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I am preparing for an exam and trying to solve this exercise about nowhere dense set. I would be very happy if someone could help me. Let $(X, \tau )$ be a topological space. A set $S\subset X$ is nowhere dense if it does not contain any internal point. Prove that a closed set $C\subset X$ is nowhere dense if and only if it is the boundary of an open set $U\subset X$. Thank you very much in advance! |
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HINT: Let $U=X\setminus C$; $U$ is of course open. If $C$ is nowhere dense, $U$ is a dense open set, and therefore $C=X\setminus U=(\operatorname{cl}U)\setminus U=\operatorname{bdry}U$. Now let $U$ be an open set, and suppose that $C=\operatorname{bdry}U$. Use the fact that $U$ is open to show that $C\cap U=\varnothing$, and then use the fact that $C\subseteq\operatorname{cl}U$ to show that $C$ cannot have an interior point. If you don’t already know this characterization, you may find it useful to show that in general $\operatorname{bdry}A=\operatorname{cl}A\cap\operatorname{cl}(X\setminus A)$. |
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