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A space $X$ is said to be extremally disconnected if every open set has an open closure. A subset $S$ of a space $X$ is called a zero-set if there is a continuous function $f:X\to \mathbb{R}$ such that $S=f^{-1}(0)$, and a subset $S'$ of $X$ is called a cozero-set if it is the complement of a zero-set. A space $X$ is basically disconnected if every cozero-set has an open closure. Hence any extremally disconnected space is basically disconnected.

Are the following statements correct?

  1. The space $X$ is basically disconnected if and only if every pair of disjoint open sets have disjoint closures.
  2. The closure of every cozero-set is a zero-set.
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  • $\begingroup$ You shouldn't change you question after getting and answer to the original one, but rather post a new question possibly linking to the original question (i.e. giving context). $\endgroup$
    – user87690
    Dec 24, 2017 at 10:43

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The condition that pairs of disjoint open sets have disjoint closures is actually equivalent to extremal disconnection, not basic disconnection. (Also note that “$^*$extremely disconnected” should be replaced with “extremally disconnected.”)

Necessity$\phantom{--}$ Suppose that $X$ is extremally disconnected. Let $U,V\subseteq X$ be disjoint open sets. For the sake of contradiction, assume that $x\in(\operatorname{cl} U)\cap(\operatorname{cl}V)$. Since $V^{\mathsf c}$ is closed and $U\subseteq V^{\mathsf c}$, it follows that $\operatorname{cl}U\subseteq V^{\mathsf c}$. Similarly, $\operatorname{cl} V\subseteq U^{\mathsf c}$. Therefore, $x$ is neither in $U$ nor in $V$. But since $x\in\operatorname{cl} U$, it must be an accumulation point of $U$. Given that $\operatorname{cl} V$ is an open neighborhood of $x$, there must exist some $y\in U\cap(\operatorname{cl} V)\subseteq U\cap U^{\mathsf c}=\varnothing$, a contradiction.

Sufficiency$\phantom{--}$ For the other direction, suppose that $X$ is not extremally disconnected. Then, there exists an open set $S\subseteq X$ such that $\operatorname{cl} S$ is not open. This means that $\operatorname{int}(\operatorname{cl}S)$ is a strict subset of $\operatorname{cl}S$, so that there exists some $x\in\operatorname{cl}S$ such that $x\notin \operatorname{int}(\operatorname{cl}S)$. But then $$x\in[\operatorname{int}(\operatorname{cl}S)]^{\mathsf c}=\operatorname{cl}[\operatorname{int}(S^{\mathsf c})],$$ so that $S$ and $\operatorname{int}(S^{\mathsf c})$ are disjoint open sets, yet their closures intersect.

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  • $\begingroup$ Is there a relationship between completely separated and extremally (basically )disconnected? $\endgroup$
    – Jak
    Dec 24, 2017 at 5:24
  • $\begingroup$ Is it true ? "The space $X$ is basically disconnected if and only if every pair of disjoint cozero-sets have disjoint closures". $\endgroup$
    – Jak
    Dec 24, 2017 at 5:28
  • $\begingroup$ @Jak: about the relationship: extremal disconnectedness boosts up Hausdorff to totally separated, and regular to zerodimensional. $\endgroup$
    – user87690
    Dec 24, 2017 at 10:41

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