A filter is the intersection of the ultra filter refining it 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!
 A: 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.
