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.

  • $\begingroup$ Have you proved that every ultrafilter is prime? $\endgroup$ – Asaf Karagila May 21 '13 at 2:53
  • $\begingroup$ yes, this follows easily from the condition at the end of my question. $\endgroup$ – Camilo Arosemena-Serrato May 21 '13 at 2:55
  • $\begingroup$ 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.) $\endgroup$ – Asaf Karagila May 21 '13 at 4:09
  • $\begingroup$ 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. $\endgroup$ – Camilo Arosemena-Serrato May 21 '13 at 4:42
  • $\begingroup$ @CamiloArosemena This property of disjoint zero-sets is called (in papers I've seen) "screenability". It's used in the Wallman compactifications, IIRC. $\endgroup$ – Henno Brandsma May 21 '13 at 6:51

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.


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