Proof by contradiction: $ \forall \epsilon \in \mathbb{R}^{>0}(a< b+\epsilon) \to a \leq b$

Question

Let $a,b \in \mathbb{R}$, I must proof:

$ \forall \epsilon \in \mathbb{R}^{>0}(a< b+\epsilon) \to a \leq b$

Proof by contradiction: I have by negation of thesis "$a>b$ (or $b \leq a \wedge a \neq b$), but if $a>b$ then $a-b>0$ and by hypothesis $a<b+(a-b)=a$ therefore $a<a$ is absurd! Is correct?

Thanks in advance!

P.S.= $ \forall \epsilon \in \mathbb{R}^{>0}(a< b+\epsilon) \to a < b$ is true?

Does this answer your question? Intuition: If $a\leq b+\epsilon$ for all $\epsilon>0$ then $a\leq b$? — user1110341, May 27 '23 at 15:19

Thomas · Accepted Answer · 2013-12-30T17:03:06.213

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Maybe the problem is: You are given two real numbers $a,b$ such that for all $\epsilon > 0$, you have $a < b+ \epsilon$, then prove that $a\leq b$.

In this case your solution is correct. So you can assume by contradiction that $a > b$. Then $a - b > 0$, and so $b + (a-b) = a < a$ is the contradiction.

edited Dec 30 '13 at 17:03

answered Dec 30 '13 at 16:56

Thomas

44,491

I edited my post, is correct? – mle Dec 30 '13 at 17:00
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@Soviet: Not quite. I think you want $(\forall \epsilon \in \mathbb{R}^{>0}, a< b+\epsilon )\to (a \leq b)$ or something like that. – Thomas Dec 30 '13 at 17:01

Proof by contradiction: $ \forall \epsilon \in \mathbb{R}^{>0}(a< b+\epsilon) \to a \leq b$

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