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

joeshipman@aol.com joeshipman at aol.com
Wed Oct 24 12:06:48 EDT 2007

>-----Original Message-----
>From: John Baldwin <jbaldwin at uic.edu>
>It was suggested at a logic dinner this evening that many 
>set theorists, especially those interested in cardinal invariants in 
>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.

For example, under Martin's axiom, all sets of reals of cardinality 
less than continuum are measure zero, so the graph of a well-ordering 
of [0,1] of the lowest possible order type gives a set which is 
measure-0 on horizontal lines and co-measure-0 on vertical lines, while 
under RVM there are small nonmeasurable sets and "strong Fubini 
theorems" are true in all dimensions (any iterated integrals of 
non-negative functions which exist must be equal).

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)?

-- JS

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