The negation of a limit condition, Spivak style.

Question

I got stuck on this bit from Spivak's Calculus:

When proving that $f$ does not approach $l$ at $a$, be sure to negate the definition correctly:

If it is not true that

for every $\varepsilon>0$ there is some $\delta>0$ such that, for all $x$, if $0<|x-a|<\delta$, then $|f(x)-l|<\varepsilon,$

then

there is some $\varepsilon>0$ such that for every $\delta>0$ there is some $x$ which satisfies $0<|x-a|<\delta$ but not $|f(x)-l|<\varepsilon.$

I don't understand why the negation of the top sentence is given by the bottom sentence.

Thank you so much!

EDIT

My problem is not why "continuous functions are those that let you interchange limits". It's more just a question about how you can negate sentences in general.

Answer 1

Break it down:

The statement is, 'for all $\epsilon>0$, $Q$ must be true'. Where $Q$ stands for the statement 'there is some $\delta>0$ such that, for all $x$, if $0<|x-a|<\delta$, then $|f(x)-l|<\varepsilon$'.

Clearly the negation of this is: 'there is some $\epsilon>0$ for which $Q$ is not true'.

What is the negation of $Q$? The statement $Q$ is, 'there is some $\delta>0$ such that $R$ is true' where $R$ is the rest of the statement. The negation of $Q$ is then, 'there is no $\delta>0$ such that $R$ is true' or equivalenty, 'for all $\delta>0$, $R$ is not true'.

What is the negation of $R$? The statement $R$ is, 'for all $x$, $S$ is true' where $S$ is the rest of the statement. Its negation clearly is, 'there is some $x$ for which $S$ is not true'.

Finally, the statement $S$ is, if $0<|x-a|<\delta$, then $|f(x)-l|<\epsilon$. How can this be false? This can be false only when $0<|x-a|<\delta$ is true and at the same time $|f(x)-l|<\epsilon$ is not true.

Put it all together:

There is some $\epsilon>0$ for which $Q$ is not true

OR

There is some $\epsilon>0$ such that for all $\delta>0$, $R$ is not true

OR

There is some $\epsilon>0$ such that for all $\delta>0$, there is some $x$ for which $S$ is not true.

OR

There is some $\epsilon>0$ such that for all $\delta>0$, there is some $x$ for which $0<|x-a|<\delta$ but not $|f(x)-l|<\epsilon$.

---

I suggest that you read a book about discrete mathematics. It will explain you how to write these statements using logical symbols and how to easily and immediately negate them. Your statement in logical symbols is: $$\forall \epsilon >0\,\,\exists \delta >0\,\, \forall x(0<|x-a|<\delta \implies |f(x)-l|<\epsilon)$$ Using facts such as: $\neg\forall x\,\, P(x)$ is equivalent to $\exists x\,\, \neg P(x)$, $\neg\exists x\,\,P(x)$ is equivalent to $\forall x\,\,\neg P(x)$ and $\neg(P \implies Q$) is equivalent to $P \land \neg Q$, you can easily arrive at the negation of the above statement which is, in logical symbols:

$$\exists \epsilon >0\,\,\forall \delta >0\,\,\exists x\,\,(0<|x-a|<\delta \land |f(x)-l|\geq \epsilon)$$

The symbols $\forall$, $\exists$, $\land$, $\implies$ and $\neg$ respectively mean, 'for all', 'there exists', 'and', 'implies' and 'negation of'.

Answer 2

First, look at why the negation of "for all" is "there exists". How can the statement that says "for all $x$, it is true that $P(x)$" fail? Well, it suffices that I show you one $x$ such that $P(x)$ is false, that is, a so called counterexample. And this means that $$\forall x:P(x)$$ is negated as $$\exists x:\neg P(x)$$

Now, there is another part of the limit definition that uses another logical symbol, namely $$|x-a|<\delta \implies |f(x)-f(a)|<\epsilon$$

This is a statement of the form $P\to Q$. And how can this be false? We're saying that $P$ implies $Q$. Note that then if $P$ is true but $Q$ is false, $P\to Q$ does not hold. Also, it is a nice exercise to see that $P\to Q$ is the same as $\neg P\text{ or } Q$. Thus, we see that the negation of $P\to Q$ is $P\text{ and } \neg Q$, that is $P$ holds, yet $Q$ doesn't. Now, the statement of continuity is $$\forall \epsilon >0,\exists \delta >0(\forall x,|x-a|<\delta \implies |f(x)-f(a)|<\epsilon)$$

"For each $\epsilon >0$ there exists $\delta>0$ such that for every $x$, $|x-a|<\delta$ implies $|f(x)-f(a)|<\epsilon$."

Thus, we can negate this as $$\exists \epsilon >0,\forall \delta >0(\exists x,|x-a|<\delta \text{ yet }|f(x)-f(a)|\geq \epsilon)$$ "There exists $\epsilon>0$ such that for any $\delta >0$ there exists $x$ with $|x-a|<\delta$ yet $|f(x)-f(a)|\geq \epsilon$".

(Note that the negation of $a<b$ is $a\geq b$)

Answer 3

The first one, to be true, has to work for every epsilon. As such, a single epsilon that doesn't work will disprove it: if even one counterexample exists, the top statement can't be true.

Similarly, for the top statement to work I just have to be able to provide some delta that does the job. Others might not, but I only need to reliably provide one to make my case. So the negation means that I can't find any at all.

The negation of "there is at least one that works" is "there are none that work", and the negation of "it works for all" is "it doesn't work for at least one". To say otherwise (i.e. "it works for all" becomes "it works for none") would leave out the cases where some work and others don't. It's like claiming that the opposite of OR is AND, rather than NOR. You miss out on the middle cases and don't end up with anything logically consistent.