# Intersection of two filters on a poset

Fix a poset.

A filter on the poset is its nonempty subset which is both a down-directed set and an upper set.

Conjecture Intersection of two filters is also a filter.

I have proved this conjecture for:

• meet-semilattices with greatest element (I denote it $\top$)
• join-semilattices

Can it be generalized further? Particularly, does it hold for all posets?

Note that I denote binary meets and joins as $\sqcap$ and $\sqcup$.

Proof for meet-semilattices with greatest element Let $P$, $Q$ be filters. $P\cap Q$ is nonempty because $\top\in P\cap Q$. That $P\cap Q$ is an upper set is obvious. Let $A,B\in P\cap Q$. Then $A,B\in P$ and thus $A\sqcap B\in P$, $A,B\in Q$ and thus $A\sqcap B\in Q$. Thus $A\sqcap B\in P\cap Q$ that is our set is down-directed.

Proof for join-semilattices Let $P$, $Q$ be filters. $P\cap Q$ is nonempty because $X\sqcup Y\in P\cap Q$ for some $X\in P$, $Y\in Q$. That $P\cap Q$ is an upper set is obvious. Let $A,B\in P\cap Q$. Then $A,B\in P$, thus there exists $C\in P$ such that $C\le A$ and $C\le B$. Analogously there exists $D\in P$ such that $D\le A$ and $D\le B$. Let $E=C\sqcup D$. Then $E\in P$ and $E\in Q$, thus $E\in P\cap Q$ and $E\le A$ and $E\le B$. So $P\cap Q$ is down-directed.

## 1 Answer

It certainly doesn’t hold for all posets. Let $P=\{0,1\}\times\{0,1\}$, and set $\langle a_0,a_1\rangle\le\langle b_0,b_1\rangle$ iff $a_0=b_0$ and $a_1\le b_1$. Then the filters $\{0\}\times\{0,1\}$ and $\{1\}\times\{0,1\}$ have empty intersection.