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

James Hirschorn James.Hirschorn at univie.ac.at
Thu Oct 25 14:36:40 EDT 2007

On Wednesday 24 October 2007 12:06, joeshipman at aol.com wrote:
> >continuum `believed' that  2^{\aleph_0} = 2^{\aleph_1}.
> ****
> This statement of cardinal arithmetic is considered plausible because
> it is true in the most natural models violating CH -- it follows from
> Martin's axiom, and also from the existence of a real-valued measurable
> cardinal, even though models of these axioms differ strongly in many
> other ways.

Recall that the cardinal characteristic t is the smallest cardinality of a 
maximal well ordered subfamily of (P(N)/Fin,\supseteq^*). Thus any "tower" 
(a_alpha:alpha < kappa), i.e. a_alpha's are infinite subsets of N, 
and alpha < beta implies a_beta \subseteq^* a_alpha, of height kappa < t, 
has an infinite b such that b \subseteq^* a_alpha for all alpha.

t is a relevant cardinal characteristic because of the equation 
(i.e. theorem of ZFC): 

2^{<t} = 2^{aleph-0}.  

Martin's axiom implies t > aleph-1 (and hence the equality in question), 
whereas RVM implies t = aleph-1. Because of this, and also since adding say 
aleph-(omega-1) many random reals to a model of GCH is a basic way of 
obtaining 2^{aleph-0} < 2^{aleph-1} + not-CH, I was surprised to hear that 
RVM -> 2^{aleph-0} = 2^{aleph-1}. Is there a simple proof of this?

> Can anyone point to an example in the literature of a model in which
> 2^{\aleph_0} < 2^{\aleph_1} (or an axiom which implies this) that was
> not explicitly "cooked up" to satisfy this inequality (that is, where
> the inequality follows from other properties used to define the model,
> not directly related to cardinal arithmetic)?

Doesn't V=L qualify? Perhaps you meant an axiom consistent with the negation 
of CH?

James Hirschorn

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