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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.

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The negation of AC is next to useless as an axiom, one reason being that the lowest-rank set violating AC can be arbitrarily high in the cumulative hierarchy. You may get interesting mathematics by adopting some axioms that contradict AC (such as the axiom of determinacy). – Emil Jeřábek Jan 22 '13 at 12:50
There are lots of "theorems" in this new system which are not provable in ZF, such as "There is a vector space without a basis", or "There is a partial order without a maximal chain", or "GCH does not hold". But I think your question is upside-down: in my opinion, it is more interesting (though logically equivalent) to ask for statements implying AC, rather than for consequences of non-AC. – Goldstern Jan 22 '13 at 13:21
You should be asking: what interesting results arise when we use axioms which imply the negation of AC. The answer is: many, even in computer science. The negation of AC by itself is a bit too weak to prove much of interest. – Andrej Bauer Jan 22 '13 at 14:25
There are a few questions that I think are related to "What if we didn't have AC?". I asked one about Hahn-Banach some time ago that you may be interested in:… – Loop Space Jan 22 '13 at 15:00
up vote 13 down vote accepted

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.

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"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." An extreme version of this is precisely the situation in NF. – Adam Epstein Jan 22 '13 at 15:08

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.

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While you are certainly correct, you are also missing quite a huge chunk of what we do know about models of $\lnot$AC. Perhaps you should familiarize yourself with the notion of relative constructibility introduced by Levy, denoted by $L(A)$, this is the same construction as Godel's only we start with the transitive closure of $A$ instead of the empty set. It is not uncommon to find $A$ such that the resulting model is not a model of AC. For example add $\omega_1$ Cohen reals to $L$ and consider $L(\mathbb R)$ of the generic extension. – Asaf Karagila Jan 22 '13 at 16:44

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