13 Jun 2007 (updated 15 Jun 2007 at 15:19 UTC)
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Bram's case against the Axiom of ChoiceBram Cohen wrote:
The axiom of choice is presented as 'obviously'
true, an axiom which can be simply thrown on the pile in the interests of
convenience, without raising any notable philosophical hackles.
...which is a rather strange thing to say, given how it is one of the
foremost
hackles-raising topic in the philosophy of mathematics. In fact, set theorists
don't claim that AC is obviously true of any theory of sets, but instead they
say there is a class of mathematical concpetions of set on which the axiom of
choice (AC) comes out true, most importantly the cumulative hierarchy.
There are very natural axiomatisations of set theory, particularly Fraenkel-
Mostowski set theory, which validate the Zermelo-Fraenkel (ZF) axioms but
not ZFC (ZF+AC).
The same game can be played with other axioms: if you think
the
sets of
ZFC
are "too large", you can drop the power set axiom and choose something like
Kripke-Platek set theory, with no sets of cardinality greater than the set of
real numbers. If you don't like the foundation axiom, you can choose to
replace it by Aczel's AFA axiom, a set theory which is based on bisimulation,
and so is useful for reasoning about event structures.
Bram goes on to
discuss a puzzle, via Lance Fortnow,
involving probabalistic reasoning over infinite sets. These sort of puzzles are
a little like Zeno's paradox: as Bram says, the problem is trying to extend
finitistic reasoning to the infinite case. However, AC is not to blame here: the
problem is trying to naively extend the reasoning about conditional
expectations to the infinite case: there is no right way to do this. In
particular, observe that the 'strategy' is nonrecursive, and indeed even with
the benefit of an oracle, cannot be applied in a finite number of steps.
Observe further, that the equivalence classes are not going to be measurable.
Postscript: Changed title, fixed discussion of Bram's
problem
A commentor in the thread to Bram's post has
nailed it:
Yup, nonmeasurable sets.
Proof: for each of the countably many finite subsets A of N, consider the
function fA that flips the colours of the hats of people in A. This function is
obviously measurable (i.e., inverse images of measurable sets are
measurable), and indeed measure-preserving and bijective. Now let S be your
set of representatives. Then the fA(S), as A runs over the finite subsets of N,
are disjoint, and their union is all of 2N. If any of them is measurable then
they all are and all have the same measure (because fA is measure-
preserving). But then by countable additivity the measure of the whole space
is either 0 (if |S|=0) or infinite (if |S|>0), but in fact the measure of the whole
space is 1, contradiction.
Summary of proof: consistently flipping any finite set of bits turns our set of
representatives into another set that's "obviously the same size"; there are
countably many of these, and they partition 2N, so whatever the probability p
of having no wrong guesses, we have to have infinity * p = 1, which is
impossible.
I expect this can be tweaked to yield a proof that the event "at least n errors"
is non-measurable for each n -- i.e., that the probability of getting at least n
errors is undefined -- but I'm too lazy to try. :-)