# Čech-Stone compactification of extremally disconnected space

For reference:

A space is extremally disconnected iff the closure of every open set is clopen.

I am trying to understand a simple proof for the fact that the Čech-Stone compactification $\beta X$ of an extremally disconnected space $X$ is again extremally disconnected. The proof is:

Let $U\subseteq \beta X$ be open. Then $U \cap X$ is open in $X$ (for the subspace topology), so $\text{cl}_X(U\cap X)$ is clopen in X and $\text{cl}_{\beta X}(U) = \text{cl}_{\beta X}(\text{cl}_X(U\cap X))$ is clopen in $\beta X$. Q.E.D

My question is, why is $\text{cl}_{\beta X}(\text{cl}_X(U\cap X))$ clopen?

Maybe it is useful to point out why the last equation holds. One inclusion is clear: $$\text{cl}_{\beta X}(U) \supseteq \text{cl}_X(U\cap X) \implies \text{cl}_{\beta X}(U) \supseteq \text{cl}_{\beta X}(\text{cl}_X(U\cap X))$$ For the other inclusion we use the density of $X$ in $\beta X$ $$\text{cl}_{\beta X}(\text{cl}_X(U\cap X)) \supseteq \text{cl}_{\beta X}(U\cap X) \supseteq U \cap \text{cl}_{\beta X}(X) = U \cap \beta X = U$$ Sadly I don't see how the clopenness of $\text{cl}_X(U\cap X)$ would imply that $\text{cl}_{\beta X}(\text{cl}_X(U\cap X))$ is clopen. All input is appreciated.

## 1 Answer

The missing fact is

If $A$ is clopen in $X$ then $\operatorname{cl}_{\beta X}(A)$ is clopen in $\beta X$.

Proof: $f_A: X \to [0,1]$ defined by $f_A(x) = 1$ for $x \in A$, $f_A(x) = 0$ otherwise, is continuous on $X$ (for clopen $A$) so can be extended to $g: \beta X \to [0,1]$. Then $g^{-1}[\{1\}]$ contains $A$ and is closed in $\beta X$ so $\operatorname{cl}_{\beta X}(A) \subseteq g^{-1}[\{1\}]$. Similarly $\operatorname{cl}_{\beta X} (X\setminus A) \subseteq g^{-1}[\{0\}]$. And $$\beta X = \operatorname{cl}_{\beta X}(X) = \operatorname{cl}_{\beta X}(A \cup (X \setminus A)) = \operatorname{cl}_{\beta X}(A) \cup \operatorname{cl}_{\beta X}(X\setminus A)$$

we see that $g$ is also $\{0,1\}$-valued, and so $\operatorname{cl}_{\beta X}(A) = g^{-1}[\{1\}]$ is clopen.

In your proof you end with the equality which shows that $\operatorname{cl}_{\beta X}(U)$ is the closure of a clopen subset of $X$, so is clopen by the extra fact.

• I think you want $f_A$ to have $X$ as domain. Jun 11, 2018 at 4:06
• @quantum so I do, thx. Jun 11, 2018 at 4:09