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I'm reading Matthew Dirk's The Stone Representation Theorem for Boolean Algebras, and am trying to follow the proof of Proposition 3.4 on p.6:

Proposition 3.4. Let $A$ be a Boolean algebra, and consider the set $\mathcal{S}(A) \subset 2^A$ of homomorphisms from $A$ to $2$. Then $\mathcal{S}(A)$ is a Stone space.

Proof. Fix $a \in A$. For all $x \in 2^A$, $\{ x_a \}$ is open in $A$, since $2$ has the discrete topology, and sets of the form $\{ x \in 2^A : x_a = 1 \}$ and $\{ x \in 2^A : x_a = 0 \}$ are open in $2^A$ by definition. Hence the value of $x_a$ depends continuously on $x$.

Now, if $f : X \to Y$ and $g : X \to Y$ are continuous functions into a Hausdorff space $Y$, then the set $\{ x : f(x) = g(x) \}$ is closed. (To prove this, consider the map $(f,g) : X \to Y \times Y$ and the diagonal set $\Delta = \{ (y,y) : y \in Y \}$. Then $\Delta^\text{c}$ is open iff $Y$ is Hausdorff, so $\Delta$ is closed iff $Y$ is Hausdorff.)

Therefore, for $p \in A$, the set $\{ x : x(\neg p ) = \neg x(p) \}$ is closed in $2^A$. Hence the intersection of all complement-preserving functions in $2^A$ form a closed subset of $2^A$.

Similarly the set of all functions in $2^A$ that preserve meets and joins are closed subsets of $2^A$.

Intersect these three sets to get the set of function that preserve meets, joins, and complements, that is, the set $H$ of homomorphisms $p : A \to 2$. The intersection of closed subsets is closed, so $H$ is closed and hence compact. Because $2^A$ is Hausdorff and totally disconnected, so is $H$, and we are done. $\Box$

I want to show that all sets of the form $$\{x\in 2^A :\text{x preserves the complements for a fixed } p\in A\}, \\ \{x\in 2^A :\text{x preserves the meets}\}$$ are closed. I fail to see how to prove that those are closed subsets, even using the fact that when $f,g:X\longrightarrow Y$ are continuous functions into an Hausdorff space $Y$, then the set $\{x\, :\, f(x)=g(x)\}$ is closed. I do not see how to use this fact!

Any hints?

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Note that as $2^A$ is given the product topology considering $2 = \{ 0 , 1 \}$ discrete, a subbase for it consists of all sets of the form $$U_{p\to b} = \{ x \in 2^A : x(p) = b \}$$ for $p \in A$ and $b \in 2$. As $$U_{p \to b} = 2^A \setminus U_{p \to (1-b)}$$ it follows that such sets are closed, too.

For the family $\{ x \in 2^A : x( \neg p ) = \neg x (p) \}$, we can show that this is equal to $$( U_{p \to 0} \cap U_{\neg p \to 1} ) \cup ( U_{p \to 1} \cap U_{\neg p \to 0} ),$$ which is closed. As $$\{ x \in 2^A : x\text{ preserves complements} \} = \bigcap_{p \in A} \{ x \in 2^A : x(\neg p) = \neg x(p) \}$$ it follows that this set is closed, too.

For $p,q \in A$ we can similarly show that the set $\{ x \in 2^A : x( p \wedge q) = x(p) \wedge x(q) \}$ is equal to the family $$ \begin{multline} ( U_{p \to 0} \cap U_{q \to 0} \cap U_{p \wedge q \to 0} ) \cup ( U_{p \to 0} \cap U_{q \to 1} \cap U_{p \wedge q \to 0} ) \cup \\ \cup ( U_{p \to 1} \cap U_{q \to 0} \cap U_{p \wedge q \to 0} ) \cup ( U_{p \to 1} \cap U_{q \to 1} \cap U_{p \wedge q \to 1} ). \end{multline}$$ Following a similar decomposition as above, it follows that $\{ x \in 2^A : x\text{ preserves meets} \}$ is closed.

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