Real Function is Continuous at Point iff Oscillation is Zero

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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 open subset neighborhoods of $x$.

Let $\map {\omega_f} x$ be the oscillation of $f$ at $x$:

$\map {\omega_f} x = \ds \inf \set {\map {\omega_f} I: I \in N_x}$


$\map {\omega_f} I = \ds \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$

Then $\map {\omega_f} x = 0$ if and only if $f$ is continuous at $x$.


Necessary Condition

Let $\map {\omega_f} x = 0$.

Let $\epsilon > 0$.

Suppose that $\forall I: I \in N_x: \map {\omega_f} I \ge \epsilon$.

Then by definition, $\map {\omega_f} x \ge \epsilon$.

This contradicts $\map {\omega_f} x = 0$.

From this contradiction we deduce that:

$\exists I: I \in N_x: \map {\omega_f} I < \epsilon$

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

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

So for our specific $x$, if $y$ satisfies:

$\size {x - y} < \delta$


$y \in I$

and, if $y \in D$:

$\size {\map f x - \map f y} \le \map {\omega_f} I$

Since $\map {\omega_f} I < \epsilon$ 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 $y \in D$:

$\size {x - y} < \delta \implies \size {\map f x - \map f y} < \epsilon$

Let the interval $I_\delta$ be defined as:

$I_\delta := \openint {x - \delta} {x + \delta}$

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

We have:

\(\ds \map {\omega_f} {I_\delta}\) \(=\) \(\ds \sup \set {\size {\map f y - \map f z}: y, z \in I_\delta \cap D}\) Definition of $\map {\omega_f} {I_\delta}$
\(\ds \) \(=\) \(\ds \sup \set {\size {\map f y - \map f x + \map f x - \map f z}: y, z \in I_\delta \cap D}\)
\(\ds \) \(\le\) \(\ds \sup \set {\size {\map f y - \map f x} + \size {\map f x - \map f z}: y, z \in I_\delta \cap D}\) Triangle Inequality for Vectors in Euclidean Space and Supremum of Function is less than Supremum of Greater Function
\(\ds \) \(=\) \(\ds \sup \set {\size {\map f y - \map f x}: y \in I_\delta \cap D} + \sup \set {\size {\map f x - \map f z}: z \in I_\delta \cap D}\) Supremum of Sum equals Sum of Suprema
\(\ds \) \(=\) \(\ds \sup \set {\size {\map f x - \map f y}: y \in I_\delta \cap D} + \sup \set {\size {\map f x - \map f y}: y \in I_\delta \cap D}\)
\(\ds \) \(=\) \(\ds 2 \sup \set {\size {\map f x - \map f y}: y \in I_\delta \cap D}\)
\(\ds \) \(<\) \(\ds 2 \epsilon\) as $\size {\map f x - \map f y} < \epsilon$ whenever $y \in I_\delta \cap D$

This gives:

\(\ds \map {\omega_f} x\) \(=\) \(\ds \inf \set {\map {\omega_f} I: I \in N_x}\) Definition of $\map {\omega_f} x$
\(\ds \) \(\le\) \(\ds \map {\omega_f} {I_\delta}\) Definition of Infimum of Subset of Real Numbers and that $I_\delta \in N_x$
\(\ds \) \(<\) \(\ds 2 \epsilon\)

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

Thus $\map {\omega_f} x$ must be $0$.

Hence the result.