# Definition of an infinite dimensional vector space without the axiom of choice

A finite dimensional vector space is a vector space which has a basis (linearly independent spanning subset) with finite cardinality.

If we accept the axiom of choice then it can be proven that every vector space has a (not necessarily finite) basis. The following are then equivalent:

• vector space V has no basis with finite cardinality
• vector space V has a basis with infinite cardinality

My question is, in a context where we haven't necessarily assumed the axiom of choice, is there a standard for selecting one of the two above (not necessarily equivalent) characterizations to be the definition of an infinite dimensional vector space?

My preference would be to call a vector space infinite dimensional if it admits a basis with infinite cardinality. This then leaves open the possibility for vector spaces which have no defined dimensionality since they admit no basis. But I don't know if this is standard.

A related question: The dimension theorem states that all bases for a vector space have the same cardinality. Can this theorem be proven without the axiom of choice? If not then the definition of an infinite dimensional vector space as being a vector space with infinite cardinality basis is not so good because, without the dimension theorem, one could imagine both a countable and uncountable basis for the same vector space. In this case I would prefer to say the dimensionality of the vector space is undefined.

• I don't think "finite vector space" and "infinite vector space" are used in the way you suggest. Did you mean "finite-dimensional" and "infinite-dimensional"? Commented Jan 23 at 21:25
• What would be the advantage of Option 2? It seems strange to leave open the possibility of a vector space that is neither finite-dimensional nor infinite-dimensional. Commented Jan 23 at 21:26
• @AlexKruckman yes I meant infinite/finite dimensional. I’ll fix it when I get the chance Commented Jan 23 at 22:40
• @JairTaylor it also seems strange to assign a dimensionality to a vector space with no basis when dimensionality is otherwise the cardinality of the basis Commented Jan 23 at 22:41
• Well, it's the same convention with sets. If a set is in bijection with $\{1,2, \ldots, n\}$ for some $n$ then it is finite; otherwise, infinite. So in both cases being infinite is a negative claim. It's there in the name - in-finite, ie not finite. Commented Jan 23 at 22:48

Like many other things, this depends on what is the context in which you are working.

It is conceivable that you want to focus on vector spaces with a "well-defined dimension", namely, they have a basis, and all bases have the same cardinality.

Or that you want to limit yourself, at least, to vector spaces with bases (which you can at least reasonably bound in cardinality). As it is, indeed, consistent that there is a vector space with bases of different cardinality.

Or maybe you just want to refer to bases which are "not finitely dimensional", much like how an "infinite set" simply means "a set that is not finite".

• To be clear: is your third paragraph claiming that it is indeed not possible to prove that all bases have the same dimensionality without choice? Commented Jan 23 at 22:45
• Yeah. It is consistent that there is a vector space with two distinct cardinality of bases (without AC, of course). Commented Jan 24 at 2:15
• Thank you. One more question and then I can probably accept the answer. Everything you say makes sense that you might define things differently depending on what you care about. Here's the question: Are there people who actually work with these sorts of vector spaces with no basis (with no AC)? and if so, do those people have a "standard" definition for "infinite dimensional vector space" that could be referred to? Commented Jan 24 at 2:36

I believe the definition of infinite dimensional vector space is your first bullet: a vector space that does not have a finite basis.

Note however that most infinite dimensional vector space one would meet 'in daily life' are spaces of functions. I.e. they are not abstract, we can think of their elements as very concrete entities with their own characteristics and sometimes even names.

Also, in many of these cases the infinite dimensional vector space comes with a topology that allows you to conclude that certain infinite sequences of vectors have a limit, i.e. converge to another vector. As a special case of this we have that some infinite sums (series) converge, and so that vectors can be written as the sum of infinitely many others. For some of these topological vector spaces we then have the nice situation that every vector can uniquely be written as an infinite linear combination of vectors from a fixed countable set.

From an intuitive perspective this set functions as a basis, but strictly speaking it can only be called a basis if every vector can be written as a finite linear combination of elements from the set. Proving that bases in this strict sense exist generally requires choice, but showing that the more loose, more intuitive type of "basis" (allowing infinite, converging, linear combinations) exist can in concrete cases be done without the axiom of choice.

Standard example is Fourier analysis where the functions $$\cos nx$$ and $$\sin nx$$ for all $$n$$ together form a "basis" (in above loose sense) of the vector space of all sufficiently nice $$2\pi$$-periodic functions.

By contrast: any uncountable actual basis (whose existence is guaranteed by the AoC) with the property that every such function is writable as a finite linear combination of basis functions is very hard to picture and infinitely less elegant and useful than the not-actual-basis of sines and cosines. So in the end I think that there is no shame in being happy with these not-actual-bases instead.