Proof of $\forall\epsilon>0:a

Question

Let $a,b\in\mathbb{R}$ such that $\forall\epsilon\in\mathbb{R}_{>0}:a<b+\epsilon$. Then $a\le b$.

Proof by contradiction:

Assume $a>b$. Then $a-b>0$. We also have $\forall\epsilon>0:a<b+\epsilon$. This gives $0<a-b<\epsilon$. Let $\epsilon=a-b$. Then $a-b<a-b\implies a<a$. Therefore, by contradiction $a\le b$.

In the proof above I do not understand why we are allowed to fix $\epsilon$ to a value like $a-b$. I would appreciate any help.

Thank you.

The statement should hold for any $\varepsilon>0$ and hence it should also hold for $\epsilon = a-b>0$ (note that $a-b>0$ under the assumption). Clearly, this yields a contradiction. Hence, the assumption $a>b$ is false. — Cavents, Apr 02 '16 at 16:37
So, it's a proof by contradiction and counterexample, right? — johnny09, Apr 02 '16 at 16:39
I read the question title a little differently from the question itself: I thought the implication was inside the quantifier. That makes the statement a lot less interesting. — David K, Apr 02 '16 at 16:53
Does this answer your question? Intuition: If $a\leq b+\epsilon$ for all $\epsilon>0$ then $a\leq b$? — user1110341, May 27 '23 at 15:21

score 3 · Answer 1 · answered Apr 02 '16 at 16:50

If you draw a number line, and mark a and b on that line, you can see it.

The statement tells you that, independent of how much you increased b, a is smaller than that.

Then if you assume, a is greater than b (draw it on the number line), then one can argue that if you increase b, say $\epsilon= a-b$ amount (or anything smaller than that!), a is not smaller than $b+\epsilon$ .

Since you are free to choose $\epsilon$, you can run this logic and get a contradiction.

I find it helpful to draw in such proofs.

Jimmy R. · Accepted Answer · 2016-04-02T16:47:06.103

1

You did not fix $\epsilon$ to some value. What you did is the following:

You are given that: $$\forall \epsilon>0: a<b+\epsilon\tag{1}$$
You assume that (this you want to lead to a contradiction): $$a>b\tag{2}$$

Put together $(1)$ and $(2)$ and this gives you $$0<a-b<ε$$ So, this is already the contradiction. You had that $\epsilon >0$ arbitrary but here - assuming $(2)$ - you get that $\epsilon$ must be larger than $a-b$.

edited Apr 02 '16 at 16:47

answered Apr 02 '16 at 16:40

Jimmy R.

36,148

Yes, but in the proof I write "Let $\epsilon=a-b$" and I use this value to show explicitly what is going wrong. This fixes the $\epsilon$ to some value. Is this wrong by my part? – johnny09 Apr 02 '16 at 16:48
1

No, this is not wrong. It is fine. You show that there exists an $\epsilon$ namely $\epsilon=a-b$ for which $a<b+\epsilon$ does not hold (actually any $0<\epsilon\le a-b$ would do), despite the fact that is should hold by assumption. So, assuming $(2)$ led you to a contradiction to the validity of $(1)$. – Jimmy R. Apr 02 '16 at 16:51

Proof of $\forall\epsilon>0:a

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