Showing that an implication is false

Question

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?

Answer 1

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.

Answer 2

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!

Answer 3

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.