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Let $V$ be an infinite-dimensional vector space. If $B$ and $B'$ bases of $V$, then there they are bijective.

All the proofs I have seen (example) of it involve cardinal arithmetic, which I do not understand. Does there exist any proof which is different to the one linked or does not use cardinal arithmetic?

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    In problems of this kind one looks from some special structure that allows us to bypass the machinery of cardinal numbers and Axiom of Choice, building the correspondence from what that special structure provides. Here the assumption that $B$ and $B'$ are bases of V gives us that handle. – 2017-02-12
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    @hardmath Is this true without the axiom of choice? – 2017-02-12
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    @hardmath According to [Wikipedia](https://en.wikipedia.org/wiki/Dimension_theorem_for_vector_spaces), "the uniqueness of the cardinality of the basis requires only the ultrafilter lemma, which is strictly weaker". I suppose it doesn't say that the ultrafilter lemma is necessary, but I think if it was known that it was unnecessary, then Wikipedia would say that - they're usually pretty good about elucidating exactly how strong a version of the axiom of choice is necessary for foundational issues. – 2017-02-12
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    @DustanLevenstein: That's a good find. – 2017-02-12
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    Better find: under "[Statements consistent with the negation of AC](https://en.wikipedia.org/wiki/Axiom_of_choice#Statements_consistent_with_the_negation_of_AC)", "There exists a model of ZF¬C in which there is a vector space with two bases of different cardinalities." – 2017-02-12

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