Condition for maximal filter of a lattice to be completely prime

Question

Background definitions.

• A frame $(\mathbb X,\leq)$ is a partially ordered set with arbitrary joins (least upper bounds) and finite meets (greatest lower bounds).
• A filter $F\subseteq \mathbb X$ is a nonempty up-closed proper subset of $X$ that is closed under finite meets.
• A filter is maximal when it is not contained in any strictly greater filter.
• A filter $F\subseteq\mathbb X$ is completely prime when for every $X\subseteq\mathbb X$, if $\bigvee X\in F$ then $x\in F$ for some $x\in X$.

Call a completely prime filter a point.

A motivating observation.

It is not in general the case that a maximal filter is necessarily completely prime. For a counterexample, consider $\mathbb Q$ (the rationals) with their usual topology and take the frame of its open sets. This has a filter $F_\pi$, which is the set of open neighbourhoods of $\pi$ (for $\mathbb Q$ embedded in $\mathbb R$). This is not completely prime because $(,\pi) \cup (\pi,)=\mathbb Q$ is in $F_\pi$ but neither $(,\pi)$ nor $(\pi,)$ is.

Questions.

1. What is the name for the property of a frame, that every maximal filter be completely prime?
2. What property does this correspond to on point-set topologies?

A remark.

I can see that such a condition seems to have to do with being compact -- the condition says using the terminology above that every maximal filter yields a point, after all! -- but I can't quite put my finger on a precise connection. In particular, Wikipedia has a list characterisations of compactness, but the filters referred to there are in a full powerset, not in a frame open sets. So I'm missing some piece of information. I hope someone can advise.

Thank you.

Answer 1

First a few comments/corrections:

• In your definition of frame, you left out the condition that finite meets distribute over arbitrary joins.
• In any distributive lattice, every maximal filter $F$ is prime, i.e., if $a\vee b \in F$, then $a\in F$ or $b\in F$. Note that in your example, the filter $F_\pi$ is not just not completely prime, it's not even prime. So it can't be maximal as you claim.
• To see this explicitly, consider, for example, the filter generated by $F_\pi \cup \{(-\infty,\pi)\}$. This is a filter properly containing $F_\pi$: it consists of all open sets which contain an interval of the form $(a,\pi)$ for some $a<\pi$.
• It's not going to be easy to give an explicit description of a maximal filter in the frame of open sets in $\mathbb{Q}$. In particular, none of the completely prime filters corresponding to points in $\mathbb{Q}$ are maximal.

Ok, now let's consider your actual question. Let $X$ be a spatial frame, and let $S$ be its corresponding topological space of points (so $X$ is isomorphic to the frame of open sets in $S$). Note that $S$ is a sober space. I haven't thought carefully about the situation for non-spatial frames or non-sober spaces. For $U\in X$, write $[U]$ for the corresponding open set in $S$.

Suppose $F$ is a filter in $X$ which is simultaneously completely prime and maximal. Since $F$ is completely prime, it corresponds to a point $p\in S$. Since $F$ is maximal, for every $U\in X$ such that $U\notin F$, there exists $V\in F$ such that $U\wedge V = \bot$. Translating to topological language, for every open set $[U]$ with $p\notin [U]$, there is an open neighborhood $p\in [V]$ such that $[U]$ and $[V]$ are disjoint. Equivalently, the closure of $p$ has nonempty interior, i.e., $p$ is somewhere dense. Note that if $S$ is $T_1$ (every point is closed), then $p$ is somewhere dense if and only if $p$ is an isolated point.

You ask about frames with the property that every maximal filter is completely prime. I don't know if this property has a name, but I can prove that the following are equivalent:

1. Every maximal filter in $X$ is completely prime.
2. Every cover of $S$ by closed sets has a finite subcover.
3. $S$ is a finite union of irreducible closed sets.

For the equivalence of 1 and 2, note that a filter in $X$ corresponds to a family of open sets in $S$ with the finite intersection property: any finite intersection of sets in the family is non-empty. Assuming 1, since every filter extends to a maximal filter, every family of open sets in $S$ with the finite intersection property contains a point. Conversely, if every family of open sets in $S$ with the finite intersection property contains a point, then every maximal filter $M$ extends to a completely prime filter, which must be equal to $M$ by maximality.

Thus 1 is equivalent to the condition that any family of open sets in $S$ with the finite intersection property has non-empty intersection. Taking complements and the contrapositive, we get 2.

For the equivalence of 2 and 3, note that we can cover $S$ by closed sets $\{\mathrm{cl}(\{p\})\mid p\in S\}$. Assuming $2$, this cover has a finite subcover, so there are finitely many points $p_1,\dots,p_n$ such that $S = \bigcup_{i=1}^n \mathrm{cl}(\{p_i\})$, and each set $\mathrm{cl}(\{p_i\})$ is closed and irreducible. Conversely, suppose $S = \bigcup_{i=1}^n C_i$, where each $C_i$ is closed and irreducible. Since $S$ is the space of points of a frame, it is a sober space, so each $C_i$ is the closure of a single point $p_i$. Now suppose $S = \bigcup_{j\in J} V_j$ is a cover of $S$ by closed sets. For each $1\leq i \leq n$, there is some $j_i\in J$ such that $p_i\in V_{j_i}$. Then $C_i\subseteq V_{j_i}$, so $S = \bigcup_{i=1}^n V_{j_i}$.

Note that if $S$ is $T_1$, then it has a cover $S = \bigcup_{p\in S} \{p\}$ by irreducible closed sets, so under the equivalent conditions above, $S$ is a finite discrete space.