Ytukyg

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 $P_1cap P_2$ and $P_1cup P_2$ 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 $P_1in mathcal F$ and $P_1subseteq P_2in mathcal P$ we have $P_2in mathcal F$.

A $mathcal P$-filter $mathcal F$ is said to be prime if whenever $P_1$ and $P_2$ belong to $mathcal P$ and $P_1cup P_2in mathcal F$, then $P_1in mathcal F$ or $P_2in mathcal F$. 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 $Pin mathcal P$ is such that $Pcap Fneq emptyset$ for all $Fin mathcal F$, then $Pin mathcal F$. I think this must be used in the proof but I don't know how.

All hints are appreciated.

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 $mathcal{F}$.

It's false in general, and even for cases of interest to Williard.

Let ${cal P}$ be the collection of open subsets of the real line. Let ${cal F}$ be the filter of open sets containing 0. This is clearly a prime filter.

Now, every set of the form $(0,frac{1}{n})$ intersects every element of ${cal F}$, so there is an open ultrafilter ${cal F}_1$ containing ${cal F}$ and also every set of this form.

But, in the same way, every set of the form $(-frac{1}{n},0)$ intersects every element of ${cal F}$, so there is an ultrafilter ${cal F}_2$ containing ${cal F}$ and all of these.

Clearly, ${cal F}_1$ and ${cal F}_2$ are different and ${cal F}$ is contained in both.