# $f :\mathbb N \to \mathbb R$ 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 ?

• Is every Cauchy sequence in that metric convergent? – Simon S Apr 5 '15 at 12:31
• $(\mathbb{N}, d)$ is isometric to the subspace $\{\frac{1}{n} \mathrel{|} n < 0\} \cup \{0\}$ of $\mathbb{R}$ with the usual metric. As such it is a closed and bounded subset of $\mathbb{R}$ and hence compact. – Rob Arthan Apr 5 '15 at 12:58
• @RobArthan : You should post it as an answer , its awesome – user228168 Apr 5 '15 at 13:09

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.

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.

• Brilliant, don't know why I didn't come up with this sooner ;) – Math1000 Apr 5 '15 at 12:46

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.

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.

• Every open neighbourhood of $0$ contains infinitely many elements, so the space is not discrete. – Rob Arthan Apr 5 '15 at 12:55
• @RobArthan I was editing just as your comment came up :) – GPerez Apr 5 '15 at 12:57