4 typo

## Original Question:

Let $\mathcal{P}(\omega)/fin$ denote the Boolean algebra formed from $\mathcal{P}(\omega)$ by modding out by the ideal $fin$ of finite subsets of $\omega$. As a first pass at the intended question, consider the following:

Question 0: Are there two filters $F$ and $G$ in $\mathcal{P}(\omega)/fin$ such that there is a unique ultrafilter extending $F \cup G$?

The answer, of course, is yes: consider the case where $F$ is already an ultrafilter, and $G$ is some filter such that $G \subseteq F$. We might therefore ask (what seems to be) a harder question. Given an ideal $I$ in $\mathcal{P}(\omega)/fin$, notice that

$$\{a \in \mathcal{P}(\omega)/fin \: : \: a \geq I\}$$

(where $a \geq I$ iff $(\forall b \in I)[a \geq b]$) is a filter; call such filters regular filters (I made up this terminology, and I would be glad to know if there is already a word for such objects). Now we can ask:

Question 1: Are there two regular filters $F$ and $G$ such that there is a unique ultrafilter extending $F \cup G$?

Via Stone duality, this question can be rephrased (I believe) in topological terms:

Question 1$'$: Are there two regular closed subsets $C,D \subset \omega^{*}$ such that $C \cap D$ is a singleton?

Here, a regular closed set is simply a set which is equal to the closure of its interior, and $\omega^{*} = \beta \omega \setminus \omega$, the space of all non-principal ultrafilters on $\omega$ (i.e. the Stone space of $\mathcal{P}(\omega)/fin$). If $C$ and $D$ are witnesses to a positive answer to question 1$'$, then $int(C)$ and $int(D)$ must be disjoint, in which case the ideals corresponding to these open sets form a gap; this is the basis for my original interest in this question.

## Update:

Based on the suggestions given by Andreas Blass, it turns out we have the following consistency result.

Theorem: Under CH, there exist regular filters $F$ and $G$ such that $F \cup G$ extends to a unique ultrafilter.

Proof. (sketch)

Let $\{c_{\alpha} \: : \: \alpha < \omega_{1}\}$ be an enumeration of all elements of $\mathcal{P}(\omega)/fin$. Choose elements $a_{0}, b_{0}$ such that $a_{0} \land b_{0} = 0$, $a_{0} \lor b_{0} < 1$, and furthermore such that either $c_{0} \leq a_{0} \lor b_{0}$, or else $\lnot c_{0} \leq a_{0} \lor b_{0}$.

Now suppose that for $\gamma < \omega_{1}$, we have constructed increasing sequences $\{a_{\alpha} \: : \: \alpha < \gamma\}$ and $\{b_{\alpha} \: : \: \alpha < \gamma\}$ such that, for all $\alpha < \gamma$,

(a) $a_{\alpha} \land b_{\alpha} = 0$;

(b) $a_{\alpha} \lor b_{\alpha} < 1$;

(c) either $c_{\alpha} \leq a_{\alpha} \lor b_{\alpha}$, or $\lnot c_{\alpha} \leq a_{\alpha} \lor b_{\alpha}$.

First suppose that $\gamma = \eta + 1$ is a successor ordinal. Let $d \in \{c_{\gamma}, \lnot c_{\gamma}\}$ be such that $$a_{\eta} \lor b_{\eta} \lor d < 1,$$ let ${d_{a}, d_{b}}$ be a (nontrivial, if possible) partition of $d \land \lnot(a_{\eta} \lor b_{\eta})$, and set $a_{\gamma} = a_{\eta} \lor d_{a}$ and $b_{\gamma} = b_{\eta} \lor d_{a}$. Then it is easy to see that $\{a_{\alpha} \: : \: \alpha < \gamma + 1\}$ and $\{b_{\alpha} \: : \: \alpha < \gamma + 1\}$ are increasing sequences satisfying (a) through (c).

Suppose now that $\gamma$ is a limit ordinal. Observe that the sequence $\{\lnot(a_{\alpha} \lor b_{\alpha}) \: : \: \alpha < \gamma\}$ is countable and strictly decreasing. It follows that there exists a nonzero lower bound of this sequence; equivalently, there exists an $e < 1$ such that $e \geq a_{\alpha}$ and $e \geq b_{\alpha}$ for all $\alpha < \gamma$. Moreover, since $\{a_{\alpha} \: : \: \alpha < \gamma\}$ and $\{b_{\alpha} \: : \: \alpha < \gamma\}$ are both countable, they cannot form a gap; hence there exist $a, b \in \mathcal{P}(\omega)/fin$ such that, for all $\alpha < \gamma$, $a \geq a_{\alpha}$ and $b \geq b_{\alpha}$. Replacing $a$ and $b$ by $a \land e$ and $b \land e$, if necessary, we may assume that $a, b \leq e$. Now we can repeat the argument given for the case where $\gamma$ is a successor, replacing $a_{\eta}$ by $a$ and $b_{\eta}$ by $b$.

Thus we obtain $\{a_{\alpha} \: : \: \alpha < \omega_{1}\}$ and $\{b_{\alpha} \: : \: \alpha < \omega_{1}\}$. I claim that these form a gap. If not, then there is some $\beta < \omega_{1}$ such that $c_{\beta} \geq \{a_{\alpha} \: : \: \alpha < \gamma\}$ and $\lnot c_{\beta} \geq \{b_{\alpha} \: : \: \alpha < \gamma\}$; on the other hand, we know that either $c_{\beta} \leq a_{\beta} \lor b_{\beta}$, or $\lnot c_{\beta} \leq a_{\beta} \lor b_{\beta}$, each of which readily yields a contradiction.

I claim also that for every two-element partition $\{p, q\}$ in $\mathcal{P}(\omega)/fin$, one element, say $p$, is such that $\{a_{\alpha} \land p \: : \: \alpha < \omega_{1}\}$ and $\{b_{\alpha} \land p \: : \: \alpha < \omega_{1}\}$ do not form a gap. Indeed, each such partition must be of the form $\{c_{\beta}, \lnot c_{\beta}\}$ for some $\beta < \omega_{1}$. Without loss of generality, suppose we have $c_{\beta} \leq a_{\beta} \lor b_{\beta}$; then it is not difficult to see that $$a_{\beta} \geq \{a_{\alpha} \land c_{\beta} \: : \: \alpha < \omega_{1}\}$$ and likewise $$b_{\beta} \geq \{b_{\alpha} \land c_{\beta} \: : \: \alpha < \omega_{1}\},$$ from which it follows that these sequences do \emph{not} not form a gap. $\blacksquare$

So I suppose the name of the game here is consistency results, such as

Is a negative answer to Question 1 consistent with ZFC? Can a positive answer be proved under any assumptions weaker than CH?

As this seems to be a rather slippery problem, I would welcome any suggested reading on this topic. And thanks again for the helpful replies already given; they are much appreciated.

3 minor correction to proof

## Original Question:

Let $\mathcal{P}(\omega)/fin$ denote the Boolean algebra formed from $\mathcal{P}(\omega)$ by modding out by the ideal $fin$ of finite subsets of $\omega$. As a first pass at the intended question, consider the following:

Question 0: Are there two filters $F$ and $G$ in $\mathcal{P}(\omega)/fin$ such that there is a unique ultrafilter extending $F \cup G$?

The answer, of course, is yes: consider the case where $F$ is already an ultrafilter, and $G$ is some filter such that $G \subseteq F$. We might therefore ask (what seems to be) a harder question. Given an ideal $I$ in $\mathcal{P}(\omega)/fin$, notice that

$$\{a \in \mathcal{P}(\omega)/fin \: : \: a \geq I\}$$

(where $a \geq I$ iff $(\forall b \in I)[a \geq b]$) is a filter; call such filters regular filters (I made up this terminology, and I would be glad to know if there is already a word for such objects). Now we can ask:

Question 1: Are there two regular filters $F$ and $G$ such that there is a unique ultrafilter extending $F \cup G$?

Via Stone duality, this question can be rephrased (I believe) in topological terms:

Question 1$'$: Are there two regular closed subsets $C,D \subset \omega^{*}$ such that $C \cap D$ is a singleton?

Here, a regular closed set is simply a set which is equal to the closure of its interior, and $\omega^{*} = \beta \omega \setminus \omega$, the space of all non-principal ultrafilters on $\omega$ (i.e. the Stone space of $\mathcal{P}(\omega)/fin$). If $C$ and $D$ are witnesses to a positive answer to question 1$'$, then $int(C)$ and $int(D)$ must be disjoint, in which case the ideals corresponding to these open sets form a gap; this is the basis for my original interest in this question.

## Update:

Based on the suggestions given by Andreas Blass, it turns out we have the following consistency result.

Theorem: Under CH, there exist regular filters $F$ and $G$ such that $F \cup G$ extends to a unique ultrafilter.

Proof. (sketch)

Let $\{c_{\alpha} \: : \: \alpha < \omega_{1}\}$ be an enumeration of all elements of $\mathcal{P}(\omega)/fin$. Choose elements $a_{0}, b_{0}$ such that $a_{0} \land b_{0} = 0$, $a_{0} \lor b_{0} < 1$, and furthermore such that either $c_{0} \leq a_{0} \lor b_{0}$, or else $\lnot c_{0} \leq a_{0} \lor b_{0}$.

Now suppose that for $\gamma < \omega_{1}$, we have constructed strictly increasing sequences $\{a_{\alpha} \: : \: \alpha < \gamma\}$ and $\{b_{\alpha} \: : \: \alpha < \gamma\}$ such that, for all $\alpha < \gamma$,

(a) $a_{\alpha} \land b_{\alpha} = 0$;

(b) $a_{\alpha} \lor b_{\alpha} < 1$;

(c) either $c_{\alpha} \leq a_{\alpha} \lor b_{\alpha}$, or $\lnot c_{\alpha} \leq a_{\alpha} \lor b_{\alpha}$.

First suppose that $\gamma = \eta + 1$ is a successor ordinal. Let $d \in \{c_{\gamma}, \lnot c_{\gamma}\}$ be such that $$a_{\eta} \lor b_{\eta} \lor d < 1,$$ let ${d_{a}, d_{b}}$ be a (nontrivial) nontrivial, if possible) partition of $d \land \lnot(a_{\eta} \lor b_{\eta})$, and set $a_{\gamma} = a_{\eta} \lor d_{a}$ and $b_{\gamma} = b_{\eta} \lor d_{a}$. Then it is easy to see that $\{a_{\alpha} \: : \: \alpha < \gamma + 1\}$ and $\{b_{\alpha} \: : \: \alpha < \gamma + 1\}$ are strictly increasing sequences satisfying (a) through (c).

Suppose now that $\gamma$ is a limit ordinal. Observe that the sequence $\{\lnot(a_{\alpha} \lor b_{\alpha}) \: : \: \alpha < \gamma\}$ is countable and strictly decreasing. It follows that there exists a nonzero lower bound of this sequence; equivalently, there exists an $e < 1$ such that $e \geq a_{\alpha}$ and $e \geq b_{\alpha}$ for all $\alpha < \gamma$. Moreover, since $\{a_{\alpha} \: : \: \alpha < \gamma\}$ and $\{b_{\alpha} \: : \: \alpha < \gamma\}$ are both countable, they cannot form a gap; hence there exist $a, b \in \mathcal{P}(\omega)/fin$ such that, for all $\alpha < \gamma$, $a \geq a_{\alpha}$ and $b \geq b_{\alpha}$. Replacing $a$ and $b$ by $a \land e$ and $b \land e$, if necessary, we may assume that $a, b \leq e$. Now we can repeat the argument given for the case where $\gamma$ is a successor, replacing $a_{\eta}$ by $a$ and $b_{\eta}$ by $b$.

Thus we obtain $\{a_{\alpha} \: : \: \alpha < \omega_{1}\}$ and $\{b_{\alpha} \: : \: \alpha < \omega_{1}\}$. I claim that these form a gap. If not, then there is some $\beta < \omega_{1}$ such that $c_{\beta} \geq \{a_{\alpha} \: : \: \alpha < \gamma\}$ and $\lnot c_{\beta} \geq \{b_{\alpha} \: : \: \alpha < \gamma\}$; on the other hand, we know that either $c_{\beta} \leq a_{\beta} \lor b_{\beta}$, or $\lnot c_{\beta} \leq a_{\beta} \lor b_{\beta}$, each of which readily yields a contradiction.

I claim also that for every two-element partition $\{p, q\}$ in $\mathcal{P}(\omega)/fin$, one element, say $p$, is such that $\{a_{\alpha} \land p \: : \: \alpha < \omega_{1}\}$ and $\{b_{\alpha} \land p \: : \: \alpha < \omega_{1}\}$ do not form a gap. Indeed, each such partition must be of the form $\{c_{\beta}, \lnot c_{\beta}\}$ for some $\beta < \omega_{1}$. Without loss of generality, suppose we have $c_{\beta} \leq a_{\beta} \lor b_{\beta}$; then it is not difficult to see that $$a_{\beta} \geq \{a_{\alpha} \land c_{\beta} \: : \: \alpha < \omega_{1}\}$$ and likewise $$b_{\beta} \geq \{b_{\alpha} \land c_{\beta} \: : \: \alpha < \omega_{1}\},$$ from which it follows that these sequences do \emph{not} form a gap. $\blacksquare$

So I suppose the name of the game here is consistency results, such as

Is a negative answer to Question 1 consistent with ZFC? Can a positive answer be proved under any assumptions weaker than CH?

As this seems to be a rather slippery problem, I would welcome any suggested reading on this topic. And thanks again for the helpful replies already given; they are much appreciated.

2 update with consistency result and new questions

## OriginalQuestion:

Here, a regular closed set is simply a set which is equal to the closure of its interior, and $\omega^{*} = \beta \omega \setminus \omega$, the space of all non-principal ultrafilters on $\omega$ (i.e. the Stone space of $\mathcal{P}(\omega)/fin$). If $C$ and $D$ are witnesses to a positive answer to question 1$'$, then $int(C)$ and $int(D)$ must be disjoint, in which case the ideals corresponding to these open sets form a gap; this is the basis for my original interest in this question.

## Update:

Based on the suggestions given by Andreas Blass, it turns out we have the following consistency result.

Theorem: Under CH, there exist regular filters $F$ and $G$ such that $F \cup G$ extends to a unique ultrafilter.

Proof. (sketch)

Let $\{c_{\alpha} \: : \: \alpha < \omega_{1}\}$ be an enumeration of all elements of $\mathcal{P}(\omega)/fin$. Choose elements $a_{0}, b_{0}$ such that $a_{0} \land b_{0} = 0$, $a_{0} \lor b_{0} < 1$, and furthermore such that either $c_{0} \leq a_{0} \lor b_{0}$, or else $\lnot c_{0} \leq a_{0} \lor b_{0}$.

Now suppose that for $\gamma < \omega_{1}$, we have constructed strictly increasing sequences $\{a_{\alpha} \: : \: \alpha < \gamma\}$ and $\{b_{\alpha} \: : \: \alpha < \gamma\}$ such that, for all $\alpha < \gamma$,

(a) $a_{\alpha} \land b_{\alpha} = 0$;

(b) $a_{\alpha} \lor b_{\alpha} < 1$;

(c) either $c_{\alpha} \leq a_{\alpha} \lor b_{\alpha}$, or $\lnot c_{\alpha} \leq a_{\alpha} \lor b_{\alpha}$.

First suppose that $\gamma = \eta + 1$ is a successor ordinal. Let $d \in \{c_{\gamma}, \lnot c_{\gamma}\}$ be such that$$a_{\eta} \lor b_{\eta} \lor d < 1,$$let ${d_{a}, d_{b}}$ be a (nontrivial) partition of $d \land \lnot(a_{\eta} \lor b_{\eta})$, and set $a_{\gamma} = a_{\eta} \lor d_{a}$ and $b_{\gamma} = b_{\eta} \lor d_{a}$. Then it is easy to see that $\{a_{\alpha} \: : \: \alpha < \gamma + 1\}$ and $\{b_{\alpha} \: : \: \alpha < \gamma + 1\}$ are strictly increasing sequences satisfying (a) through (c).

Suppose now that $\gamma$ is a limit ordinal. Observe that the sequence $\{\lnot(a_{\alpha} \lor b_{\alpha}) \: : \: \alpha < \gamma\}$ is countable and strictly decreasing. It follows that there exists a nonzero lower bound of this sequence; equivalently, there exists an $e < 1$ such that $e \geq a_{\alpha}$ and $e \geq b_{\alpha}$ for all $\alpha < \gamma$. Moreover, since $\{a_{\alpha} \: : \: \alpha < \gamma\}$ and $\{b_{\alpha} \: : \: \alpha < \gamma\}$ are both countable, they cannot form a gap; hence there exist $a, b \in \mathcal{P}(\omega)/fin$ such that, for all $\alpha < \gamma$, $a \geq a_{\alpha}$ and $b \geq b_{\alpha}$. Replacing $a$ and $b$ by $a \land e$ and $b \land e$, if necessary, we may assume that $a, b \leq e$. Now we can repeat the argument given for the case where $\gamma$ is a successor, replacing $a_{\eta}$ by $a$ and $b_{\eta}$ by $b$.

Thus we obtain $\{a_{\alpha} \: : \: \alpha < \omega_{1}\}$ and $\{b_{\alpha} \: : \: \alpha < \omega_{1}\}$. I claim that these form a gap. If not, then there is some $\beta < \omega_{1}$ such that $c_{\beta} \geq \{a_{\alpha} \: : \: \alpha < \gamma\}$ and $\lnot c_{\beta} \geq \{b_{\alpha} \: : \: \alpha < \gamma\}$; on the other hand, we know that either $c_{\beta} \leq a_{\beta} \lor b_{\beta}$, or $\lnot c_{\beta} \leq a_{\beta} \lor b_{\beta}$, each of which readily yields a contradiction.

I claim also that for every two-element partition $\{p, q\}$ in $\mathcal{P}(\omega)/fin$, one element, say $p$, is such that $\{a_{\alpha} \land p \: : \: \alpha < \omega_{1}\}$ and $\{b_{\alpha} \land p \: : \: \alpha < \omega_{1}\}$ do not form a gap. Indeed, each such partition must be of the form $\{c_{\beta}, \lnot c_{\beta}\}$ for some $\beta < \omega_{1}$. Without loss of generality, suppose we have $c_{\beta} \leq a_{\beta} \lor b_{\beta}$; then it is not difficult to see that$$a_{\beta} \geq \{a_{\alpha} \land c_{\beta} \: : \: \alpha < \omega_{1}\}$$$$b_{\beta} \geq \{b_{\alpha} \land c_{\beta} \: : \: \alpha < \omega_{1}\},$$from which it follows that these sequences do \emph{not} form a gap. $\blacksquare$

So I suppose the name of the game here is consistency results, such as

Is a negative answer to Question 1 consistent with ZFC? Can a positive answer be proved under any assumptions weaker than CH?

As this seems to be a rather slippery problem, I would welcome any suggested reading on this topic. And thanks again for the helpful replies already given; they are much appreciated.

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