Re: Continuum hypothesis

Alan Smaill says...

george <greeneg@xxxxxxxxxx> writes:

On Aug 20, 11:39 am, stevendaryl3...@xxxxxxxxx (Daryl McCullough)
wrote:
Bell's Theorem proves that no measurable function f can possible
satisfy this constraint. However, Pitowsky proved that if one
assumes the continuum hypothesis, one can construct a nonmeasurable
function that satisfies this constraint.

One line of the truth table still has not been completed here.
If one DENIES the continuum hypothesis, can there still
exist a NON-constructible nonmeasurable function that
satisfies the constraint? Or is the truth of the CH necessary
to the existence of the non-measurable function at all (regardless
of whether it can be proven to exist)?

Since the proof given uses Martin's Axiom and not the stronger CH,
and ZF+MA is consistent with not AC (assuming ZF consistent,
presumably), the existence of such functions is consistent
with not CH.

As I understand Pitowsky's construction, what is needed for the
construction to go through is something along the lines of this axiom:

A: There exists a well-ordering < of the reals such that for every
real x, the set of all y < x has Lebesgue measure 0.

The continuum hypothesis of course implies this (because if
you well-order the reals with order type omega-1, then the
set of all y < x will always be a countable set, which always
has Lebesgue measure zero). But my axiom A doesn't imply the
continuum hypothesis.

--
Daryl McCullough
Ithaca, NY

.

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