[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
combinatorial
>set theorists, especially those interested in cardinal invariants in
the
>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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