I'm working with the metric space $(\mathbb{N}, \rho)$ where $\mathbb{N}$ is the set of natural numbers and $\rho(x,y) = |\frac{1}{x} - \frac{1}{y}|$.

I'm considering the open balls on this metric. Are there any that are finite? Infinite? All of $\mathbb{N}$?

My hunch is that there are open balls that are finite and infinite. For example, the open ball $B(1, \frac{1}{2})$ seems to be just {$1$}.

But if we make the radius larger than $1$ doesn't the open ball becoming infinite?

Am I correct? Are there any open balls that are finite? Infinite? All of $\mathbb{N}$? Any other general statements we can make about the open balls?

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    $\begingroup$ You might find it helpful to note that your space is isometric to the set $\{\frac{1}{n} : n \in \mathbb{N}\} \subset [0,1]$ (with the Euclidean metric). Try drawing a picture of this. $\endgroup$ May 11, 2012 at 3:53

2 Answers 2


$B(a,r)$ is just the solution of $\left|\dfrac1a -\dfrac1x\right|<r$, which is easily solved:

$ x>\dfrac{a}{1+a r} \qquad \qquad \qquad\text{if } r\ge\dfrac1a $

$ \dfrac{a}{1+a r}<x<\dfrac{a}{1-a r} \qquad \text{if } r<\dfrac1a $

In the first case, the ball is infinite. In the second case, the ball is finite.

In particular, the ball is infinite if $r\ge1$ and the ball centered at $1$ is all of $\mathbb N$ if $r=1$.


Your suspicion is correct. Open balls in $(\mathbb N,\rho)$ are either finite or cofinite (meaning have all but finitely many elements of $\mathbb N$). To see this, consider the open ball $B(x,r)$. If $r\leq 1/x$ then $$\begin{eqnarray}y\in B(x,r)&\implies& |1/x-1/y|<r\\ &\implies& 1/y>1/x\text{ or }1/y>1/x-r\\ &\implies& y<x\text{ or } y<\frac{1}{1/x-r} \end{eqnarray}$$ and so there are finitely many $y$ in $B(x,r)$, while if $r> 1/x$ then $$\begin{eqnarray}y>x &\implies& |1/x-1/y|<1/x<r\\ &\implies& y\in B(x,r) \end{eqnarray}$$ thus all but finitely many $y$ are in $B(x,r)$. It is indeed possible for an open ball to be all of $\mathbb N$, such as $B(x,1)$ for any $x\in \mathbb N$.


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