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

[joeshipman@aol.com]

>-----Original Message-----
>From: James Hirschorn <James.Hirschorn at univie.ac.at>
>
>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.

Thnaks for clarifying this. I was aware of the distinction, and by RVM 
I understand the same thing you do, that Lebesgue measure can be 
extended to all subsets of the real line. However, when I looked at 
Fremlin's monograph I misinterpreted the statement of Prikry's theorem, 
which does indeed require the stronger statement that the continuum is 
a real-valued measurable cardinal (any real-valued measurable cardinal 
is weakly inaccessible and no larger than c, but it could be smaller).

My main point was unaffected by this, since the model you describe is 
directly constructed to make 2^{aleph-0} be less than 2^{aleph-1}, 
while I was asking if any models of ~CH which were not specifically so 
constructed had nonetheless turned out to have this property.

 From a broader Platonistic perspective, I think the general theme that 
the continuum is "much larger" than any sets of smaller cardinality 
makes it intuitively plausible that, if an atomlessly-measurable 
cardinal exists, the continuum is one, and even that it is the first 
one. To the extent that one can imagine a countably additive measure on 
all subsets of the continuum, it seems that sets of cardinality smaller 
than the continuum "ought" to have measure 0 in that measure.

-- JS
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