[FOM] how it is that Martin's Axiom is crucial in understanding theory of the Continuum

Tom Dunion tom.dunion at gmail.com
Sat Sep 15 15:10:48 EDT 2012

Several "natural" strengthenings of Martin's Axiom make the value of
2^{\aleph_0} to be \aleph_2, which accords with Goedel's intuition,
and is a result favored by many.  I want to focus on just a few
observations that could be considered pros and cons of MA.

Pro: (besides providing the beginning of plausible trails leading to
that particular resolution of the CH).
(1) Easily simplifies the determination of many basic cardinal
characteristics of the Continuum
(2) Resolves Suslin's Hypothesis (in the affirmative)
(3) Makes all powersets 2^{\kappa} equal to 2^{\aleph_0}, for
infinite \kappa < 2^{\aleph_0}
(4) Makes all initial segments "nice" in the sense that, given any
well-ordering of [0,1] of length continuum, all sets of reals
corresponding to [0, \alpha) are Lebesgue measurable, for all
\alpha < (the initial ordinal for) 2^{\aleph_0}, in consequence of their
having cardinality less than the continuum.

Against: (1) Trivializes the theory of cardinal characteristics of the continuum
(2) Conflicts with several of Goedel's intuitions (e.g. see Martin &
Solovay's 1970 paper "Internal Cohen Extensions")
(3) Blocks strong Fubini theorems, even in just the plane.
(4) Makes those initial segments too nice. (For, recall, a well-order
of [0,1] of length continuum makes the set S in the unit square to be
nonmeasurable, where (x,y) is in S precisely when x< y in the
well-ordering).  Why then should all those sets in [0,1] be measurable?
(5) Seems to be in conflict with the spirit of the Convergence theorems in
basic theory of Lebesgue integration.  For we can have continuum many
integrals (i.e.of the characteristic function on the measure-zero set
corresponding to each [0, \alpha) all being equal to zero, but the final
integration gives an integral over [0,1] = 1.  (This is not the only situation
where "integral of the limit" does not equal "limit of the integrals"
so to speak, under MA.  The scare quotes mean I'm not suggesting
there is any actual contradiction with Lebesgue's theorems.)

Still, I wonder if MA could somehow be "not quite" true, yet still
indicate some genuine insight (more than simply consistency with ZFC)
that aims us in what might be the right direction toward resolving CH.
(Assuming CH is indeed resolvable in some fact of the matter sense.)

-- TD

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