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 $|f(x)-l|<\varepsilon,$


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!


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.


3 Answers 3


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


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


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


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'.

  • $\begingroup$ So again - the truth operators are working from left to right? $\endgroup$
    – bryanj
    Jul 19, 2013 at 4:16
  • $\begingroup$ Negating the "highest level" can correspond to negations at "lower levels" in the manner that Alraxite and Robert and Peter have already mentioned. Because of the structure of this example, it looks like the negation is churning through the sentence left-to-right, toggling $\forall$s and $\exists$s back and forth. $\endgroup$ Jul 19, 2013 at 4:25
  • $\begingroup$ @bryanj If you want to find the negation of a statement like $\forall x\,\, \exists y\,\, \forall z\,\, P(x,y,z)$ where $P(x,y,z)$ is a statement about $x,y$ and $z$, you can repeatedly use $\neg\forall x\,\, P(x)\iff\exists x\,\, \neg P(x)$ and $\neg\exists x\,\,P(x) \iff \forall x\,\,\neg P(x)$ to arrive at its negation. So $\neg(\forall x\,\, \exists y\,\, \forall z\,\, P(x,y,z)) \iff \exists x\,\, \neg(\exists y\,\, \forall z\,\, P(x,y,z))\iff \exists x\,\, \forall y\,\, \neg(\forall z\,\, P(x,y,z))\iff \exists x\,\, \forall y\,\, \exists z\,\, \neg P(x,y,z).$ $\endgroup$
    – Alraxite
    Jul 19, 2013 at 4:36
  • $\begingroup$ @bryanj So, it seems that the negation operator moves from left to right. Also for clarity: $\forall x\,\, \exists y\,\, \forall z\,\, P(x,y,z)$ is the same thing as saying $\forall x\,\, (\exists y\,\, (\forall z\,\, P(x,y,z)))$ $\endgroup$
    – Alraxite
    Jul 19, 2013 at 4:38
  • $\begingroup$ +1 for really great explanation. This is exactly how complex logic inferences can be handled. $\endgroup$
    – Paramanand Singh
    Jul 19, 2013 at 11:37

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 $a\geq b$)

  • $\begingroup$ Thank you very much. It seems to me that we are negating logical operators from left to right? Are we allowed to do it this way? And if so, why? $\endgroup$
    – bryanj
    Jul 19, 2013 at 3:51
  • $\begingroup$ Downvote? Maybe a logician angry at my sloppyness? $\endgroup$
    – Pedro
    Jul 19, 2013 at 4:40
  • $\begingroup$ @bryanj See what Alraxite commented in his answer. It is all a matter of following certain rules of logic and its sentences. The answer is yes, we put a negation sign in the left and work ourselves into the sentence. $\endgroup$
    – Pedro
    Jul 19, 2013 at 4:43

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.


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