The exercise is as follows: Given a function $f : D \mapsto \mathbb R$ Are the following conditions stronger, weaker or not comparable with continuity?
$$\forall a \in D, \exists \epsilon \gt 0, \exists \delta \gt 0, \forall x \in D: |x-a| \lt \delta \Longrightarrow |f(x) - f(a)|\lt \epsilon $$
The solution that was given is:
It is weaker because $\forall \epsilon \gt 0 P(\epsilon) \Rightarrow \exists \epsilon \gt 0 : P(\epsilon)$
As an example $f(x)= \begin{cases} 0 & x \ne 0\\ 1 & x = 0 \end{cases}$
Now (c), because $\epsilon = 2, a\in D, \delta = 1, \forall x \in D : |x-a| \lt 1 \Longrightarrow |f(x) -f(a)|\le 1 \lt 2 $
What I understand from this "solution" is that because of the quantifier $\exists$ is weaker than the quantifier $\forall$ the statement is weaker. Ok, from here on I don't understand it.
The example is a discontinuous function, ok. But I don't see the connection between the statement and the example. At least I'm not 100 % sure of it because I don't see why the statement suggests anything discontinuous.