# Clopen subspaces of Stonean spaces

A Stonean space is a compact Hausdorff space in which the closure of any open set is again open. Suppose S is Stonean and C is clopen in S, i.e. C is both open and closed. Now C, endowed with the subspace topology, is clearly compact and Hausdorff. Is it Stonean too?

Yes. Any open set in $C$ has the form $C\cap U$, where $U$ is open in $S$. It's easy to check that the closure in $C$ of $C\cap U$ is equal to $C\cap \bar{U}$, where $\bar{U}$ is the closure of $U$ in $S$. Since $S$ is Stonean, $\bar{U}$ is open, so $C\cap \bar{U}$ is relatively open in $C$.
To every Boolean algebra there corresponds the "Stonean space" of all $2$-valued homomorphisms on that algebra, with the topology of pointwise convergence of nets of such homomorphisms.
This pairs off Boolean algebras and "Stonean spaces" in a well behaved way, and the content of "well behaved" is an interesting story which I omit from this answer for the time being, and it reduces the question to this: If $a_0$ is a fixed member of a Boolean algebra $A$, then is $\{b\wedge a_0 : b\in A\}$ a Boolean algebra in its own right? If you phrase it that way, then the answer is obviously "yes".
• This is not quite right--Stonean spaces are not the same as Stone spaces, and correspond to complete Boolean algebras, not all Booolean algebras. So you need $\{b\wedge a_0:b\in A\}$ to be complete if $A$ is complete, which is also true but slightly less obvious than just that it is a Boolean algebra. – Eric Wofsey May 9 at 2:40