Let’s say I consider the Banach–Tarski paradox unacceptable, meaning that I would rather do all my mathematics without using the axiom of choice. As my foundation, I would presumably have to use ZF, ZF plus other axioms, or an approach in which sets were not fundamental.

Suppose that all I want is enough analysis to express all existing theories in physics. Is ZF enough? If not, then is there any attractive, utilitarian system of the form ZF+x, where x represents some other axiom(s), that does suffice, without allowing Banach-Tarski?

Wikipedia has a list of statements that are equivalent to choice: http://en.wikipedia.org/wiki/Axiom_of_choice#Equivalents The only one that seems obviously relevant is Blass’s result that you need choice to prove that every vector space has a basis. But if all I care about is vector spaces that would actually be used in physics (probably nothing fancier than the space of functions from Rm to Rn), does this matter? I.e., are the spaces for which you need choice to prove the existence of a basis too pathological to be of interest to a physicist? In cases of physical interest, it seems like it would be trivial to construct a basis explicitly.

Is Solovay’s theorem relevant? I’m confused about the role played by the existence of inaccessible cardinals.

I’m a physicist, not a mathematician, so I would appreciate answers pitched at the level of a dilettante, not that of a professional logician.

[EDIT] André Nicolas asks: “[…] why should Banach-Tarski be unacceptable?” Fair enough. Let me try to clarify what I had in mind. The real number system contains stuff that is physically meaningless, but (a) I have a clear idea of which of its features can’t mean anything physical (e.g., the distinction between rationals and irrationals), and (b) doing math in Q would be much less convenient than doing math in R. Similarly, I might prefer to think of my dy‘s and dx‘s as infinitesimals, and although those are unphysical, I understand what’s unphysical about them, and they’re convenient. But when it comes to choice, it’s not obvious to me how to distinguish physically meaningful consequences from physically meaningless ones, and it’s not obvious that I would lose any convenience by limiting myself to ZF.

**Answer**

An interesting question, that would take many pages to begin to answer! We make a small disjointed series of comments.

In the last few years, there has been a systematic program, initiated by Friedman and usually called Reverse Mathematics, to discover precisely how much we need to prove various theorems. The rough answer is that for many important things, we need very much less than ZFC. For many things, full ZF is vast overkill. Small fragments of second-order arithmetic, together with very limited versions of AC, are often enough.

About the Axiom of Choice, for a fair bit of basic analysis, it is pleasant to have Countable Choice, or Dependent Choice, at least for some kinds of sets. We really want, for example, sequential continuity in the reals to be equivalent to continuity. One could do this without full DC, but DC sounds not unreasonable to many people who have some discomfort with the full AC. This was amusingly illustrated in the early 20-th century. A number of mathematicians who had publicly objected to AC turned out to have unwittingly used some form of AC in their published work.

Next, bases. For finite dimensional vector spaces, there is no problem, we do not need any form of AC (though amusingly we do to prove that the Dedekind definition of finiteness is equivalent to the usual definition.)

For *some* infinite dimensional vector spaces, we cannot prove the existence of a basis in ZF (I guess I have to add the usual caveat “if ZF is consistent”). However, an algebraic basis is not usually what we need in analysis. For example, we often express nice functions as ∑∞0anxn. This is an infinite “sum.” The same remark can be made about Fourier series. True, we would use an algebraic basis for R over Q to show that there are strange solutions to the functional equation f(x+y)=f(x)+f(y). But are these strange solutions of any conceivable use in Physics?

Finally, why should the Banach-Tarski result be unacceptable to a physicist *as* physicist? It is easy to show that the sets in the decomposition cannot be all measurable. In mathematical models of physical situations, do non-measurable sets of points in R3 ever appear?

**Attribution***Source : Link , Question Author : Community , Answer Author : Cloudscape*