3 Two useful facts
We will need the following fundamental fact about Boolean algebras which is sometimes called the ultrafilter principle. It could also be deduced from the ring-theoretic fact (probably in Mathlib) that any non-unit element of a ring is in some maximal ideal, but a direct proof is not hard.
Let \(A\) be a Boolean algebra. For any \(a \in A \setminus \{ 1\} \), there exists a homomorphism \(x \colon A \to \mathbf{2}\) such that \(x(a) = 0\).
By Zorn’s lemma, let \(I\) be a maximal element of the set of proper ideals of \(A\) which contain \(a\). Define \(x(b) = 0\) iff \(b \in I\). Clearly, \(x(a) = 0\); we need to check that \(x\) is a homomorphism. The equalities \(x(0) = 0\) and \(x(b \vee b') = x(b) \vee x(b')\) are easy to check from the defining properties of an ideal. To see that \(x(\neg b) = \neg x(b)\) for any \(b \in A\), the crucial observation is that if \(b \not\in I\) and \(\neg b \not\in I\), then it is possible to enlarge \(I\) by adding \(b\) to it and generating an ideal \(I'\), and \(I'\) will still be proper because \(\neg b \not\in I\). By maximality of \(I\) this is impossible. We thus get that, for any \(b \in A\), one of \(b\) and \(\neg b\) must be in \(I\), and they can never be both in \(I\), since that would give \(1= b \vee \neg b\) in \(I\), again contradicting that \(I\) is proper. It now follows from the definitions that \(x(\neg b) = \neg x(b)\).
Recall that a space \(X\) is called totally separated if, for any distinct \(x,y \in X\), there exist disjoint open sets \(U, V \subseteq X\) such that \(x \in U\) and \(y \in V\).
Any compact Hausdorff totally disconnected space is totally separated.
Suppose that, for any distinct \(x,y \in X\), there exists a clopen set \(K \subseteq X\) such that \(x \in K\) and \(y \not\in K\). Then \(X\) is totally disconnected.
The following equivalent definition of Boolean spaces is convenient.
A compact Hausdorff space \(X\) is totally disconnected if, and only if, for any distinct \(x, y \in X\), there exists a clopen set \(K \subseteq X\) such that \(x \in K\) and \(y \not\in K\).
Finally, in any totally separated space, one may actually separate distinct points by clopens, and .