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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:

  1. [$\forall \epsilon > 0$] 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?)

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

  3. [$\forall x$] 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?)

  4. [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.

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  • $\begingroup$ Points 1 and 2 are incorrect. $\forall \epsilon > 0$ becomes $\exists \epsilon > 0$, and $\exists \delta > 0$ becomes $\forall \delta > 0$. $\endgroup$
    – user169852
    May 28 '21 at 1:41
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    $\begingroup$ I'm voting to reopen this question because although the linked duplicate is similar, OP's question really pertains to why their negation was wrong as opposed to what the correct negation is. The linked question doesn't touch on OP's specific misunderstanding. To state briefly how to fix this in case it is not reopened, the definition of $\forall \varepsilon > 0 P(\varepsilon)$ is $\forall \varepsilon (\varepsilon > 0 \implies P(\varepsilon))$. Try negating this and you'll see why you get $\exists \varepsilon > 0 \neg P(\varepsilon)$. $\endgroup$ May 28 '21 at 2:05
  • $\begingroup$ @paul-blart-math-cop Although I fully agree with you, it looks like the link to the similar question disappeared on reopening. So I will give it again in the next comment. $\endgroup$
    – J.-E. Pin
    May 28 '21 at 3:29
  • $\begingroup$ A related question. $\endgroup$
    – J.-E. Pin
    May 28 '21 at 3:29
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Your argument is essentially correct except for Points 1 and 2, where there is a big misunderstanding, as correctly pointed out by paul blart math cop in his comment. I will try to expand his comment, to understand why you do not have to change inequalities at the beginning of the statement of continuity when you negate it. There is no magic, on the contrary it is in accordance with general logical rules.

In general, the negation of a statement of the form $\forall x A(x)$ ("every $x$ has the property $A$") is a statement of the form $\exists x \lnot A(x)$ ("at least one $x$ does not have the property $A$"), as correctly stated by the OP. And dually, the negation of $\exists x A(x)$ ("at least one $x$ has the property $A$") is $\forall x \lnot A(x)$ ("no $x$ has the property $A$").

The statement of continuity of a function $f$ at point $y$ is of the form $\forall \varepsilon > 0, P(\varepsilon)$, for some property $P$. What is the logical form $\forall \varepsilon > 0, P(\varepsilon)$? This is the point that the OP is missing. To correctly negate a statement of the form $\forall \varepsilon > 0, P(\varepsilon)$, we first have to understand its real logical form.

(Notation. In logic, $\to$ represents implication (if$\dots$ then $\dots$), and $\land$ represents conjunction (and).)

Writing $\forall \varepsilon > 0, P(\varepsilon)$ is just a shorthand for $\forall \varepsilon (\varepsilon > 0 \to P(\varepsilon))$, this is its real logical form. So, the negation of $\forall \varepsilon > 0, P(\varepsilon)$ is the negation of $\forall \varepsilon (\varepsilon > 0 \to P(\varepsilon))$ and hence: \begin{align} & \ \lnot \forall \varepsilon \, (\varepsilon > 0 \to P(\varepsilon)) \\ \equiv& \ \exists \varepsilon \, \lnot (\varepsilon > 0 \to P(\varepsilon)) &&\text{(according to the law of negation of quantifiers seen above)} \\ \equiv& \ \exists \varepsilon \, (\varepsilon > 0 \land \lnot P(\varepsilon)) &&\text{(according to the law of negation of implication, as correctly explained in OP's Point 4)} \end{align}

The last statement is usually shortened as $\exists \varepsilon > 0, \lnot P(\varepsilon)$, where $P(\varepsilon)$ is a statement of the form $\exists \delta > 0, Q(\varepsilon, \delta)$, whose real logical form is $\exists \delta (\delta > 0 \land Q(\varepsilon, \delta))$. Starting from $\lnot P(\varepsilon)$, you can iterate the process of internalizing negation (according to the laws of negation for quantifiers and connectives). Finally, you get the statement \begin{align} \exists \varepsilon > 0, \forall \delta > 0, \exists x \text{ such that } |x -y|< \delta \text{ and } |f(x) - f(y)| \geq \varepsilon \end{align} whose real logical form is \begin{align} \exists \varepsilon \Big(\varepsilon > 0 \land \forall \delta \big(\delta > 0 \to \exists x \, (|x -y|< \delta \land |f(x) - f(y)| \geq \varepsilon) \big) \Big) \end{align}

Aside remark. There is also a little mistake in OP's Point 4. The negation of $|f(x) - f(y)| < \varepsilon$ is $|f(x) - f(y)| \not< \varepsilon$. Since the order $>$ on $\mathbb{R}$ is total (that is, for all $x,y \in \mathbb{R}$, either $x < y$ or $x > y$ or $x = y$), claiming that $|f(x) - f(y)| \not< \varepsilon$ is equivalent to $|f(x) - f(y)| \geq \varepsilon$, and not to $|f(x) - f(y)| > \varepsilon$, as wrongly written in Point 4 by the OP.

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  • $\begingroup$ Why did I receive one downvote? Just to understand. $\endgroup$ May 28 '21 at 6:34
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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.

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