Boolean algebra is isomorphic to algebra of clopens in a compact space

Question

I'm working with the "abstract" definition of boolean algebra (without presupposing Stone's representation theorem).
The hint I'm given is to consider a morphism $ f: B \rightarrow X $ where $B$ is the boolean algebra and $X$ is the topological space of ultrafilters on $B$ with the topology given by the base of the sets of ultrafilters of the form $[b] = \lbrace \mathcal{U} \mid b \in \mathcal{U} \rbrace$, where $b$ ranges over the elements of $b$. Now, $f$ is defined as $ b \mapsto [b] $. I have already proven that the map is a morphism and is injective. Now I have to prove that it is surjective, which is to say, for any clopen $C$ in the topology generated by that base, I can find $b \in B$ such that $ b \mapsto C$.

Answer 1

Let $C$ be clopen in $X$. So for every ultrafilter $U \in B$ we can find $b_U \in B$ such that $U \in [b_U]$ (i.e. $b_U \in U$) such that

$$[b_U]\subseteq C$$

(because $C$ is open and these sets form a base for $X$).

As $C$ is closed in the compact $X$, it's itself compact so finitely many cover $C$ too, i.e. there are finitely many $b_{U_i}, 1 \le i \le n$ for some $U_i \in C$ so that $C = \bigcup_{i=1}^n [b_{U_i}]$. From the fact that we're working with ultrafilters we know that $[b] \cup [b'] = [b \cup b']$ for all $b,b' \in B$ and it follows that

$$C = [\bigvee_{i=1}^n b_{U_i}]$$

which is a basis set, as required.

Bonus: proof that $X$ is actually compact, as I use this fact.

Let $\{[b_i], i \in I\}$ be a cover of $X$ by basic open subsets of $X$, we need to find a finite subcover. I'll denote by $b'$ the complement of $b$ in $B$. Consider $\mathcal{B}:=\{b_i' \mid i \in I\} \subseteq B$. Suppose that $0 = \bigwedge_{i \in F} b'_i$ for some finite subset $F$ of $I$. By De Morgan's law for Boolean algebras:

$$1 = (\bigwedge_{i \in F} b'_i)' = \bigvee_{i \in F} b_i, \text{ so } X=[1]= [\bigvee_{i \in F} b_i] = \bigcup_{i \in F} [b_i]$$ and we have a finite subcover for our basic cover. So $\mathcal{B}$ has the f.i.p. (or we'd be done) but then there is an ultrafilter $u$ in $B$ that contains $\mathcal{B}$ by Zorn or some other maximal principle.

But then we have a contradiction: $b'_i \in \mathcal{B} \subseteq u$ so $u \notin [b_i]$ for all $i$, contradicting that we had a cover in the first place. So the fip cannot always hold and so our cover did have a finite subcover.