[FOM] free ultrafilters
Thomas Forster
T.Forster at dpmms.cam.ac.uk
Fri Oct 13 17:51:14 EDT 2006
I think what Bill was after (his query arose in a conversation we were
having) was the cutest/simplest (pair of) constructions of two
nonprincipal ufs over N - using AC of course. As it might be: "Here is
how to build a Ramsey ultrafilter; here is how to build a p-point that
isn't Ramsey." That sort of thing. The cardinality argument is cute, but
i think he was after something more informative
Thomas
On Fri, 13 Oct 2006, Andreas
Blass wrote:
> Bill Taylor asked:
>
> > Is it possible/easy to describe in some way, how to have two non-
> > isomorphic
> > free ultrafilters on the integers, where isomorphism is defined
> > with respect to permutations on the integers.
>
> A lot depends here on the meaning of "have". If you want a
> *definition* of two non-isomorphic free ultrafilters, then the answer
> is negative. In fact, you can't expect to get a definition for even
> one free ultrafilter on the integers. It is consistent, relative to
> ZFC, that no free ultrafilter on the integers is definable, even with
> real and ordinal numbers allowed as parameters in the definition.
> (Of course definitions become possible if one assumes G"odel's axiom
> of constructibility, which provides a definable well-ordering of the
> universe and thus allows one to define things, like free
> ultrafilters, that usually come from the axiom of choice.)
> If, on the other hand, you just want a proof that two non-isomorphic
> free ultrafilters exist, then this can be done by a counting
> argument. The number of permutations of the integers is the
> cardinality c of the continuum, so any one free ultrafilter is
> isomorphic to at most c others (in fact, exactly c, but that's not
> needed here). But it is known that there are 2^c free ultrafilters
> on the integers. So they can't all be in one isomorphism class; in
> fact there have to be 2^c isomorphism classes.
>
> Sincerely yours,
> Andreas Blass
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