For filters on sets defined equivalence relation being isomorphic
.
Posed some open problems like this: are every two nontrivial ultrafilters isomorphic?
Definition.
For any function we’ll define the function
by the formula (for any set
):
Definition.
I call filters and
directly isomorphic when there are a bijection
such that
is a bijection from
to
.
Definition.
Let ,
are sets. Let
and
are the lattices of all subsets of these sets. Let
and
are filters on the lattices
and
correspondingly. I will call filters
and
isomorphic when there exist sets
and
such that filters
and
are directly isomorphic.
Obvious. Filters and
are isomorphic iff there exist sets
and
such that there are a bijection
such that
is a bijection from
to
.
Remark.
It seems that there exist a simpler definition of isomorphic filters in terms of reloids, but that stumbles on some open problems about reloids.
Proposition.
If two filters are directly isomorphic then they are isomorphic.
Proof.
Take ,
. Then
and
.
Example.
Let is the lattice of all subsets of some set
. Then there are isomorphic filters on
which are not directly isomorphic.
Proof.
Consider two filters on :
,
. There are no bijection from
to
because
. So
and
are not directly isomorphic.
Now let ,
and
is defined by the formula
. Then
and
;
is a bijection from
to
and
is a bijection from
to
. So
and
are directly isomorphic, that is
and
are isomorphic filters.
Theorem.
Being directly isomorphic
for filters is an equivalence relation.
Proof.
- Reflexivity
- Let
is a filter on some set
. Then the identity function
is a bijection from
to
. Evidently
is a bijection from
to
. So
and
are directly isomorphic.
- Symmetry
- Let filters
and
are directly isomorphic. Then exists a bijection
such that
is a bijection from
to
.
is a bijection. To finish the proof it is enough to show that
is a bijection from
to
. First
is a function with domain
. Let
; if
then
because
is a bijection, consequently
. So
is an injection. Let
; then
that is
where
;
; that is
is a function onto
. So
is a bijection from
to
. Hence
and
are directly isomorphic.
- Transitivity
- Let filters
and
are directly isomorphic and filters
and
are directly isomorphic. Then exist bijections
and
such that
is a bijection from
to
and
is a bijection from
to
. The function
is a bijection from
to
. We have
is a bijection as a composition of two bijections. So filters
and
are directly isomorphic.
Lemma.
If and
are directly isomorphic, then for any
there exists
such that
and
are directly isomorphic.
Proof.
Let and
are directly isomorphic; let
. Then there are a bijection
such that
is a bijection from
to
.
; hence
is an injection defined on the set
. Let
. We have
. Let
; then
; by definition of
directly isomorphic
exists such that
. We have
because
and because
is a bijection. So
. Thus
is a function onto
. So
is a bijection.
Theorem.
Being isomorphic
for filters is an equivalence relation.
Proof.
- Reflexivity
- Let
is a filter. Let
. Then
is directly isomorphic to itself. Consequently
is isomorphic to itself.
- Symmetry
- Let filters
and
are isomorphic. Then there exist sets
and
such that filters
and
are directly isomorphic. By proved above
and
are directly isomorphic. But this means that
and
are isomorphic.
- Transitivity
- Let filters
and
are isomorphic and filters
and
are isomorphic. Then
and
are directly isomorphic and
and
are directly isomorphic where
,
,
. Let
. We have
. By the lemma (taking in account also symmetry proved above) there exist sets
and
such that
and
are directly isomorphic and
and
are directly isomorphic. By transitivity of being direct isomorphic we have that
and
are directly isomorphic. Hence
and
are isomorphic.
Proposition.
- Principal filters, generated by sets of the same cardinality, are isomorphic.
- If a filter is isomorphic to a principal filter, then it is also a principal filter with the same cardinality of the sets which generate these principal filters.
Proof.
- Let
and
are principal filters. Let they are generated by the sets
and
correspondingly where
. Enough to prove that
and
are directly isomorphic. But this follows from
and
and
, because any bijection from
to
sends
to
.
- Let
is a principal filter generated by a set
, let
is a filter isomorphic to
. We shall prove that
is principal and is generated by a set
such that
. Let
and
and
is a bijection from
to
such that
is a bijection from
to
. We have
So
has the smallest element
. Consequently
has the smallest element
and so
is a principal filter generated by the set
whose cardinality is
.
Proposition.
A filter isomorphic to a nontrivial ultrafilter is a nontrivial ultrafilter.
Proof.
Let is a nontrivial ultrafilter and let
is isomorphic to
. Then there are sets
and
and a bijection
such that
is a bijection from
to
.
The filter cannot be a trivial ultrafilter because otherwise
would be also a principal filter. So it’s enough to prove that
is an ultrafilter. We will prove that for any set
either
or
is an element of
.
Let . Then
;
for some
. Because
is an ultrafilter, either
or
. If
then
and so
and consequently
. If
then
; consequently
;
;
;
And as the culmination of this passage about isomorphic filters, several (related) open problems:
Question. Are any two nontrivial ultrafilters isomorphic?
Question. Are any two nontrivial ultrafilters on the same set directly isomorphic?
Question. Are any two nontrivial ultrafilters on isomorphic?
Question. Are any two nontrivial ultrafilters on directly isomorphic?
(These are open problems for me, the author of this text. If someone knowing the solutions of these problems, please contact me.)
In this newsgroup thread is proved that there are non-isomorphic non-trivial ultrafilters on
. This solves all four open problems.