Let X be a Tychonoff space such that for any closed set A there exist a continuous function f: X to R such that A=cl(X-Z(f)). Is this space X discrete?
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1$\begingroup$ What is Z(f)?${}$ $\endgroup$– WojowuCommented Feb 13, 2017 at 13:05
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$\begingroup$ @Wojowu I assume Z(f) is the zero set of f. $\endgroup$– Yemon ChoiCommented Feb 13, 2017 at 14:16
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1$\begingroup$ The answer is yes and this is a trivial exercise. $\endgroup$– YCorCommented Feb 13, 2017 at 14:46
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Here I assume $Z(f)$ indeed means the zero set.
Let $x\in X$ be arbitrary. As $X$ is Tychonoff, $A=\{x\}$ is closed, so by assumption there is an $f$ such that $\{x\}=cl(X\setminus Z(f))$. But this clearly implies $X\setminus Z(f)=\{x\}$. Since $Z(f)$ is closed as the preimage of a closed set $\{0\}$, $\{x\}$ is open so $X$ is discrete.
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$\begingroup$ I do not unterstand how {x}=cl(X-Z(f)) implies X- Z(f)={x}. $\endgroup$– S.BCommented Feb 13, 2017 at 18:29
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$\begingroup$ @Wojowu why are you answering what is clearly someones pointset topology homework? $\endgroup$ Commented Feb 13, 2017 at 19:14
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$\begingroup$ @StevenGubkin To be fair I didn't even notice this is MO, I was answering this from mobile... $\endgroup$– WojowuCommented Feb 13, 2017 at 19:16