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 $x$ be a point in $D$.

Let $N_x$ be the set of neighborhoods of $x$ where we use the usual topology for the reals.

Let $\omega_f \left({x}\right)$ be the oscillation of $f$ at $x$:
 * $\omega_f \left({x}\right) = \displaystyle \inf \left\{ \omega_f \left({I}\right): I \in N_x \right\}$

where:
 * $\omega_f \left({I}\right) = \displaystyle \sup \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$

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

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

Let $\epsilon > 0$.

Suppose that $\forall I: I \in N_x: \omega_f \left({I} \right) \ge \epsilon$.

Then by definition, $\omega_f \left({x}\right) \ge \epsilon$.

This contradicts $\omega_f \left({x}\right) = 0$.

From this contradiction we deduce that:
 * $\exists I: I \in N_x: \omega_f \left({I}\right) < \epsilon$

For this particular $I$, there is an open set $O \subset I$ by the definition of neighborhood.

Therefore, a $\delta \in \R_{>0}$ exists such that $\left({x - \delta \,.\,.\, x + \delta}\right)$ is a subset of $O$.

So for our specific $x$, if $y$ satisfies:
 * $\left \vert {x - y} \right \vert < \delta$

then:
 * $y \in I$

and, if $y \in D$:
 * $\left\vert {f \left({x}\right) - f \left({y}\right)} \right\vert \le \omega_f \left({I}\right)$

Since $\omega_f \left({I}\right) < \epsilon$ it follows by the definition of continuity that $f$ is continuous at $x$.

Sufficient Condition
Let $f$ be continuous at $x$.

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

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

$I_\delta$ is an element of $N_x$ as $I_\delta$ is a neighborhood of $x$.

We have:

This gives:

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

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

Hence the result.