[FOM] free ultrafilters
K. P. Hart
K.P.Hart at tudelft.nl
Tue Oct 17 04:35:01 EDT 2006
On Mon, Oct 16, 2006 at 03:11:48PM +0200, Freek Wiedijk wrote:
> Bill Taylor:
>
> >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.
>
> I think I remember having been told that if you put the
> Cech-Stone topology on the collection of free ultrafilters
> over the integers, then that topological space -- called
> "\beta\omega - \omega" -- is _not_ homogeneous (in the sense
> that you can't map any point to any other point using an
> automorphism of the space), which is a little bit surprising
> as of course the space of the integers that the Cech-Stone
> compactification starts from _is_ homogeneous.
>
> Some googling to find a reference for this led me to:
>
> $\beta\omega - \omega$ is not first order homogeneous" by
> Eric K. van Douwen and Jan van Mill, Proceedings of the
> American Mathematical Society, Vol. 81, No. 3 (Mar., 1981),
> pp. 503-504.
>
> The abstract of this "shorter note" is:
>
> "We find a first order property shared by some but not all
> points of $\beta\omega - \omega$."
This is not the type of homogeneity that was asked for, though it is
much much stronger.
The Van Douwen-Van Mill paper establishes a property of points in the
*space* $\beta\omega-\omega$ that can be formulated in first-order terms
using only closed sets as variables/constants and that, as the abstract
puts it, some points do have and others do not.
The original question treats ultrafilters as families of subsets of $\omega$
and asks for two that are *not* mapped to each other by any permutation of
$\omega$. This was answered by Andreas Blass (twice) in previous posts.
If $\pi$ is a permutation of $\omega$ that transforms an ultrafilter $u$
into an ultrafilter $v$ then its \v{C}ech-Stone extension induces an
autohomeomorphism $\pi^*$ of $\beta\omega-\omega$ that maps the point $u$
to the point $v$.
The points constructed by Van Douwen and Van Mill therefore also answer
the original question but their constructions (or at least of one of them)
are rather more involved than that given by Blass.
KP Hart
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