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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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Your collection $\mathcal P$ is known as a distributive lattice. It is known that the following statement is equivalent to the Axiom of Choice: "Given any distributive lattice, each filter on the lattice is contained in some ultrafilter on the lattice." Given that, I've added the (axiom-of-choice) tag, since it seems that your question may be closely related. –  Cameron Buie May 20 '13 at 22:44
    
@Cameron: I disagree with the tag addition. This is not a question about where does the axiom of choice enters the proof, or how to infer the axiom of choice from the principle which you have mentioned. If we start adding the AC tag to every question which is closely related to a principle equivalent to some choice principle, the AC tag will rival [homework] fairly soon! –  Asaf Karagila May 20 '13 at 22:47
    
@Asaf: Fair point. I guess I should have read the tag description first. :-P –  Cameron Buie May 20 '13 at 22:49
    
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

1 Answer 1

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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