Closure of a set is not necessarily compact

Question

Let $(X,d)$ be a metric space and suppose $(x_n)$ is a Cauchy sequence in $(X,d)$. Is $\overline{\{x_n : n \in \mathbb{N} \}}$ necessarily compact?

The answer is obviously no, consider $x_n = 1/n$ in $X = (0,1]$ with the usual metric. $(x_n)$ is Cauchy in $X$ and $\overline{\{x_n : n \in \mathbb{N} \}} \subset (0,1]$. However, no subsequence of $(x_n)$ can converge in $(0,1]$ because 0 is the only possible limit, but $0 \not \in (0,1]$.

This is fine, however, now I have to show that if $\overline{\{x_n : n \in \mathbb{N} \}}$ is complete, then it is compact, and specify under what conditions $\overline{\{x_n : n \in \mathbb{N} \}}$ is complete/compact.

I've made little progress .

Answer 1

Note that $\{x_n\}$ has a limit $x$ in the completion of $X$, because it is a Cauchy sequence. Since any subsequence of this sequence converges to the same limit we have two options for the set $\overline{\{x_n\}}$. If $x\notin X$, then $\overline{\{x_n\}} = \{x_n\}$. If $x\in X$, then $\overline{\{x_n\}} = \{x_n\} \cup \{x\}$. In the first case, as you showed in your example we see that $\overline{\{x_n\}}$ is not compact. However, if it is complete, then it must contain $\{x\}$ and thus we are in the second case. Note that any sequence in $\{x_n\} \cup \{x\}$ has a subsequence which either converges to $x_n$ for some $n$, or it converges to $x$.

Answer 2

HINT: Let $Y=\operatorname{cl}\{x_n:n\in\Bbb N\}$.

• Use the fact that $\langle x_n:n\in\Bbb N\rangle$ is Cauchy to show that $Y$ is totally bounded.

Then recall that a metric space is compact if and only if it’s complete and totally bounded. (If you’ve not seen this result, you’ll find a sketch of a proof of the required direction here.)