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My question is, if we suppose the Axiom of Choice or Zorn's Lemma, are any infinite dimensional vector spaces isomorphic as vector spaces? If not, are all countable vector spaces isomorphic as vector spaces? Or are all uncountable (or, say, $\aleph_1$ cardinality) vector spaces isomorphic as vector spaces?

Edit: I guess I've sort of answered part of my own question here, because an isomorphism must be a bijection, and there is no bijection between countable and uncountable sets. But I'm still curious about the other parts of the question . . .

Rasputin
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  • By countable vector space, do you mean a vector space that is countable as a set, or a vector space that is of countable dimension over its base field? – Servaes May 10 '17 at 21:18
  • This theorem holds if all vector spaces have bases "two vector spaces are isomorphic if and only if their bases have the same cardinality". – Maxime Ramzi May 10 '17 at 21:19
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    In general, vector spaces over the same field are isomorphic iff they have basis of equal cardinality. Period. Now, to prove that any vector space over some field has a basis you need AC (in fact, both claims are equivalent in FZ). – DonAntonio May 10 '17 at 21:20
  • @Servaes, sorry, I should fix that ambiguity in the question – Rasputin May 10 '17 at 21:21
  • @DonAntonio, I see. Does the conception of cardinality in this statement differentiate between different kinds of uncountability, or only between countable and uncountable? – Rasputin May 10 '17 at 21:22
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    @Rasputin It distinguishes between different kinds of uncountability. E.g. by Cantor's diagonal argument, the set of sets of real numbers is of strictly greater cardinality than the set of real numbers. In fact there are lots of levels of uncountable cardinality - a proper class of them! – Noah Schweber May 10 '17 at 21:24
  • @NoahSchweber, got it, that makes sense. Is the existence of these different levels of uncountability (e.g. $\mathbb{R}$, the power set of $\mathbb{R}$, the power set of the power set of $\mathbb{R}$, etc.) at all dependent on whether we accept the axiom of choice? – Rasputin May 10 '17 at 21:28
  • @Rasputin Nope, it's provable without using choice or any other complicated axioms of set theory at all. To prove that a set $A$ has strictly smaller cardinality than its powerset $\mathcal{P}(A)$, all you need is a tiny amount of the axiom of Separation; so separation + powerset (+ the axiom of infinity) gives you tons of distinct uncountable cardinalities. – Noah Schweber May 10 '17 at 21:28
  • Ahh ok got it, thanks! – Rasputin May 10 '17 at 21:30

2 Answers2

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Assuming the axiom of choice, two vector spaces over a fixed field are isomorphic if and only if they have bases of the same cardinality.

So if $V$ and $W$ are two vector spaces over some field $\Bbb F$, then they are isomorphic if and only if they they bases of the same cardinality.

Now, even with or without the axiom of choice, not all countable vector space are isomorphic. Note that $\Bbb Q^n$ and $\Bbb Q^m$ are isomorphic if and only if $m=n$, this is because finite dimensional vector spaces have the "usual theory" without appealing to choice. So you might want to be more careful in that formulation.

Asaf Karagila
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    +1. Note to the OP that the reason choice is invoked for the first sentence is because without choice, some vector spaces need not have bases. (Clearly if $V$ and $W$ have bases of the same cardinality, they're isomorphic, and this doesn't use choice.) – Noah Schweber May 10 '17 at 21:23
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There are infinitely many cardinal numbers for a base, all $\aleph_\alpha$, $\alpha \in \operatorname{Ord}$. If we have two vector spaces $V$ and $W$ over some field $\mathbb{K}$, Zorn tells us they both have a base. They are isomorphic iff these bases have the same cardinality. This is clearly necessary and sufficient.

Henno Brandsma
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