# Correct proof idea? Metric spaces and closed sets.

Let $(X,d)$ be an arbitrary complete metric space and suppose $S\subseteq(X,d)$. Show that $S$ is closed if and only if every Cauchy sequence in $S$ converges to a point in $S$.

I did the forward direction, is it correct?

Suppose $S$ is closed. Let $x\in S$ be an Cauchy sequence, $x=(x_n)_{n \in \mathbb N}$. Then, since $S$ is closed, it contains all of its limit points. Therefore, $\lim_{n\to\infty} (x_n)$ will converge to an element in $S$. So every Cauchy sequence in $S$ converges to an element in $S$.

For the backward direction, would I just let $S\subseteq(X,d)$ and suppose that every Cauchy sequence in $S$ converges to a point in $S$. Then show that a limit point $x$ is in $S$?

Any feedback is appreciated, thanks.

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The first direction is almost ok, but you have to argue why $(x_n)_n$ converges at all. Here you need, that $(x_n)_n$ is also a C-sequence in $(X,d)$ which is complete.
To finish the proof, by the assumption "every Cauchy sequence in S converges to a point in S", then the limit point of the convergent sequence is in $S$. Hence $S$ is closed. –  Mhenni Benghorbal Nov 27 '12 at 7:36
Prove the contrapositive for the other direction. Assume $S$ is not closed and find a Cauchy sequence in $S$ which does not converge in $S$ (not closed means that one limit point is not there in the set, so construct a Cauchy sequence which converge to this limit point).