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Prove that in a metric space $(X,d)$ for each $x \in X$ and for each open neighbourhood $U$ of $x$, $\exists$ an open set $V$ such that $x \in V \subseteq \overline V \subseteq U$. Here $\overline V$ is the closure of $V$.

My attempt: Suppose $x \in X$ is arbitrary and $U$ be a open nbd of $x$. Then $\exists\; r > 0$ such that $x\;\in\; B(x,r)\;\subset\;U$. And $B(x,r)\;\subseteq\;\overline{B(x,r)}$. If I can now show that $\overline{B(x,r)}\;\subseteq\; U$ then the proof is done. But how to prove this step. Here $B(x,r)$ means a open ball centered at $x$ and radius $r$.

ビキ マンダル
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If $B(x,r)\subset U$, then$$B\left(x,\frac r2\right)\subset\overline{B\left(x,\frac r2\right)}\subset B(x,r)\subset U.$$

José Carlos Santos
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