TITLE: A unique ultrafilter extending a union of filters?
QUESTION [12 upvotes]: 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 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.

REPLY [6 votes]: The following is due to Alan Dow:
In any model obtained by adding $\aleph_2$ many Cohen reals to a model
of $\mathsf{CH}$ the statement is false.
We force with $\mathbb{P}=\operatorname{Fn}(\omega_2,2)$ and we let
$\dot{\mathcal{I}}$, $\dot{\mathcal{J}}$ and $\dot u$ be 
$\mathbb{P}$-names such
that $\dot{\mathcal{I}}$ and  $\dot{\mathcal{J}}$ are forced to be ideals
and $\dot u$ is forced to be the unique ultrafilter that extends the
two associated regular filters.
Now let $M$ be an elementary substructure of a suitable large $H(\theta)$
that has cardinality $\aleph_1$ and that is closed under $\omega$-sequences.
Let $\delta=M\cap\omega_2$ and 
$\mathbb{P}_M=\operatorname{Fn}(\delta,2)$.
By elementarity the $\mathbb{P}_M$-names
$\dot{\mathcal{I}}\cap M$ and  $\dot{\mathcal{J}}\cap M$ are forced to be 
ideals and $\dot u\cap M$ is forced to be the unique ultrafilter that extends 
the two associated regular filters.
Work in $V[G\cap M]$.
We write $\mathcal{I}_M$, $\mathcal{I}_M$ and $u_M$ for the interpretations
of $\dot{\mathcal{I}}\cap M$, $\dot{\mathcal{J}}\cap M$ and $\dot u\cap M$
in this model.
Note that every element of $u_M$ meets elements of $\mathcal{I}_M$ 
and $\mathcal{J}_M$.
On the other hand if $\mathcal{I}'$ and $\mathcal{J}'$ are countable 
subfamilies of $\mathcal{I}_M$ and $\mathcal{J}_M$ respectively then some
infinite subset, $x$, of $\omega$ separates the elements of $\mathcal{I}'$ from
those of $\mathcal{J}'$.
This implies that, without loss of generality, for every countable subfamily
$\mathcal{J}'$ of $\mathcal{J}_M$  there is an element $x\in u_M$ such that
$x\cap y$ is finite for all $y\in\mathcal{J}'$.
Using this one can construct a sequence 
$\langle b_\alpha:\alpha<\omega_1\rangle$
in $\mathcal{J}_M$
such that every element of $u_M$ contains all but countably many of 
the $b_\alpha$. 
In  $V[G]$ consider the next Cohen real $c$, added by 
$\operatorname{Fn}([\delta,\delta+\omega,2)$, say and assume, 
without loss of generality, that $c\notin u$.
There must be a set $Y\subseteq c$ that separates 
$\lbrace x\cap c:x\in\mathcal{I}\rbrace $ from $\lbrace y\cap c:y\in\mathcal{J}\rbrace $.
In $V[G\cap M]$ we take names, $\dot c$ and $\dot Y$, for $c$ and $Y$,
note that these are $\operatorname{Fn}(C,2)$-names for some countable set $C$.
For every $\alpha<\omega_1$ it is forced that $\dot Y\cap b_\alpha$ is finite;
by pigeon-holing there will be one $p\in\operatorname{Fn}(C,2)$ and one 
$n\in\omega$ such that $p$ forces $\dot Y\cap b_\alpha\subseteq n$ for 
uncountably many $\alpha$.
Still in $V[G\cap M]$ let 
$Y_p=\lbrace k:(\exists q\le p)(q \Vdash k\in\dot Y)\rbrace $.
Then $b_\alpha\cap Y_p\subseteq n$ for uncountably many $\alpha$ so that
$Y_p\notin u_M$.
On the other hand for every $x\in\mathcal{I}_M$ it is forced that 
$x\cap\dot c\subseteq \dot Y$ (mod finite); this implies that 
$x\setminus Y_p$ is finite for all $x\in\mathcal{I}_M$, so that $Y_p\in u_M$.