If $B_{1},B_{2}$ are two bases for a vector space $V$, prove that there exists a bijection $f:B_{1}\rightarrow B_{2}$.
It is known to all that the statement is true for finite-dimensional vector spaces, with quite a lot of elegant proofs. I suppose there at least should be one or two similarly elegant proofs for the infinite-dimensional version, assuming that the Axiom of Choice is correct.
Clearly, if any one basis is finite then they are both finite. So we may assume they are both infinite.
For each vector $v$ in $B_1$ there is a finite subset $S_v$ of $B_2$ that suffices to express $v$ as a linear combination. Moreover, every vector in $B_2$ must occur in one of the $S_v$, lest a proper subset of $B_2$ span $B_1$ and hence $V$. Therefore, $$|B_2| = \left|\bigcup_{v\in B_1}S_v\right|\leq \sum_{v\in B_1}|S_v| \leq \sum_{v\in B_1}\aleph_0 = \aleph_0|B_1| = |B_1|.$$ Symmetrically, $|B_1|\leq |B_2|$.
