[FOM] free ultrafilters

[Andreas Blass]

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