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Unlike the reals considered as a vector space over the rationals, I know of a number of nice examples of vector spaces with uncountable dimension that have a nice basis.

For example, the space of rational functions on the real or complex numbers with complex coefficients whose numerator degree does not exceeds the denominator degree has as basis the fractional functions $1/(1+sx)^k$ with arbitrary complex $s$ and positive integers $k$.

Another example is the space of linear combinations of exponential functions $e^{sx}$ on the real or complex numbers.

Is there a common generalization? Or even some sort of general theory behind these examples?

Edit: Given the discussion so far, let me make the question more precise: Is there a fairly general way of proving that certain nice uncountable families of nice functions on some nice domain have the property that any finite subset is linearly independent? One kind of niceness criterion (but not the only relevant one) could be that the space spanned by the functions from the family can be characterizd in a basis-free way. Another niceness criterion is that the proof of linear independence should be not completely trivial.

Arnold Neumaier
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    It sounds as though the common generalization is that we are looking at the set of all finite linear combinations of an uncountable set. – Anthony Quas Sep 02 '15 at 16:18
  • @AnthonyQuas: But this has in general no easily described basis, and the basis might be finite or countable. One needs the space of all finite linear combinations of an uncountable set where every finite subset is linearly independent. This is a nontrivial condition not easily established. For example, the uncountable set of all polynomials in a variable $x$, or all linear functions does not qualify. – Arnold Neumaier Sep 02 '15 at 16:21
  • An example where there exists an explicit basis but it is quite nontrivial to find is a field of rational functions over $\mathbb{Q}$ in an arbitrary set of variables, considered as a vector space over $\mathbb{Q}$; see this answer for details. – Eric Wofsey Sep 02 '15 at 16:27
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    If you take any set, finite, countable or uncountable, then a standard construction produces the free vector space over this set and the latter is, by the very definition, a basis thereof. As a soecial case, if you have family of, say functions in a function space, then if they are linearly independent the free vector space is just their linear hull. Both of your examples are of this form. By the way, in analysis the free locally locally convex space over a topological space (or variants thereof) are perhaps more interesting. One still gets a basis, but not in the algebraic sense. – priel Sep 02 '15 at 17:03
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    A very pretty way of exhibiting an explicit continuum size $\mathbb{Q}$-linearly independent set of reals was given by François Dorais here: http://mathoverflow.net/a/23206/78711 – Todd Trimble Sep 02 '15 at 17:47
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    I think the idea of constructing the free vector space on an arbitrary uncountable set may be missing OP's point. If I understand it correctly, what's wanted is a nice vector space with a nice basis; that is, the vector space has to have some description (such as, rational functions with that condition on degrees) other than just "the vector space generated by the elements of the nice set $B$." – Gerry Myerson Sep 03 '15 at 00:34
  • I think we are talking at cross purposes. Thus concerning the second sentence, how do express $\dfrac 1{(x-1)^2}$ as a linear combination of such functions? – priel Sep 03 '15 at 07:33
  • and @Gerry Myerson, how is the third sentence not the kind of description that you decry?. Also I can't parse the third line of the edit of the Op in any way that makes sense to me. Finally, on the positive side, is the space of finite sequences indexed over an uncountable set the kind of example you are looking for? (If I have misunerstood your query, please just ignore this). – priel Sep 03 '15 at 07:48
  • @priel: I corrected my statements. I consider the two examles given as being nice in different ways, the first one satisfying my explicitly suggested possible niceness criterion. I consider the second nice, too, though I don't know of an independent characterization of the space it generates. – Arnold Neumaier Sep 03 '15 at 09:56

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The linearly independent set $\{ e^{sx} \}$ is generated by a simple mechanism: namely, it consists of eigenvectors for an operator $\frac{d}{dx}$ acting on a vector space all of whose eigenspaces are $1$-dimensional. The rational functions, I think, don't naturally appear in this way but they are all annihilated by some polynomial differential operator in the Weyl algebra $k \left[ x, \frac{d}{dx} \right]$.

In general if you have an algebra $A$ acting on a vector space $V$ it's interesting to look at the vectors $v \in V$ such that the $A$-submodule $Av$ generated by them is simple. (This is one of a few possible natural generalizations of being an eigenvector.) Then if $v_i, i \in I$ are vectors such that the corresponding $A$-submodules $A v_i$ are both simple and nonisomorphic, the $v_i$ are linearly independent.

Qiaochu Yuan
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  • Thanks. Could you please give a reference containing a bit more details about ''If the corresponding A-submodules are simple and nonisomorphic then the corresponding vectors are linearly independent''? – Arnold Neumaier Nov 23 '15 at 14:25
  • @Arnold: just observe that two simple and nonisomorphic submodules of a module must have trivial intersection. – Qiaochu Yuan Nov 23 '15 at 17:06
  • But wouldn't you have to show that the intersection of a submodule and the direct sum of finitely many other submodules have trivial intersection? – Arnold Neumaier Nov 23 '15 at 17:58
  • @Arnold: yes, but that's true too, as long as all of the submodules are simple. – Qiaochu Yuan Nov 23 '15 at 21:18
  • Then i am lacking knowledge about some basic properties of these concepts. Where can I inform myself? – Arnold Neumaier Nov 23 '15 at 21:40
  • @Arnold: maybe a ring theory textbook? Simplicity is doing all of the work here; none of the statements I'm making are particularly deep. – Qiaochu Yuan Nov 25 '15 at 02:30