I have to do this exercise for my math study, and I'm having trouble with doing the second part of it.
Let $x, y, \epsilon \in \mathbb{R}$ and $\epsilon > 0$. Prove: $|x - y| \leqslant \epsilon$ $\forall \epsilon > 0 \Leftrightarrow x = y$
I think I have the left implication $\Leftarrow$:
Assume $x = y \Rightarrow x - y = 0 \Rightarrow|x - y| = 0 = |0| \Rightarrow|x - y| = 0 < \epsilon \Rightarrow |x - y| \leqslant \epsilon$ $\forall \epsilon > 0 \in \mathbb{R}$
Is this argument correct?
For the right implication, I do only have an idea of how to prove it. I think I have to assume first that $x > y$ and then $x < y$ and get a contradiction form both.
Could you please explain me the right implication and tell me if my left implication is correct?
Thanks in advance!