# Can two filters be “separated”?

Let $A$ and $B$ be filters on some set. (Note that I do include improper filter into the set of filters.)

Do necessarily exist filters (on the same set) $X\supseteq A$ and $Y\supseteq B$ such that $X\cap Y = A\cap B$ and $X$ and $Y$ are separated (that is there are non-intersecting sets $P$, $Q$ such that $P\in X$ and $Q\in Y$)?

• Not necessarily. A trivial example is when $A=B$ is a principal ultrafilter. I suggest you to add the hypothesis "distinct filters". – Crostul Feb 6 '18 at 21:48
• @Crostul I've corrected the question: Note that I do include improper filter into the set of filters. – porton Feb 6 '18 at 21:49
• @Crostul Now your example is no more a counter-example – porton Feb 6 '18 at 21:50

This is perhaps easier to think about via Stone duality: if $S$ is the Stone space of the power set Boolean algebra, there is an inclusion-reversing correspondence between closed subsets of $S$ and filters. So in terms of closed subsets of $S$, your question is, if $A$ and $B$ are closed sets, do there exist closed sets $X\subseteq A$ and $Y\subseteq B$ such that $X\cup Y=A\cup B$ and $X\cap Y=\emptyset$?

Let me first demonstrate that a counterexample exists in any topological space $S$ which is not extremally disconnected. If $S$ is not extremally disconnected, that means there is an open set $U\subset S$ whose closure is not open. Let $A=\overline{U}$ and let $B=S\setminus U$; then $A$ and $B$ are closed and $A\cup B=S$. Suppose there exist disjoint closed sets $X\subseteq A$ and $Y\subseteq B$ such that $X\cup Y=S$. Then $X$ must contain $A\setminus B=U$, so $X$ must be all of $A=\overline{U}$ since it is closed. But $X$ is also open, since its complement $Y$ is closed. This is a contradiction, since we assumed $\overline{U}$ was not open.

Now, the Stone space $S$ of a power set Boolean algebra is extremally disconnected, so this does not immediately give a counterexample. However, if your set is infinite, then $S$ has a closed subspace $T$ which is not extremally disconnected (namely, the subspace of nonprincipal ultrafilters, since the power set algebra modulo the ideal of finite subsets is not complete). Taking a counterexample inside $T$ as above, that counterexample will still work for $S$.

Here's the same argument, translated concretely into specific filters. Take a set $E$ which is partitioned into infinitely many infinite subsets $E_i$. Let $B$ be the filter of subsets of $E$ which contain $E_i$ for all but finitely many $i$. Let $A$ be the filter of subsets of $E$ which contain all but finitely many points of $E_i$ for each $i$. Note that $A\cap B$ is the cofinite filter.

Now suppose $X$ and $Y$ are separated filters extending $A$ and $B$ such that $X\cap Y$ is still the cofinite filter. If $X\neq A$, then there is some element $P\in X$ which omits infinitely many points of $E_i$ for some $i$. But then $P\cup (E\setminus E_i)\in X\cap B\subseteq X\cap Y$ and is not cofinite, which is a contradiction. Thus $X=A$.

Now, since $X$ and $Y$ are separated, let $P\in X=A$ and $Q\in Y$ be disjoint. Since $P\in A$ and $Q$ is disjoint from $P$, $Q\cap E_i$ must be finite for all $i$. We can enlarge $Q$ to a set $R$ which contains all but one element of $E_i$ for each $i$. Then $R\in X\cap Y$ but $R$ is not cofinite, which is a contradiction.

• What is the difference between $\overline{U}$ and $S\setminus U$? They seem to denote the same thing! – porton Feb 6 '18 at 23:12
• By $\overline{U}$ I mean the closure, not the complement. – Eric Wofsey Feb 6 '18 at 23:15