I am self-studying Royden's Real Analysis; Exercise 58 of Section 9.5, "Compact Metric Spaces", asks:
Let $E$ be a subset of the compact metric space $X$. Show that the subspace $E$ is compact if and only if $E$ is a closed subset of $X$.
The reverse implication is easy: if $E$ is closed, given an open cover $\mathcal{O}$ of $E$, $\mathcal{O} \cup \{X \setminus E\}$ is an open cover of $X$, which, by compactness of $X$, contains a finite subcover $\mathcal{O}'$ of $X$; removing $X \setminus E$ from $\mathcal{O}'$ if necessary, we arrive at a finite subcover of $\mathcal{O}$ of $E$, so $E$ is compact. (In fact, the same argument holds in general for any topological space, not just metric spaces.)
However, I am stumped in trying to prove the forward implication, namely that $E$ being compact in the compact metric space $X$ implies that $E$ is closed in $X$.
This section of the text does contain what it calls the Characterization of Compactness for Metric Spaces, which states:
For a metric space $X$, the following three assertions are equivalent:
(i) $X$ is complete and totally bounded;
(ii) $X$ is compact;
(iii) $X$ is sequentially compact.
I had the thought of using this to say that if $E$ is compact in $X$, then it is complete and totally bounded in $X$, and to show that $E$ being complete implies that $E$ is closed in $X$. However, using the definition of "complete", all we can say is that every Cauchy sequence in $E$ converges in $E$. In contrast to the case of $\mathbb{R}^n$, (according to my understanding) in a general metric space not every convergent sequence is Cauchy, so we cannot use completeness of $E$ to say that $E$ contains all of it limit points in order to prove it to be closed.
I feel as though I am missing something obvious here. Direction would be appreciated.
Best Answer
Assume that $E$ is a compact set. Assume that $(x_n)$ be a sequence in $E$ such that $x_n\to x\in X$. In order to prove that $E$ is closed set, we need to prove $x\in E$. In fact, since $(x_n)\subset E\subset X$, there is a sub-sequence $(x_{n_{k}})$ such that $x_{n_k}\to y\in X$ as $k\to\infty$. On the other hand, we also have $x_{n_k}\to x$ as $k\to\infty$. So $x=y$ from the uniquely of the limitations. Since $y\in E$, we deduce $x\in E$.