(I'm not good at English, so please tell
me if there is something wrong!)
I came up with the following conjecture but I can neither prove it nor give a counterexample. I want to know which it is true or not true.
Conjecture:
Let $S$ be a set and $F, F'$ be two filters on $S$.
For $X\subset S$, define $F_{X} :=\{X'\in F \mid X' \supset X\}$. ($F_X$ is also a filter.)
Then, is there always $X\subset S$ such that $F_X\subset F'$ and $F'_{S\setminus X}\subset F$ ???
For example, if there is $X\in F'$ such that $S\setminus X \in F$, then obviously $F_X\subset F'$ and $F'_{S\setminus X}\subset F$ because of the definition of filters. So, the conjecture is true.
Thanks!!
This is perhaps easier to think about via Stone duality: if $Y$ is the Stone space of the power set Boolean algebra $P(S)$, there is an inclusion-reversing correspondence between closed subsets of $Y$ and filters on $S$. So in terms of closed subsets of $Y$, your question is, if $C$ and $D$ are closed sets, does there exist a clopen subset $A$ such that $A\cup C\supseteq D$ and $A^c\cup D\supseteq C$? (Here $A$ corresponds to the principal filter generated by $X$, and unions of closed sets correspond to intersections of filters.)
These two inclusions are equivalent to $A\supseteq D\setminus C$ and $A^c\supseteq C\setminus D$. In other words, we want to be able to separate $D\setminus C$ and $C\setminus D$ by clopen sets. Two subsets of a Stone space are separated by clopen sets iff their closures are disjoint, so equivalently we want $D\setminus C$ and $C\setminus D$ to have disjoint closures.
Restricting to the closed subspace $Z=C\cup D$ of $Y$, we can think of $C$ and $D$ as an arbitrary pair of closed sets which cover $Z$, and so their complements $D\setminus C$ and $C\setminus D$ are an arbitrary pair of disjoint open subsets of $Z$. So, finally, our question is: in any closed subspace $Z$ of $Y$, do disjoint open sets always have disjoint closures? That is, is every closed subspace of $Y$ extremally disconnected?
Finally, we conclude the answer is no: for instance, if $S$ is an infinite set and $Z$ is the closed subspace of $Y$ consisting of the nonprincipal ultrafilters, then $Z$ is not extremally disconnected, since its algebra of clopen sets is $P(S)/\mathrm{fin}$ which is not complete.
Here's an explicit example of filters you can extract from this argument. Take a set $S$ which is partitioned into infinitely many infinite subsets $S_i$. Let $F$ be the filter of subsets of $S$ which contain $S_i$ for all but finitely many $i$. Let $F'$ be the filter of subsets of $S$ which contain all but finitely many points of $S_i$ for each $i$. Note that $F\cap F'$ is the cofinite filter.
Now suppose $X$ is such that $F_X\subseteq F'$ and $F'_{S\setminus X}\subseteq F$. Since $F_X\subseteq F'$, we must have $X\in F'$ (if $X$ is missing infinitely many points of $S_i$ for some $i$, then $X\cup(S\setminus S_i)$ is in $F_X$ but not $F'$). But then $S\setminus X$ contains only finitely many points of each $S_i$ so we can easily enlarge it to get a set which is in $F'_{S\setminus X}$ but not in $F$. Thus no such $X$ can exist.