0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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) $\endgroup$ – b00n heT Jun 30 '16 at 5:40
  • $\begingroup$ I haven't constructed the reals yet. $\endgroup$ – the_dude Jun 30 '16 at 5:41
  • $\begingroup$ Sure. Makes sense :) $\endgroup$ – b00n heT Jun 30 '16 at 5:41
  • $\begingroup$ 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. $\endgroup$ – the_dude Jun 30 '16 at 5:44
  • $\begingroup$ Yes and no: You are claiming that this is a contradiction without giving a proof of it, thus it is mathematically incorrect. $\endgroup$ – b00n heT Jun 30 '16 at 5:46
1
$\begingroup$

...But since $\delta < \epsilon$ , $\forall \epsilon > 0, \epsilon \in \mathbb Q $...

This part is not right.

Make sure to use the right argument with the right inequality!

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

$\endgroup$
  • $\begingroup$ Yeah right. I get the absurdity. $\endgroup$ – the_dude Jun 30 '16 at 5:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.