Condition for Continuity on Interval

Theorem
Let $$f$$ be a real function defined on an interval $$\mathbb{I}$$.

Then $$f$$ is continuous on $$\mathbb{I}$$ iff $$\forall x \in \mathbb{I}: \forall \epsilon > 0: \exists \delta > 0: y \in \mathbb{I} \land \left|{x - y}\right| < \delta \Longrightarrow \left|{f \left({x}\right) - f \left({y}\right)}\right| < \epsilon$$.

Proof
Let $$x \in \mathbb{I}$$ such that $$x$$ is not an end point.

Then the condition $$y \in \mathbb{I} \land \left|{x - y}\right| < \delta$$ is the same as $$\left|{x - y}\right| < \delta$$ provided $$\delta$$ is small enough.

The criterion given therefore becomes the same as the statement $$\lim_{y \to x} f \left({y}\right) = f \left({x}\right)$$, that is, that $$f$$ is continuous at $$x$$.

Now suppose $$x \in \mathbb{I}$$ and $$x$$ is a left hand end point of $$\mathbb{I}$$.

Then the condition $$y \in \mathbb{I} \land \left|{x - y}\right| < \delta$$ reduces to $$x \le y < x + \delta$$ provided $$\delta$$ is small enough.

The criterion given therefore becomes the same as the statement $$\lim_{y \to x^+} f \left({y}\right) = f \left({x}\right)$$, that is, that $$f$$ is continuous on the right at $$x$$.

Similarly, if $$x \in \mathbb{I}$$ and $$x$$ is a left hand end point of $$\mathbb{I}$$, then the criterion reduces to the statement that $$f$$ is continuous on the left at $$x$$.

Thus the assertions are equivalent to the statement that $$f$$ is continuous at all points in $$\mathbb{I}$$, that is, that $$f$$ is continuous on $$\mathbb{I}$$.