# Quantification notation and discontinuous function

I have to describe what it means, when a function is discontinuous in a. I also have to use quantification notation.

I have tried following:

$$\exists \varepsilon > 0, \exists \delta > 0 : |x-a| < \delta \Rightarrow |f(x)-f(a)| \geq \varepsilon$$

But when I went to the classes they used a different notation. But is this wrong?

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Unfortunately it's wrong, irrespective of the notation. The function $$f(x) = \begin{cases} 0 & x < 0 \\ 1 & x \geq 0 \end{cases}$$ is certainly discontinuous at $0$. – Dylan Moreland Sep 7 '11 at 16:17
In essence, this is the same question as this one and my answer fits here as well. In this question $P(x,y)=|x-y|<\delta$ and $Q(x,y)=|f(x)-f(y)|<\varepsilon$ (these are the notations from the linked question, of course). – Asaf Karagila Sep 7 '11 at 16:22

I'm afraid that your negation of continuity is incorrect. First, let us remember the definition of "continuous at $a$":

$\forall \epsilon\gt 0\Biggl( \exists \delta\gt 0\biggl(\forall x\bigl( |x-a|\lt \delta \Rightarrow |f(x)-f(a)|\lt \epsilon\bigr)\biggr)\Biggr)$.

Now, how do we write down the negation? (I'm assuming that "discontinuous at $a$" means "defined at $a$ but not continuous, by the way)?

Let's break it down: what is the negation of a "For all" statement?

The negation of $\forall x (P(x))$ is $\exists x(\neg P(x))$.

That is, the negation of "For every $x$, $P(x)$ is true" is "There is at least one $x$ for which $P(x)$ is false". So, the negation of the definition of continuity will be:

$\exists \epsilon\gt 0\Biggl(\neg\Biggl(\forall \delta\gt 0 \biggl(\forall x\bigl( |x-a|\lt \delta \Rightarrow |f(x)-f(a)|\lt \epsilon\bigr)\biggr)\Biggr)\Biggr)$.

Now we need to find the negation of the statement inside the "For all", which is a statement of the form "There exists..." What is the negation of a "There exists" statement?

The negation of $\exists y(Q(y))$ is $\forall y(\neg Q(y))$.

That is: the negation of "There exists a $y$ such that $Q(y)$ is true" is "For every $y$, $Q(y)$ is false". So, to negate that "There exists" statement, we have:

$\exists \epsilon\gt 0\forall\delta\gt 0\Biggl(\neg\Biggl(\forall x\bigl( |x-a|\lt \delta\Rightarrow |f(x)-f(a)|\lt \epsilon\bigr)\Biggr)\Biggr)$.

Now we need to negate a "for all" statement. We already know how to do that:

$\exists \epsilon\gt 0 \forall\delta\gt 0 \Biggl( \exists x\Bigl( \neg\bigl( |x-a|\lt\delta\Rightarrow |f(x)-f(a)\lt\epsilon\bigr)\Bigr)\Biggr)$.

Now we need to find the negation of the implication. What is the negation of an implication?

The negation of $A\Rightarrow B$ is $A\text{ and }\neg B$.

That is, the negation of "If $A$, then $B$" is "$A$, and not $B$." So:

$\exists \epsilon\gt 0 \forall \delta\gt 0\Biggl( \exists x\bigl( |x-a|\lt\delta \text{ and }\neg(|f(x)-f(a)|\lt\epsilon)\bigr)\Biggr)$.

Finally, we need to find the negation of "$|f(x)-f(a)|\lt\epsilon$". That's $|f(x)-f(a)|\geq \epsilon$. So we finally get:

$\exists \epsilon\gt 0\forall\delta \gt 0 \Biggl( \exists x\bigl( |x-a|\lt\delta\text{ and }|f(x)-f(a)|\geq \epsilon\bigr)\Biggr)$.

In words: "there is an $\epsilon\gt 0$ such that, no matter what $\delta\gt 0$ you pick, there is an $x$ which is $\delta$-close to $a$, but with $f(x)$ not $\epsilon$-close to $f(a)$."

Note. Some people say that $f$ is discontinuous at $a$ if and only if it is defined at $a$ but not continuous at $a$. Less common is to say that $f$ is discontinuous at $a$ if and only if it is not continuous at $a$. Under the latter (in my experience uncommon) definition, "$f$ is discontinuous at $a$" would be a disjunction between the formula above, and "$f$ is not defined at $a$".

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Thank you so much. Your explanation is really good, better than my teacher assistant's explanation :) – Brugerfugl Sep 7 '11 at 16:46

Your attempt is lacking a quantifier for $x$. It will be wrong no matter where you place it, but in different ways. What you should have done is take the definition for being continuous at a: $$\forall \epsilon > 0\; \exists \delta > 0\; \forall x\;\big( |x-a|<\delta \Rightarrow |f(x)-f(a)|<\epsilon \big)$$ and then negated that, moving the negation inwards $$\neg \forall \epsilon > 0\; \exists \delta > 0\; \forall x\;\big( |x-a|<\delta \Rightarrow |f(x)-f(a)|<\epsilon \big)$$ $$\exists \epsilon > 0\; \neg\exists \delta > 0\; \forall x\;\big( |x-a|<\delta \Rightarrow |f(x)-f(a)|<\epsilon \big)$$ $$\exists \epsilon > 0\; \forall \delta > 0\; \neg\forall x\;\big( |x-a|<\delta \Rightarrow |f(x)-f(a)|<\epsilon \big)$$ $$\exists \epsilon > 0\; \forall \delta > 0\; \exists x\;\neg\big( |x-a|<\delta \Rightarrow |f(x)-f(a)|<\epsilon \big)$$ $$\exists \epsilon > 0\; \forall \delta > 0\; \exists x\;\big( |x-a|<\delta \land |f(x)-f(a)|\ge\epsilon \big)$$

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