FOM: reply to Joe Shipman
Matt Insall
montez at rollanet.org
Tue Feb 22 09:27:14 EST 2000
Joe Shipman already replied to this (when I replied, I expected my reply to
go to the list, but it went only to him), but I am hoping someone else
familiar with NSA can explain my conundrum about AC and Tychonoff's theorem
(see below). Also, I wonder if my reason for a strong denial of CH is
familiar to others.
Professor Shipman said:
> Yes, although to a non-analyst CH may appear more
> fundamental than RVM, being a simpler statement.
Agreed. I find it so, although I have other (non-analysis) reasons for
preferring as strong as possible a denial of CH. For one thing, most
properties we ascribe to sets in general are derived from our intuition
about how properties of finite sets generalize. When this is considered in
regard to the status of CH, consider the finite case: between n and 2^n,
there is a very fast-growing cardinality of the set {n,...,2^n}. Why should
this property not continue in the infinite cardinals?
> AC seems a VERY clear application of the "MAXIMIZE"
> principle, because it is equivalent to "a product of
> nonempty sets is nonempty", and denying it would
> clearly be minimizing the size of the cartesian product
> set.
I am certainly not averse to AC. In fact, I have struggled with myself to
even understand why AC is independent of the other axioms of set theory.
(Perhaps most FOMers have no trouble seeing sets as being
non-well-orderable, but not me.) And I concur that the product of nonempty
sets should be nonempty. I would like to point out, though that it seems to
me that in some famed applications of AC, or ``equivalents of AC'',
especially certain ones which use this version you mention, appear to be
avoidable. In particular, consider the Tychonoff theorem: ``A product of
compact spaces is compact.'' Now, there is a very nice, short proof of this
result using nonstandard analysis. (cf. Hurd and Loeb, or Albeverio, et.
al.). The first line of the proof applies AC in the cartesian product form,
but of course, in the case that the product space is empty, the conclusion
holds anyway. That is, even if the product of some compact spaces is empty,
it is compact, because the empty space is a compact space. The remainder of
the proof uses infinitesimals, which are available in any set theory with
the Boolean Prime Filter theorem (BPF, or equivalently, BPI), to show that,
when the product of compact spaces is nonempty, it is compact. Now, BPF is
strictly weaker than AC (cf. Rubin and Howard, and many others), so one does
not need, IMHO, the full strength of AC for the Tychonoff Theorem. (Perhaps
I am overlooking something. If there are some others in NSA out there to
show me where I am missing something, I'd appreciate it.)
> Yes, but c=aleph_c is not as strong as RVM, because c
> could simply be the aleph_oneth (aleph_first?) fixed
> point of the aleph function, you don't need to go
> beyond ZFC to show this is consistent.
>
I am glad to learn that c = \aleph_c is not as strong as RVM. I had not yet
seen that this was the case. I also, as I said before, had no reason to
believe that c = \aleph_c was inconsistent. I am a little confused by your
comment above, though, that c could simply be the \aleph_1'st fixed point of
the aleph function. How could c=\aleph_c and c=\aleph_{\aleph_1}? Either I
misunderstood your comment or this implies that \aleph_1=\aleph_{\aleph_1},
it seems. [I hope I'm not just getting myself confused. :-)]
> It is an interesting question how much stronger the
> assumption of a measure is than the assumption that
> there are weak inaccessibles <= c. I'm pretty sure
> that it is consistent for c to be weakly inaccessible
> but for no measure to exist. Can anyone confirm this?
>
I hope someone does confirm it. It does sound interesting.
Name: Matt Insall
Position: Associate Professor of Mathematics
Institution: University of Missouri - Rolla
Research interest: Foundations of Mathematics
More information: http://www.umr.edu/~insall
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