# Showing that if a subset of a complete metric space is closed, it is also complete

Let $(X, d(x,y))$ be a complete metric space. Prove that if $A\subseteq X$ is a closed set, then $A$ is also complete.

My attempt: I tried to prove that every Cauchy sequence $(b_n)$ of points of $A$ converges to a point $b\in A$. However could not figure out the exit way. Maybe I am on the wrong track. Could you please help me?

edit: More from my attempt:

Suppose $A$ is a closed set and let $(x_n)$ be a sequence of points $A$ such that $\lim_n x_n\to b$.

Suppose now that $A$ has the property that $b\in A$, whenever $x_n$ converges to $b$. We know that every element of $x_n$ which is convergent in $X$ also converges to a point in $A$. Since $x_n$ is a Cauchy sequence in $A$, it must converge to a point $y\in A$. But the limit of a convergent sequence is unique.

Take $x\in A$ and select an appropriate $n$, which enables $x_n$ to converge to a point $x$ in $X$. Since the limit is unique, it must follow that $x=y$. Thus $x\in A$ and $A$ is closed.

If $A$ is a closed subset of $X$, then any Cauchy sequence of a point in $A$ is convergent in $X$ and hence converges to a point in $A$. Thus $A$ is complete.

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To show this one must start with a Cauchy sequence, not a convergent sequence. Let $(x_n)$ be a Cauchy sequence in $A$. Then $(x_n)$ is a Cauchy sequence in $X$. Since $X$ is complete, $x_n\to x$ for some $x\in X$. Since $A$ is closed, $x\in A$. Hence $A$ is complete.

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 was that all? :D – Amadeus Bachmann Nov 26 '12 at 2:37

Let $(M,d)$ be a complete metric space, and let $A \subseteq M$. Suppose $A$ is closed.

Claim. $(A,d)$ is complete.

Proof. Let $(x_n)$ be a Cauchy sequence in $A$. Then $(x_n)$ is a Cauchy sequence in $(M,d)$ (trivial to verify). So $x_n \to x \in M$. But $A$ is closed, so it contains all of its limit points. So $x \in A$. Hence, $(A,d)$ is complete.

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Hint: You are on the right track, I believe. Any Cauchy sequence converges to some $b\in X$.

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 I am extremely tired so that it has become really hard to think clear, but I added more details about what I have done so far. Could you please take a look and help me to put this proof in a correct shape? Thanks! – Amadeus Bachmann Nov 26 '12 at 2:26

Here it is understood that the metric used on the subset say $N$ is the same as that used on $M$, the super set. Let $(x_n)$ be a sequence (Cauchy) in $N$. Regarding it as a sequence in $M$, it is still a Cauchy sequence in $M$ , and so since $M$ is complete, it converges in $M$ (say to $x$). Since, $x_n$ belongs to $N$ and $x_n \to x$ and $N$ is closed (all accumulation/cluster points are contained in $N$), $x$ belongs to $N$. Thus $(x_n)$ converges in $N$, and so $N$ is complete.

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