Is the set of extended natural numbers compact? Is the metric space $(\overline{\mathbb{N}}, d)$ compact? Here, $\overline{\mathbb{N}}=\mathbb{N} \cup \{\infty\}$ and $$d(m,n)=\left|\frac{1}{m} - \frac{1}{n}\right|\ \ \ \text{ for }  m,n \in \mathbb{N}\text{ and }$$
$$d(n,\infty)=\frac{1}{n}\text{ for }n\in\Bbb N.$$
Let $m \in \mathbb{N}$. We claim that the set $\lbrace m \rbrace$ is open in this metric. We want to find an $\epsilon>0$ such that $B(m,\epsilon) \subset \{m\}$. Notice that for each natural number $n$, there is such a thing as the closest point, namely, $n+1$. Thus, if we choose $\epsilon = \frac{1}{m(m+1)}$, then $B(m,\epsilon) \subset \{m\}$. Hence, each $\lbrace m \rbrace$ is open. $\lbrace \infty \rbrace$ however is not open, for each $\epsilon>0$ there is a point with distance less than $\epsilon$.
I considered using this to come up with an open cover that did not admit a finite subcover, like $U= \bigcup\limits_{n=1}^{\infty} B\left(n, \frac{1}{n(n+1)}\right)$, but ran into trouble with $\infty$, because any $\epsilon$ radius around it would contain infinitely many points and thus give us a finite cover.
Any help would be appreciated.
 A: Let $\{U_\alpha\}$ be an open cover of $\overline{ \Bbb N}$ and let $\infty\in U_{\alpha_0}$. So, we have an open ball centered at $\infty$ and contained in $U_{\alpha_0}$, let's say $\big\{x\in \overline{\Bbb N}:d(x,\epsilon)<\epsilon\big\}=B(\infty,\epsilon)\subseteq U_{\alpha_0}$ for some $\epsilon>0$. Then, $$B(\infty,\epsilon)=\bigg\{n\in\Bbb N\bigg| \frac{1}{n}<\epsilon\bigg\}\cup\{\infty\}\subseteq U_{\alpha_0}$$Since, there are only finitely many $n\in\Bbb N$ with $\frac{1}{n}\geq \epsilon$ we are done.
That is to say, let $n_0:=\left\lfloor\frac{1}{\epsilon}\right\rfloor+1$, then $n_0\in \Bbb N$, and $n\geq n_0$ implies $n>\frac{1}{\epsilon}\implies \frac{1}{n}<\epsilon$ i.e. $n\geq n_0\implies n\in B(\infty,\epsilon)\subseteq U_{\alpha_0}$.
Now, let $U_1,..., U_{n_0-1}\in \{U_\alpha\}$ be such that $k\in U_k$ for all $k=1,...,(n_0-1)$. Then, $U_1,...,U_{n_0-1}, U_{\alpha_0}$ is a finite sub-cover of the open cover $\{U_\alpha\}$.
A: If you have an open cover $\mathscr{U}$ of $\overline{\mathbb{N}}$, then $\infty\in U_0\in\mathscr{U}$. Therefore $U_0$ contains a ball centered at $\infty$, say $B(\infty,r)$. Take $k\in\mathbb{N}$, $k>0$, such that $1/k<r$. Then the set
$$
S=\{n\in\mathbb{N}:n\notin B(\infty,r)\}
$$
is finite because for all $n>k$ we have $n\in B(\infty,r)$.
