Every singleton set is open. Prove that in the metric space $(\Bbb N ,d)$, where we define the metric as follows: let $m,n \in \Bbb N$ then, $$d(m,n) = \left|\frac{1}{m} - \frac{1}{n}\right|.$$ Then show that each singleton set is open.
I am facing difficulty in viewing what would be an open ball around a single point with a given radius?
Can I take the open ball around an natural number $n$ with radius $\frac{1}{2n(n+1)}$??
 A: Consider $$K=\left\{ \frac 1 n \,\middle|\, n\in\mathbb N\right\}$$
equipped with the standard metric $d_K(x,y) = |x-y|$. Then $(K,d_K)$ is isometric to your space $(\mathbb N, d)$ via $\mathbb N\to K, n\mapsto \frac 1 n$. Open balls in $(K, d_K)$ are easy to visualize, since they are just the open balls of $\mathbb R$ intersected with $K$. This should give you an idea how the open balls in $(\mathbb N, d)$ look.
A: I'll let $d_1$ denote the usual metric, and $d_2$ your metric.
Let $a \in \Bbb N$. We have:
{$a$} $= B_{d_1}(a,\frac12)$, because:
If $x \in$ {$a$}, then $x = a$ and we are done.
If $x \in B_{d_1}(a,\frac12)$, then $d(a,x) < \frac12$, so $|a - x| < \frac12$, but this means that $|a - x| = 0$, since $a,x \in \Bbb N$. Thus, $x = a$, and so $x \in$ {$a$}.
Therefore for any $a \in \Bbb N$, {$a$} is an open ball in $(\Bbb N, d_1)$, and every open ball is an open set.
On the other hand, $d_1$ and $d_2$ are topologically equivalent. Therefore singletons are also open in $(\Bbb N, d_2)$.
