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
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 $B(x,\frac{1}{x(x+1)})$