$f :\mathbb N \to \mathbb R$ be the function $f(0)=0 , f(n)=\dfrac 1 n , \forall n >0$;is $\mathbb N$ induced with the metric $|f(x)-f(y)|$ compact? Let $\mathbb N$ be the set of non-negative integers and $f :\mathbb N \to \mathbb R$ be the function $f(0)=0 , f(n)=\dfrac 1 n , \forall n >0$ , then obviously $f$ is injective , so $d : \mathbb N \times \mathbb N \to \mathbb R $ defined as $d(x,y)=|f(x)-f(y)|$ induces a metric on $\mathbb N$ . Then , is it true  that $(\mathbb N , d)$ is compact ? 
 A: Yes it is, because it's sequentially compact:
Let $x_n$ be a sequence in $\mathbb{N}$. Case 1: it takes on finitely many values. Then there is a constant subsequence which obviously has a limit in $\mathbb{N}$. Case 2: it takes on infinitely many values. Then it has a strictly increasing(in the usual sense) subsequence. This subsequence has to converge to $0$ with respect to your metric.
A: By construction, $(\mathbb{N}, d)$ is isometric to the subspace $\{\frac{1}{n} \mathrel{|} n > 0\} \cup \{0\}$ of $\mathbb{R}$ with the usual metric. This subspace is closed and bounded and hence compact by the Heine-Borel theorem.
A: Assume the space is covered with open sets $U_i$, $i\in I$. Then $0$ must be in one of these sets, so there must be one open set V among the $U_i$ cotaining $0$. So we can find  a natural number $m$  s.t $0 \in B(0, 1/m) \subset V$ where $B$ is the open ball in this space. Now the numbers $\{m+1, m+2, \dots\}$ are in this ball. Take $V$ and the open sets from the collection $U_i$ containing the numbers upto $m$ and you have a finite cover.
A: Take $m\in \Bbb N$. Now, observe that the set of numbers $$\left\{\left|\frac1m-\frac{1}{x}\right|:x\in\Bbb N\right\}$$
is bounded below by $l_m = \left|\frac1m-\frac{1}{m+1}\right|$. Knowing this, by taking any $\varepsilon <l_m$, we obtain $$B_\varepsilon(m) = m$$
Thus, the space is discrete and infinite, so it cannot be compact.
Edit: I did not notice that $0$ was included in $\Bbb N$ in this case, for which the above fails. What's more, in other answers it can be seen that the presence of $0$ is exactly what makes the space compact.
