Infimum of Set of Oscillations on Set is Arbitrarily Close

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

Let $x$ be a point in $D$.

Let $S_x$ be a set of real sets that contain (as an element) $x$.

Let:


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

where $\map {\omega_f} I$ is the oscillation of $f$ on a real set $I$:


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

Let $\epsilon \in \R_{>0}$.

Let $\map {\omega_f} x \in \R$.

Then an $I \in S_x$ exists such that:
 * $\map {\omega_f} I - \map {\omega_f} x < \epsilon$

Proof
Let $\epsilon \in \R_{>0}$.

Let $\map {\omega_f} x \in \R$.

We need to prove that an $I \in S_x$ exists such that:
 * $\map {\omega_f} I - \map {\omega_f} x < \epsilon$

We have that $\map {\omega_f} I \in \overline \R_{\ge 0}$ for every $I \in S_x$ by Oscillation on Set is an Extended Real Number.

Therefore:


 * $\set {\map {\omega_f} I: I \in S_x}$ is a subset of $\overline \R$

We have also:


 * $\ds \inf \set {\map {\omega_f} I: I \in S_x} \in \R$ as $\ds \inf \set {\map {\omega_f} I: I \in S_x} = \map {\omega_f} x$

Therefore, an $I \in S_x$ exists such that: