Real Function is Continuous at Point iff Oscillation is Zero

Theorem
Let $f: D \to \R$ be a real function where $D \subseteq \R$.

Let $\omega_f$ be the oscillation of $f$, that is:


 * $\omega_f \left({I}\right) = \sup \left\{{\vert f \left({x}\right) - f \left({y}\right) \vert: x, y \in I}\right\}$


 * $\omega_f \left({x}\right) = \inf \left\{{\omega_f \left({I}\right): x \in I}\right\}$

Then $\omega_f \left({x}\right) = 0$ iff $f$ is continuous at $x$.

Necessary Condition
Suppose $\omega_f \left({x}\right) = 0$.

Let $\epsilon > 0$.

We have that $\exists I$ such that $x \in I$ and $\omega_f \left({I}\right) < \epsilon$.

Let $\delta = \sup \left\{{\left \vert {u - v} \right \vert: u, v \in I}\right\}$

So if:
 * $\left \vert {x - y} \right \vert < \delta \implies x, y \in I$

then:
 * $\left \vert {f \left({x}\right) - f \left({y}\right)} \right \vert < \sup \left\{{\left \vert {f \left({x}\right) - f \left({y}\right)} \right \vert: x, y \in I}\right\} = \omega_f \left({x}\right) = 0 < \epsilon$

So from the definition of continuity, we have that $f$ is continuous at $x$.

Sufficient Condition
Suppose $f$ is continuous at $x$.

Then $\forall \epsilon > 0: \exists \delta \in \R$ such that:
 * $\left \vert x-y \right \vert < \delta \implies \left \vert f \left({x}\right)-f \left({y}\right) \right \vert < \epsilon$

Let the interval $I_\delta$ be defined as:
 * $I_\delta := \left({x - \delta . . x + \delta}\right)$

Then:

This holds true for any value of $\epsilon$.

Thus $\omega_f \left({x}\right)$ must be $0$.

Hence the result.