Real Analysis – Prove |x – y| ? ? ? ? > 0 iff x = y

Question

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!

Answer 1

Your left implication is OK. For the right implication, if $x\neq y$, just take $\epsilon=\frac{|x-y|}{2}$ to obtain a contradiction.