Why is the Stone space of a Boolean algebra compact?

Question

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.

Answer 1

An ultrafilter is a subset of $\mathscr{B}$. View a subset of $\mathscr{B}$ as a $\{0,1\}$-valued function on $\mathscr{B}$. If you put the discrete topology on $\{0,1\}$, then $\{0,1\}^{\mathscr{B}}$ 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 $\{0,1\}^{\mathscr{B}}$ 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.

Answer 2

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 $I' \subseteq I$ , $\bigcup_{p \in I'} U_p \neq S$, 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).