Proof Check: $x \leq y+ \epsilon$ for all $\epsilon >0$ iff $x \leq y$.

Question

Synopsis

I want to be sure I'm utilizing proof by contradiction correctly, so please check my proof of the exercise below. It's relatively simple, so it shouldn't take you too much time.

Exercise

Let $x$ and $y$ be real numbers. Show that $x \leq y + \epsilon$ for all real numbers $\epsilon > 0$ if and only if $x \leq y$.

Proof

Suppose $x \leq y + \epsilon$ for all $\epsilon > 0$ and $x > y$. Then $x - y > 0$ and $x - y + \epsilon > \epsilon$ for all $\epsilon > 0$. But if $x \leq y + \epsilon$, then $y - x + \epsilon \geq 0$. So $0 \leq y-x+\epsilon < y-x+(x-y+\epsilon) = \epsilon$, a contradiction. For the converse, suppose $ x \leq y$. Then it is obvious that $x \leq y+\epsilon$. This concludes our proof.

Update

This proof is obviously wrong. It is not a contradiction that $\epsilon >0$. For some reason, I deluded myself that my conclusion stated that $\epsilon < 0$, but that's just due to my occasional stupidity and habitual lack of double checking. Instead, consider some $\epsilon$ such that $0 < \epsilon < x-y$. Then $x \leq y + \epsilon < y+x-y < x$, a contradiction. Thank you to the various people who commented on the issues with my proof. This was a very stupid mistake, and I don't even know how I overlooked what I did.

I think this post should be deleted by its author instead of being closed as a duplicate. It doesn't seem useful to future readers. — Anne Bauval, Jun 10 '23 at 17:13

score 2 · Accepted Answer · answered Jul 12 '20 at 07:28

2

0≤y−x+ϵ<y−x+(x−y+ϵ)=ϵ, a contradiction.

Why is that a contradiction? $0$ is $< \epsilon$.

....

Instead Note $\epsilon$ is not fixed. If $x-y >0$ then we can let $\epsilon$ be some value $0 < \epsilon < x-y$ and .... then what happens?

$x \le y + \epsilon < y+(x-y) = x$ so $x < x$. Which certainly is a contradiction!

answered Jul 12 '20 at 07:28

fleablood

130,341

Thank you for your answer. I can't believe I made such a silly mistake ;-; – alvnya Jul 12 '20 at 07:39

score 1 · Answer 2 · answered Jul 12 '20 at 07:12

No, it is not correct. I tried to prove it by contradiction, and you got the conclusion that 0<\varepsilon, claiming that that's a contradiction. Why? There is no contradiction there.

If $x>y$, let $\varepsilon=\frac12(x-y)$. Then $\varepsilon>0$ and therefore $x\leqslant y+\varepsilon$. But this means that$$x\leqslant y+\frac12(x-y)=\frac12(x+y)<\frac12(x+x)=x.$$This gives you a contradiction: $x>x$.

Proof Check: $x \leq y+ \epsilon$ for all $\epsilon >0$ iff $x \leq y$.

2 Answers2