# Closed subset of complete metric space…don't understand last part of theorem.

A closed subset of a complete metric space is a complete subspace. Proof. Let $S$ be a closed subspace of a complete metric space X. Let $(x_n)$ be a Cauchy sequence in $S$. Then $(x_n)$ is a Cauchy sequence in $X$ and hence it must converge to a point $x$ in $X$. But then $x \in \bar{S} = S$. Thus $S$ is complete.

I have seen this theorem in several places but I never know why they are able to say "But then $x \in \bar{S} = S$"...Why is $x$ in $\bar{S}$?

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Because S is a closed subset, hence $S = \bar{S}$ – amWhy Nov 1 '12 at 20:48
Do you know what $\bar S$ means? – Chris Eagle Nov 1 '12 at 20:48
Yes, its the closure of $S$, ie. the smallest closed subset containing $S$ or alternatively the set $S \cup$ the limit points of $S$. I don't see how they can say $x$ is in the closure of $S$ though. – sonicboom Nov 1 '12 at 20:53

Let $U$ be any open nbhd of $x$. There is an $r>0$ such that $B(x,r)\subseteq U$, where $B(x,r)$ is the open ball of radius $r$ centred at $x$. Since $\langle x_n:n\in\Bbb N\rangle\to x$, there is an $m\in\Bbb N$ such that $d(x_n,x)<r$ whenever $n\ge m$. In particular, $d(x_m,x)<r$, so $x_m\in B(x,r)\subseteq U$. Moreover, $x_m\in S$, so $x_m\in U\cap S$. This shows that $U\cap S\ne\varnothing$ for every open nbhd $U$ of $x$, and it follows at once that $x\in\operatorname{cl}S$.

Added: Here in one place are some definitions that seem to be giving you a bit of trouble. Let $\langle X,d\rangle$ be a metric space, $S\subseteq X$ and $x\in X$.

• $x\in\operatorname{cl}S$: for each $r>0$, $B(x,r)\cap S\ne\varnothing$. In words, every open ball at $x$ contains at least one point of $S$.
• $x$ is a limit point (or cluster point) of $S$: for each $r>0$, $B(x,r)\cap(S\setminus\{x\})\ne\varnothing$. In words, every open ball at $x$ contains at least one point of $S$ other than $x$.
• $x\in\operatorname{int}S$: there is an $r>0$ such that $B(x,r)\subseteq S$.
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Cheers, I am going to have to write that out and draw some diagrams to make sure I understand whats going on. – sonicboom Nov 1 '12 at 21:00
When they say "Let $(x_n)$ be a Cauchy sequence in $S$" are they saying that all values of $x_n$ will be in $S$ and the only question is then where the limit of the sequence lies? – sonicboom Nov 1 '12 at 21:07
I get it now...no matter how small we make $r$ there will be an open ball with center $x$, radius $r$ that contains a point of $S$ other than $x$ itself, hence $x$ is a limit point. – sonicboom Nov 1 '12 at 21:14
@sonicboom: Actually, it’s possible that $x\in S$ and that the sequence is eventually constant at $x$, so $x$ might not be a limit point. But either it’s a limit point or it’s in $S$, and either way it’s in $\operatorname{cl}S$. – Brian M. Scott Nov 1 '12 at 23:23
@sonicboom: No, points of $S$ are not necessarily limit points of $S$: $x$ is a limit point of $S$ if every open neighborhood of $x$ contains a point of $S\setminus\{x\}$. Thus, $3$ is not a limit point of $S=[0,1]\cup\{3\}$: $(2,4)$ is an open nbhd of $3$ that contains no point of $S\setminus\{3\}$. If $x$ has an open nbhd that lies entirely within $S$, then $x$ is in the interior of $S$. – Brian M. Scott Nov 2 '12 at 11:11