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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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    Is it true that in general, for any open $U$, $U\cap T\ne\emptyset$? As from the definition, it is for any open ball of $x$ instead of using any open set. Although intuitively, it seems trivial2012-09-29
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    In a metric space, you ultimately rely on open balls, yes. My argument works for arbitrary topological spaces. In a metric space, if $U$ is open and contains $x$ then there is also a ball around $x$ contained in $U$. Or simply replace "open $U\ni x$" with "open ball $B(x,r)$ with $r>0$" throughout the argument.2012-09-29
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    actually if $S$ is open, can we conclude that there must be a open ball lying in $S\cap T$2012-10-10
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    and is it a must for S to be open2012-10-10