Ytukyg

The question is quite simple: if $mathbb{V}$ is a vector space and $B$ and $B'$ are basis for $mathbb V$, then do $B$ and $B'$ have the same cardinality?

I've tried to answer the question as follows: suppose that $B'$ is bigger than $B$. Then there is a map $f:B'to B$ surjective but non-injective. Extending this to a linear map $f':mathbb Vtomathbb V$ we have that $mathbb V/ker f'cong mathrm{Im} f'=mathbb V$, with $ker f'$ being non-trivial. But is this really a contradiction? (of course it is in finite dimension, since we know that all basis have the same cardinality in that case, but this is exactly what we are trying to prove here...)

Your argument doesn't work, since - as mentioned in the comments - given any infinite set $A$ there is a map $f:Arightarrow A$ which is surjective but not injective (or injective but not surjective, if you prefer).

However, the statement is true, and the proof is simpler:

• Suppose $B, B'$ are infinite bases of $mathbb{V}$ with $vert Bvert<vert B'vert$.




• For each $bin B$, there is a finite $F_bsubseteq B'$ with $binlangle F_brangle$.




• Since the union of $vert Bvert$-many finite sets has cardinality $vert Bvert$, the set $$hat{B}=bigcup_{bin B}F_b$$ has cardinality $vert Bvert$.




• Since $vert Bvert<vert B'vert$, there must therefore be some $ain B'setminushat{B}$.




• But since $B$ spans $mathbb{V}$, $a$ can be written as a linear combination of elements of $B$, and hence can be written as a linear combination of elements of $hat{B}$. This contradicts the linear independence of $B'$.

As a quick pedagogical aside, here's an attempted proof which doesn't quite work:

Since $B$ is a basis for $mathbb{V}$, we have that every element of $B'$ can be represented as a finite linear combination of elements of the field of scalars $F$. Using the fact that the set of finite subsets of an infinite set has the same cardinality as the original set, the cardinality of the set of linear combinations of elements of $B$ is $vert Bverttimesvert Fvert$, and so we get $$vert B'vertlevert Bverttimesvert Fvert.$$

In case $F$ is "small" - that is, in case $vert Fvertlevert Bvert$ - then we get $vert B'vertlevert Bvert$ (since multiplication of two infinite cardinalities just results in the larger of the two). But in case $F$ is large, we don't get anything. E.g. this argument doesn't rule out the possibility of a vector space over a field of cardinality $aleph_{omega^3+omegacdot 842+17}$ with one basis of cardinality $aleph_{omega^2+9}$ and another basis of cardinality $aleph_0$.

Note that we've used the axiom of choice crucially in the above when we calculated the cardinality of $hat{B}$ (which we needed in order to conclude the existence of $a$). Without the axiom of choice, the argument above breaks down. This is the only essential use of choice - note that the second bulletpoint does not require choice, since there is a unique minimal choice of $F_b$.

The full axiom of choice is not needed, and the general study of how much choice is needed to prove various mathematical results - around vector spaces and other topics - is quite rich; I just want to point out the reliance on something beyond set theory without choice, here.