# Isomorphic ultrafilters - open problems

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### Victor Porton

Jun 10, 2009, 2:25:46 PM6/10/09
to nhin...@aol.com
Hello mathematicians,

I have defined the relation "being isomorphic" for two filters and
proved that this relation is an equivalence relation as well as some
other simple theorems.

I published it at this blog entry:
http://portonmath.wordpress.com/2009/06/10/isomorphic-filters/

For me it remains an open problems: 1. whether any two non-trivial
ultrafilters are isomorphic; 2. whether any two non-trivial filters on
the set of natural numbers are isomorphic.

Maybe you know the answers for my open problems?

You may post comments for the above mentioned blog post.

### hagman

Jun 10, 2009, 4:54:33 PM6/10/09
to

If I got this right, given two non-trivial ultrafilters, you want
to find subsets in these filters such that the restriction of the
filters to these subsets are directly isomorphic (i.e. via a bijection
of these underlying sets).

Let N be the set of the naturals, R the set of the reals.
Let a_0 = { x c N | #(N\x) < oo } and refine this to an ultrafilter
a (using Axiom of Choice, of course).
Let b_0 = { x c R | #(R\x) <= #N } and refine this to an
ultrafilter b.
If a and b were directly isomorphic, then there should exist
sets A e a, B e b and a bijection f:A->B such that ...
But B is uncountable (else R\B e b_o c b) hence no such bijection
exists.

Hagen

### Butch Malahide

Jun 10, 2009, 6:43:16 PM6/10/09
to
On Jun 10, 1:25 pm, Victor Porton <por...@narod.ru> wrote:
> Hello mathematicians,
>
> I have defined the relation "being isomorphic" for two filters and
> proved that this relation is an equivalence relation as well as some
> other simple theorems.
>
> I published it at this blog entry:http://portonmath.wordpress.com/2009/06/10/isomorphic-filters/
>
> For me it remains an open problems: 1. whether any two non-trivial
> ultrafilters are isomorphic; 2. whether any two non-trivial filters on
> the set of natural numbers are isomorphic.
>
> Maybe you know the answers for my open problems?

No, there are many nonisomorphic nontrivial ultrafilters on the set N
of natural numbers.

There are 2^(2^{aleph_0}) ultrafilters on N. Each isomorphism class
contains at most 2^(aleph_0) ultrafilters. Therefore the number of
isomorphism classes is 2^(2^{aleph_0}). One of these isomorphism
classes consists of all the trivial (i.e. principal) ultrafilter on N.

### Victor Porton

Jun 11, 2009, 6:34:41 AM6/11/09
to

Why each isomorphism class contains at most 2^(aleph_0) ultrafilters?

I see only that the number of ultrafilters in a class is not more than
the number of bijection between subsets of N. What is this number?

### Victor Porton

Jun 11, 2009, 11:42:48 AM6/11/09
to

Oh, it seems that I've got it:

The number of bijections between any two subsets of N is no more that
(aleph_0)^(aleph_0) = 2^(aleph_0).

The number of all bijections between any two subsets of N is no more
than
2^(aleph_0) * 2^(aleph_0) = 2^(aleph_0).

Therefore Each isomorphism class contains at most 2^(aleph_0)
ultrafilters.

Q.E.D.