# Do Stone-Čech compactifications have property that disjoint closed subsets of $X$ have disjoint closures in $\beta X$?

I came across this in Van Douwen´s paper, Characterizations of $$\beta \mathbb{Q}$$ and $$\beta \mathbb{R}$$, Proposition 4.

Van Douwen writes: "We show that $$\gamma H$$ = $$\beta H$$ by showing that disjoint closed subsets of $$H$$have also disjoint closures in $$\gamma H$$." ($$\gamma H$$ is just an arbitrary compactification of the space $$H$$).

Is this a special property of Stone-Čech compactifications? Why does this imply that the compactification is Stone-Čech? I know the Stone-Čech compactification is compact and Hausdorff, but that is not anything extra about Stone-Čech particurarly.

What do you think? Thank you.

• Yes, it's in fact the unique compactification with that property. Aug 16, 2021 at 21:48

## 1 Answer

Yes that is a special property of Stone-Cech compactifications. In fact, the property that any two disjoint zero-sets in $$X$$ have disjoint closures in $$\beta X$$ characterizes the Stone-Cech compactification. What I will write here is based on Gillman and Jerison's book, "Rings of Continuous Functions".

To see this, suppose is a completely regular space $$X$$ dense in $$T$$ with the property that any continuous $$\tau : X \to Y$$ with $$Y$$ compact has a continuous extension $$\bar{\tau} : T \to Y$$. If two zero-sets $$A$$ and $$B$$ of $$X$$ are disjoint, then there exists a function $$f \in C_b(X)$$ such that $$f|_{A} = 0$$ and $$f|_{B} = 1$$. Then $$f$$ is a continuous mapping into the compact space $$\operatorname{cl} (f(X))$$, so it has a continuous extension $$\bar{f} \in C(T)$$. But then clearly $$\bar{f}|_{A} = 0$$ so $$\bar{f}|_{\operatorname{cl}(A)} = 0$$, and similarly $$\bar{f}|_{B} = 1$$ so $$\bar{f}|_{\operatorname{cl}(B)} = 1$$. Therefore $$\operatorname{cl}(A)$$ and $$\operatorname{cl}(B)$$ are disjoint.

Conversely suppose any two disjoint zero-sets of $$X$$ have disjoint closures in $$T$$. Let $$\tau : X \to Y$$ be a continuous mapping from $$X$$ into a compact space $$Y$$. Since $$X$$ is dense in $$T$$, for each $$p \in T$$ there is a $$z$$-ultrafilter $$\mathscr{A}_p$$ on $$X$$ with limit $$p$$. Moreover, because of the assumption that disjoint zero-sets of $$X$$ have disjoint closures on $$T$$, this $$z$$-ultrafilter is unique (because distinct $$z$$-ultrafilters contain disjoint zero-sets). Therefore for each $$p \in T$$ we may define a set $$\tau^\# \mathscr{A}_p := \{ E \subset Y : E \text{ is a zero-set and } \tau^{-1}(E) \in \mathscr{A} \}$$. One can quickly verify that $$\tau^\# \mathscr{A}_p$$ is a prime $$z$$-filter on $$Y$$. But $$Y$$ is compact, so $$\tau^\# \mathscr{A}_p$$ has a cluster point, and by primality converges to that cluster point, which we define to be $$\overline{\tau}(p)$$. Thus we have defined a mapping $$p \mapsto \bar{\tau}(p)$$ which clearly extends $$\tau$$, since for any $$p \in X$$, $$p \in \bigcap \mathscr{A}$$ so $$\tau(p) \in \bigcap \tau^\# \mathscr{A}_p$$. So we just need to show $$\bar{\tau}$$ is continuous. Let $$p \in T$$, and $$F$$ be a zero-set neighbourhood of $$\bar{\tau}(p)$$. Let $$F'$$ be a zero-set whose complement is a neighbourhood of $$\bar{\tau}(p)$$ contained in $$F$$. Therefore $$F \cup F' = Y$$, so letting $$Z =\tau^{-1}(F)$$ and $$Z' = \tau^{-1}(F')$$, we have that $$Z \cup Z' = X$$, and so $$\operatorname{cl}(Z) \cup \operatorname{cl}(Z') = T$$, and in particular $$T - \operatorname{cl}(Z') \subset \operatorname{cl}(Z)$$. Now $$p \notin Z'$$ because $$\bar{\tau}(p) \notin F'$$, so $$T - \operatorname{cl}(Z')$$ is a neighbourhood of $$p$$, contained in $$\operatorname{cl}(Z)$$, so $$\overline{\tau}(q) \in F$$ for every $$q \in T - \operatorname{cl}(Z')$$. This proves that $$\overline{\tau}$$ is continuous.

• Thank you, just a few questions. 1. What is $C_b(X)$? The b confuses me. 2. What is a "zero set"? 3. In the latter "Conversely, suppose..." part, you are showing the other direction - that $T$ is Stone-Čech, from the assumption about the disjoint closures, right? (I am not familiar with filters so far, so I have to study that for more time.) Aug 16, 2021 at 21:34
• $C_b(X)$ denotes the continuous bounded (real valued) functions on $X$. A zero-set is a subset $A$ of $X$ such that $A = \{x \in X : f(x) = 0\}$ for some $f \in C(X)$. In the conversely part, I am indeed showing that $T$ satisfies the universal property of the Stone-Cech compactification based on the assumption about disjoint closures.
– jl00
Aug 16, 2021 at 21:46
• Thank you, I really appreciate your answer. Aug 16, 2021 at 21:46