Free ultrafilter generated by a collection I'm doing an exercise that asks me to show that if $\mathcal{A}$ is a collection of subsets of $X$, then $\mathcal{A}$ has the $\omega$-intersection property iff it can be extended to a free ultrafilter. Here, the $\omega$-intersection property means that any finite subset of $\mathcal{A}$ has infinite intersection, and free means that the filter has no finite subset of $X$ as a member. 
I proceeded as follows: let $\Sigma$ be the set of free filters that contain $\mathcal{A}$, ordered by inclusion. Then $\Sigma \neq \emptyset$, because the filter generated by $\mathcal{A}$ belongs to $\Sigma$. I then proved that each chain has an upper bound (in this case, a supremum) and used Zorn's Lemma to conclude that $\Sigma$ must have a $\subseteq$-maximal element. I'm having trouble, however, showing that such maximal free filter containing $\mathcal{A}$ is an ultrafilter. Is there anyone with some insight that can help me?
 A: The $\Rightarrow$ part is pretty straightforward. If $\mathcal{A}$ doesnt have the $\omega$-intersection property, then there is a finite subcolection $\mathcal{A_f}$ such that $\bigcap \mathcal{A_f}$ is finite and, if $F$ is a filter and $\mathcal{A} \subset F$ then $F$ is closed by finite intersections and then has a finite subset ($\bigcap \mathcal{A_f}$), hence $F$ is not free.
Now, by your construction, $\Sigma$ is free and it is already maximal in $\subset$ order. Let $A$ be a subset of $X$. Now we show that either $\{A\} \cup \Sigma$ or $\{X\setminus A\} \cup \Sigma$ has $\omega$-intersection property. 
Suppose the contrary, then there are $U$ and $V \in \Sigma$ such that $U \cap A$ is finite and $V \cap (X\setminus A)$ is finite. But then $U\cap V$ is finite because $U\cap V = ((U\cap V) \cap A) \cup ((U\cap V) \cap (X\setminus A))$, the union of two finite sets. But this is a contradiction because $\Sigma$ has $\omega$-intersection property.
That is, either $\{A\} \cup \Sigma$ or $\{X\setminus A\} \cup \Sigma$ is a free filter extending $\Sigma$ that has $\mathcal{A}$, but, by construction, that is exactly $\Sigma$ and thus $\Sigma$ is an ultrafilter.
A: As you have already recognized, one can use Zorn's lemma to construct a suitable free ultrafilter. However, and this may be of interest to some, it is also possible to use already known theorems about filters and ultrafilters.

Ultrafilter lemma. Let $X$ be a nonempty set and $\mathcal F$ a filter of $X$, then $\mathcal F$ is contained in an ultrafilter $\mathcal U$.

The proof of the ultrafilter lemma already uses Zorn's lemma.

Extension to filters. Let $X$ be a nonempty set and $\mathcal S$ a collection of subsets of $X$. The collection $\mathcal S$ fulfills the finite intersection property iff every finite subcollection has a nonempty intersection. Is this the case, there exists a filter $\mathcal F$ of $X$ with $\mathcal S\subseteq\mathcal F$.


Theorem about free ultrafilters. Let $X$ be an infinite set and $\mathcal U$ an ultrafilter of $X$, then $\mathcal U$ is free iff the Fréchet-Filter $\mathcal F_{\rm F}$ (consisting of all cofinite sets) is contained in $\mathcal U$.

With these tools, the more difficult part "$\mathcal A$ has $\omega$-intersection property $\Rightarrow$ there exists a free ultrafilter containing $\mathcal A$" will look like this:

*

*Show that $\mathcal A\cup\mathcal F_{\rm F}$ fulfills the finite intersection property. This is the only step where the $\omega$-intersection property is used.

*Then there exists a filter $\mathcal F$ containing $\mathcal A\cup\mathcal F_{\rm F}$.

*From the ultrafilter lemma follows, that there exists an ultrafilter containing $\mathcal F$.

*By construction, the ultrafilter $\mathcal U$ contains the Fréchet-Filter $\mathcal F_{\rm F}$ as well as $\mathcal A$. Hence, $\mathcal U$ is a free ultrafilter containing the collection $\mathcal A$.

