If $X$ is complete and $K$ is relatively compact, then the closure $\overline{K}$ is compact.

Question

Currently I'm self studying functional analysis, namely compact operators. In the text, the author gives the following definitions:

Definition 1: A set $K\subseteq X$ is called a compact set if and only if for every sequence $x_n\in K$ there exists an $x\in K$ and a subsequence $x_{n_k}$ of $x_n$ so that $x_{n_k}\to x$.

Definition 2: A set $K\subseteq X$ is called a relatively compact set if and only if every sequence $x_n\in K$ has a Cauchy subsequence $x_{n_k}$.

The author then gives the following remark:

Remark: If $X$ is complete and $K$ is relatively compact, then the closure $\overline{K}$ is compact.

I decided to try to show this myself, but I'm having trouble. Any help would be appreciated.

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My attempt, though unfinished, goes as follows:

Proof. Let $x_n\in\overline{K}$. To show that $\overline{K}$ is a compact set we need to show that there exists an $x\in\overline{K}$ and a subsequence $x_{n_k}$ of $x_n$ so that $x_{n_k}\to x$. By defining of the closure operation, $x_n\in \overline{K}$ implies there exists a sequence $y_m\in K$ so that $y_m\to x_n$. Since $K$ is relatively compact, the sequence $y_m$ has a Cauchy subsequence $y_{m_K}$. Since $X$ is complete $y_{m_k}\to x\in\overline{K}$, as $\overline{K}$ is closed.

Answer 1

The argument is a little more subtle - the sequence $\{y_m\}$ you construct in the question converges to $x_n$ ($n$ fixed), so you can't use that to get at a subsequence of $\{x_n\}$.

For each $n \in \mathbb{N}$, since $x_n \in \overline{K}$, there is a point $y_n \in K$ such that $d(y_n,x_n) < 1/n$. Now $\{y_n\}$ has a Cauchy subsequence $\{y_{n_k}\}$. Now, use the triangle inequality to show that $\{x_{n_k}\}$ is also Cauchy.