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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?

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  • $\begingroup$ Many people define "ultrafilter" as "$\subset$-maximal filter". $\endgroup$ – DanielWainfleet Nov 29 '19 at 7:33
  • $\begingroup$ 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 $\endgroup$ – Nuntractatuses Amável Nov 30 '19 at 0:46
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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.

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