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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?

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2 Answers 2

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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 $(\operatorname{cl}A)\cap\operatorname{cl}B=\varnothing$.

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

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