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As the topics, Prove that $S\cap \cl(T)\subset \cl(S\cap T)$, if $S$ is an open sets,where $\cl$ is the closure of a sets. $S,T\in \mathbb{R^n}$

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If $x\in S\cap \operatorname{cl}(T)$ then $x\in S$ and for all open $U\ni x$, we have $U\cap T\ne\emptyset$. Given an open $U\ni x$, we see that $S\cap U$ is also an open neighbourhood of $x$, hence $(S\cap U)\cap T$ is nonempty. We conclude that $U\cap (S\cap T)$ is nonempty for every open $U\ni x$, i.e. $x\in \operatorname{cl}(S\cap T)$. In summary, $S\cap \operatorname{cl}(T)\subseteq \operatorname{cl}(S\cap T)$.

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    and is it a must for$S$to be open2012-10-10