Infimum of Set of Oscillations on Set is Arbitrarily Close

From ProofWiki
Jump to navigation Jump to search


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) $x$.


$\omega_f \left({x}\right) = \displaystyle \inf \left\{{\omega_f \left({I}\right): I \in S_x}\right\}$

where $\omega_f \left({I}\right)$ is the oscillation of $f$ on a real set $I$:

$\omega_f \left({I}\right) = \displaystyle \sup \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$

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

Let $\omega_f \left({x}\right) \in \R$.

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

$\omega_f \left({I}\right) - \omega_f \left({x}\right) < \epsilon$


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

Let $\omega_f \left({x}\right) \in \R$.

We need to prove that an $I \in S_x$ exists such that:

$\omega_f \left({I}\right) - \omega_f \left({x}\right) < \epsilon$

We have that $\omega_f \left({I}\right) \in \overline{\R}_{\ge 0}$ for every $I \in S_x$ by Oscillation on Set is an Extended Real Number.


$\left\{{\omega_f \left({I}\right): I \in S_x}\right\}$ is a subset of $\overline{\R}$

We have also:

$\displaystyle \inf \left\{{\omega_f \left({I}\right): I \in S_x}\right\} \in \R$ as $\displaystyle \inf \left\{{\omega_f \left({I}\right): I \in S_x}\right\} = \omega_f \left({x}\right)$

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

\(\displaystyle \omega_f \left({I}\right) - \displaystyle \inf \left\{ {\omega_f \left({I'}\right): I' \in S_x}\right\}\) \(<\) \(\displaystyle \epsilon\) by Infimum of Subset of Extended Real Numbers is Arbitrarily Close
\(\displaystyle \iff \ \ \) \(\displaystyle \omega_f \left({I}\right) - \omega_f \left({x}\right)\) \(<\) \(\displaystyle \epsilon\) as $\omega_f \left({x}\right) = \displaystyle \inf \left\{ {\omega_f \left({I'}\right): I' \in S_x}\right\}$