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Hi: Suppose that $E$ is a compact of $\mathbb{R}^{n}$, with $n>1$. Why $$E^{\prime}:=\bigcup_{x\in E}\overline{B(x,r)}$$ is a compact? Here $B(x,r)$ is the ball of center $x$ and radius $r$. Thanks for your reply.

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Why do you not allow $n = 1$? You use $r$ without saying anything about it (e.g., it is a fixed positive number). –  KCd Jun 12 at 16:07
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Yes, if you mean the usual metric. Heine-Borel theorem will help you. –  ThePortakal Jun 12 at 16:08

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up vote 3 down vote accepted

I assume that $r$ is fixed. By definition, you have $$E'=\{ x\in\mathbb R^n;\; {\rm dist}(x,E)\leq r\}\, .$$ Since the function $x\mapsto {\rm dist}(x,E)$ is continuous, it follows that $E'$ is closed in $\mathbb R^n$. Moreover, $E'$ is also bounded because $E$ is bounded (this is easy to check). Hence, $E'$ is compact.

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