Infimum of Set of Oscillations on Set is Arbitrarily Close

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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:

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

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

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


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

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


Then an $I \in S_x$ exists such that:

$\map {\omega_f} {I \cap D} - \map {\omega_f} x < \epsilon$


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:

$\map {\omega_f} {I \cap D} - \map {\omega_f} x < \epsilon$


By Oscillation on Set is an Extended Real Number:

$\forall I \in S_x: \map {\omega_f} {I \cap D} \in \overline \R_{\ge 0}$


Therefore:

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


We have also:

$\inf \set {\map {\omega_f} {I \cap D}: I \in S_x} \in \R$

as $\inf \set {\map {\omega_f} {I \cap D}: I \in S_x} = \map {\omega_f} x$


Therefore:

\(\ds \exists I \in S_x: \, \) \(\ds \map {\omega_f} {I \cap D} - \inf \set {\map {\omega_f} {I' \cap D}: I' \in S_x}\) \(<\) \(\ds \epsilon\) Infimum of Subset of Extended Real Numbers is Arbitrarily Close
\(\ds \leadstoandfrom \ \ \) \(\ds \map {\omega_f} {I \cap D} - \map {\omega_f} x\) \(<\) \(\ds \epsilon\) as $\map {\omega_f} x = \inf \set {\map {\omega_f} {I' \cap D}: I' \in S_x}$

$\blacksquare$