Axiom Of Choice

The Axiom of Choice states that given a set X of non-empty sets, there exists a function that when it is applied to any of the sets in X gives a member of that set (∃ f such that f(x) ∈ x ∀ x ∈ X). That is to say, one can choose one element from each of the sets in X. The axiom of choice has been proved to be independent of the Zermelo-Fraenkel set theory axioms, and so it can be used with them without fear of introducing a contradiction. It is generally regarded as being true, and its truth is equivalent to the truth of many useful propositions. However, it causes the BanachTarskiParadox?. -- RobertFurber?

Here's a dissenting view:

Experience with the RussellParadox taught us that we have to be very careful how we specify sets. A mere EnglishLanguage predicate test for set membership is not enough to define a set, or else you invite deep dark problems. The domain of a function is a set, and so is the range of a function F. The domain of F the axiom of choice is the set X, a set of sets. The range is another set Y, given by the chosen members of each set in X. Now let's think carefully: If an EnglishLanguage proposition is inadequate to infer the existence of a set, should it be enough to infer the existence of a function? Besides making the bald-faced claim that Y exists, what limits can you place on it? Can you make a non-trivial test of reasonability for an assertion about what Y might be or contain?

I seem to recall reading a "proof" that a gold ball could be duplicated by sufficiently arcane cutting and pasting. The basic structure of the proof was that you could use AxiomOfChoice to make a 1::1 correspondence between the points in one sphere and those in two separate spheres. Now common sense tells you that this is preposterous, and it is preposterous.

I claim that the function F exists at the same level of existence as the set of all sets not containing themselves: It's a clear paradox, and it admits paradoxical results, and it should not be allowed.

Except that AxiomOfChoice has been proved to have no paradoxical result. (Unless Zermelo-Fraenkel already has paradoxical results by itself, that is.)

Can you please point to this proof that the AxiomOfChoice has no paradoxical result? I'd like to know how "paradoxical" is defined there. By most people's standards, the BanachTarskiParadox? is plenty paradoxical. See http://en.wikipedia.org/wiki/Banach-Tarski_paradox. -- BenKovitz

I would not be so certain that the BanachTarskiParadox? is as paradoxical as you think. For example, it does _not_ say that a gold ball, or any physical thing, could be broken into parts and reassembled in two copies. The "pieces" it refers to are unmeasurable. Physical things, like gold balls, are made up of finite numbers of discrete clumps. Finite sites of discrete clumps do not admit unmeasureable sets. I am not trying to be pedantic but rather to point out that applying intuition about finite, discrete sets like clumps of matter does not always hold up when thinking about mathematical entities like a continuum. They are simply different things and behave differently. Nor is it reasonable to say that the only "correct" math is that which describes atoms in the physical world. A pure mathematician might point out that math is beautiful in its own right, much like a piece of music. As an applications philistine, I would point out fields like financial analysis which are not constrained to quanta of physical matter.

CategoryMath