What is the difference between $\implies$ and $\land$ in definition of continuity

Question

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 $\exists x, x_o \in D, |x-x_o| < \delta \land |f(x) - f(x_o)| \geq \epsilon$

Answer 1

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.

Answer 2

$\forall \epsilon > 0, \exists \delta$ such that $\forall x, x_o \in D, |x-x_o| < \delta \implies |f(x) - f(x_o)| < \epsilon$

Says, "For any $\epsilon$, there must be a delta such that if any 2 points, $x$ and $x_0$ are 'close' (less than $\delta$ apart), then their images, $f(x), f(x_0)$, must also be 'close' (less than $\epsilon$ apart)." In less technical language, this means that for a function to be continuous you cannot have points which are some finite distance apart from each other in the domain have images in the codomain that are arbitrarily far apart. If you consider points in the domain that are arbitrarily far apart, then this definition has nothing to say about such points, because the antecedent of the 'if' is always false.

$\forall \epsilon > 0, \exists \delta$ such that $\forall x, x_o \in D, |x-x_o| < \delta \land |f(x) - f(x_o)| < \epsilon$

This says that in order for a function to be continuous, there must be a $\delta$ such that ANY two points in $D$ are within $\delta$ of each other (so $D$ must be bounded). Oh, and by the way the images of EVERY 2 points in $D$ must all be within $\epsilon$ of each other (for any $\epsilon$).

On this second definition, the only functions that are continuous are those that map the entire bounded set $D$ into a single point $y_0$.