So i haven't done this stuff in a really long time and i need some help to complete this proof. I want to show: Let $(X,d)$ be a complete metric space and ${I}_{n}\subset X$, $n\in \mathbf{N}$ ($\mathbf{N}$ are the natural Numbers) be a arbitrary sequence of finite sets and let $K:=\bigcap _{ n\in \mathbf{N}} \bigcup_{ x\in {I }_{ n }} {\overline{{B}_{\frac{1}{n}}(x)}}$
Show that $K$ is compact. (The above should denote closed balls, my edititng is bad)
So in a Metric space the compactness of $K$ is equivalent to K being complete and totally bounded which is equivalent to saying $K$ needs to be closed and totally bounded since closed subsets of complete spaces are complete. $K$ is obviously closed since the union is finite and finite unions of closed sets are closed and intersection of closed sets is closed. Now the tricky part for me is showing that $K$ is totally bounded how can i find a finite cover of $K$ consisting of open Balls with radius Epsilon and center in $K$??
We know that a metric space $(X,d)$ is compact if and only if it is complete and totally bounded. That is for every $\varepsilon>0$ there is a finite covering of $X$ by (open) balls of radius $\varepsilon$. Since
$$K = \bigcap_{n\in \mathbf{N}}\underbrace{\bigcup_{x\in I_n}\overline{B_{\frac{1}{n}}(x)}}_{\text{finite union}}$$
it is, as you say, clear that $K$ is closed and therefore also complete. We are now only left to verify total boundedness. We note that for every $n$
$$K\subseteq\bigcup_{x\in I_n}\overline{B_{\frac{1}{n}}(x)}.$$
That is $K$ is covered by finitely many closed balls of radius $1/n$. Show now that it is covered by finitely many OPEN balls of radius $2/n$. Given $\varepsilon$ can we cover $X$ by finitely many balls of radius $\varepsilon$?