Mathematics with the negation of AC - MathOverflow

Mathematics with the negation of AC

2013-01-22T12:41:58Z

Clearly Very important results in Math require the Axiom of choice, for example "any vector space has a base". But in the absence of AC (i.e., only in ZF) it is possible that a vector space has no basis.

In another direction append the negation of AC to ZF. What happens to algebra or analysis now ? If you know any Theorem in this new system (other than those that can be drived in ZF alone) please let me know. A fine reference would also be helpful.

Answer by arsmath for Mathematics with the negation of AC

2013-01-22T13:58:54Z

It's tempting to think that if you add the negation of axiom of choice, then you can prove things like "all sets of reals are Lebesgue measurable," but it's not quite that easy. To really get a grip on what AC adds to ZF, I suggest familiarizing yourself with Gödel's constructible universe, at least in broad outlines. Gödel iteratively constructs (using only ZF) a hierarchy of sets, where (roughly) at each stage he adds the sets that set theory requires. This is a model of ZF. He then proves a surprising result: in this model, the axiom of choice holds.

That means that if the negation of the axiom of choice is true, then floating out there, somewhere, outside this model, is a collection of sets that lacks a choice function. Nothing tells us where this mysterious collection is -- if you stick to the types of constructions Gödel uses, you'll miss it.

What's interesting, then, is not so much the negation of AC, but the addition of axioms that let you do things that imply that the axiom of choice is false. One example would be the axiom "all sets of reals are Lebesgue measurable". Another, more dramatic one is the axiom of determinacy, which implies that all sets of reals are measurable, and much else besides.

Answer by Asaf Karagila for Mathematics with the negation of AC

2013-01-22T14:11:22Z

The negation of the axiom of choice only allows us to prove that there is some set which cannot be well-ordered. There is some family of non-empty sets whose product is empty. There is some partially ordered set in which every chain is bounded, but there is no maximal element. And so on.

Of course, from a family without a choice function we can easily construct a set which cannot be well-ordered, and with it a partial order witnessing the failure of Zorn's lemma, and other examples. But we cannot really say much about this family. Is it a family of finite sets? Is it a family of countable sets? Is this family well-orderable? And so on.

It turns out that mathematics is not just "you either love someone or you hate them". If you have no choice, you can still have plenty of degrees of choice, and without pointing out how much choice you have, or don't have, it's very hard to say much.

Furthermore it is possible that the axiom of choice fails so very very far up the cumulative hierarchy that no set used by any mathematician (except set theorists, maybe) is a witness for this failure. In such universe it is true that some theorems will fail (e.g. there will be a commutative unital ring without a maximal ideal, and there will be a vector space without a basis), but their failure occurs so far beyond our interest that it's just as well possible to assume that it doesn't happen.

All we can say, in case we assume the negation of AC, that all those principles equivalent to the axiom of choice fail somewhere, but we cannot possible give an intelligent answer about where these failures occur.

You may also be interested in this math.SE answer of mine, which discusses a similar question.