Let $U$ be a non-empty open set in a metric space. I want to show that there is an open ball $B(x,r)$ such that the closure $\overline{B(x,r)}\subseteq U$. It is clear that there is an open ball contained in $U$ but why its closure will lie inside $U$. I am not able to justify my self. What is the problem if we assume that for all open ball $B(x,r)$ in $U$, $\overline{B(x,r)}\not\subseteq U$?
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Since $U$ is open an non-empty, there is some $x\in U$ and there is some $r>0$ such that $B(x,r)\subset U$. Since $B(x,r)\subset U$, and since $\overline{B\left(x,\frac r2\right)}\subset B(x,r)$, $\overline{B\left(x,\frac r2\right)}\subset U$.
On the other hand, you cannot deduce that $\overline{B(x,r)}\subset u$ just from the fact that $B(x,r)\subset U$. Suppose that you are working on $\mathbb{R}$ with its usual metric and that $U=(-1,1)$. Then $B(0,1)\subset U$, but $\overline{B(0,1)}\varsubsetneq U$.
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