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## Homework Statement

Let ##M## be a metric space. Prove that ##\overline{\overline{S}} = \overline{S}## for ##S\subseteq M##.

## Homework Equations

## The Attempt at a Solution

First we know that ##\overline{S} \subseteq \overline{\overline{S}}## is true (just take this for granted, since I know how to do it and it's the other direction that I am interested in).

Second, we want to show that ##\overline{\overline{S}} \subseteq \overline{S}##. So let ##x## be a closure point for ##\overline{S}##. So there exists a sequence ##\{x_n\} \subseteq \overline{S}## such that ##\lim_{n\to\infty}x_n = x##. So by the definition of convergence there exists ##N_1## such that ##d(x_{N_1},x) < \epsilon /2##, where ##\epsilon## is an arbitrary positive real number. Note that in particular that ##x_{N_1}\in \overline{S}##, so there exists a sequence ##\{y_m \}\subseteq S## such that ##\lim_{m\to\infty}y_m = x_{N_1}##. So by the definition of convergence there exists ##N_2\in \mathbb{N}## such that ##m\ge N_2## implies ##d(y_m,x_{N_1}) < \epsilon /2##. So if ##m \ge N_2## then ##d(y_m,x) \le d(y_m,x_{N_1}) + d(x_{N_1},x) < \epsilon /2 + \epsilon /2 = \epsilon##. Hence ##\lim_{m\to\infty}y_m = x##, and so that ##x## is a limit point for ##S##.

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