[FOM] Does 2^{\aleph_0} = 2^{\aleph_1}?

James Hirschorn James.Hirschorn at univie.ac.at
Sun Oct 28 16:20:56 EDT 2007

On Friday 26 October 2007 13:01, joeshipman at aol.com wrote:
> 2^{\aleph_1}. I don't know who first proved this, but the stronger
> result that RVM implies no cardinal smaller than the continuum has a
> power set larger than the continuum was proved by Prikry (Bulletin of
> the AMS 81 (1975), 907-909). You can find a short proof online in
> section 5E of Fremlin's monograph "Real-Valued Measurable Cardinals",
>
> http://www.essex.ac.uk/maths/staff/fremlin/rvmc/index.htm

I think there was some confusion about the statement RVM. Please correct me if
I am wrong, but I believe that by RVM you mean that "2^{aleph-0} is a
real-valued measurable cardinal". There is apparently some precedence for
this, but I'm not sure when or by whom the acronym RVM was first used.

By RVM I meant "Lebesgue measure can be extended to all subsets of the real
line", or equivalently "there exists an atomlessly measurable cardinal". This
was the definition of RVM I used in my recent paper:

http://arxiv.org/abs/math/0604085v3

I was unaware of the precedent for RVM when I wrote it (even though I had
heard of the acronym), and the latter RVM seemed to be the best candidate for
an axiomatization of random forcing.

While Prikry's result entails that the former RVM implies 2^{aleph-0} =
2^{aleph-1}, clearly the latter RVM does not: Starting with GCH and a
measurable cardinal kappa, and adding kappa^{+aleph-(omega-1)} (i.e. the
aleph-(omega-1)'th successor of kappa) many random reals gives a
counterexample.

James Hirschorn