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For a vector space $V$, let $BS(V)$ be the set of cardinalities (not necessarily $\aleph$s) of bases of $V$. Of course, in ZFC each $BS(V)$ is a singleton, but supposing the axiom of choice fails, it is consistent to have $BS(V)=\emptyset$ or $\vert BS(V)\vert>1$.

Now let $\mathfrak{BS}$ be the class $\{BS(V): V\text{ a vector space}\}$. My question is:

What restrictions on $\mathfrak{BS}$ can be proved in ZF?

(The question I want to ask is "What are the possible values for $\mathfrak{BS}$?", but since without choice there is no canonical picture of what the cardinalities 'are' I'm not even sure how to phrase that.)

Of course, the question can be asked in general for generating sets of arbitrary algebraic structures. One more question I would love to know the answer to, but which I think is too broad, is How different is the situation when we ask the same question in this much greater generality?

I am asking this question for vector spaces over all fields, but I would also be interested in answers for vector spaces over specific (classes of) fields, especially (classes of) well-orderable fields.

To help make this question less hopelessly broad, let me ask a specific sub-question, the answer to which I'm sure is "no" (but I can't prove it):

Is it consistent with ZF that $\mathfrak{BS}$ is closed under finite intersections?

In general, what are the "best" (i.e., most structurally rich - closed under lots of operations) kinds of algebraic structure $\mathfrak{BS}$ can have?

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  • $\begingroup$ If you are able to talk about the subclass of script BS which arise from well-ordered bases, you should be able to show that subclass is closed under intersection. You might then explore what weakenings (larger subclasses) you have in which you can show things like closed under finite intersection. I recommend going through a universal algebra text and seeing which basic constructions are choiceless or require a minimum amount (e.g. Galois connections, congruence lattices). You may end up with something CS-theory people are doing. Gerhard "Giving Noah S's HS Theorem" Paseman, 2014.04.21 $\endgroup$
    – Gerhard Paseman
    Commented Apr 21, 2014 at 18:20

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Sizes of bases of vector spaces without the axiom of choice shows that assuming $\sf BPI$ we have that every two bases have the same cardinality. This means that $BS(V)$ is either empty, or a singleton.

So in a model of $\sf ZF+\lnot AC+BPI$ we have that $\frak BS$ is singletons and the empty set, which is definitely closed under finite (or otherwise) intersections.

As for a general structure, note that there is a strong restriction on the elements of $BS(V)$, when $V$ is a vector space over $F$. Namely, whenever $B\in BS(V)$ then $$|V|=|[F\times B]^{<\omega}|.$$

This means that there are bijections between the various $[F\times B]^{<\omega}$ sets. So even if $BS(V)$ has several different elements, they are all bounded below $|V|$.

So over a fixed field, if $B$ is a basis for a vector space of one cardinality, and $B'$ is a vector space of a different cardinality, then $\{B,B'\}$ cannot possible be a member of $\frak BS$.

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  • $\begingroup$ @Noah: Does that answer the second question? $\endgroup$
    – Asaf Karagila ♦
    Commented Apr 21, 2014 at 18:16
  • $\begingroup$ That's nice! Yes, it does. (I've deleted my previous comments since they're no longer relevant.) I wonder if we can make it closed under finite intersections and finite unions? In general, I'm curious what's the "best" algebraic structure it can have? $\endgroup$
    – Noah Schweber
    Commented Apr 21, 2014 at 20:05
  • $\begingroup$ Noah, it cannot possibly be closed under finite unions. It is impossible to have a vector space with a basis of $1$ element, and another of $2$ elements. $\endgroup$
    – Asaf Karagila ♦
    Commented Apr 21, 2014 at 20:16
  • $\begingroup$ Oh, doy. OK, maybe I ought to be restricting attention to infinite cardinalities. :) $\endgroup$
    – Noah Schweber
    Commented Apr 21, 2014 at 20:19
  • $\begingroup$ Also good. If $\kappa$ and $\lambda$ are two distinct $\aleph$'s then $\{\kappa,\lambda\}$ is not there. $\endgroup$
    – Asaf Karagila ♦
    Commented Apr 21, 2014 at 20:21

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