Is there anything wrong with this definition of discontinuity? Is there anything wrong with this definition of discontinuity for a function y = f(x)?
$\forall \delta>0\, \exists \varepsilon>0$ such that $\vert x-c\vert < \delta$, but $\vert f(x) - f(c)\vert > \epsilon$.
 A: If you take the usual definition of continuity and negate it, you get
$$
\exists \epsilon > 0 \forall \delta > 0 \exists c\in \mathbb{R}:     |x - c| < \delta \text{ and } |f(x) - f(c)| > \epsilon.
$$
Your definition is missing quantification on $c$, but more importantly, you've switched the orders of the quantifiers on $\epsilon$ and $\delta$, which is not legitimate. As explained in the comments, with the quantifiers reversed, your definition is implied by but does not imply discontinuity.
A: EDIT:  the answer below is incorrect, but I'm leaving it up to show the possible pitfalls of alternative approaches.  If time allows, I'll try to refine and correct it at a later date.

It's almost true;  it needs a slight refinement:

$f(x)$ is discontinuous at $x$ when $\forall \delta > 0, \exists \epsilon > 0$ such that for all $x$ satisfying $0 < |x - c| < \delta$, $|f(x) - f(c)| > \epsilon$.

Without the "for all" I've bolded above, it doesn't work. For example, take $f(x) = x$ and $c = 0$.  For a given $\delta > 0$, choose $\epsilon = \delta/2$.  Then for at least some $x$ such that $|x| < \delta$, $|f(x)| = x \ngtr \epsilon = \delta /2$.
A: Here's one negation for the definition of continuity
$\exists \varepsilon>0:\nexists\delta>0$ for which $|x-c|<\delta\implies|f(x)-f(c)|<\varepsilon$ holds.
A: As explained in the comments

This is actually a definition of not being locally constant. – Rahul  Feb 4 at 17:49 

The proposed definition
$\forall \delta>0\, \exists \varepsilon>0$ such that $\vert x-c\vert < \delta$, but $\vert f(x) - f(c)\vert > \epsilon$
is equivalent to
$\forall \delta>0\, \exists c$ such that $\vert x-c\vert < \delta$, but $f(x) \neq f(c)$,
which is the statement that the function is non-constant on arbitrarily small neighborhoods of $x$.  Requiring the same condition for all $x$ is the statement that the function is not locally constant near any point.
