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

Question

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 05:40
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 05:44
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 05:46
I don't see why $\delta < \epsilon$ gives a contradiction so you should point out why it is so. — Jeff L., Jun 30 '16 at 05:47
If the main purpose of your post is checking your proof (as opposed to asking for any solution), you should probably use the ([tag:proof-verification]) tag, see the tag-info. — Martin Sleziak, Jun 30 '16 at 06:51
Does this answer your question? Intuition: If $a\leq b+\epsilon$ for all $\epsilon>0$ then $a\leq b$? — user1110341, May 27 '23 at 15:22

score 1 · Accepted Answer · answered Jun 30 '16 at 05:38

1

...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.

answered Jun 30 '16 at 05:38

Daniel

7,185

Yeah right. I get the absurdity. – the_dude Jun 30 '16 at 05:53

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

1 Answers1