# [FOM] free ultrafilters

Andreas Blass ablass at umich.edu
Thu Oct 19 08:58:01 EDT 2006

As K.P. Hart pointed out, the paper

> $\beta\omega - \omega$ 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.

does considerably more than was asked for here on fom.  It provides
two free ultrafilters on the integers that are not only not
isomorphic via a
permutation of the integers but can be explicitly distinguished by their
topological behavior in the Stone-Cech remainder of the integers and
indeed
by a first-order property over that topology.  But the
reference to the paper reminded me of another way to answer the original
question, i.e., to produce two non-isomorphic free ultrafilters on the
integers.  This way avoids Kunen's method of independent sets; it's
essentially Frolik's proof that the Stone-Cech remainder of the
integers is
not homogeneous ("Sums of Ultrafilters," Bull. A.M.S. 73 (1967) 87-91).
First construct (using the axiom of choice) some free
ultrafilter U
on the set Z of integers.  Then observe that the following is a free
ultrafilter on the set ZxZ of ordered pairs of integers:
V = { X subset of ZxZ : {a in Z :  {b in Z : (a,b) in X} in
U} in U}.
[To decipher this rather convoluted definition, read it as "A statement
S(a,b) about a and b is true for V-almost-all (a,b) iff, for U-almost-
all a,
for U-almost-all b, S(a,b)."]  Use any bijection from ZxZ to Z, to
transfer
V to an ultrafilter V' on Z.  Then V' and U are non-isomorphic free
ultrafilters on Z.
It's not immediately obvious that these ultrafilters aren't
isomorphic, but it has been known for a long time.  For a proof (in the
appropriate generality), see Theorem 9.2(b) of "The Theory of
Ultrafilters"
by Comfort and Negrepontis.

Andreas Blass