Quick introduction: I left mathematics about $30$ years ago to begin law school and pursue a career as a lawyer. I've reacquired the itch, and I've been very slowly going through Set Theory: An Introduction to Independence Proofs by Kenneth Kunen ($1980$ edition).

After about $18$ months, I'm working my way through Chapter $2$ and I've reached the discussion of Martin's Axiom. I'm looking specifically at Theorem $3.4$, which asserts a number of equivalent conditions to Martin's Axiom.

Kunen sets up the proof by using the Stone space of a complete Boolean algebra $\mathscr B$. The Stone space is a new concept for me. I think the definition Kunen states is standard -- the points of the space are the algebra's ultrafilters, and a basis for the topology is $\{ U_p \mid p \in \mathscr B \setminus \{ \mathbb 0 \} \}$, where $U_p$ is the set of all ultrafilters containing $p \in \mathscr B$.

I see why this set is a basis for a topology. Kunen asserts that it's a compact, totally disconnected Hausdorff space. I see why it's Hausdorff and that each set in the basis is clopen. I'm having trouble seeing why the Stone space is compact. (Kunen asserts that we don't need the algebra to be complete but I'm willing to use the fact if it helps.)

I'm trying to prove that a collection of closed sets with the finite intersection property has non-empty intersection. I think that any closed set turns out to be an arbitrary intersection of basic open sets, so we can assume without loss of generality that the closed sets in our collection are in fact basic open sets, $U_{p_\alpha}$. Therefore, if the collection has the f.i.p., the various $p_\alpha$ must be pairwise compatible. But I don't think that gets me anywhere because the infimum of the $p_\alpha$ still can be $\mathbb 0$.

So why does the Stone space of a (complete) Boolean algebra have to be compact? Thanks for helping me understand this.

  • $\begingroup$ It suffices that every finite subset has non-zero meet. Then we can extend the family to an ultrafilter (that's necessary and sufficient). Not all Boolean algebras are complete, only relatively few are; these correspond to the extremally disconnected Stone spaces. $\endgroup$ Aug 29 '20 at 13:06
  • $\begingroup$ @HennoBrandsma Thanks for your answer below. I'm still working through how we have satisfied the conditions of Zorn's Lemma. I may be back later with a follow-up question. Please don't tell me yet -- I need the practice so I want to see whether I can work through it myself. $\endgroup$ Aug 29 '20 at 20:54

You only need to consider a cover of $S$ (the Stone space) by basic open subsets, so a cover of the form $U_p, p \in I$ where $I$ is some subset of $\mathcal{B}$. Suppose it does not have a finite subcover, so for any finite $I' \subseteq I$ , $\bigcup_{p \in I'} B_p \neq S$, so there is some ultrafilter $F_{I'} \subseteq \mathcal{B}$ such that $F_{I'} \notin U_p$ or $p \notin F_{I'}$, so $p' \in F_{I'}$ (where $p'$ denotes the complement in the B.A.) for all $p \in I'$. So $\land_{p \in I'} p'\in F_{I'}$ and so $\land_{p \in I'} p' \neq \Bbb 0$ and it follows that the set $\{p'\mid p \in I\}$ has the finite intersection property (and finite subset has non-zero meet). So Zorn (or another AC-implied "maximal principle") implies that there is an ultrafilter $F$ that contains all $p'$ for $p \in I$, and this $F$ would not have been covered by the original basic cover $U_p, p \in I$, which is a contradiction. So there must be a finite subcover after all and $S$ is compact. Note that I use that for ultrafilters $F$ in a B.A. we have that $p \notin F$ iff $p' \in F$, which is a standard fact. This also implies that $S\setminus U_p = U_{p'}$ so all basic open sets are also closed (clopen) which explains the total disconnectedness of $S$ (it's even zero-dimensional, but those notions are equivalent for compact Hausdorff spaces).

We also argue that the $U_p$ form a base for the closed sets and every family of $U_p$ with f.i.p. corresponds to a family of $p$ with the f.i.p etc. But personally I like the cover approach somewhat better as it is more direct (vis a vis the definition of compactness).

  • $\begingroup$ I think I see it now. In the terms I used in my question, the union of any chain of finite collections of $U_{p_\alpha}$ is also a filter, so any chain is bounded above in the set of filters and we have satisfied the conditions of Zorn's Lemma and there must be a maximal filter, which is an ultrafilter. It must contain all finite intersections of $U_{p\alpha}$ or it would not be maximal. This ultrafilter is therefore in the intersection of all $U_{p_\alpha}$, so that intersection is non-empty and we have proved compactness. $\endgroup$ Aug 30 '20 at 7:35
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    $\begingroup$ @RobertShore If $Q$ is a subset of $\mathcal{B}$ such that every finite subset has a non-zero meet, define a filter $\mathcal{F}$ by the set $\{b \in \mathcal{B}\mid \exists Q' \subseteq Q \text{ finite }: \land Q' \le b \}$. This is a well-defined non-trivial filter by the unite-meet-property (meaning that $\Bbb 0 \notin \mathcal{F}$) and then Zorn implies that we can extend it to an ultrafilter on $\mathcal{B}$, which thus contains all of $Q$. $\endgroup$ Aug 30 '20 at 17:17
  • $\begingroup$ Ah, it was simpler than I thought. Just note that a subset of a Boolean algebra with the f.i.p. generates a non-trivial filter. Then Zorn's Lemma guarantees the existence of an ultrafilter containing that filter. $\endgroup$ Aug 30 '20 at 19:07

An ultrafilter is a subset of the powerset $\mathcal{P}(\mathscr{B})$ of $\mathscr{B}$. View a subset of $\mathcal{P}(\mathscr{B})$ as a $\{0,1\}$-valued function on $\mathcal{P}(\mathscr{B})$. If you put the discrete topology on $\{0,1\}$, then $\{0,1\}^{\mathcal{P}(\mathscr{B})}$ is compact by Tychonoff. It is not hard to show that the Stone space (i.e. the set of functions that correspond to ultrafilters) is closed subset of $\{0,1\}^{\mathcal{P}(\mathscr{B})}$ in this topology, and the induced topology on the Stone space is precisely the one defined by Kunen. So the topology on the Stone space is compact since it is a closed subspace of a compact space.

Here is also a direct proof. Let $S$ be the Stone space. Suppose we have an open cover of $S$ with no finite subcover. We can assume this open cover is of the form $\{U_{p_i}:i\in I\}$ for some index set $I$ and $p_i\in\mathscr{B}$. For a finite $X\subseteq I$, let $p_X=\bigcup_{i\in X}p_i\in\mathscr{B}$. Then $\bigcup_{i\in X}U_{p_i}=U_{p_X}$. So $U_{p_X}\neq S$ for any finite $X$ by assumption. It follows that $p_X\neq 1$ for all finite $X$. In other words $\{\neg p_i:i\in I\}$ has the finite intersection property. So there is an ultrafilter $\mathcal{U}\in S$ containing $\neg p_i$ for all $i\in I$. So $\mathcal{U}\not\in U_{p_i}$ for all $i\in I$, a contradiction.

Note that both proofs involve something like the Axiom of Choice through either Tychonoff's Theorem or extending a filter to an ultrafilter.

  • $\begingroup$ I think the fact that all Stone spaces are compact implies "almost AC". So it's quite necessary to use it. (it implies that all products of compact Hausdorff spaces are compact and full AC is equivalent to the same fact without the Hausdorff). $\endgroup$ Aug 29 '20 at 13:00
  • $\begingroup$ Yes, "all Stone spaces are compact" is equivalent to the ultrafilter lemma (or Boolean prime ideal theorem) which is weaker than AC. For example BPI is equivalent to "products of compact Hausdorff spaces are compact", while AC is equivalent to "products of compact spaces are compact". $\endgroup$ Aug 29 '20 at 13:09
  • $\begingroup$ @halrankard2 Thanks for your answer. I think I see why the set of functions that aren't ultrafilters has to be open. It's $$\{ f \mid f(\mathbb 0)= 1 \} \cup \{ f \mid f(p)=f(p') \} \cup \{ f \mid f(p)=f(q) = 1, f(p \land q) = 0 \} \cup \{ f \mid p \leq q, f(p) = 1, f(q)=0 \}$$ and each of these components (taking unions in one case over all $p, q \in \mathscr B$ and in the other case over all $p \leq q \in \mathscr B$) is open in the product topology. Am I right? I'm still working through how we've satisfied the conditions of Zorn's Lemma, which both you and Henno Brandsma suggested. $\endgroup$ Aug 29 '20 at 21:02
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    $\begingroup$ The argument for showing the set of ultrafilters is closed is right. The use of Zorn's Lemma to extend a filter to an ultrafilter is separate argument. Given a filter $F$ on $\mathscr{B}$, consider the collection of all filters on $\mathscr{B}$ that contain $F$. This collection is nonempty since $F$ is in it, and the union of any increasing chain of filters containing $F$ is a filter containing $F$. So the collection satisfies the hypotheses of Zorn, and a maximal element is an ultrafilter extending $F$. (Of course, Zorn is equivalent to AC, and thus stronger than the ultrafilter lemma.) $\endgroup$ Aug 30 '20 at 11:29

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