# I need to show the inequality between $|x|<$ all positive real numbers $\epsilon$ iff $x=0$.

Let $x ∈ \mathbb R$. Prove that $|x| < \epsilon$ for all $\epsilon > 0$ iff $x = 0$

I understand that for all real numbers $\epsilon>0$ can only be greater than $|x|$ because taking the inequality of anything other than $0$ will result in a real number that can be greater than $\epsilon$. Therefore contradiction. I just need help showing that. I am new to analysis.

Suppose $|x| < \epsilon$ for all $\epsilon>0$. If $x \neq 0$, then take $\epsilon = {|x| \over 2}$ which gives $|x| < {|x| \over 2}$, or $|x| <0$, which is impossible. Hence $x=0$.
As $|x|=0$, one direction is trivial.
For the other let $|x|<\epsilon$ for all $|\epsilon>0|$. Assume $x\ne 0$. Then one of $x,-x$ is $>0$, hence $|x|:=\max\{x,-x\}>0$. So if we let $\epsilon=|x|$, we do not have $|x<<\epsilon$. Hence the assumption that $x\ne 0$ must be false and we conclude $x00$.
The set $\{\epsilon\mid \epsilon>0\}$ is simply $\Bbb R_{>0}$ which has $0$ as greatest lower bound. Now by hypothesis we have $|x|\le \epsilon$ for all $\epsilon>0$ so $|x|\le0$ but also $|x|\ge0$ hence $|x|=0\iff x=0$.