# A topological space is extremally disconnected iff every two disjoint open sets have disjoint closures

Show that for any topological space $X$ the following are equivalent:

• $X$ is extremally disconnected
• Every two disjoint open sets in $X$ have disjoint closures.

My attempt at a solution

Assume every two disjoint open sets in $X$ have disjoint closures. Let A open in $X$, $A$ and $X \setminus \overline{A}$ disjoint open set $\overline{A} \cap (\overline{X \setminus \overline{A}} )= \emptyset$,

$(\overline{X \setminus \overline{A}} ) \subset X \setminus \overline{A}$, which implies that $X \setminus \overline{A}$ is closed in $X$. Thus $\overline{A}$ is open in $X$. So we obtain $X$ is extremally disconnected.

Otherwise Let $X$ is extremally disconnected. Then the closure of every open set is open. So $X$ is completely separated. Let $A$, $B$ two disjoint open sets. Then there exist a continuous function $g:X \rightarrow [0,1]$ such that $g(A)=0$, $g(B)=1$

How can I continue?

• Sorry I guess I deleted your answer accidentally. Can you write again please, graydad? – NicolA Sep 13 '14 at 22:27
• You can't really delete someone else's answer (not without substantially more privileges than you have for now, anyway). He deleted it himself (maybe because he figured he shouldn't do it for you?). – tomasz Sep 13 '14 at 23:00
• @NicolA I deleted it because I'm fixing the huge hole you pointed out haha. I had to leave my computer for a while but I will see if I can quickly make a fix and repost. – graydad Sep 14 '14 at 2:00

## 2 Answers

The first half of your argument is fine. For the second half you don’t need the function $g$. You have the disjoint open sets $A$ and $B$. If $x\in B$, then $B$ is an open nbhd of $x$ disjoint from $A$, so $x\notin\operatorname{cl}A$. Thus, $B\cap\operatorname{cl}A=\varnothing$. Now suppose that $x\in\operatorname{cl}A$; $\operatorname{cl}A$ is open, so it’s an open nbhd of $x$ disjoint from $B$, and $x\notin\operatorname{cl}B$. It now takes just one small step to finish the argument to show that $(\operatorname{cl}A)\cap\operatorname{cl}B=\varnothing$.

Look at the open set $\overline{A}\cap \overline{B}$. Is it possible for it to meet $A$?