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? 
 A: 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
$$\neg \forall \epsilon > 0\; \exists \delta > 0\; \forall x\;\big( |x-a|<\delta \Rightarrow |f(x)-f(a)|<\epsilon \big)$$
$$\exists \epsilon > 0\; \neg\exists \delta > 0\; \forall x\;\big( |x-a|<\delta \Rightarrow |f(x)-f(a)|<\epsilon \big)$$
$$\exists \epsilon > 0\; \forall \delta > 0\; \neg\forall x\;\big( |x-a|<\delta \Rightarrow |f(x)-f(a)|<\epsilon \big)$$
$$\exists \epsilon > 0\; \forall \delta > 0\; \exists x\;\neg\big( |x-a|<\delta \Rightarrow |f(x)-f(a)|<\epsilon \big)$$
$$\exists \epsilon > 0\; \forall \delta > 0\; \exists x\;\big( |x-a|<\delta \land |f(x)-f(a)|\ge\epsilon \big)$$
A: 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 $a$". 
