## Wednesday, April 1, 2015

### An Axiom of Choice strong enough to puzzle

All the main puzzles that follow from the Axiom of Choice (AC)--nonmeasurable sets, Banach-Taski and guessing future coin tosses--need only a weaker version of AC. One weaker version that suffices is this:

(*) There is a choice function for any partition of the interval (0,1) into non-empty countable sets.

Now imagine worlds with point-sized particles that never move, but can perish and come into existence. The world starts at time 0. Each particle has a lifetime between 0 and 1, exclusive. Some locations in the world are never occupied by a particle. Call these "vacant". At all other locations, a particle comes into existence at time 0. Two particles never occupy the same location at the same time. Call such worlds p-worlds.
For each non-vacant location x in a p-world w, there is an associated set L(w,x) of numbers in (0,1), where a number y is in L(w,x) iff some particle at x has lifetime of length y. I now need a crucial metaphysical plenitude assumption:

(**) For any set S such that (a) every member of S is a countable non-empty collection of members of (0,1) and (b) the cardinality of S is at most that of the continuum, there is a p-world w such for each A in S there is a unique location x in w such that L(w,x)=A.
In other words, any set S satisfying (a) and (b) is the set of sets of lifetime lengths for non-vacant locations in some p-world, without duplication.

Given the plenitude assumption, I get the version of AC needed for the paradoxes. For given a partition S of (0,1) into countable sets, there will be a p-world as in (**). Given a member A of S, there will be a unique location x such that L(w,x)=A. Let f(A) be the lifetime of the first particle at x in w. This is our choice function.