I was working with an exercise about topology but this involucres boolean algebras. I don't have much experience working with boolean algebras and that's the reason because I can't solve. First some definitions:
Let $X$ be a zero-dimensional space. A family $E\subseteq B(X)$ (here $B(X)$ is the collection of all clopen sets of $X$) is called a Boolean base for $X$ if $E$ is a Boolean subalgebra of $B(X)$ and $E$ is a base for $X$. Let $BB(X)=\left\{E\subseteq B(X)\mid E \ \text{is a Boolean base for} \ X \right\}$. (Here $(B(X),\cap,\cup,\emptyset,X)$ is the boolean algebra).
Now, suposse that $X$ is a zero-dimensional and locally compact Hausdorff space and take $E=\left\{C\in B(X)\mid C \ \text{or} \ X\setminus C \ \text{is compact} \right\}$ and suposse that $E$ is a boolean base for $X$. Prove that $E$ is the $\subseteq$-minimum of $BB(X)$. Here $BB(X)$ is partially ordered by inclusion.
My attempt:
Given $D$ another boolean base, our goal to prove that $E\subseteq D$. Let $C\in E$ and we can suposse w.l.g. that $C$ is compact. By the hausdorfness of $X$ and the fact that $D$ is a basis, for all $x\in X\setminus C$ there exists $U_x$ and $V_x$ elements of $D$ such that $C\subseteq U_x$ and $x\in V_x$. Clearly the family $$\mathscr{U}=\{U_x\mid x\in X\setminus C\}$$ is an open cover of $C$ and because $C$ is a compact set, there exists $X_0\subseteq X\setminus C$ a finite subset $$\mathscr{U}_0=\{U_x \mid x\in X_0\}$$ such that $C\subseteq \bigcup \mathscr{U}_0$. We claim that $\bigcap \mathscr{U}_0=C$. Clearly $C\subseteq \bigcap \mathscr{U}_0$. But, the other contention still holds? I know and is easy to prove that $\bigcap \mathscr{U}=C$ but, is it true that $\bigcap\mathscr{U}_0=C$?. I thought in this teqnique because $D$ being a boolean subalgebra is closed under finite intersections (here the infimum and supremum are equal to intersection and union respectively). I think that I'm close to the solution. A thing that I have noticed is the fact that I never used the locally compactness. Any hint? Thanks.
Regarding your argument, each $U_x$ covers $C$ by definition, so passing to a finite subcover of $\mathscr{U}$ is kind of trivial, seeing as how any member by itself will do.
I'll change some of the formatting, because I find using capital Roman letters for things of different types to be a bit confusing. In what follows, $B(X)=\mathscr B (X)$, $E=\mathscr E$, and $D=\mathscr D$.
The missing ingredient is that $C$ is open (because $\mathscr{E}\subset \mathscr{B}(X)$). Argue as follows: for each $x\in C$, pick $U_x\in \mathscr D$ such that $x\in U_x\subset C$. Then we have $$ C=\bigcup_{x\in C}U_x $$ In particular, $\{U_x:x\in C\}$ is an open cover of $C$. Since $C$ is compact, there are $x_1,\dots,x_n\in C$ such that $$ C\subset \bigcup_{i=1}^n U_{x_i} $$ Since the reverse inclusion also holds, it follows that $$ C=\bigcup_{i=1}^n U_{x_i}\in \mathscr D $$ because $\mathscr D$ is a Boolean algebra.