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.

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    $\begingroup$ It sounds as though the common generalization is that we are looking at the set of all finite linear combinations of an uncountable set. $\endgroup$ Sep 2, 2015 at 16:18
  • $\begingroup$ @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. $\endgroup$ Sep 2, 2015 at 16:21
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    $\begingroup$ 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. $\endgroup$
    – priel
    Sep 2, 2015 at 17:03
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    $\begingroup$ A very pretty way of exhibiting an explicit continuum size $\mathbb{Q}$-linearly independent set of reals was given by François Dorais here: mathoverflow.net/a/23206/78711 $\endgroup$
    – Todd Trimble
    Sep 2, 2015 at 17:47
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    $\begingroup$ 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 $B$." $\endgroup$ Sep 3, 2015 at 0:34

1 Answer 1


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.

  • $\begingroup$ 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''? $\endgroup$ Nov 23, 2015 at 14:25
  • $\begingroup$ @Arnold: just observe that two simple and nonisomorphic submodules of a module must have trivial intersection. $\endgroup$ Nov 23, 2015 at 17:06
  • $\begingroup$ 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? $\endgroup$ Nov 23, 2015 at 17:58
  • $\begingroup$ @Arnold: yes, but that's true too, as long as all of the submodules are simple. $\endgroup$ Nov 23, 2015 at 21:18
  • $\begingroup$ Then i am lacking knowledge about some basic properties of these concepts. Where can I inform myself? $\endgroup$ Nov 23, 2015 at 21:40

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