Assuming the axiom of choice does not hold we have that there is a vector space without a basis. The situation can be, in some sense, worse. It is consistent that there are vector spaces that have two bases with completely different cardinalities. 


Is anything known on when a vector space is spanned by sets of different cardinalities, and on the relation between those cardinalities? 


Is there a known relation between common choice principles (BPIT, DC, etc.) and possible cardinalities of a vector space? (For example, does BPIT implies that every two bases have the same size?)



Sizes of bases of vector spaces without the axiom of choice
Yes, ZF+BPIT implies that vector space dimension is well-defined.
[Edit: some Googling shows that James Halpern gave the same answer [back in the 1960s](http://www.ams.org/journals/proc/1966-017-03/S0002-9939-1966-0194340-1/S0002-9939-1966-0194340-1.pdf).]


Working in ZF+BPIT, fix a field $F$ and an $F$-vector space $V$ and bases $A$ and $B$ of V. That is, each element of $V$ is a unique $F$-linear combination of elements of $A$; likewise for $B$. For each $a\in A$, let $S\_a$ be the minimal subset of $B$ such that $a$ is spanned by $S\_a$. Each $S\_a$ is finite; give it the discrete topology. Let $X=\prod\_{a\in A}S\_a$, which is nonempty by BPIT (and is compact Hausdorff). By Schroeder-Bernstein, it suffices to show that some $f\in X$ is injective. By compactness, it suffices to show that for every finite subset $K$ of $A$, there is an $f\in X$ that is injective on $K$. Since each $\prod\_{a\in A\setminus K}S\_a$ is nonempty by BPIT, it suffices to show that there is an injection in every $\prod\_{a\in K}S\_a$. That is a nice little linear algebra exercise you can solve in ZF using the finite case of Hall's marriage theorem.


