limit ordinal, because it is the union of its elements.

Transfinite induction has two canonical statements. The "strong" statement, $$ (\forall \alpha)((\forall \beta)((\beta<\alpha) \rightarrow P(\beta)) \rightarrow P(\alpha))\rightarrow (\forall \alpha)P(\alpha), $$ doesn't split anything into cases. The version used most frequently in proofs says that any property preserved under unions and successors holds for all ordinals. Zero should rarely be a special case.

Also, "limit ordinals" should totally be called "colimit ordinals". The term "limit ordinal" refers to limit points in the order topology, thus excluding zero, but this is silly.

Answer 19

The empty product in a group $G$ is the unit of $G$. This is the only way to avoid mistake in calculations.

Set theory begins by the construction of finite ordinals. The first one is $\emptyset$ and is denoted $0$. The next one is $\{\emptyset\}$, which is not empty ! It is denoted $1$. More generally, every finite ordinal is defined only in terms of the empty set recursively: $n+1:=n\cup\{ n \}$. Physicists (or administrators, politicians, whoever is asked to fund mathematics) might find this pedantic, but it is actually powerful.

Answer 20

Zsbán Ambrus brings up an interesting example in the comments: the degree of the zero polynomial. The first time I was told about this issue I was told that it is largely a matter of convention. Well, maybe. Here is some evidence suggesting that $\deg 0 = \infty$:

• The most basic one: the zero polynomial has infinitely many roots in an algebraic closure.

• If one wants the degree of a polynomial to be a valuation, then we must define $\deg 0 = \infty$. This is the unique choice consistent with the requirements that $\deg fg = \deg f + \deg g$ and $\deg (f+g) \ge \text{min}(\deg f, \deg g)$, and it is necessary in order to make the corresponding absolute value nondegenerate.

• One way to say the above geometrically when $F = \mathbb{C}$ is that the degree should describe the order of the pole of $f$ at infinity on the Riemann sphere. The function $0$ decays faster than the reciprocal of any polynomial in the neighborhood of infinity. In fact, the sequence of functions $x^n$ converges uniformly to it in a neighborhood of infinity as $n \to \infty$.

• Another way to say the above is that, in the natural topology on $F[[x]]$, we have $x^n \to 0$. This can be appreciated even if you are, for example, a combinatorialist, because it allows you to say natural things about generating functions like $\frac{1 - x^n}{1 - x} \to \frac{1}{1 - x}$ as $n \to \infty$.

• One can also define the degree of $f$ as $[F[x]/(f(x)) : F]$, in which case again we find that $\deg 0 = \infty$. This is just a fancier version of the first reason.

Edit: As James Borger points out, the middle ideas are mistaken. Corrected, they actually suggest that $\deg 0 = -\infty$:

• $\deg 0 = -\infty$ is the unique choice consistent with the requirements that $\deg fg = \deg f + \deg g$ and $\deg (f + g) \le \max(\deg f, \deg g)$. With this definition, $|f| = 2^{\deg f}$ is now an absolute value.

• Geometrically, when $F = \mathbb{C}$ the function $0$ has a zero of infinite order at infinity, hence a pole of order $-\infty$.

• The relevant local ring here is really $F[[ \frac{1}{x} ]]$, and in the natural topology on this ring we have $\frac{1}{x^n} \to 0$.

• Another reason to like this definition is that it gives a uniform statement of the division algorithm on $F[x]$.

Answer 21

Recall that an abstract simplicial complex consists of a family of finite sets $K\subseteq 2^V$ such that $\sigma\in K$ and $\tau\subseteq\sigma$ implies $\tau\in K$. (Sometimes it is also assumed that $\{v\}\in K$ for all $v\in V$.) Elements of $K$ are called faces of $K$. Dimension of a face is its cardinality minus 1.

According to the above definition, if $S\not=\emptyset$, then the empty set $\emptyset$ is a face of the complex (of dimension $-1$, since it has $0$ elements). Applying the usual definition of simplicial homology gives us what is called reduced homology, which is often much better-behaved than the nonreduced one (obtained when we forget about the empty face).

In this context it is important to distinguish between the empty simplicial complex $K_e=\{\emptyset\}$ and the void simplicial complex $K_v=\{\}$, which may be understood as the $(-1)$-sphere and the $(-1)$-disk. (In particular $H_{-1}(K_e)=\mathbb{Z}$, $H_n(K_e)=0$ for all $n\not=-1$, and $H_m(K_v)=0$ for all $m$.)

Answer 22

One is not a prime number, but zero is! It's different than the other prime numbers in $\mathbb{Z}$, though, because it has height zero rather than height one.

Zero is prime in any integral domain. Remember that the trivial ring is not an integral domain.

Answer 23

I'm always amused (and unpopular for it) when mathematicians claim that some terminology or notation is "true" rather than "convention", as if there is some God of Mathematics out there who hands us definitions that us mere humans are forbidden to tamper with. Actually all mathematical notation is convention.

We could, if we agreed as a body to do so, define "$m+n$" to mean what it used to mean, unless $m=n=1$ in which case it means 3. It isn't forbidden, because we invented "$+$"; it belongs to us and we can choose what it means. We won't do that, though, not because it is wrong according to some objective source of truth, but because it would cause us a whole heap of trouble. For example we would have to rewrite a huge number of theorems to add exceptional cases, and we hate exceptional cases. It isn't a question of "truth" since all of modern mathematics could be correctly stated using the new definition.

In the same way, we choose to set the empty product equal to 1, and the empty sum equal to 0, because those are the conventions that make our symbol manipulations so much easier and simpler than any alternative conventions would make them. It isn't really different, except in degree, to excluding 1 from being a prime. We do that because otherwise we would have to spend all day writing "let $p$ be a prime other than 1". Qualitatively the reason is the same: we like our theorems to be cleaner, with fewer preconditions and fewer subcases. Simplicity and generality is a fundamental aesthetic of mathematics.

Answer 24

Pfister's local–global theorem has an interesting "vacuous" instance:

Remember that quadratic forms over a field $F$ of characteristic $\neq 2$ are diagonalizable, and hence a nondegenerate quadratic form can be represented by a sequence $\langle a_1, \dotsc, a_m \rangle$ of elements $a_i \in F^*$.

One can "add" diagonal quadratic forms by declaring $\langle a_1, \dotsc, a_m \rangle \oplus \langle b_1, \dotsc, b_k \rangle :=\langle a_1, \dotsc, a_m, b_1, \dotsc, b_k \rangle$ and one can multiply quadratic forms with natural numbers by declaring $n\cdot\langle a_1, \dotsc, a_m \rangle:=\underbrace{\langle a_1, \dotsc, a_m \rangle\oplus \dotsb \oplus \langle a_1, \dotsc, a_m \rangle}_{n \text{ times}}$.

If "$<$" is an ordering on $F$ (compatible with $+$ and $\cdot$), then to a quadratic form $f=\langle a_1, \dotsc, a_m\rangle$ one can assign its signature $\sigma(f)(<):=\#\{ a_i \mid 0 < a_i\} - \#\{a_i \mid a_i < 0\}$. That is, the signature is the number of positive minus the number of negative diagonal entries, according to the ordering "$<$". One can show that this signature is independent of the choice of diagonalization.

Thus, if we denote by $X(F)$ the set of all orderings of $F$, from a fixed quadratic form $f$ we obtain a well-defined map $X(F)\to \mathbb{Z}$, ${<} \mapsto \sigma(f)(<)$.

Pfister's local–global theorem says something about how much the map $\sigma(f)$ determines $f$:

Let $f$, $g$ be quadratic forms of the same rank over a field $F$. If $\sigma(f)=\sigma(g)$, then there exists $\ell \in \mathbb{N}$ such that $2^\ell\cdot f \cong 2^\ell \cdot g$ (i.e. the multiplied forms are isometric). [The theorem says more, but this is the relevant part.]

When $F$ is not orderable at all (e.g. if $F$ is algebraically closed or of characteristic $>0$, or $F=\mathbb{Q}_p$), then $X(F)=\emptyset$. Therefore there exists only one map $S\colon X(F) \to \mathbb{Z}$, i.e. any two quadratic forms $f$, $g$ satisfy $\sigma(f)=\sigma(g)$. Thus, for a non-orderable field, any two quadratic forms have isometric multiples: $2^\ell\cdot f \cong 2^\ell \cdot g$ for some $\ell$.

I learned this from a wonderful expository note Introduction to the real spectrum by Pete L. Clark.

Answer 25

Let $\bigotimes_{i \in I} M_i$ denote the tensor product of $R$-modules $M_i$. Then $\bigotimes_{i \in \emptyset} M_i$ is $R$.

(Reason: $\prod_{i \in \emptyset} M_i$ is the terminal object in the category of sets (see the answer of Eivind Dahl), i.e. a point, and multilinear maps on this to $N$ are just elements of $N$, i.e. homomorphisms $R \to N$.)

Another one: I know this is really silly and already contained somehow in the other answers, but anyway:

$$\prod_{i \in \emptyset} 0 = 1$$

Answer 26

The zero ring has Krull dimension $-\infty$. Why? It was already explained why $\sup(\emptyset)=-\infty$ in another answer, and the zero ring has no prime ideal chain, so that's it. Also notice that the proof of $\dim(A[T]) \geq \dim(A)+1$ works for every commutative ring $A$ (with equality for Noetherian $A$), including $A=0$. You cannot have $\dim(0)=0$ or even $\dim(0)=-1$ (Wikipedia states this as an option, which is absurd).

Accordingly, a commutative ring in which every prime ideal is maximal does not have to be of Krull dimension $0$. It is just $\leq 0$.

Answer 27

The full rank factorization of the 3x2 zero matrix is the product of a 3x0 matrix times a 0x2 matrix. There exist empty matrices with n>0 rows and 0 column. The 0x0 empty matrix is the only non nonsingular zero matrix.

Answer 28

A "convention" floating around in (co)homology theory is that given a space $X$ the quotient with the empty set is $X/\emptyset = X_+ = X \sqcup *.$ This seems rather arbitrary if one defines $X/A$ to be given by coequalizing the inclusion $A \subseteq X$ and a constant map $\smash{A \rightarrow * \underset{a}{\rightarrow} X}$ to some point $a \in A$.

But there is another possible definition: $X/A$ is the pushout $$\begin{array}{cc} A & \rightarrow & X\\ \downarrow & & \downarrow\\ * & \rightarrow & X/A \end{array}$$ Setting $A=\emptyset$ we can recover $X/\emptyset = X \sqcup * = X_+$. In particular this specializes to the counterintuitive identity $\emptyset / \emptyset = *$.

Answer 29

Let $p,q$ be Hilbert space projections. If $pq+qp$ is a non-zero projection then, where $\varphi$ is the golden ratio, $$\lVert pq\rVert=\frac{1}{\varphi}.$$

It follows very quickly from Walters - Anticommutator Norm Formula for Projection Operators… but is it vacuous? This answer to Is $p q + q p$ ever a projection? on MSE seemed to resolve the issue but has an issue.

Answer 30

$\forall M\in\mathsf{Set}\setminus\{\emptyset\} : \emptyset^M=\emptyset$, but

$\forall N\in\mathsf{Set}: N^\emptyset = \{\emptyset\}$.

Answer 31

Maybe it is not serious, but then in a sense none of the examples in this thread are.

Jordan decomposition represents every operator on a finite-dimensional space in a unique way as $S+N$ where $S$ is a diagonalizable operator, $N$ is a nilpotent operator, and $SN=NS$.

The Jordan decomposition of the zero operator is $0+0$. It thus is the only operator which is diagonalizable and nilpotent at the same time.

Similarly, the zero Lie algebra is the only one which is both semisimple and nilpotent. And one might also argue that it is not simple. Or is it?...

Answer 32

A The empty set is a covering map of any topological space. More generally, a covering map needn't be surjective (although many books claim just that). For example the inclusion of a closed and open subset of a space is a covering. Strangely, I would argue that this entails that the empty topological space, although connected, is not simply connected.

B Dually, given a field $K$, the zero algebra over $K$ is diagonal and in particular étale: the morphism of affine schemes $\varnothing \to \operatorname{Spec}(K) $ is étale. In the same vein, a nonzero constant polynomial over $K$ is separable (its nonexistent roots in an algebraic closure of $K$ are certainly distinct) . We may then say without any exception that the $K$-algebra $K[X]/(f(X))$ is étale iff $f(X)$ is a separable polynomial.

Answer 33

If we need to define the value of the Euler's function $\varphi$ at infinity, the best choice will be $\varphi(\infty)=2$ This is because $\varphi(n)$ is the number of generating elements in the cyclic group of order $n$, and so if $n\to \infty$, the the cyclic group tends to $\mathbb{Z}$ which has just two generating elements.

$0^0=1$ is another example, which is easy to prove for the natural zero, but it is not true for the real zero!

Answer 34

There are a number of examples from logic.

Many-sorted first-order logic with 0 sorts is propositional logic. A 0-ary function symbol is a constant symbol, and a 0-ary relation symbol is a proposition. These “degenerate” cases are actually quite interesting and important.

Answer 35

I suppose another example would be those proofs by induction in which you don't need a basis, although no nice examples come to mind instantly. Here I mean proofs in which you show that if $P(m)$ for all $m < n$, then $P(n)$. You don't need to show $P(n)$ holds for the smallest value of $n$, since it is vacuously true that $P(m)$ holds for all smaller values, and therefore the thing proved in the inductive step entails the smallest instance as a special case.

This works not only for natural numbers, but for infinite ordinals.

Answer 36

Is the span of the empty set in a vector space equal to $\lbrace 0\rbrace$, or does it have no span? The "correct" answer in my opinion is the latter. See Example 3.10.3 of https://web.archive.org/web/20201101123724id_/http://math.mit.edu/~rstan/ec/ec1.pdf for a reason:

3.10.3 Example. We give one simple example of the use of equation (3.34). We wish to count the number of spanning subsets of $V_n$. For the purpose of this example, we say that the empty set $\emptyset$ spans no space, while the subset $\{0\}$ spans the zero-dimensional subspace $\{0\}$.* If $W \in B_n(q)$, then let $f(W)$ be the number of subsets of $V_n$ whose span is $W$, and let $g(W)$ be the number whose span is contained in $W$. Hence $g(W) = 2^{q^{\dim W}} − 1$, since $\emptyset$ has no span. Clearly $$ g(W) = \sum_{T \le W} f(T) $$ so by Möbius inversion in $B_n(q)$, $$ f(W) = \sum_{T \le W} g(T) \mu(T, W) $$ Putting $W = V_n$, there follows $$ \begin {aligned} f (V_n) &= \sum_{T \in B_n(q)} g(T) \mu(T, V_n) \\ &= \sum_{k=0}^n {\boldsymbol n \choose \boldsymbol k} (-1)^{n-k} q^{n - k \choose 2} ( 2^{q^k} - 1 ). \end {aligned} $$

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* The standard convention is that the empty set spans $\{0\}$. If we wish to retain this convention, then we need to enlarge $B_n(q)$ by adding $\emptyset$ below $\{0\}$.

On the other hand, a reason (which I find unconvincing) for the span to be $\lbrace 0\rbrace$ is given by PBRMEASAP at https://www.physicsforums.com/archive/index.php/t-84017.html. This site has a discussion of whether the empty set is a vector space. The correct answer is that it isn't, because one of the axioms is the existence of an additive identity 0.

Update. I agree with the comments that the span of the empty set is $\lbrace 0\rbrace$. What I said above was foolish.