1
$\begingroup$

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 $Q\in Y$)?

$\endgroup$
  • $\begingroup$ Not necessarily. A trivial example is when $A=B$ is a principal ultrafilter. I suggest you to add the hypothesis "distinct filters". $\endgroup$ – Crostul Feb 6 '18 at 21:48
  • $\begingroup$ @Crostul I've corrected the question: Note that I do include improper filter into the set of filters. $\endgroup$ – porton Feb 6 '18 at 21:49
  • $\begingroup$ @Crostul Now your example is no more a counter-example $\endgroup$ – porton Feb 6 '18 at 21:50
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ What is the difference between $\overline{U}$ and $S\setminus U$? They seem to denote the same thing! $\endgroup$ – porton Feb 6 '18 at 23:12
  • 1
    $\begingroup$ By $\overline{U}$ I mean the closure, not the complement. $\endgroup$ – Eric Wofsey Feb 6 '18 at 23:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.