I have the following proof, but I don't understand one of the steps:

Theorem 4.4. A Boolean algebra is complete iff its Stone space is exlremally disconnected.

Proof. Identify the given Boolean algebra $B$ with the clopen algebra of $\mathsf{S}B$. Suppose that $B$ is complete, and let $U$ be an open set in $\mathsf{S}B$. Let $\mathscr{A}$ be the family of all members of $B$ included in $U$; then, since $B$ is a base for $\mathsf{S}B$, we have $U = \bigcup \mathscr{A}$. Since $B$ is complete, $\mathscr{A}$ has a supremum $V$ in $B$ which must by definition be a clopen subset of $\mathsf{S}B$. We claim that $\overline{U} = V$. Since $V$ is an upper bound for $\mathscr{A}$, certainly $U = \bigcup \mathscr{A} \subseteq V$, so that $\overline{U} \subseteq V$ since $V$ is closed. If $V - \overline{U} \neq \emptyset$, then $V—\overline{U}$ is a non-empty open set which must, since $\mathsf{S}B$ is a Boolean space, include a non-empty clopen set $W$. But then $V-W$ is a clopen set which includes $\bigcup \mathscr{A}$ and is properly included in $V$. This contradicts the choice of $V$ as the supremum of $\mathscr{A}$. Therefore $\overline{U}=V$ as claimed, so a fortiori $V$ is open.

Conversely, suppose that $\mathsf{S}B$ is extremaily disconnected, and let $\mathscr{A}$ be a subfamily of $B$. Then $U = \bigcup\mathscr{A}$ is an open subset of $\mathsf{S}B$ (recall that we are identifying $B$ with $\mathsf{CS}B$!) and so, since $\mathsf{S}B$ is extremally disconnected, $\overline{U}$ is clopen and hence in $B$. We claim that $\overline{U} = \bigvee \mathscr{A}$ in $B$. Certainly $\overline{U}$ is an upper bound for $\mathscr{A}$ in $B$; on the other hand, if $V$ is a member of $B$ which includes each member of $\mathscr{A}$, then $U = \bigcup \mathscr{A} \subseteq V$ so that $\overline{U} \subseteq V$ since $F$ is clopen. Therefore $\overline{U} = \bigvee \mathscr{A}$ as claimed, and $B$ is complete. $\blacksquare$

I do not understand the argument that why $V - \overline{U}$ is open set. and why should it include $W$ clopen set.


Since $V$ is clopen and the closure of $U$, $\overline U$, is closed, then as $V-\overline U=V\cap(SB-\overline U)$ by definition of set difference, then $V-\overline U$ is the finite intersection of two open sets, which is open by definition. In a topological space with an clopen base, any open set is the union of clopen sets, and hence a non-empty open set will contain a non-empty clopen set, W say, as a subset.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.