From: Fred Galvin <galvin@math.ukans.edu>
Subject: Re: A real thick problem
Date: Sun, 5 Aug 2001 13:59:20 -0500
Newsgroups: sci.math
Summary: What values of cardinal arithmetic are consistent with ZFC?

On Sat, 4 Aug 2001, denis-feldmann wrote:

> Is it possible that 2^(aleph_1)=2^(aleph_0)??

Yes. You can "force" a model of ZFC in which 2^(aleph_0) and
2^(aleph_1) are both equal to aleph_2. More interestingly, if you
assume that Lebesgue measure can be extended to a countably additive
measure defined for all subsets of R, then it follows that
(a) 2^k = 2^(aleph_0) for every infinite cardinal k < 2^(aleph_0), and
(b) 2^(aleph_0) is much bigger than aleph_1, in fact, there are
2^(aleph_0) different alephs below 2^(aleph_0).
==============================================================================

From: Fred Galvin <galvin@math.ukans.edu>
Subject: Re: A real thick problem
Date: Sun, 5 Aug 2001 15:00:33 -0500
Newsgroups: sci.math

On Sun, 5 Aug 2001, Fred Galvin wrote:

> On Sat, 4 Aug 2001, denis-feldmann wrote:
> 
> > Is it possible that 2^(aleph_1)=2^(aleph_0)??
> 
> Yes. You can "force" a model of ZFC in which 2^(aleph_0) and
> 2^(aleph_1) are both equal to aleph_2.

I got that right.

> More interestingly, if you assume that Lebesgue measure can be
> extended to a countably additive measure defined for all subsets
> of R, then it follows that
> (a) 2^k = 2^(aleph_0) for every infinite cardinal k < 2^(aleph_0), and
> (b) 2^(aleph_0) is much bigger than aleph_1, in fact, there are
> 2^(aleph_0) different alephs below 2^(aleph_0).

But I got this wrong. The conclusions (a) and (b) follow from the
assumption that 2^(aleph_0) is a so-called "real-valued measurable
cardinal", which means that Lebesgue measure can be extended to an
everywhere-defined 2^(aleph_0)-additive measure, i.e., the union of
fewer than 2^(aleph_0) measure zero sets has measure zero. I hope I
got it right this time! I should leave such technical stuff to the
experts on set theory.