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

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.

• @Jose answered, keep in mind that convergent $\implies$ Cauchy'' always. – b00n heT Sep 29 '19 at 9:31
• Erwin Kreyszig. Introductory functional analysis with applications. Page 30. Theorem 1.4-7. – Neil hawking Sep 29 '19 at 9:45

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, $$s=s^\ast\in S$$.
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.
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.