A topological space $X$ is an $EF$-space if if for any two collections $\mathcal{U}$ and $\mathcal{V}$ of clopen subsets of $X$ with $\bigcup \mathcal{U}\cap \bigcup \mathcal{V}=\emptyset$, we have $\bigcup \mathcal{U}$ and $\bigcup \mathcal{V}$ are completely separated.

By deffinition, it is essy to see the following. $(a)$ If $X$ is an $EF$-space and $\mathcal{U}$ and $\mathcal{V}$ are to collection of clopen subsets of $X$ with $\bigcup \mathcal{U}\cap \bigcup \mathcal{V}=\emptyset$, then $cl(\bigcup \mathcal{U})\cap cl(\bigcup \mathcal{V})=\emptyset.$

In Normal space the converse of $(a)$ is true.

Question: For a completely regular space $X$ is the convesre of $(a)$ true?


Partial answer: if the space is zero-dimensional then any two disjoint open sets have disjoint closures, so the space is extremally disconnected. And then closures of open sets are clopen, hence your implication holds. At the other extreme, is the space is connected then both conditions are true vacuously.

  • $\begingroup$ Dear Hart, Yes in Zero-dimensional space the following are equivalent. (1) $X$ ia an EF-spac. (2) $X$ is an extremally disconnected. (3) Any union of clopen subsets of $X$ is $C^{*}$-embedded. (4) If $\mathcal{U}$ and $\mathcal{V}$ are two collections of clopen subsets of $X$ with $\bigcup \mathcal{U}\cap \bigcup \mathcal{V}=\emptyset$, then $cl(\bigcup \mathcal{U})\cap cl(\bigcup \mathcal{V})=\emptyset$. $\endgroup$ – Ali Jan 26 '13 at 15:13

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