Proof that $\varepsilon$-arguments are valid

Question

A lot of times when proving for example inequalities like $$x \leq y$$ for real numbers $x,y$ the argument looks like $$x \leq y + \varepsilon$$ for all $\varepsilon > 0$, hence $x \leq y$.

Now this is obviously very intuitive, but is there a "proof" that this conclusion is correct? And is it always sufficient in order to proof $x \leq y$ to show $x \leq y + \epsilon$ for all $\varepsilon > 0$?

I'd appreciate any explanations!

NOTE: I know that these kinds of arguments are correct when dealing with sequences. But here we have no sequences so I wanted to understand this too.

To conclude $x \le y$, it suffices to know $x < y + \varepsilon$ for all $\varepsilon$. — mlc, Mar 30 '17 at 21:45
Does this answer your question? Intuition: If $a\leq b+\epsilon$ for all $\epsilon>0$ then $a\leq b$? — user1110341, Jun 10 '23 at 06:53
@user1110341 I don't think it does: that post was about intuition, not formal proof. — Anne Bauval, Jun 10 '23 at 17:19
Does this answer your question? Let $x$ and $y$ be real numbers. Prove that if $x\le y + \epsilon$ for every positive real number $\epsilon$, then $x \leq y$. (found using Approach$0$) — Anne Bauval, Jun 10 '23 at 17:20

Olivier · Answer 1 · 2017-03-30T21:55:13.410

4

Suppose $x < y + \varepsilon$ for all $\varepsilon >0$, but that $x>y$. Then, taking $\varepsilon = x-y >0$, you obtain $x<x$, a contradiction.

edited Mar 30 '17 at 21:55

answered Mar 30 '17 at 21:49

Olivier

4,003

score 2 · Answer 2 · answered Mar 30 '17 at 21:46

2

Suppose $x \leqslant y + \varepsilon$ for every positive $\varepsilon$, and for the sake of contradiction, suppose $x > y$.

For the specific error $\varepsilon := \frac{1}{2}(x - y) > 0$ we have that $$x \leqslant y + \varepsilon < y + 2\varepsilon = y + (x - y) = x,$$ which is a desired contradiction.

answered Mar 30 '17 at 21:46

joeb

2,828

That is exactly what I was looking for. However, is there also a direct proof for this? Just for more clarification that $(\forall \varepsilon > 0: x < y + \varepsilon) \implies (x \leq y)$. – Staki42 Mar 30 '17 at 21:50
I meant a proof that does not use contradiction, i.e. $A \implies B$ rather than $\neg B \implies \neg A$ – Staki42 Mar 30 '17 at 22:00

score 1 · Accepted Answer · answered Mar 30 '17 at 22:08

1

Consider

$$t\le\epsilon$$ for all $\epsilon>0$.

Clearly,

$$t\le0$$ is compatible, while

$$t>0$$ is not because

$$0<t\le\epsilon$$ cannot hold for all $\epsilon>0$.

Rewrite with $t:=x-y$.

answered Mar 30 '17 at 22:08

That made it clear! Thanks! – Staki42 Mar 30 '17 at 22:10

score 1 · Answer 4 · answered Mar 30 '17 at 23:37

A silly direct proof (for those who don't) like proofs by contradiction).

$x\le y +\epsilon$ for all $\epsilon > 0$.

Let $\tau =y-x $.

$y= x +\tau $ so $\tau \le 0$.

So $y \le x $

$-\tau \ge 0$. So as $x\ge y $ we have $x+(-\tau) \ge y+(-\tau) =y-y+x = x $. So $-\tau \le 0$.

Well, I guess technically that is a proof by contradict. Still... I thought it was cute.

So $\tau = y-x=0$. And $x=y $.

Proof that $\varepsilon$-arguments are valid

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