Theorem 1   Given a Boolean algebra $B$ there exists a totally disconnected compact Hausdorff space $X$ such that $B$ is isomorphic to the Boolean algebra of clopen subsets of $X$ .

Proof. [Still very rough sketch of proof, slowly filling in details] Let the set $X$ be defined as follows: $$X=\left\{f\co B\to \{0,1\} \mid f \text{ is a homomorphism}\right\}$$ Alternatively, $X$ may be defined as the set of all ultrafilters of $B$ . (Recall that a subset of a Boolean algebra is the kernel of a homomorphism if and only if it is an ultrafilter.) We may endow $X$ with the subspace topology induced by the product topology on ${\{0,1\}}^B$ . Recall that a basis for this topology is given by the sets $O_{i} (x)$ where $i = 0,1$ and $x \in B$ . These sets are defined as follows: $$O_i (x) = \{ f \in X \mid f(x) = i \}$$

Let us note that $X$ is a closed subspace of ${\{0,1\}}^B$ . Suppose that $g \notin X$ . By definition, $g$ is not a homomorphism, so there must exist $x,y \in B$ such that $g (\sim (x \cup y)) \neq \sim (g(x) \cup g(y))$ . Then no element of the open set $O_{g(x)} (x) \cap O_{g(y)} (y) \cap O_{g (\sim (x \cup y))} (\sim (x \cup y))$ will be a homomorphism because, for every element $h$ of this set, $$h (\sim (x \cup y)) = g (\sim (x \cup y)) \neq \sim (g(x) \cup g(y)) = \sim (h(x) \cup h(y))$$ By Tychonoff's theorem, ${\{0,1\}}^B$ is compact. Hence, as a closed subset of a compact topological space, $X$ is compact.

Then $X$ is a totally disconnected Hausdorff space. Let $\mathrm{Cl}(X)$ denote the Boolean algebra of clopen subsets of $X$ , then the following map $$T\co B \to \mathrm{Cl}(X),\quad T(x)=\{f\in X\,|\, f(x)=1\}$$ is well defined (i.e. $T(x)$ is indeed a clopen set), and an isomorphism.