1
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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 $C$.

$\endgroup$
0
$\begingroup$

The following may seem like a moderately complicated way to think about it, but if you're fluent in it, you see instantaneously that the answer to the question posed here is "yes".

To every "Stonean space" there corresponds the Boolean algebra of all clopen subsets of the space.

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".

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Eric Wofsey May 9 at 2:40

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.