# $\mathbb N$ when given the metric $d(m,n)=\frac{1}{m}-\frac{1}{n}$

How to show that $\mathbb N$ when given the metric $d(m,n)=\dfrac{1}{m}-\dfrac{1}{n}$ and when given the subspace topology as inherited from $\mathbb R$ are equivalent.

When $\mathbb N$ is given the subspace topology then $\mathbb N$ becomes the discrete space.But I cant show that in the first case single point sets are open.

I know to show equivalent we have to proceed as

take any $x\in B(x,r)$ and then find $s>0$ such that $x\in B(x,s)\subset B(x,r)$

But I cant find the $s>0$ in the metric given .Please help

• @AdamHughes: Are you sure about that? I fail to see how your inequality is accomplished. For example for $m = 1$ we have $d(1,n) = 1 - \frac{1}{n}$ which is fairly large for $n > 1$. – Matthias Klupsch Feb 25 '15 at 8:19
• Oh rats, @MatthiasKlupsch thanks, I got the direction reversed. Silly mistake. – Adam Hughes Feb 25 '15 at 8:21

I think you mean $d(x,y) = |\frac{1}{x} -\frac{1}{y}|$ for symmetry reasons. So, I'm going from there.
Since the $\mathbb N$ when seen as a subspace of $\mathbb R$ is a discrete space, any other topology (choice of open subsets) $\mathcal X$ on $\mathbb N$ is contained in this one, so take $\mathcal X$ to be this metric topology, all you need to do now is to provide, for each $x\in \mathbb N$ a basic open set $U_x$ that only contains $x$. So, without loss of generality let $y > x$, this means that $d(x,y) = \frac{1}{x} - \frac{1}{y}$, which is minimized when $y = x+1$ so just take $U_x$ to be $B(x,\frac{1}{x(x+1)})$