4
$\begingroup$

We define a filter on a given set $X$ as a non-empty collection $\mathcal{F}$ of subsets of $X$ such that

  • $\emptyset \notin \mathcal{F}$

  • $F_1, F_2 \in \mathcal{F} \Rightarrow F_1 \cap F_2 \in \mathcal{F} $

  • $F \in \mathcal{F}, F \subset E \Rightarrow E \in \mathcal{F} $

If $\mathcal{F}$ is a filter, we call it an ultrafilter if for every $A \subset X$, we have that either $A \in \mathcal{F}$ or $A^c \in \mathcal{F}$. (Equivalently, we could have defined it as a filter which is maximal in the poset of all filters on $X$ partially ordered by inlcusion).

What I don't know how to prove is the following:

For every filter $\mathcal{F}$ on $X$, we have that $\mathcal{F} = \cap \{ \mathcal{U}: \mathcal{U} \textit{ is an ultrafilter and } \mathcal{F} \subset \mathcal{U} \}$

I do understand that the fact that the we are not taking an intersection of an empty collection (i.e. that there is always at least an ultrafilter containing $\mathcal{F}$) requires the AC.

The result I am seeking to prove is taken for granted here: Is it true in general that a filter is given by the intersection of the ultrafilters refining it?

Thank you!

$\endgroup$
  • $\begingroup$ This doesn't say a filter is the intersection of "the ultrafilter" that refines it; rather it says a filter is the intersection of all ultrafilters that refine it. $\endgroup$ – Michael Hardy Jun 12 '18 at 1:14
4
$\begingroup$

Well as you mentioned you need AC to always have $\{\mathcal{U}, \mathcal{U}$ is an ultrafilter and $\mathcal{F} \subset \mathcal{U} \} \neq\emptyset$ (actually you can do it with weaker statements but let's not worry about it here).

So in the following I assume AC (for those who are interested, the Ultrafilter lemma is sufficient).

Now it's obvious that $\mathcal{F}$ is a subset of the filter of the mentioned intersection (which is also a filter, we let $\mathcal{G} := \bigcap \{\mathcal{U}, \mathcal{U}$ is an ultrafilter and $\mathcal{F} \subset \mathcal{U} \}$). Therefore it suffices to show that $\mathcal{G}\subset \mathcal{F}$.

Let $A\in \mathcal{G}$ and assume $A\notin \mathcal{F}$. Let $B= A^c$. If there existed $C\in \mathcal{F}$ with $C\cap B = \emptyset$, then $C\subset B^c = A$, and therefore $A\in\mathcal{F}$, a contradiction.

Therefore for all $C\in \mathcal{F}$, $B\cap C \neq \emptyset$.

Therefore $\mathcal{F}\cup \{B\}$ is a filter basis, which means there exists a filter $\mathcal{U'}$ that contains it. Now using the ultrafilter lemma (either as an axiom or as a consequence of AC - you can prove it using Zorn's lemma for instance) there exists an ultrafilter $\mathcal{U}$ containing $\mathcal{U'}$.

But $\mathcal{U}$ contains $\mathcal{F}$ and so it contains $\mathcal{G}$, and so $A\in\mathcal{U}$. But $B=A^c \in \mathcal{U}$, and so $\emptyset \in \mathcal{U}$, a contradiction.

Therefore $A\in \mathcal{F}$, and so $\mathcal{G}\subset \mathcal{F}$, and so $\mathcal{G} = \mathcal{F}$, which is what we wanted.

$\endgroup$
  • $\begingroup$ I have trouble seeing why $\mathcal{F}\cup \{B\}$ is a filter basis... $\endgroup$ – Rlos Nov 15 '17 at 3:48
  • $\begingroup$ Well, I just figured that one can't really tell whether $\mathcal{F}\cup \{B\}$ is a filter basis, but the set $\omega = \{ C \cap B : C \in \mathcal{F} \}$ is a filter basis (by your observation) whose spanned filter contains both $\mathcal{F}$ and $B$, which will lead to the desired contradiction... $\endgroup$ – Rlos Nov 15 '17 at 4:14
  • $\begingroup$ @Rlos : Sorry, what I meant by "filter basis" here was "has the finite intersection property" (there are differences in different languages of what we mean by this, maybe my translation was the right one). But what you said is definitely what works $\endgroup$ – Max Nov 15 '17 at 6:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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