Interior of $\{\frac{1}{n}\mid n\in\Bbb{N}\}$ is empty

Question

Prove that the interior of $S=\left\{\frac{1}{n}\mid n\in\Bbb{N}\right\}$ is empty.

What I tried: We want to prove that for all $x\in S$ and all $\varepsilon>0$, $B(x,\varepsilon)\not\subseteq S$.

Choose $x=\frac{1}{m}$. Let $\frac{1}{m}-\varepsilon>\frac{1}{m+1}$. Choose $y=\frac{1}{m}-\frac{\varepsilon}{2}\in B(x,\varepsilon)$, then $\frac{1}{m+1}<y<\frac{1}{m}$ therefore $y\not\in S$.

How do we proceed for other $\varepsilon$?

Answer 1

You don't have to do all that. The set $B(x,\varepsilon)$ is just the interval $(x-\varepsilon,x+\varepsilon)$, which is uncountable. Since $S$ is countable, $(x-\varepsilon,x+\varepsilon)\varsubsetneq S$.