Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $\mathcal{F}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.