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?

• Many people define "ultrafilter" as "$\subset$-maximal filter". Nov 29 '19 at 7:33
• yes, that is my definition. however, being maximal in $\Sigma$ is not the same thing (actually, it is, but that's what I wanted to prove) as being maximal in the set of all filters Nov 30 '19 at 0:46

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.