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Let $A,B\subseteq\mathbb R^d$, $A$ closed and $B$ open. If $x\in A\cap B\neq\emptyset$ does there exist $\varepsilon>0$ such that $B_\varepsilon(x)\subseteq A\cap B$?

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If $d=1$, $A=\{0\}\subset (-1,1)=:B$, there isn't any open ball centered at $0$ and contained in $A$. – Davide Giraudo Nov 6 '12 at 9:34
ah yes. good example – trotinett Nov 6 '12 at 9:37

as $x\in B$, and $B$ is open so there exist $\epsilon>0$ such that $B_{\epsilon}(x)\subseteq B$, but our $B_{\epsilon}$ may not be in the intersection. So it is ingeneral not true, for example in Real line, $A=\{0\}$, $B=(-1,1)$

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