Infimum of Set of Oscillations on Set

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

where $\omega_f \left({I}\right)$ is the oscillation of $f$ on the set $I$:
 * $\omega_f \left({I}\right) = \sup \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$

Then:
 * $\omega_f \left({x}\right) \in \R$ if and only if $\left\{{\omega_f \left({I}\right): I \in S_x}\right\}$ contains a real number.

Proof
Let:
 * $S = \left\{{\omega_f \left({I}\right): I \in S_x}\right\}$

We observe that:
 * $\inf S = \omega_f \left({x}\right)$

Necessary Condition
Let $\inf S \in \R$.

We need to prove that $S$ contains a real number.

Note that $S$ is non-empty as the empty set does not admit an infimum (in $\R$).

Therefore, $S$ has at least one element.

Accordingly, there is an $I \in S_x$ such that $\omega_f \left({I}\right) \in S$.

Let $I \in S_x$.

Therefore, $x \in I$.

From this follows by Oscillation on Set is an Extended Real Number that $\omega_f \left({I}\right)$ is an extended real number.

Therefore, $S$ is a set of extended real numbers as $S = \left\{{\omega_f \left({I}\right): I \in S_x}\right\}$.

Accordingly, $S$ contains a real number by Infimum of Subset of Extended Real Numbers is Arbitrarily Close as $\inf S \in \R$.

Sufficient Condition
Let $S$ contain a real number.

We need to prove that $\inf S \in \R$.

Let:
 * $SR = S \cap \R$

We have:
 * $SR$ is not empty as $S$ contains a real number

Let $I \in S_x$.

Therefore, $x \in I$.

From this follows by Oscillation on Set is an Extended Real Number that $\omega_f \left({I}\right) \in \overline \R_{\ge 0}$.

Therefore:
 * $S$ is a subset of $\overline \R_{\ge 0}$ as $S = \left\{{\omega_f \left({I}\right): I \in S_x}\right\}$

Accordingly:
 * $S$ is bounded below

From this follows that:
 * $SR$ is bounded below as $SR$ is a subset of $S$

We have:
 * $SR$ is bounded below
 * $SR$ is not empty

Therefore:
 * $\inf SR \in \R$ Continuum Property

We have:
 * $S$ is a set of extended real numbers as $S$ is a subset of $\overline \R_{\ge 0}$
 * $S$ is bounded below

Therefore:
 * $\inf S \in \R$ by Infimum of Real Subset as $\inf SR \in \R$