# Sizes of bases of vector spaces without the axiom of choice

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?)

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## 1 Answer

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.]

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.

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Very interesting! How strong is the assertion that the dimension is well defined? Does it imply BPIT or a weak version of it? – Asaf Karagila Apr 7 '12 at 10:01
My guess is that it's a very weak assertion. What does it buy you when the dimension is $\infty$ (that is, when there is no basis)? I conjecture that you could have all your favorite "pathological" sets in a model where vector space dimension is well defined; you'd just need to "protect" the pathologies by ensuring that all vector spaces into which they inject have dimension $\infty$. – David Milovich Apr 9 '12 at 16:49
If I was looking for some kind of reversal, I would play with the $F_2$-vector space of all functions from a given set $S$ to $F_2$. There a basis is exactly a minimal family of subsets of $S$ such that every subset of $S$ is a symmetric sum of finitely many sets from the family. In any case, I wouldn't be surprised if the literature has already answered the questions in your comment. I'm just not very familiar with this literature. – David Milovich Apr 9 '12 at 17:03
I can think of one pathology where you can't "avoid" a basis: the power set $W$ of an amorphous set $S$ has an $F_2$-basis: the singletons and $S$ itself. Try this: can you prove $dim(W)=|S|+1$ from the amorphousness of $S$? – David Milovich Apr 9 '12 at 17:28
I actually think of no basis as "no dimension". Take for example $\ell_p$ spaces in the presence of "All sets of reals have Baire property", this makes $\ell_p$ and $\ell_q$ non-isomorphic as vector spaces. Both have no Hamel basis. Dimension is only defined for vector spaces which already have some basis, and any other basis have the same cardinality. I am not sure that I understand your example about the amorphous set. Do you suggest that there is a well defined notion of dimension or that there is none? – Asaf Karagila Apr 9 '12 at 17:41