# Proof that a subspace $A$ of a complete metric space $X$ is complete iff $A$ is closed

Here's my proof in my own words, does it stack up?

1. Showing $A$ is complete implies $A$ is closed. Let $(x_n)$ be a convergent sequence in $A$. $A$ is complete $\implies (x_n) \to p \in A$. Hence $A$ is closed.

2. Showing $A$ is closed $\implies$ $A$ is complete. Let $(x_n)$ be a convergent sequence in $A$. $A$ is closed $\implies (x_n) \to p \in A$. As every convergent sequence is a Cauchy sequence, $(x_n)$ is a Cauchy sequence in $A$ that converges to $p \in A$ and hence $A$ is complete.

That sound ok?

As an aside, it seems to me that the definition of completeness and closed are basically identical, why are there two definitions for the same thing? Am I missing something here?

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– Martin Sleziak Nov 23 '12 at 14:59

Your part 2 is slightly wrong. We need to show that $(x_n)$ $\in A$ Cauchy implies it converges to a limit in $A$. (This is different to what you have said, see my next paragraph) We do this by saying $X$ is complete, so $(x_n) \rightarrow x \in X$ and then use the fact that $A$ is closed in $X$ to say that since $(x_n) \rightarrow x \in X$, that $x\in A$.