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

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.

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.

• Another (more standard?) definition of relatively compact is en.wikipedia.org/wiki/Relatively_compact_subspace Jun 22 '21 at 4:23
• @copper.hat . In the context of metric spaces, some authors have used "$K$ is pre-compact" to mean $\overline K$ is compact. Jun 22 '21 at 5:45

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.
• Triangle inequality (norm notation): $$||x_{n_k}-x_{n_j}||=||x_{n_k}-y_n+y_n-x_{n_j}||\leq||y_n-x_{n_k}||+||y_n-x_{n_j}||<2/n\to0.$$ Correct? To me, my argument seems off as I don't see where I made use of $\{y_{n_k}\}$ being a Cauchy subsequence... Jun 22 '21 at 1:23
• What you need is $\|x_{n_k} - x_{n_j}\| \leq \|x_{n_k} - y_{n_k}\| + \|y_{n_k} - y_{n_j}\| + \|y_{n_j} - x_{n_j}\|$. Jun 22 '21 at 1:32
• Just to add, so by the completeness of $X$, there exists an $x\in X$ for which $x_{n_k}\to x$. By the closedness of $\overline{K}$, and $\{x_{n_k}\}\in\overline{K}$, it follows that $x\in\overline{K}$, and hence $\overline{K}$ is compact. Correct? Jun 22 '21 at 1:40