$\forall \epsilon>0\ \exists x,y\ |x−y|=\epsilon/2.$ Why isn't it a counterexample to $\forall \epsilon>0\ (|x−y|<\epsilon) \implies x=y$

Question

I'm studying $\forall \epsilon>0(|x−y|<\epsilon) \implies x=y$ at the moment and there is something I can not figure out. I really understand the logic behind the contrapositive demonstration in this post: Prove : $|x - y| \leqslant \epsilon$ $\forall \epsilon > 0$ iff $x = y$ but I also can not understand why the logic below is false:

$\forall \epsilon>0\ \exists x,y\ (x\neq y \land |x−y|=\epsilon/2<\epsilon),$ which looks like a counter example of the initial statement.

I'm I making a mistake?

(I'm French so sorry in advance for any potential grammar or spelling mistake)

Thanks in advance!

In the post that you are referring to, $x$ and $y$ are given and can not be chosen depending on $\epsilon$. — Martin R, Sep 21 '21 at 11:26
Thank you. I finally managed to see the mistake in my argumentation. — Idoia, Sep 21 '21 at 11:39
Does this answer your question? Intuition: If $a\leq b+\epsilon$ for all $\epsilon>0$ then $a\leq b$? — user1110341, Jun 10 '23 at 06:55

score 3 · Answer 1 · answered Sep 21 '21 at 11:27

3

Let $\varepsilon>0$. If you take $x,y\in\Bbb R$ such that $|x-y|=\frac\varepsilon2$, then indeed $|x-y|<\varepsilon$ and $x\ne y$. But it's not true that $|x-y|$ is smaller than every number greater than $0$ (which is what the assertion $(\forall\varepsilon>0):|x-y|<\varepsilon$ means). Just note that, for instance, $|x-y|>\frac\varepsilon3$.

answered Sep 21 '21 at 11:27

José Carlos Santos

440,053

Thank you very much. I finally understand the mistake in my argumentation ! – Idoia Sep 21 '21 at 11:37
This https://math.stackexchange.com/q/1468292/42969 is very close, is it good enough as a duplicate target? – Martin R Sep 21 '21 at 11:43
@MartinR It's not clear-cut to me but I will not oppose it. – José Carlos Santos Sep 21 '21 at 11:58

score 1 · Answer 2 · answered Jun 10 '23 at 16:55

Your $$\forall \epsilon>0(|x−y|<\epsilon) \implies x=y\tag1$$ is ambiguous:

You seem to interpret $(1)$ as$$\forall x,y\quad\forall \epsilon>0\quad\left(|x−y|<\epsilon\implies x=y\right)\tag2$$or equivalently$$\forall \epsilon>0\quad\forall x,y\quad\left(|x−y|<\epsilon\implies x=y\right),\tag3$$the negation of which is$$\exists\epsilon>0\quad\exists x,y\quad\left(|x−y|<\epsilon\text{ and }x\ne y\right).\tag4$$ You correctly prove $$\forall\epsilon>0\quad\exists x,y\quad\left(|x−y|<\epsilon\text{ and }x\ne y\right),$$ which implies $(4)$ and therefore disproves $(3)$ and $(2).$
The intended meaning of $(1)$ is however the proposition proved in the post you linked to: $$\forall x,y\quad\left[\left(\forall \epsilon>0\quad|x−y|<\epsilon\right)\implies x=y\right].\tag5$$ The parenthesis make a big difference between $(2)$ (false) and $(5)$ (true).

Pat · Answer 3 · 2021-09-21T12:07:57.403

The correct original statement is: "if $\lvert x-y \rvert \lt \epsilon$ holds for all $\epsilon \gt 0$, then $x=y$". When this statement is true, your proposed counter-example cannot be valid, you cannot suggest a positive number equal to $\lvert x-y \rvert$ because $\lvert x-y \rvert$ is smaller than any such number (i.e. any $\epsilon \gt 0$).

I am guessing that your misunderstanding comes from reading the original statement as "if for every $\epsilon \gt0$ some $x,y$ are such that $\lvert x-y \rvert \lt \epsilon$ then they are $x=y$", which obviously cannot be true, as proven by your counter-example. But the correct original statement is that after you have taken some $x,y$ then you examine whether $\lvert x-y \rvert \lt \epsilon$ for every positive $\epsilon$. If so, $x=y$.

$\forall \epsilon>0\ \exists x,y\ |x−y|=\epsilon/2.$ Why isn't it a counterexample to $\forall \epsilon>0\ (|x−y|<\epsilon) \implies x=y$

3 Answers3