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Some backround:

Let $\mathcal P$ be a class of subsets of a topological space such that if $P_1$ and $P_2$ are sets from $\mathcal P$ then $P_1\cap P_2$ and $P_1\cup P_2$ belong to $\mathcal P$. A $\mathcal P$-filter $\mathcal F$ is a collection of nonempty elements of $\mathcal P$ closed for finite intersections and such that for any $P_1\in \mathcal F$ and $P_1\subseteq P_2\in \mathcal P$ we have $P_2\in \mathcal F$.

A $\mathcal P$-filter $\mathcal F$ is said to be prime if whenever $P_1$ and $P_2$ belong to $\mathcal P$ and $P_1\cup P_2\in \mathcal F$, then $P_1\in \mathcal F$ or $P_2\in \mathcal F$. A $\mathcal P$-ultrafilter is just a maximal $\mathcal P$-filter.

My question is, is every prime $\mathcal P$-filter contained in a unique $\mathcal P$-ultrafilter?; this is exercise 12E.6 of Willard's General Topology.

I have proved that if $\mathcal F$ is a $\mathcal P$-ultrafilter and $P\in \mathcal P$ is such that $P\cap F\neq \emptyset$ for all $F\in \mathcal F$, then $P\in \mathcal F$. I think this must be used in the proof but I don't know how.

All hints are appreciated.

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Have you proved that every ultrafilter is prime? – Asaf Karagila May 21 '13 at 2:53
    
yes, this follows easily from the condition at the end of my question. – Camilo Arosemena May 21 '13 at 2:55
    
I know it does, I asked if you proved that. Did you also prove the slightly stronger proposition: If $\cal F$ is a filter, and $P$ is such that for all $F\in\cal F$, $P\cap F\neq\varnothing$, then there exists $\cal F'$ extending $\cal F$ such that $P\in\cal F'$? (The statement about ultrafilters follows trivially form this.) – Asaf Karagila May 21 '13 at 4:09
    
yes, I did. I've also proved the question for when $\mathcal P$ is the set of all zero-sets of the space. In there you have to use the following property: Given any two disjoint zero-sets $A,B$, there are zero-sets $C,D$ such that $A\subseteq C^c$, $B\subseteq D^c$ and $C^c\cap D^c=\emptyset$. Perhaps the exercise is false, and we can come up with a counterexample with this. – Camilo Arosemena May 21 '13 at 4:42
    
@CamiloArosemena This property of disjoint zero-sets is called (in papers I've seen) "screenability". It's used in the Wallman compactifications, IIRC. – Henno Brandsma May 21 '13 at 6:51
up vote 5 down vote accepted

In general it's false, I think, even for finite collections.

The question is really about distributive lattices, as the commenters already said, and there are standard examples of distributive lattices such that a prime filter need no be extendible to a unique ultrafilter, I just have to instantiate such an example as a collection of subsets of a topological space..:

Let $X = \mathbb{R}$, say, and $\mathcal{P} = \{[0,9],[0,6],[3,9],[3,6],[3,5],[4,6],[4,5]\}$ (the lattice diagram is a "double diamond").

Then $\mathcal{F} = \{[0,9], [3,9]\}$ is a prime filter, but both $\mathcal{U} = \mathcal{P}\setminus \{[3,5],[4,5]\}$ and $\mathcal{U}' = \mathcal{P} \setminus \{[4,6],[4,5]\}$ are ultrafilters extending $\mathcal{F}$.

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