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.
 A: 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.
