TITLE: The negation of a limit condition, Spivak style.
QUESTION [8 upvotes]: 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.

REPLY [22 votes]: 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'.