From: kramsay@aol.commangled (Keith Ramsay)
Subject: Re: How Many Transcendental Numbers?
Date: 05 Mar 2001 02:13:07 GMT
Newsgroups: sci.math
Summary: Upper bounds on the order type of 2^{Aleph_0}

In article <stevel-11766C.01214527022001@news.dnai.com>,
Steve Leibel <stevel@bluetuna.com> writes:
|Now the question is, which of the higher cardinalities describes the 
|real numbers?  Is it the next one up from the countable ones, or is it a 
|higher cardinality than that?
|
|And the answer is -- nobody knows.  And not only does nobody know, but 
|nobody knows in sort of a strange way.
|
|It turns out that you can just assume it's ANY ONE of the uncountable 
|cardinals (called the "alephs").  You can take as an axiom that the 
|number of reals is any aleph that you like, and you still get a 
|consistent theory and you can do conventional mathematics.

This is not quite true. It's not possible, for example, for the
continuum to be Aleph_omega, where omega is the smallest infinite
ordinal. If I remember correctly, the continuum isn't the union of
a countable collection of sets of smaller cardinality. A set of
cardinality Aleph_omega is the union of subsets X1, X2, X3,...
where Xn has cardinality Aleph_n.

It's also not possible for the continuum to be Aleph_{c^+}, where
c^+ is the smallest cardinality greater than c, or any other Aleph_x
where x>c.

The result uses some trick to convert a model M of set theory with an
ordinal x which is not the limit of countably many smaller ordinals
into a model M' in which Aleph_x is the continuum. Of course, if we
do this trick to an ordinal x which is greater than the continuum in
M, then in M' it's no longer true that x is greater than the
continuum. Thus x may have lost its identity as c^+ or something
else which might get moved by the trick.

The result does imply, though, that for any integer n>0, the statement
c=Aleph_n is consistent with ZFC, and so is c=Aleph_{omega+n}, and so
on, which leaves the continuum problem pretty wide open. At different
points in time, Goedel leaned toward c=Aleph_1 and c=Aleph_2 at least
somewhat, although I don't think he ever was convinced of either one.

Keith Ramsay