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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.

Don't hesitate ask me questions about this.

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

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?

> 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.

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?

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.

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