How shall we prove that the converse statement of "epsilon-delta definition of continuity" is true?

Question

The epsilon-delta definition of continuity:

A function $f(x)$ from $\mathbf{R}$ to $\mathbf{R}$ is continuous at point $x_0 \in \mathbf{R}$ if for every $\epsilon > 0$ there exists a $\delta > 0$ such that whenever $|x–x_0| < \delta$ then $|f(x)–f(x_0)| < \epsilon$

Now from this, how shall we deduce that the converse statement is true?

Converse statement:

If a function $f(x)$ from $\mathbf{R}$ to $\mathbf{R}$ is continuous at point $x_0 \in \mathbf{R}$, then for every $\epsilon > 0$ there exists a $\delta > 0$ such that whenever $|x–x_0| < \delta$ then $|f(x)–f(x_0)| < \epsilon$

Answer 1

Long comment

See T.Tao, Analysis, I, page 227 :

Definition 9.4.1 (Continuity). Let $X$ be a subset of $\mathbb R$, and let $f : X → \mathbb R$ be a function. Let $x_0$ be an element of X. We say that $f$ is continuous at $x_0$ iff we have :

$\lim_{x → x_0;x∈X} f(x) = f(x_0).$

Thus, there is nothing to prove.

Answer 2

This is not the inverse statement. You need to change the quantifiers as well. That is if the original statement had an "for every" or "for all" the negation would have "there exists" and vice versa. Moreover if you have a statement of the form $A$ implies $B$ the negation would be $A$ and (the inverse of $B$). So for instance the negation of "$|x-x_0|<\delta\Rightarrow |f(x)-f(x_0)|<\varepsilon$" is "$|x-x_0|<\delta$ and $|f(x)-f(x_0)|\geq\varepsilon$".

In other words the negation of

"for every $\varepsilon>0$ there exists $\delta>0$ such that whenever $|x-x_0|<\delta$ then whenever $|x-x_0|<\delta$ we have $|f(x)-f(x_0)|<\varepsilon$"

would be

"There exists $\varepsilon>0$ such that for all $\delta>0$ there exists $x$ such that $|x-x_0|<\delta$ and $|f(x)-f(x_0)|\geq \varepsilon$."

Tip: It is the best to write the statement with the quantifiers $\forall$ and $\exists$ instead of words. That way it is much easier to replace each $\forall$ with $\exists$ and vice-versa.