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There is this homework assignment that I seem to keep getting wrong. The question is:

Negate the following statement:

"For every positive number $\epsilon$, there is a positive number $\delta$ such that |x-a| < $\delta$ implies |f(x)-f(a)| < $\epsilon$".

My answer was:

"There exists a positive number $\epsilon$, such that for every positive number $\delta$, there exists an x such that |x-a| < $\delta$ and |f(x)-f(a)| $\geqslant \epsilon$".

In math symbols this is

$\exists \epsilon > 0, \forall \delta >0, \exists x, (|x-a|<\delta) \wedge(|f(x)-f(a)|\geqslant\epsilon)$

But then the answer I got back from my teacher, was "for which x,a? Is this true for all x,a such that |x-a|<$\delta$ or just for one set of x.a?"

I just can't seems to figure out exactly what I'm missing. A quantifier for a?

Can anybody help me, please?

Caroline
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  • In the first statement, there is no quantifier for $a$. So $a$ is a free variable in that sentence. In its negation $a$ should be still a free variable. – ajotatxe Oct 19 '18 at 12:49
  • @Caroline You negated the statenent. It seems from their comment that the wanted you yo do more, that is to specify $x $ and $a $. Was the function given? Did you know which $a $ it was? – AnyAD Oct 19 '18 at 12:52
  • Why did you write $\exists x$ but not $\exists x,a$ in your negation? What made you treat $x,a$ differently? – Michal Adamaszek Oct 19 '18 at 12:54
  • See Continuous function : definition. – Mauro ALLEGRANZA Oct 19 '18 at 13:16
  • In the def you have missed "for every $x$": this is why you need $\exists x$ in the negated version. – Mauro ALLEGRANZA Oct 19 '18 at 14:39
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    I don't understand your teacher's note either. They seem to be asking you for more quantification of $x$ and $a,$ but the fact is that you have given too much quantification. Also, it is curious that the statement you are supposed to negate is so close to the definition of "$f$ continuous at $a,$" yet is not the correct definition. My question would be, was the quantifier "for all $x$" left out of the problem statement on purpose? – David K Oct 19 '18 at 15:13

4 Answers4

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A function $f:D\to\Bbb R$ is continuous at $a\in D$ if $$\forall \epsilon\in\Bbb R^+~\exists \delta\in\Bbb R^+~\forall x\in D~~(\lvert x-a\rvert<\delta~\to~\lvert f(x)-f(a)\rvert <\epsilon)\tag 1$$

So a function $f:D\to\Bbb R$ is discontinuous at $a\in D$ if $$\exists \epsilon\in\Bbb R^+~\forall \delta\in\Bbb R^+~\exists x\in D~~(\lvert x-a\rvert<\delta~\land~\lvert f(x)-f(a)\rvert \geqslant\epsilon)\tag 2$$

However, while (2) is the negation of (1), (1) is not the statement which you quoted:$$\forall \epsilon\in\Bbb R^+~\exists \delta\in\Bbb R^+~~(\lvert x-a\rvert<\delta~\to~\lvert f(x)-f(a)\rvert <\epsilon)\tag 3$$

$x$ and $a$ are free variables in that statement. They should still be free variables in its negation.

Graham Kemp
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The original statement is

$$\forall \epsilon > 0\quad \exists \delta >0 :\quad P(x,a)$$

and the negation is

$$\exists\epsilon > 0\quad \forall \delta >0: \quad \lnot P(x,a)$$

user
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I could be mistaken, but I think your confusion lies in the following.

You have "... $\delta$ such that $|x - a| < \delta$ implies $|f(x) - f(a)| < \epsilon$". Implicitly, this means "... $\delta$ such that, for all $x$, ($|x - a| < \delta \implies |f(x) - f(a)| \le \epsilon$)" -- note the additional "for all $x$" that wasn't there before. This is definition of "continuous at $a$". That's why in the negation of "continuous at $a$" you have a "there exists $x$" but nothing to do with $a$; $a$ is some fixed number at the start.

In general (but not always), if one says something like this with no quantifier for $x$, it means "all $x$ which satisfy $\texttt{statement1}$, we have $\texttt{statement2}$", which in this case is "for all $x$ which satisfy $|x - a| < \delta$, we have $|f(x) - f(a)| < \epsilon$.

Other answers show the correct solution, but hopefully this helps you understand why you were confused! :)

Sam OT
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As several people have commented: did you include the complete question? Was there anything else about $x$ and $a$? Which are fixed, and which are free variables? e.g. if $a$ were fixed but $x$ could vary, it would be the definition of $f$ being continuous at $a$. But if $x$, $a$ were both free variables, it would be the definition of $f$ being uniformly continuous on whatever set it's defined on. These are two very different things, and there is simply no way to know which is intended without more information about $a$.

(Of course, most beginners find uniform continuity very confusing, so I'm guessing that you wouldn't have covered that yet, and so continuity at the point $a$ is intended, in which case the question should have said that $a$ is fixed at the beginning).

Dr Z
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  • The question is from the book "Book of Proof" by Richard Hammack. (It is available as a PDF on the internet.) It is from chapter 2, section 10, question 4. That what I have written in the original post, is all there is to the question. No further info on quantifiers is given. – Caroline Oct 20 '18 at 13:04