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My question is about the proof of compactness of the Lemma 3.1(page 5) in this paper.

Let $\beta \mathbb{N}$ be the set of all the ultrafilters on $\mathbb{N.}$

For each $A\subseteq \mathbb{N}$, we define $A^*=\{\mathcal{F}\in \beta\mathbb{N}\colon A \in \mathcal{F} \}$ and claim that $\mathcal{B}=\{A^*\colon A\subseteq \mathbb{N}\}$ is a basis for a compact Hausdorff topology.

To show the compactness, he uses the FIP (finite intersection property). By taking a collection of closed sets with the FIP, he claims that

the FIP (of those closed sets) is equivalent to the statement that for every finite collection of $\mathcal{F}_i$’s, there is an ultrafilter that extends $\cup_i \mathcal{F_i}$.

But why is this statement true? I tried to show that any element (an ultrafilter) in the finite intersection of some closed sets can serve as an extension of $\cup_i \mathcal{F_i}$ but failed.

Thank you in advance!

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    $\begingroup$ I think you may be referring to the fact that any collection of subsets such that the intersection of finitely many is non-empty can be extended to an ultrafilter. This is straight Zorn's Lemma, there is a maximal collection with the FIP extending our collection. $\endgroup$
    – André Nicolas
    Nov 7, 2015 at 6:59
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    $\begingroup$ By the way, the paper has a typo in the paragraph preceding the one you're asking about. It states each basic open set is also closed, since $A^* = (A^c)^*$, but that should be $(A^*)^c= (A^c)^*$. This probably didn't help. $\endgroup$
    – BrianO
    Nov 7, 2015 at 7:21
  • $\begingroup$ @AndréNicolas that is the finish of the proof, after he notes the statement that the OP is asking about. It's what eventually proves it, but not what the question is about, I think. $\endgroup$
    – Henno Brandsma
    Nov 7, 2015 at 7:26
  • $\begingroup$ Do you mean $A\in\Bbb N$ or $A\subseteq\Bbb N$? $\endgroup$
    – Asaf Karagila ♦
    Nov 7, 2015 at 7:28
  • $\begingroup$ @AsafKaragila $A\subseteq \mathbb{N}$. And now I see that the question needs a few corresponding edits :) $\endgroup$
    – BrianO
    Nov 7, 2015 at 7:31

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Quoting from the paper (slightly modifying notation): "now let $\{F_i: i \in I\}$ be a collection of closed sets with the FIP. Each $F_i$ consists of all ultrafilters that include all of a collection $\mathcal{F}_i$ of subsets of $\mathbb{N}$". (this is hopefully clear, this is argued in the previous paragraph).

So we know $F_i = \{ \mathcal{F} \in \beta\mathbb{N}: \mathcal{F_i} \subset \mathcal{F} \}$, for some collection $\mathcal{F_i} \subset \mathcal{P}(\mathbb{N})$. This is the really essential fact.

"The FIP is equivalent to the statement that for every finite collection of $\mathcal{F}_i$'s, there is an ultrafilter that extends $\cup \mathcal{F_i}$".

It's more exact to say that for every finite subset $J$ of $I$, there is an ultrafilter that extends $\cup_{i \in J} \mathcal{F_i}$. This is clear, because if $\mathcal{F}$ (a point in $\beta \mathbb{N}$, so an ultrafilter) is in the intersection of the $F_i, i \in J$, it means that $\mathcal{F}$ contains all the sets from $\mathcal{F_i}$, for $i \in J$, because that is exactly what it means to be in $F_i$, and so the ultrafilter contains all sets of $\cup_{i \in J} \mathcal{F_i}$, i.e. it extends that union (as is the common parlance). And if such an ultrafilter exists it is in the finite intersection, reasoning the other way round. It just follows from the definition of the topology, plus the general form of the $F_i$.

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