Isomorphic filters – open problems | Math Research of Victor Porton

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 $f : A \rightarrow B$ we’ll define the function $\left\langle f \right\rangle : \mathscr{P} A \rightarrow \mathscr{P} B$ by the formula (for any set $X$ ): $\left\langle f \right\rangle X = \left\{ f x | x \in X \right\}.$

Definition.
I call filters $a$ and $b$ directly isomorphic when there are a bijection $f : \bigcup a \rightarrow \bigcup b$ such that $\left\langle f \right\rangle |_a$ is a bijection from $a$ to $b$ .

Definition.
Let $U$ , $W$ are sets. Let $\mathscr{P} U$ and $\mathscr{P} W$ are the lattices of all subsets of these sets. Let $a$ and $b$ are filters on the lattices $\mathscr{P} U$ and $\mathscr{P} W$ correspondingly. I will call filters $a$ and $b$ isomorphic when there exist sets $A \in a$ and $B \in b$ such that filters $a \cap \mathscr{P} A$ and $b \cap \mathscr{P} B$ are directly isomorphic.

Obvious. Filters $a$ and $b$ are isomorphic iff there exist sets $A \in a$ and $B \in b$ such that there are a bijection $f : A \rightarrow B$ such that $\left\langle f \right\rangle |_{a \cap \mathscr{P} A}$ is a bijection from $a \cap \mathscr{P} A$ to $b \cap \mathscr{P} B$ .

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 $A = \bigcup a$ , $B = \bigcup b$ . Then $a \cap \mathscr{P} A = a$ and $b \cap \mathscr{P} B = b$ .

Example.
Let $\mathscr{P} U$ is the lattice of all subsets of some set $U$ . Then there are isomorphic filters on $\mathscr{P} U$ which are not directly isomorphic.

Proof.
Consider two filters on $\mathbb{N}$ : $a = \left\{ \mathbb{N} \right\}$ , $b = \left\{ \mathbb{N} \setminus \left\{ 0 \right\}, \mathbb{N} \right\}$ . There are no bijection from $a$ to $b$ because $\mathrm{card} a \neq \mathrm{card} b$ . So $a$ and $b$ are not directly isomorphic.

Now let $A = \mathbb{N}$ , $B = \mathbb{N} \setminus \left\{ 0 \right\}$ and $f : A \rightarrow B$ is defined by the formula $f x = x + 1$ . Then $a \cap \mathscr{P} A = \left\{ \mathbb{N} \right\}$ and $b \cap \mathscr{P} B = \left\{ \mathbb{N} \setminus \left\{ 0 \right\} \right\}$ ; $f$ is a bijection from $A$ to $B$ and $\left\langle f \right\rangle |_{a \cap \mathscr{P} A} = \left\langle f \right\rangle |_{\left\{ \mathbb{N} \right\}}$ is a bijection from $a \cap \mathscr{P} A$ to $b \cap \mathscr{P} B$ . So $a \cap \mathscr{P} A$ and $b \cap \mathscr{P} B$ are directly isomorphic, that is $a$ and $b$ are isomorphic filters.

Theorem.
Being directly isomorphic for filters is an equivalence relation.

Proof.

Reflexivity: Let $a$ is a filter on some set $A$ . Then the identity function $\mathrm{Id}_A$ is a bijection from $\bigcup a$ to $\bigcup a$ . Evidently $\left\langle \mathrm{Id}_A \right\rangle |_a$ is a bijection from $a$ to $a$ . So $a$ and $a$ are directly isomorphic.
Symmetry: Let filters $a$ and $b$ are directly isomorphic. Then exists a bijection $f : \bigcup a \rightarrow \bigcup b$ such that $\left\langle f \right\rangle |_a$ is a bijection from $a$ to $b$ . $f^{- 1} : \bigcup b \rightarrow \bigcup a$ is a bijection. To finish the proof it is enough to show that $\left\langle f^{- 1} \right\rangle |_b$ is a bijection from $b$ to $a$ . First $\left\langle f^{- 1} \right\rangle |_b$ is a function with domain $b$ . Let $X, Y \in b$ ; if $X \neq Y$ then $\left\langle f^{- 1} \right\rangle X \neq \left\langle f^{- 1} \right\rangle Y$ because $f^{- 1}$ is a bijection, consequently $\left\langle f^{- 1} \right\rangle |_b X \neq \left\langle f^{- 1} \right\rangle |_b Y$ . So $\left\langle f^{- 1} \right\rangle |_b$ is an injection. Let $X \in a$ ; then $\left\langle f \right\rangle X \in b$ that is $\left\langle f \right\rangle X = Y$ where $Y \in b$ ; $X = \left\langle f^{- 1} \right\rangle \left\langle f \right\rangle X = \left\langle f^{- 1} \right\rangle Y = \left\langle f^{- 1} \right\rangle |_b Y$ ; that is $\left\langle f^{- 1} \right\rangle |_b$ is a function onto $a$ . So $\left\langle f^{- 1} \right\rangle |_b$ is a bijection from $b$ to $a$ . Hence $b$ and $a$ are directly isomorphic.
Transitivity: Let filters $a$ and $b$ are directly isomorphic and filters $b$ and $c$ are directly isomorphic. Then exist bijections $f : \bigcup a \rightarrow \bigcup b$ and $g : \bigcup b \rightarrow \bigcup c$ such that $\left\langle f \right\rangle |_a$ is a bijection from $a$ to $b$ and $\left\langle g \right\rangle |_b$ is a bijection from $b$ to $c$ . The function $g \circ f$ is a bijection from $\bigcup a$ to $\bigcup c$ . We have $\left\langle g \circ f \right\rangle |_a = \left\langle g \right\rangle \circ \left\langle f \right\rangle |_a = \left\langle g \right\rangle |_b \circ \left\langle f \right\rangle |_a$ is a bijection as a composition of two bijections. So filters $a$ and $c$ are directly isomorphic.

Lemma.
If $a \cap \mathscr{P} A$ and $b \cap \mathscr{P} B$ are directly isomorphic, then for any $A' \in a \cap \mathscr{P} A$ there exists $B' \in b \cap \mathscr{P} B$ such that $a \cap \mathscr{P} A'$ and $b \cap \mathscr{P} B'$ are directly isomorphic.

Proof.
Let $a \cap \mathscr{P} A$ and $b \cap \mathscr{P} B$ are directly isomorphic; let $A' \in a \cap \mathscr{P} A$ . Then there are a bijection $f : A \rightarrow B$ such that $\left\langle f \right\rangle |_{a \cap \mathscr{P} A}$ is a bijection from $a \cap \mathscr{P} A$ to $b \cap \mathscr{P} B$ . $a \cap \mathscr{P} A' \subseteq a \cap \mathscr{P} A$ ; hence $\left\langle f \right\rangle |_{a \cap \mathscr{P} A'}$ is an injection defined on the set $a \cap \mathscr{P} A'$ . Let $B' = \left\langle f \right\rangle A'$ . We have $B' \subseteq B$ . Let $Y \in b \cap \mathscr{P} B'$ ; then $Y \in b \cap \mathscr{P} B$ ; by definition of directly isomorphic exists $X \in a \cap \mathscr{P} A$ such that $\left\langle f \right\rangle X = Y$ . We have $X \subseteq A'$ because $Y \subseteq B'$ and because $f$ is a bijection. So $X \in a \cap \mathscr{P} A'$ . Thus $\left\langle f \right\rangle |_{a \cap \mathscr{P} A'}$ is a function onto $b \cap \mathscr{P} B'$ . So $\left\langle f \right\rangle |_{a \cap \mathscr{P} A'}$ is a bijection.

Theorem.
Being isomorphic for filters is an equivalence relation.

Proof.

Reflexivity: Let $a$ is a filter. Let $A = \bigcup a$ . Then $a = a \cap \mathscr{P} A$ is directly isomorphic to itself. Consequently $a$ is isomorphic to itself.
Symmetry: Let filters $a$ and $b$ are isomorphic. Then there exist sets $A \in a$ and $B \in b$ such that filters $a \cap \mathscr{P} A$ and $b \cap \mathscr{P} B$ are directly isomorphic. By proved above $b \cap \mathscr{P} B$ and $a \cap \mathscr{P} A$ are directly isomorphic. But this means that $b$ and $a$ are isomorphic.
Transitivity: Let filters $a$ and $b$ are isomorphic and filters $b$ and $c$ are isomorphic. Then $a \cap \mathscr{P} A$ and $b \cap \mathscr{P} B_1$ are directly isomorphic and $b \cap \mathscr{P} B_2$ and $c \cap \mathscr{P} C$ are directly isomorphic where $A \in a$ , $B_1, B_2 \in b$ , $C \in c$ . Let $B_0 = B_1 \cap B_2$ . We have $B_0 \in b$ . By the lemma (taking in account also symmetry proved above) there exist sets $A_0 \in a \cap \mathscr{P} A$ and $C_0 \in c \cap \mathscr{P} C$ such that $a \cap \mathscr{P} A_0$ and $b \cap \mathscr{P} B_0$ are directly isomorphic and $b \cap \mathscr{P} B_0$ and $c \cap \mathscr{P} C_0$ are directly isomorphic. By transitivity of being direct isomorphic we have that $a \cap \mathscr{P} A_0$ and $c \cap \mathscr{P} C_0$ are directly isomorphic. Hence $a$ and $c$ 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 $a$ and $b$ are principal filters. Let they are generated by the sets $A$ and $B$ correspondingly where $\mathrm{card} A = \mathrm{card} B$ . Enough to prove that $a \cap \mathscr{P} A$ and $b \cap \mathscr{P} B$ are directly isomorphic. But this follows from $a \cap \mathscr{P} A = \left\{ A \right\}$ and $b \cap \mathscr{P} B = \left\{ B \right\}$ and $\mathrm{card} A = \mathrm{card} B$ , because any bijection from $A$ to $B$ sends $\left\{ A \right\}$ to $\left\{ B \right\}$ .
Let $a$ is a principal filter generated by a set $A$ , let $b$ is a filter isomorphic to $a$ . We shall prove that $b$ is principal and is generated by a set $B$ such that $\mathrm{card} B = \mathrm{card} A$ . Let $A \in a$ and $B \in b$ and $f$ is a bijection from $A$ to $B$ such that $\left\langle f \right\rangle |_{a \cap \mathscr{P} A}$ is a bijection from $a \cap \mathscr{P} A$ to $b \cap \mathscr{P} B$ . We have $b \cap \mathscr{P} B = \left\{ \left\langle f \right\rangle |_{a \cap \mathscr{P} A} X | X \in a \cap \mathscr{P} A \right\} = \left\{ \left\langle f \right\rangle X | X \in a \cap \mathscr{P} A \right\}.$ So $b \cap \mathscr{P} B$ has the smallest element $\left\langle f \right\rangle A$ . Consequently $b$ has the smallest element $\left\langle f \right\rangle A$ and so $b$ is a principal filter generated by the set $B = \left\langle f \right\rangle A$ whose cardinality is $\mathrm{card} A$ .

Proposition.
A filter isomorphic to a nontrivial ultrafilter is a nontrivial ultrafilter.

Proof.
Let $a$ is a nontrivial ultrafilter and let $b$ is isomorphic to $a$ . Then there are sets $A \in a$ and $B \in b$ and a bijection $f : A \rightarrow B$ such that $\left\langle f \right\rangle |_{a \cap \mathscr{P} A}$ is a bijection from $a \cap \mathscr{P} A$ to $b \cap \mathscr{P} B$ .

The filter $b$ cannot be a trivial ultrafilter because otherwise $a$ would be also a principal filter. So it’s enough to prove that $b$ is an ultrafilter. We will prove that for any set $Y \in \mathscr{P} \bigcup b$ either $Y$ or $( \bigcup b) \setminus Y$ is an element of $b$ .

Let $Y \in \mathscr{P} \bigcup b$ . Then $Y' = Y \cap B \in \mathscr{P} B$ ; $Y' = \left\langle f \right\rangle X'$ for some $X' \in \mathscr{P} A$ . Because $a$ is an ultrafilter, either $X' \in a$ or $( \bigcup a) \setminus X' \in a$ . If $X' \in a$ then $X' \in a \cap \mathscr{P} A$ and so $Y' = \left\langle f \right\rangle X' = \left\langle f \right\rangle |_{a \cap \mathscr{P} A} X' \in b \cap \mathscr{P} B$ and consequently $Y \in b$ . If $( \bigcup a) \setminus X' \in a$ then $A \setminus X' \in a \cap \mathscr{P} A$ ; consequently $\left\langle f \right\rangle |_{a \cap \mathscr{P} A} (A \setminus X') \in b \cap \mathscr{P} B$ ; $\left\langle f \right\rangle |_{a \cap \mathscr{P} A} (A) \setminus \left\langle f \right\rangle |_{a \cap \mathscr{P} A} (X') \in b \cap \mathscr{P} B$ ; $B \setminus Y' \in b$ ; $( \bigcup b) \setminus Y \in b.$