Finding a basis of an infinite-dimensional vector space?

Question

The other day, my teacher was talking infinite-dimensional vector spaces and complications that arise when trying to find a basis for those. He mentioned that it's been proven that some (or all, do not quite remember) infinite-dimensional vector spaces have a basis (the result uses an Axiom of Choice, if I remember correctly), that is, an infinite list of linearly independent vectors, such that any element in the space can be written as a finite linear combination of them. However, my teacher mentioned that actually finding one is really complicated, and I got a sense that it was basically impossible, which reminded me of Banach-Tarski paradox, where it's technically 'possible' to decompose the sphere in a given paradoxical way, but this cannot be actually exhibited. So my question is, is the basis situation analogous to that, or is it actually possible to explicitly find a basis for infinite-dimensional vector spaces?

Answer 1

It's known that the statement that every vector space has a basis is equivalent to the axiom of choice, which is independent of the other axioms of set theory. This is generally taken to mean that it is in some sense impossible to write down an "explicit" basis of an arbitrary infinite-dimensional vector space. On the other hand,

• Some infinite-dimensional vector spaces do have easily describable bases; for example, we are often interested in the subspace spanned by a countable sequence $v_1, v_2, ...$ of linearly independent vectors in some vector space $V$, and this subspace has basis $\{ v_1, v_2, ... \}$ by design.
• For many infinite-dimensional vector spaces of interest we don't care about describing a basis anyway; they often come with a topology and we can therefore get a lot out of studying dense subspaces, some of which, again, have easily describable bases. In Hilbert spaces, for example, we care more about orthonormal bases (which are not Hamel bases in the infinite-dimensional case); these span dense subspaces in a particularly nice way.

Answer 2

The "hard case" is essentially equivalent to this one:

Find a basis for the real numbers $\mathbb{R}$ over the field of the rational numbers $\mathbb{Q}$.

The reals are obviously an extension field of the rationals, so they form a vector space over $\mathbb{Q}$. It should be clear that such a basis has to be uncountable (for if it were countable, the reals would likewise also be countable).

It should also be clear that such a basis is a subset of $\{1\} \cup \mathbb{R} \setminus \mathbb{Q}$. The trouble is, that the power set of the reals is "so big" that it's not even clear how to name the sets we need to apply the axiom of choice TO. Linearly independent subsets however, DO satisfy the requirements for Zorn's Lemma, a form of the Axiom of Choice.

A relatively easy-to-follow proof of the existence of a basis for any vector space using Zorn's Lemma can be found here: Link