How to show that $ \inf \{ |x-a| : a \in A\} > 0 $

Question

I was trying to prove something and part of it was to show that

$ \inf \{ |x-a| : a \in A\} > 0 $ where

$ A= \{ 1, \frac{1}{2}, \frac{1}{3}, \dots, \frac{1}{n}, \dots \} $

It is clear that

$ \inf \{ |x-a| : a \in A\} \geq 0 $

I know it seems easy, however, I'm just stuck.

Answer 1

As you have noticed, $x\in A\cup\{0\}$ if and only if $\inf\{|x-a|:a\in A\}$. More generally, for any subset $A$ of $\mathbb{R}$, for any $x\in\mathbb{R}$, the number $d(x,A):=\inf\{|x-a|:a\in A\}$ is called the distance between $x$ and $A$. We have $d(x,A)=0$ iff for any $n>0$, there is $a_n\in A$ such that $|x-a_n|<\frac{1}{n}$ iff $\displaystyle\lim_{n\to\infty} a_n = x$ for some sequence $(a_n)\subseteq A$ iff $x\in\overline{A}$. In your case, $\overline{A}=A\cup \{0\}$.