Real analysis question $ a = b$ if and only if for every $\epsilon \gt 0, \left\lvert a-b \right\rvert \lt \epsilon$

Question

I've encountered the proof that BWOC, $|a-b| \lt \epsilon$, we assume that $a \ne b.$

Without loss of generality, $a\gt b, a-b\gt 0$, let $\epsilon = a-b.$ Then $|a-b| \lt a-b$, which is a contradiction. Thus $a=b$.

I do not get which part of it is a contradiction.

PS. I'm completely new to stack exchange, so if my wording is weird, please let me know how I should phrase the questions next time so that you understand them better! Thank you!

Answer 1

Since $a-b > 0$, $|a-b|=a-b < a-b$.

The same number can't be less than itself. This is a contradiction.

Answer 2

If $|d| < \epsilon$ for all $\epsilon>0$, then we must have $d=0$, because if not, then setting $\epsilon = {1 \over 2} |d|$ will yield $1 < {1 \over 2}$ which is a contradiction.

Set $d=a-b$ in the above.