Negating a statement: is there indeed a quantifier missing? 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?
 A: 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.
A: The original statement is
$$\forall \epsilon > 0\quad  \exists \delta >0 :\quad P(x,a)$$
and the negation is
$$\exists\epsilon > 0\quad  \forall \delta >0: \quad \lnot P(x,a)$$
A: 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! :)
A: 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).
