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