Theorem: Oscillation on Set is an Extended Real Number

Let $f: D \to \R$ be a real function where $D \subseteq \R$.
Let $x$ be a point in $D$.
Let $I$ be a real set that contains (as an element) $x$.
Let $\map {\omega_f} {I \cap D}$ be the oscillation of $f$ on $I \cap D$:
:$\map {\omega_f} {I \cap D} = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$
Then:
:$\map {\omega_f} {I \cap D} \in \overline \R_{\ge 0}$
and:
:<nowiki>$\map {\omega_f} {I \cap D} = \begin{cases}
\text{a positive real number} & \set {\size {\map f y - \map f z}: y, z \in I \cap D} \text{is bounded above} \\
\infty & \set {\size {\map f y - \map f z}: y, z \in I \cap D} \text{is not bounded above}
\end{cases}$</nowiki>

Proof:
We observe that $\size {\map f y - \map f z} = 0$ for $y = z = x$.
Therefore:
:$0 \in \set {\size {\map f y - \map f z}: y, z \in I \cap D}$ as $x \in I \cap D$
Accordingly:
:$\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ is non-empty
We have that $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ is a real set as $\size {\map f y - \map f z} \in \R$ for every $y, z \in D$.
Every real number is less than $\infty$.
Therefore:
:$\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ is bounded above in $\overline \R$
There are two cases: either $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ is bounded above (in $\R$), or it is not.
First, assume that:
:$\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ is bounded above (in $\R$)
We have that $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ is bounded above and non-empty.
Therefore:
:$\sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$ exists as a real number by the Continuum Property
We know that $\sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$ is greater than or equal to every element of $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$.
Also $0 \in \set {\size {\map f y - \map f z}: y, z \in I \cap D}$.
Therefore, $\sup \set {\size {\map f y - \map f z}: y, z \in I \cap D} \ge 0$.
We also have that $\sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$ exists as a real number.
Therefore, $\sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$ is a positive real number.
In other words:
:$\map {\omega_f} {I \cap D}$ is a positive real number as $\map {\omega_f} {I \cap D} = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$
Next, assume that:
:$\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ is not bounded above (in $\R$)
We have that $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ is bounded above in $\overline \R$.
Therefore, $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ has $\infty$ as an upper bound.
A number that is less than $\infty$ is a real number or equal to $-\infty$.
No real number is an upper bound for $\set {\size {\map f y - \map f z} y, z \in I \cap D}$ as $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ is not bounded above in $\R$.
$-\infty$ is not an upper bound for $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ as this set contains $0$.
Therefore, $\infty$ is the only upper bound for $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$.
Accordingly, $\infty$ is the least upper bound of $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$.
In other words:
:$\map {\omega_f} {I \cap D} = \infty$ as $\map {\omega_f} {I \cap D} = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$
In either case, whether $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ is bounded above or not:
:$\map {\omega_f} {I \cap D} \in \overline \R_{\ge 0}$.
{{qed}}
Category:Real Analysis
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