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Say we have $A \subset X$, given $(X, d)$ a metric space. I want to prove that the closure of $A$, that is

$$\overline{A} = \{x \in X \ | \ \forall \varepsilon > 0, \ B(x, \varepsilon) \cap A \neq \emptyset \}$$ is equivalent to $$\overline{A} = \{x \in X \ | \inf_{a \in A} d(x, a) = 0 \}.$$

What I've done is the following: by manipulating the first definition

$$\overline{A} = \{x \in X \ | \ \forall \varepsilon > 0, \ B(x, \varepsilon) \cap A \neq \emptyset \} $$ $$\ \ = \{x \in X \ | \ \forall \varepsilon > 0\ , \exists a \in A, \ d(a, x) < \varepsilon \}$$

All help is welcomed.

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    $\begingroup$ By definition, if $\inf d(x,a)=0$, then $\exists \{a_n\} s.t. d(x,a_n)\to 0$. Therefore, $\forall \varepsilon >0$... $\endgroup$
    – bartgol
    Jul 19, 2016 at 14:37

1 Answer 1

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Let $B=\{x\in X\mid \inf\limits_{a\in A}d(x,a)=0\}$.

We want to prove that $B=\bar{A}$.

  • Let $b\in B$ and let $\varepsilon>0$. By definition, $\inf\limits_{a\in A}d(b,a)=0$, so there is $a\in A$ with $d(b,a)<\varepsilon$. Thus $a\in B(b,\varepsilon)\cap A$. Therefore $B\subseteq\bar{A}$.

  • Suppose $c\in X\setminus B$, so that $\varepsilon=\inf\limits_{a\in A}d(c,a)>0$ (it can't be $<0$, of course). If $a\in A$, then $d(c,a)\ge\varepsilon$, so $a\notin B(c,\varepsilon/2)$. Therefore $B(c,\varepsilon/2)\cap A=\emptyset$ and so $c\notin\bar{A}$. Hence $\bar{A}\subseteq B$.

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