# Isomorphic filters – open problems

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.

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