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Suppose $a$ is a real number and that $f(x)$ is a real-valued function. In calculus, the statement "$f(x)$ is continuous at a" means that for every positive real number $E$, there exists a positive real number $\sigma$ such that for every real number x in the open interval $(a-\sigma, a+\sigma)$ we have $|f(x)-f(a)| < E$. In plain English, what does it mean for a real-valued function $f(x)$ to not be continuous $a$?

I guess I need to negate $f(x)$ is continuous at $a$.

there exists a positive real number $\sigma$ such that for every real number $x$ in the open interval $(a-\sigma, a+\sigma)$ we have $|f(x)-f(a)| < E$.

Here is my negation. Does it correct? (This does sound weird to me)

For all positive real number, there exists a real number $x$ in the open interval $(a-\sigma, a+\sigma)$, we have $|f(x)-f(a)| \geq E$.

The Phenotype
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merry-go-round-one
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4 Answers4

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1) Well, If being continuous means that for all positive numbers $E$ there is something that is true, then then negation is:

It is not true that for all positive numbers there is something that true

Which means there must be some positive number $E$ where that thing is false.

2) The thing that was true was that there is some positive number $\sigma$ where something happens.

The negation if that there isn't any positive number where something happens.

That means for all positive numbers $\sigma$ that something never happens.

3) The thing that happens is that for all $x \in (a - \sigma,a + \sigma)$ that something can be said about the $x$.

If that is negated then it is not the case that for all $x \in (a - \sigma,a + \sigma)$ something can be said.

That means there is at least one $x \in (a-\sigma, a + \sigma)$ where that something isn't true.

4) Finally the something about $x$ is that $|f(x) - f(a)| < E$.

The negation of that is simple. $|f(x)-f(a)| \ge E$.

Soo.... putting in all together:

The exists some positive number $E$ where for every possible positive $\sigma$ there will always be some $x \in (a - \sigma,a + \sigma)\in (a - \sigma,a + \sigma)$ where $|f(x) -f(a)| \ge E$.

(Intuitively $f(x)$ "makes a jump" of size $E$ at $a$ and you can't find get close enough no matter how small a $\sigma$ you pick where all the $x$ within $\sigma$ have $f(x)$ closer to $f(a) $ than E$.)

In general "not all" = "there exists at least one" and "not there exists" = "for all".

fleablood
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The negation is correct. You switch the 'for all' to 'exists' and vice versa, leave the sets alone, and do the complement of the inequality sign.

Here is a link for negating inequalities

and here for negating quantifiers

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How about: For at least one positive real number $\epsilon$, there is no positive real number $\delta$, where $|f(x)-f(a)| \lt \epsilon$ for all $|x-a| \lt \delta$.

herb steinberg
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$f$ is not continuous in $a$ iff:

$\exists \sigma \in R,\space \exists E \in R, \space \forall x \in [a - \sigma, a+ \sigma]\setminus \lbrace{a}\rbrace \space |f(x)- f(a)| \geq E$ .

Which barely translate as "somewhere near $a$, you cannot have f(x) as near of f(a) as you would like".

Tal Singrist
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