# Prove that $\mathbb{N}$ with the metric $d(m,n)=\lvert m^{-1} - n^{-1}\rvert$ is a discrete metric space

Prove that $\mathbb{N}$, along with the metric $d(m,n)=\lvert m^{-1}-n^{-1}\rvert$, is a discrete metric space.

I am stuck with this one, I don't know how to proceed ?

Any help will be appreciated.

• As before, can you show that singletons are open here as well? – Prahlad Vaidyanathan Oct 4 '13 at 13:57
• @PrahladVaidyanathan How do i do that here ? the distance between any two different natural numbers is always going to be greater than zero but less than 1. So, what should be the radius of the ball that i choose ? – johny Oct 4 '13 at 14:07

Fix a point $m\in \mathbb{N}$, and you want to show that $\{m\}$ is an open set. In other words, you want to find $r > 0$ such that $$B(m,r)\cap \mathbb{N} = \{m\}$$ So for any $n\in \mathbb{N}$, consider $$\left | \frac{1}{m} - \frac{1}{n} \right |$$ If $n\to \infty$, this distance approaches $|1/m| > 0$. In particular, there is some $n_0\in \mathbb{N}$ such that $$d(m,n) > \frac{1}{2|m|} \quad\forall n \geq n_0, n\neq m$$ Now consider the set $$S = \{d(m,n) : n < n_0, n\neq m\}$$ This is a finite set, and every number inside is positive, hence its minimum is a positive number. Take $$r = \frac{1}{2} \min\{ \min(S), |1/m|\} > 0$$ Then, for any $n \in \mathbb{N}, n\neq m$, it follows that $$d(m,n) > r$$ Hence, $$B(m,r)\cap \mathbb{N} = \{m\}$$ Does this help?