# AA epsilon>0,EEdelta in R,AAx in R(|x−c|<delta=>|f(x)−f(c)|<epsilon) is false?

Let the function $f:\mathbb{R}\to \mathbb{R}$ be discontinuous at c. Then the statement: $\mathrm{\forall }ϵ>0,\mathrm{\exists }\delta \in \mathbb{R},\mathrm{\forall }x\in \mathbb{R}\left(|x-c|<\delta \phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}|f\left(x\right)-f\left(c\right)|<ϵ\right)$ is false. The negation of the statement: $\mathrm{\exists }ϵ>0,\mathrm{\forall }\delta \in \mathbb{R},\mathrm{\exists }x\in \mathbb{R}\left(|x-c|<\delta \phantom{\rule{thickmathspace}{0ex}}and\phantom{\rule{thickmathspace}{0ex}}|f\left(x\right)-f\left(c\right)|⩾ϵ\right)$ is false because whenever $\delta$ is negative, $|x-c|<\delta$ is false. Is anything wrong here? Thank you!
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Your original statement is meant to express "continuity at $c$", but it's not written precisely enough! In fact, the original statement you gave is true, because you can pick $\delta =0$, regardless of the value of $ϵ$.
Here's the corrected version of the original statement:
$\mathrm{\forall }ϵ>0\phantom{\rule{thickmathspace}{0ex}}\mathrm{\exists }\delta >0\phantom{\rule{thickmathspace}{0ex}}\mathrm{\forall }x\in \mathbb{R}\left(|x-c|<\delta \phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}|f\left(x\right)-f\left(c\right)|<ϵ\right)$
This statement is actually false (by definition), assuming that $f$ is not continuous at $c$. The negation is

This statement is actually true.