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Let $a,b\in \mathbb{R}$ be two real numbers. Suppose that $$\forall \epsilon \gt0: a\le b+\epsilon$$ holds. I'm aware that the correct implication is $$\forall \epsilon \gt0: a\le b+\epsilon \implies a\le b,$$ but why is the implication $$\forall \epsilon \gt0: a\le b+\epsilon \implies a\lt b \tag{1}$$ false? What's the general strategy for showing implications like $(1)$ are false?

S.H.W
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    The general strategy is to find a counterexample. Here, if we let $a=b=0$, then $a\leq b+\varepsilon$ for all $\varepsilon>0$, but we do NOT have $a<b$. – Feng Dec 27 '23 at 07:43
  • @Feng Do you even need to set $a=b=0$? It seems rather obvious that $a\le a+\varepsilon$ for all $\varepsilon\gt 0$ and all $a$ – Divide1918 Dec 27 '23 at 07:53
  • @Feng Thanks. I think the important detail was $\forall a,b\in \mathbb{R}$. So the complete implication is $$\forall a,b\in \mathbb{R}\ \ \forall \epsilon \gt0: a\le b+\epsilon \implies a\lt b,$$ right? – S.H.W Dec 27 '23 at 07:59
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    For two mathematical statements $P$ and $Q$, the general strategy for showing that $P \Rightarrow Q$ is false is to demonstrate a case where $P$ is true and $Q$ is false. – AlkaKadri Dec 27 '23 at 08:47
  • @AnneBauval Thanks for the comment. If you like you can add your comment as an answer so that I can accept it. – S.H.W Dec 27 '23 at 09:05
  • Well... okay. You have a $P \implies Q \text{ or } R$ and you want to know why $P \implies Q$ might not be true. Find a case where $P$ is true but $Q$ is not. As we know $P\implies Q\text{ or } R$ and we want $Q$ to be false so we will have $R$ true. So consider $a =b$. Then $a \le a+\epsilon$ for all $\epsilon$ and ... whoa! baby we are done. $a\le a + \epsilon$ for all $\epsilon > 0$ but $a \not < a$ so our hypothesis is false. – fleablood Jan 09 '24 at 21:56

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Summary of the various comments: $$\forall a,b\in \mathbb{R}\ \ [(\forall \epsilon \gt0: a\le b+\epsilon) \implies a\lt b]$$ is false "because" $$\exists a,b\in \mathbb{R}\ \ [(\forall \epsilon \gt0: a\le b+\epsilon) \text{ and } a\ge b]$$ is true. To prove it, take for instance $a=b=$ any number.

Anne Bauval
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Yes, you are righ about your intuition.

One logic observation: The $\varepsilon$ does not occur in $p: a<b$, therefore, the correct statement would be: $\forall \varepsilon >0, a\le b+\varepsilon \implies a\le b.$

Let's prove that $\forall \varepsilon >0, a\le b+ \varepsilon \implies a\le b$.

Proof: Suppose that $\forall \varepsilon >0, a\le b+\varepsilon$ but $a>b.$ Then $\exists p>0$ such that $a=b+p.$ Thus, $\forall \varepsilon >0, b+p\le b+ \varepsilon$, i.e. $\forall \varepsilon, p\le \varepsilon,$ ABSURD!

Gonçalo
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MATH VICTHOR
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Try to find a counterexample, so that the statement does NOT hold. If we plug in $0$ for $a$ and $b$, we see that $a < b + \varepsilon$ holds, whereas $a<b$ (that is: $0<0$) does not hold.

Alex M.
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alena2000
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