Suppose I want to show that two infinite-dimensional vector spaces are not isomorphic to each other. This is easy if my vector spaces are finite-dimensional as I just find a basis for each and show they are of different size, since finite-dimensional vector spaces are isomorphic if and only if they have the same dimension.

If the vector spaces have different cardinality then of course we would not be able to find a bijection between them, so they are not isomorphic, but what if the cardinality is the same? I know I need to show that every bijection is not linear, but that sounds like a lot of work. What is the best way to go about it generally?


1 Answer 1


Two vector spaces (over the same field) are isomorphic iff they have the same dimension - even if that dimension is infinite.

Actually, in the high-dimensional case it's even simpler: if $V, W$ are infinite-dimensional vector spaces over a field $F$ with $\mathrm{dim}(V), \mathrm{dim}(W)\ge \vert F\vert$, then $V\cong W$ iff $\vert V\vert=\vert W\vert$. In particular, if $F=\mathbb{Q}$, two infinite-dimensional vector spaces over $F$ are isomorphic iff they have the same cardinality.

So, for example:

  • As vector spaces over $\mathbb{Q}$, $\mathbb{R}$ and $\mathbb{C}$ are isomorphic (this assumes the axiom of choice).

  • The algebraic numbers $\overline{\mathbb{Q}}$ are not isomorphic to the complex numbers $\mathbb{C}$ as vector spaces over $\mathbb{Q}$, since the former is countable while the latter is uncountable.

EDIT: All of this assumes the axiom of choice - without which, the idea of "dimension" doesn't really make sense. See the comments for a bit more about this.

  • $\begingroup$ Ah, this wasn't covered in lectures. How do I in general find the dimension of an infinitely dimensional vector space? I heard it is not always possible to construct a basis, e,g for the set of continuous functions from [a,b] to R. $\endgroup$
    – user85798
    Commented Oct 26, 2015 at 19:19
  • $\begingroup$ Assuming the axiom of choice, it is always possible to construct a basis (or rather, a basis always exists; "construct" is a funny word where the axiom of choice is involved :D). Without the axiom of choice, infinite-dimensional vector spaces are indeed extremely strange. Surprisingly, the statement "Every vector space has a basis" is equivalent to the axiom of choice; this was proved by Andreas Blass. $\endgroup$ Commented Oct 26, 2015 at 19:20
  • $\begingroup$ The most straightforward way to show that bases exist is via Zorn's lemma: given a vector space $V$ over a field $F$, let $\mathbb{P}$ be the partial order of all linearly independent subsets of $V$, ordered by inclusion. Then by Zorn, $\mathbb{P}$ has a maximal element; and it's easy to see that a maximal linearly independent set must be a basis. Of course, you now really ought to prove Zorn's lemma from the axiom of choice, which is nontrivial, but usually in early algebra classes you are allowed to just assume Zorn. $\endgroup$ Commented Oct 26, 2015 at 19:22
  • $\begingroup$ Meanwhile, the proof that two vector spaces with bases $B_0$ and $B_1$ of the same cardinality are isomorphic is completely straightforward, and doesn't use choice at all. By the way, it is consistent with ZF (=set theory without choice) that $\mathbb{R}$ and $\mathbb{C}$ are not isomorphic as $\mathbb{Q}$-vector spaces, but this is quite hard to show. $\endgroup$ Commented Oct 26, 2015 at 19:24
  • $\begingroup$ By the way, note that without a basis, the word "dimension" makes little sense - at best, we can use the phrase "infinite-dimensional" as slang for "does not have a finite basis." But you should be careful about this - using the word "dimension" in the context of vector spaces may lead people to assume you're assuming choice. $\endgroup$ Commented Oct 26, 2015 at 19:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .