Simpson’s book “Subsystems of Second-Order Arithmetic” shows how a large amount of mathematics can be expressed in a language that talks only about integers and sets of integers. This language he calls Z_2, or second-order arithmetic (though strictly speaking it’s a two-sorted first-order language). Moreover, even with a fairly weak set of axioms, one can prove a large number of standard mathematical theorems. In particular, the axiom of choice is not used. Anything provable in Simpson’s book is provable in ZF.

There is a slight catch, in that you do lose some generality in the translation from ordinary mathematical language into the language of Z_2. The first sacrifice is that because we’re only allowed to talk about sets of integers, “intrinsically uncountable” statements will not be expressible. So for example, you can prove a version of the Hahn-Banach theorem, but it will only be for separable spaces.

The second sacrifice, which is more relevant to the current discussion, is that you can’t talk about “arbitrary sets,” even finite ones. Sets are always sets of integers, which have a canonical ordering on them among other things. Returning to the example of the algebraic closure of Q, in Z_2 the uniqueness problem mentioned above is “defined out of existence.” That is, the whole problem of arbitrary sets of roots of polynomials with no canonical ordering on them doesn’t come up, because you can’t express such things in Z_2. You can only talk about an algebraic closure of Q that is built up from sets of integers, and for such things you can prove uniqueness up to isomorphism.

The point is that these sacrifices sound like ones that Tim is willing to make. The answer to the question “How far can you get?” is therefore “Very far indeed.” It’s quite difficult to find examples of theorems that are not provable in , the strongest set of axioms considered in Simpson’s book, other than by “cheating” and picking intrinsically uncountable statements. The one example I know of a well-known theorem not provable in is the Robertson-Seymour graph minor theorem. This uses some fancy transfinite induction at some point. Even then it’s conceivable that the axiom of choice isn’t really needed, but I’m not familiar enough with the details to be able to say for sure.

But for the sets that matter we can find ways of making the choices explicit.

There are counterexamples to this in analysis, at least. For example, to construct a finitely-additive measure on c_0 that is not countably additive requires something the axiom of choice. To construct a non-trivial element of the Stone-Cech compactification of the integers requires something like the axiom of choice.

Going back to your main question, there is a sense of “matters” that I don’t pretend to be able to define precisely, in which it doesn’t matter to me that not all algebraic closures of the rationals are isomorphic. The sense is something like this. For convenience we like to talk about “arbitrary” objects, but what actually matters is explicit objects. When we talk about the arbitrary ones we can’t get hold of their elements (because nothing is described), so we’re often forced to use the axiom of choice. But for the sets that matter we can find ways of making the choices explicit.

I don’t necessarily believe that that last sentence always holds, but it’s there to give some idea of what I’m talking about (and which I’m sure set theorists have thought about far more deeply than I am ever likely to).

In the paper by Hodges that I cited above, he demonstrates some other “weird” facts: In ZF, one cannot prove any of the following statements.

1. Let L be an algebraic closure of Q. Then Gal(L/Q) is non-trivial.
2. Let L be an algebraic closure of Q. Then there is a non-trivial absolute value on L.
3. Every principal ideal domain either has prime elements or is a field.

The argument would be like this. Let be any algebraic closure of the rationals. I shall build inside a subset that’s countable and algebraically closed, which, by the minimality of , will prove that is countable.

The construction of is the obvious one. I start with the rationals. Then I …

Actually, I’m even more confused. I was about to build up in stages, but can’t one do it in just one stage? The algebraic numbers form an algebraically closed field, so they’re obviously the algebraic closure of the rationals. And that seems to be true even if I define “algebraic number” to mean “solution in of a polynomial with integer coefficients”.

I can see how one might need Zorn’s lemma to build an isomorphism from the usual algebraic numbers to the algebraic numbers in (since as you build up you find roots that are algebraically indistinguishable so you have to make arbitrary choices about which one corresponds to which), but I don’t see how the algebraic numbers in could fail to be countable.

Looking at Wikipedia I see that it says that the algebraic numbers are the algebraic closure of the rationals, but that Zorn’s lemma is needed to prove that every field has an algebraic closure.

These exotic algebraic closures have some interesting features. See for example W. Hodges, “Lauchli’s algebraic closure of Q,” Math. Proc. Camb. Phil. Soc. 79 (1976), 289-297.

ZF also doesn’t show that every infinite finitely-branching tree has an infinite path (Koenig’s lemma). Again it might be the case that if you’re studying (say) the game tree of some specific combinatorial game, you might be able to construct infinite paths explicitly whenever you need them. I imagine, though, that maintaining that level of logical hygiene might get tiresome after a while. Our intuition that there has to be an infinite path is so strong that it would be easy to forget to define explicitly every infinite path we needed.

You need countable choice to show that an infinite set has a countable subset. I don’ t know if that’s the kind of result you had in mind.

for all n, there exists m, R(n,m)

then T also proves

there exists f, for all n, R(n,f(n))

[These results can be found for instance in Troelstra and van Dalen - Constructivism in Mathematics]

So, you will not find any problems in the constructive parts of mathematics, e.g. in the theory of the rationals.

I think there are extensions of this to parts of mathematics with classical logic, but I will have to check.