# Prove that for rational $x$,$y$ and $\epsilon$ if $|x-y| \leq \epsilon$ , $\forall \epsilon > 0$ then $x=y$

I know that this is a repeated question, but I wanted to show my attempt.

Suppose $x \neq y$ and (wlog) $x > y$, then $x$ can be written as

$x = y + \delta$, for some ($\delta > 0$ and $\delta \in \mathbb Q$)

Therefore $|x-y| = \delta$.

But since $\delta < \epsilon$ , $\forall \epsilon > 0, \epsilon \in \mathbb Q$ , and we assumed that $\delta > 0$ ; therefore contradiction arises. Hence $x=y$.

Is this a correct proof? I want to know what is missing or how to add more rigour in the last statement.

• The step "$\delta<\epsilon,\forall \epsilon >0$ [...] and therefore a contradiction arises" could be improved, by simply choosing explicitly $\epsilon=\frac{\delta}{2}<\delta$. For the rest your proof is correct (although I don't see why you need the rationality assumption here) – b00n heT Jun 30 '16 at 5:40
• I haven't constructed the reals yet. – the_dude Jun 30 '16 at 5:41
• Sure. Makes sense :) – b00n heT Jun 30 '16 at 5:41
• But is it wrong / immature the way I wrote it? That since $\delta < \epsilon$ for all $\epsilon > 0$, and also that $\delta$ is positive; Hence the contradiction. – the_dude Jun 30 '16 at 5:44
• Yes and no: You are claiming that this is a contradiction without giving a proof of it, thus it is mathematically incorrect. – b00n heT Jun 30 '16 at 5:46

...But since $\delta < \epsilon$ , $\forall \epsilon > 0, \epsilon \in \mathbb Q$...
Since $\delta >0$, there exists $\epsilon \in \Bbb Q$ such that $0 < \epsilon < \delta$ and therefore $\epsilon < |x-y| = \delta$, which is absurd.