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.
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.
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.