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I am trying to solve the following exercise in Kolmogorov's real analysis textbook.

Prove that a subspace of a complete metric space $R$ is complete if and only if it is closed.

I think I must not fully understand the concept of completeness, because I almost see complete and closed as synonyms, which is surely not the case. With that said, here is my attempt at a proof.

($\implies$)

Suppose $S \subset R$ is complete. Then, by definition, any Cauchy sequence in $S$ converges to a limit in $S$. So let $(s_n)$ be a Cauchy sequence in $S$, where $s_n \to s$. Since $S$ is complete $s \in S$, hence, $(s_n)$ contains all of its limit points, and is thus closed.

I don't think I have figured out this first implication. If a sequence converges, it clearly only has one limit point. It seems straightforward to show that a limit is contained in $S$, but how would I deal with limit points of sequences that do not converge? Do these not make a difference here?

Now, for the opposite implication, the proof of which I believe I am more confident about.

($\implies$)

Let $S \subset R$ be closed. Let $(s_n)$ be a Cauchy sequence of elements in $S$. But, since $S \subset R$, $s_n \in S$ implies that $s_n \in R$, and since $R$ is a complete space, $(s_n) \to s$, where $s \in R$. Since $S$ is closed, though, it contains all of its limit points, so $\lim s_n = s \in S$, meaning that $S$ is complete, as $(s_n)$ was an arbitrary Cauchy sequence in $S$.

Any help would be greatly appreciated.

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  • $\begingroup$ @Jose answered, keep in mind that ``convergent $\implies$ Cauchy'' always. $\endgroup$ – b00n heT Sep 29 '19 at 9:31
  • $\begingroup$ Erwin Kreyszig. Introductory functional analysis with applications. Page 30. Theorem 1.4-7. $\endgroup$ – Neil hawking Sep 29 '19 at 9:45
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In order to prove the first implication, you should prove that if a sequence $(s_n)_{n\in\mathbb N}$ converges to some $s\in R$, then you actually have $s\in S$. But since $(s_n)_{n\in\mathbb N}$ converges, it is a Cauchy sequence. And therefore, since $S$ is complete, it must converge to an element of $s^\ast\in S$. Since a sequence cannot converge two distinct limits, $s=s^\ast\in S$.

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First implication: You don't have to worry about sequences that don't converge. You just need to show that every convergent sequence converges to a point in $\mathcal S$. And it will because $\mathcal S$ is complete.

Because of the result, you have to look outside complete metric spaces to find closed subspaces that are not complete. Your intuition probably has a lot to do with the real numbers.

For instance, let $\mathcal R$ be the open interval $(0,2)\subset\Bbb R$. Then $\mathcal S=[1,2)\subset\mathcal R$ is closed but not complete. That is possible because $\mathcal R$ is not complete.

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