What is the difference between $\implies$ and $\land$ in definition of continuity Can someone please explain to me what would happen if instead:
Given $D \subset \mathbb{R}$
$\forall \epsilon > 0, \exists \delta$ such that $\forall x, x_o \in D, |x-x_o| < \delta \implies |f(x) - f(x_o)| < \epsilon$
I wrote:
$\forall \epsilon > 0, \exists \delta$ such that $\forall x, x_o \in D, |x-x_o| < \delta \land |f(x) - f(x_o)| < \epsilon$?
And also, why is it that we can use $\land$ in the negation of continuity but not $\implies$?
Update: Negation
$\exists \epsilon > 0, \forall \delta > 0$ such that $\exists x, x_o \in D, |x-x_o| < \delta \land |f(x) - f(x_o)| \geq \epsilon$

It is not necessarily the case that there exists a $\delta$ such that every $x,x_0$ you have $|x-x_0|<\delta$.  In particular if the space you are working in is infinite in measure.  The "and" of the implication will fail because the first part is untrue when $|x-x_0|\geq \delta$ (which will happen in particular for $x=\delta+x_0$)
For trivial example, consider the constant function $f(x)=0$.
Although it is always true that for any $\epsilon>0$ you have $|f(x)-f(x_0)|<\epsilon$  (it will be identically zero), it will not be true that there exists a $\delta$ such that for every $x,x_0$ you have $|x-x_0|<\delta$
Suppose there was such a delta.  Then that would imply that for every $x,x_0$ you have $|x-x_0|<\delta$.  But when $x=\delta+x_0$, one has $|x-x_0|=|\delta+x_0-x_0|=|\delta|\not\lt \delta$, a contradiction.
Since the first part of the and statement is false, the entire and statement is false for that choice of $x,x_0$.  Since this is false for all $\delta$, the entire statement is false.
