TITLE: Negation of the definition of continuity
QUESTION [5 upvotes]: I would like to find the negation of this statement of continuity of a function $f \colon \mathbb{R} \to \mathbb{R}$ at a point $y$:

$\forall \epsilon > 0$, $\exists \delta > 0$, such that $\forall x$ if |$x-y$| < $\delta$ then |$f(x)- 
    f(y)$| < $\epsilon$

So, when considering the negation of this statement I would want to consider negating the following components of the statement:

[$\forall \epsilon > 0$] becomes [$\exists \epsilon < 0$]? (I am a little confused because not($\forall x, P(x)$ becomes $\exists \epsilon$ such that not$(P(x))$, but would the statement that $\epsilon > 0$ be considered the predicate P($\epsilon$) in this instance? If it is then I assume what I wrote above is correct then?)

[$\exists \delta > 0$] becomes [$\forall \delta < 0$] (similar reasoning as above)

[$\forall x$] becomes [$\exists x$] (here I am a little confused because I see the universal quantifier and want to negate it into the existential quantifier but the definition of the negation of the quantifiers includes a not statement of the predicate, but there is no predicate in this component. $x$ is used in a subsequent implication, which I am assuming is a separate negation on its own. Is this correct thinking?)

[if $|x-y|$ < $\delta$ then $|f(x)-f(y)| < \epsilon$] would become [$|x-y| < \delta$ and $|f(x)-f(y)| > \epsilon$]. Here I am just using the definition of the negation of the implication, which is $\text{not}(A \Rightarrow B) = (A \text{ and not}(B))$.


so that my negation of the statement of continuity here is:

$\exists \epsilon < 0, \forall \delta < 0$ such that $\exists x,  |x-y| < \delta$ and $|f(x)-f(y)| > \epsilon$

Looking at the statement as a whole it seems like the first two components should only have their quantifiers changed and not their inequalities, but the rest of the statement seems to make sense to me as describing a point of discontinuity  on the function. Is my reasoning here correct? Or am I missing something entirely.
I appreciate any help with this problem.

REPLY [2 votes]: You're quite correct that the negation of $\ \forall  x,P(x)\ $ is $\ \exists x,\neg P(x)\ $, and that $\ \epsilon\ $ is a free variable in the predicate you're quantifying over.  Strictly speaking, $\ y\ $ is also a free variable in that predicate, so it should be $\ P(\epsilon,y)\ $ rather than just $\ P(\epsilon)\ $, but there's no harm here treating $\ y\ $ as if it were a constant rather than a free variable and referring to it as just $\ P(\epsilon)\ $ as you've done.
Expanding on paul blart math cop's comment, the expression $\ \forall \epsilon>0,P(\epsilon)\ $ is synonymous with $\ \forall \epsilon, (\epsilon>0)\Rightarrow P(\epsilon)\ $, and $\ \exists\epsilon>0,P(x)\ $ is synonymous with $\ \exists\epsilon, (\epsilon>0)\wedge P(x)\ $.  So applying the rule you know to the first expression, you get
\begin{align}
\neg\forall\epsilon>0,P(\epsilon)&\equiv\neg\forall \epsilon, (\epsilon>0)\Rightarrow P(\epsilon)\\
&\equiv\exists\epsilon,\neg\big((\epsilon>0)\Rightarrow P(\epsilon)\big)\\
&\equiv\exists\epsilon,\neg\big(\neg(\epsilon>0)\vee P(\epsilon)\big)\\
&\equiv\exists\epsilon,\neg\neg(\epsilon>0)\wedge \neg P(\epsilon)\\
&\equiv\exists\epsilon,(\epsilon>0)\wedge \neg P(\epsilon)\ ,
\end{align}
from De Morgan's laws and the definition of material implication.