Oscillation at Point (Infimum) equals Oscillation at Point (Epsilon-Neighborhood)

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

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

Let $N_x$ be the set of open subset neighborhoods of $x$.

Let $E_x$ be the set of $\epsilon$-neighborhoods of $x$.

Let $\omega_f \left({x}\right)$ be the oscillation of $f$ at $x$ based on $N_x$:
 * $\omega_f \left({x}\right) = \inf \left\{{\omega_f \left({I}\right): I \in N_x}\right\}$

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

Let $\omega^E_f \left({x}\right)$ be the oscillation of $f$ at $x$ based on $E_x$:
 * $\omega^E_f \left({x}\right) = \inf \left\{{\omega_f \left({I}\right): I \in E_x}\right\}$

Then:
 * $\omega_f \left({x}\right) \in \R$ if and only if $\omega^E_f \left({x}\right) \in \R$

and, if $\omega_f \left({x}\right)$ and $\omega^E_f \left({x}\right)$ exist as real numbers:
 * $\omega_f \left({x}\right) = \omega^E_f \left({x}\right)$

Necessary Condition
Let:
 * $\omega_f \left({x}\right) = \inf \left\{{\omega_f \left({I}\right): I \in N_x}\right\}$
 * $\omega^E_f \left({x}\right) = \inf \left\{{\omega_f \left({I}\right): I \in E_x}\right\}$
 * $\omega_f \left({I}\right) = \sup \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$

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

We need to prove:
 * $\omega^E_f \left({x}\right) \in \R$
 * $\omega^E_f \left({x}\right) = \omega_f \left({x}\right)$

First, we intend to prove that $\omega^E_f \left({x}\right) \in \R$.

We have $\inf \left\{{\omega_f \left({I}\right): I \in N_x}\right\} = \omega_f \left({x}\right)$.

Therefore, $\inf \left\{{\omega_f \left({I}\right): I \in N_x}\right\} \in \R$ as $\omega_f \left({x}\right) \in \R$.

From this follows by Lemma 12 that $\left\{{\omega_f \left({I}\right): I \in N_x}\right\}$ contains a real number.

Accordingly, an $I \in N_x$ exists such that $\omega_f \left({I}\right)$ is a real number.

$I$ is an open subset neighborhood of $x$ as $I \in N_x$.

This means that $I$ contains an open subset that contains (as an element) $x$.

Therefore, an $h \in \R_{>0}$ exists such that $\left({x - h \,.\,.\, x + h}\right)$ is a subset of $I$.

Observe that $\left({x - h \,.\,.\, x + h}\right) \in N_x$.

We have:
 * $I \in N_x$
 * $\omega_f \left({I}\right) \in \R$
 * $\left({x - h \,.\,.\, x + h}\right) \in N_x$
 * $\left({x - h \,.\,.\, x + h}\right) \subset I$

This gives by Lemma 4:
 * $\omega_f \left({\left({x - h \,.\,.\, x + h}\right)}\right) \in \R$

We observe that $\left({x - h \,.\,.\, x + h}\right)$ is an $\epsilon$-neighborhood of $x$.

Therefore, $\left({x - h \,.\,.\, x + h}\right) \in E_x$.

Accordingly, $\omega_f \left({\left({x - h \,.\,.\, x + h}\right)}\right) \in \left\{{\omega_f \left({I}\right): I \in E_x}\right\}$.

Therefore, $\left\{{\omega_f \left({I}\right): I \in E_x}\right\}$ contains a real number as $\omega_f \left({\left({x - h \,.\,.\, x + h}\right)}\right) \in \R$.

From this follows by Lemma 12 that $\left\{{\omega_f \left({I}\right): I \in E_x}\right\}$ admits an infimum (in $\R$).

In other words:
 * $\omega^E_f \left({x}\right) \in \R$ as $\omega^E_f \left({x}\right) = \inf \left\{{\omega_f \left({I}\right): I \in E_x}\right\}$

This is the first statement that we intended to prove.

Next, we need to prove that $\omega^E_f \left({x}\right) = \omega_f \left({x}\right)$.

This result follows by Lemma 1 as $\omega_f \left({x}\right)$ and $\omega^E_f \left({x}\right)$ exist as real numbers.

Sufficient Condition
Let:
 * $N = \left\{{\omega_f \left({I}\right): I \in N_x}\right\}$
 * $E = \left\{{\omega_f \left({I}\right): I \in E_x}\right\}$
 * $\omega_f \left({x}\right) = \inf N$
 * $\omega^E_f \left({x}\right) = \inf E$
 * $\omega_f \left({I}\right) = \sup \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$

Let $\omega^E_f \left({x}\right) \in \R$.

We need to prove:
 * $\omega_f \left({x}\right) \in \R$
 * $\omega_f \left({x}\right) = \omega^E_f \left({x}\right)$

First, we intend to prove that $\omega_f \left({x}\right) \in \R$.

We have $\omega^E_f \left({x}\right) \in \R$.

Therefore, $\inf E \in \R$ as $\inf E = \omega^E_f \left({x}\right)$.

Accordingly:
 * $E$ contains a real number by Lemma 12

We observe by the definitions of $E_x$ and $N_x$ that every $I$ in $E_x$ is also an element of $N_x$.

Therefore, $E_x$ is a subset of $N_x$.

Accordingly:
 * $E$ is a subset of $N$ by the definitions of $E$ and $N$

We have:

This is the first statement that we intended to prove.

Next, we need to prove that $\omega_f \left({x}\right) = \omega^E_f \left({x}\right)$.

This result follows by Lemma 1 as $\omega^E_f \left({x}\right)$ and $\omega_f \left({x}\right)$ exist as real numbers.

Lemma 1
Let $f: D \to \R$ be a real function where $D \subseteq \R$.

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

Let $N_x$ be the set of open subset neighborhoods of $x$.

Let $E_x$ be the set of $\epsilon$-neighborhoods of $x$.

Let $\omega_f \left({x}\right)$ be the oscillation of $f$ at $x$ based on $N_x$:
 * $\omega_f \left({x}\right) = \inf \left\{{\omega_f \left({I}\right): I \in N_x}\right\}$

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

Let $\omega^E_f \left({x}\right)$ be the oscillation of $f$ at $x$ based on $E_x$:
 * $\omega^E_f \left({x}\right) = \inf \left\{{\omega_f \left({I}\right): I \in E_x}\right\}$

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

Let $\omega^E_f \left({x}\right) \in \R$.

Then $\omega_f \left({x}\right) = \omega^E_f \left({x}\right)$.

Proof
We have:
 * $\omega_f \left({x}\right) \in \R$
 * $\omega^E_f \left({x}\right) \in \R$

We need to prove that:
 * $\omega_f \left({x}\right) = \omega^E_f \left({x}\right)$

Let:
 * $N = \left\{{\omega_f \left({I}\right): I \in N_x}\right\}$
 * $E = \left\{{\omega_f \left({I}\right): I \in E_x}\right\}$
 * $\omega_f \left({x}\right) = \inf N$
 * $\omega^E_f \left({x}\right) = \inf E$
 * $\omega_f \left({I}\right) = \sup \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$

We have:
 * $\inf N \in \R$ as $\omega_f \left({x}\right) \in \R$
 * $\inf E \in \R$ as $\omega^E_f \left({x}\right) \in \R$

Let:
 * $NR = N \cap \R$

Let $I \in N_x$.

Therefore, $x \in I$.

Lemma 7 gives that $\omega_f \left({I}\right)$ is an extended real number.

Therefore:
 * $N$ is a set of extended real numbers as $N = \left\{{\omega_f \left({I}\right): I \in N_x}\right\}$

Also, we have that $N$ is bounded below (in $\R$) as $\inf N \in \R$.

This gives by Lemma 9:
 * $\inf NR \in \R$ as $\inf N \in \R$
 * $\inf NR = \inf N$ as $\inf NR \in \R$ and $\inf N \in \R$

Let:
 * $ER = E \cap \R$

We observe by the definitions of $E_x$ and $N_x$ that every $I$ in $E_x$ is also an element of $N_x$.

Therefore, $E_x$ is a subset of $N_x$.

Accordingly:
 * $E$ is a subset of $N$ by the definitions of $E$ and $N$

Therefore:
 * $ER$ is a subset of $NR$ by Set Intersection Preserves Subsets/Corollary

We have that $N$ is a set of extended real numbers.

Also, we have that $E$ is a subset of $N$.

Therefore, $E$ is a set of extended real numbers.

Also, we have that $E$ is bounded below (in $\R$) as $\inf E \in \R$.

This gives by Lemma 9:
 * $\inf ER \in \R$ as $\inf E \in \R$
 * $\inf ER = \inf E$ as $\inf ER \in \R$ and $\inf E \in \R$

Assume that:
 * $s$ is a real number in $N$

Then an $I \in N_x$ exists such that:
 * $\omega_f \left({I}\right) = s$ as $N = \left\{{\omega_f \left({I'}\right): I' \in N_x}\right\}$

We have that $\omega_f \left({I}\right) \in \R$ as $s \in \R$.

The real set $I$ is an open subset neighborhood of $x$ as $I \in N_x$.

This means that $I$ contains an open subset that contains (as an element) $x$.

Therefore, an $h \in \R_{>0}$ exists such that $\left({x - h \,.\,.\, x + h}\right)$ is a subset of $I$.

Observe that $\left({x - h \,.\,.\, x + h}\right) \in N_x$.

We have:
 * $I \in N_x$
 * $\omega_f \left({I}\right) \in \R$
 * $\left({x - h \,.\,.\, x + h}\right) \in N_x$
 * $\left({x - h \,.\,.\, x + h}\right) \subset I$

This gives by Lemma 4:
 * $\omega_f \left({\left({x - h \,.\,.\, x + h}\right)}\right) \in \R$
 * $\omega_f \left({\left({x - h \,.\,.\, x + h}\right)}\right) \le \omega_f \left({I}\right)$

We have that $\left({x - h \,.\,.\, x + h}\right) \in E_x$ as $\left({x - h \,.\,.\, x + h}\right)$ is an $\epsilon$-neighborhood of $x$.

Therefore:
 * $\omega_f \left({\left({x - h \,.\,.\, x + h}\right)}\right) \in E$ as $E = \left\{{\omega_f \left({I'}\right): I' \in E_x}\right\}$

Let:
 * $t = \omega_f \left({\left({x - h \,.\,.\, x + h}\right)}\right)$

We have $t \in E$ as $\omega_f \left({\left({x - h \,.\,.\, x + h}\right)}\right) \in E$.

Also, $t \in \R$ as $\omega_f \left({\left({x - h \,.\,.\, x + h}\right)}\right) \in \R$.

Accordingly:
 * $t \in ER$ as $ER = E \cap \R$

We have $t = \omega_f \left({\left({x - h \,.\,.\, x + h}\right)}\right)$ and $s = \omega_f \left({I}\right)$.

Therefore:
 * $t \le s$ as $\omega_f \left({\left({x - h \,.\,.\, x + h}\right)}\right) \le \omega_f \left({I}\right)$

We have assumed that $s$ is a real number in $N$.

Therefore:
 * $s \in NR$ as $NR = N \cap \R$

We have:

Lemma 4 (Oscillation on Subset)
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 $\omega_f \left({I}\right)$ be the oscillation of $f$ on a set $I$ in $S_x$:
 * $\omega_f \left({I}\right) = \sup \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$

Let $I \in S_x$.

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

Let $J \in S_x$ be a subset of $I$.

Then:
 * $\omega_f \left({J}\right) \in \R$
 * $\omega_f \left({J}\right) \le \omega_f \left({I}\right)$

Proof
Let:
 * $I, J \in S_x$
 * $J \subset I$
 * $\omega_f \left({I}\right) \in \R$

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

We need to prove:
 * $\omega_f \left({J}\right) \in \R$
 * $\omega_f \left({J}\right) \le \omega_f \left({I}\right)$

We intend to prove that $\omega_f \left({J}\right) \in \R$.

We start by proving that $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in J \cap D}\right\}$ is bounded above and non-empty.

We have that $J$ is a subset of $I$.

Therefore:
 * $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in J \cap D}\right\}$ is a subset of $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$

The statement $\omega_f \left({I}\right) \in \R$ means that $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ admits a supremum.

Therefore, $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ is bounded above.

Accordingly:
 * $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in J \cap D}\right\}$ is bounded above as $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in J \cap D}\right\}$ is a subset of $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$

We observe that $\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert = 0$ for $y = z = x$.

Therefore, $0 \in \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in J \cap D}\right\}$ as $x \in J \cap D$.

Accordingly:
 * $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in J \cap D}\right\}$ is non-empty

We have that $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in J \cap D}\right\}$ is a real set as $\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert \in \R$ for every $y, z \in D$.

We have shown that $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in J \cap D}\right\}$ is non-empty and bounded above.

Therefore, $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in J \cap D}\right\}$ admits a supremum by the Continuum Property.

In other words:

We finished proving that $\omega_f \left({J}\right) \in \R$.

It remains to prove that $\omega_f \left({J}\right) \le \omega_f \left({I}\right)$.

We have:
 * $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in J \cap D}\right\}$ is a subset of $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$
 * $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ admits a supremum
 * $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in J \cap D}\right\}$ admits a supremum

Then:

Lemma 7 (Oscillation on Set is an Extended Real Number)
Let $f: D \to \R$ be a real function where $D \subseteq \R$.

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

Let $I$ be a real set that contains (as an element) $x$.

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

Then:
 * $\omega_f \left({I}\right) \in \overline \R_{\ge 0}$

and:
 * $\omega_f \left({I}\right) = \begin{cases}

\text{a positive real number} & \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\} \text{is bounded above} \\ \infty & \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\} \text{is not bounded above} \end{cases}$

Proof
We observe that $\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert = 0$ for $y = z = x$.

Therefore:
 * $0 \in \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ as $x \in I \cap D$

Accordingly:
 * $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ is non-empty

We have that $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ is a real set as $\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert \in \R$ for every $y, z \in D$.

Every real number is less than $\infty$.

Therefore:
 * $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ is bounded above in $\overline \R$

There are two cases: either $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ is bounded above (in $\R$), or it is not.

First, assume that:
 * $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ is bounded above (in $\R$)

We have that $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ is bounded above and non-empty.

Therefore:
 * $\sup \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ exists as a real number by the Continuum Property

We know that $\sup \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ is greater than or equal to every element of $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$.

Also $0 \in \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$.

Therefore, $\sup \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\} \ge 0$.

We also have that $\sup \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ exists as a real number.

Therefore, $\sup \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ is a positive real number.

In other words:
 * $\omega_f \left({I}\right)$ is a positive real number as $\omega_f \left({I}\right) = \sup \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$

Next, assume that:
 * $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ is not bounded above (in $\R$)

We have that $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ is bounded above in $\overline \R$.

Therefore, $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ has $\infty$ as an upper bound.

A number that is less than $\infty$ is a real number.

No real number is an upper bound for $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ as $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ is not bounded above in $\R$.

Therefore, $\infty$ is the least upper bound of $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$.

In other words:
 * $\omega_f \left({I}\right) = \infty$ as $\omega_f \left({I}\right) = \sup \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$

In either case, whether $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ is bounded above or not:
 * $\omega_f \left({I}\right) \in \overline \R_{\ge 0}$.

Lemma 9 (Infimum of Real Subset)
Let $S$ be a set of extended real numbers.

Let $S$ be bounded below (in $\R$).

Let $T = S \cap \R$.

Then:
 * $S$ admits an infimum (in $\R$) if and only if $T$ admits an infimum (in $\R$)

and, if $\inf S$ and $\inf T$ exist as real numbers:
 * $\inf S = \inf T$

Proof
We observe that $T$ constitutes the real numbers of $S$.

Since there is a real number that is a lower bound for $S$, $-\infty$ is not an element of $S$.

Accordingly, $\infty$ is the only possible element of $S \setminus T$.

Therefore:
 * $S$ is a subset of $T \cup \left\{{\infty}\right\}$

First, we show that $S$ and $T$ have the same set of lower bounds.

Let $b$ be a lower bound (in $\R$) for $S$.

Then $b$ is a lower bound for $T$ as $T$ is a subset of $S$.

Therefore:
 * the set of lower bounds for $S$ is a subset of the set of lower bounds for $T$

Assume that $c$ is a lower bound (in $\R$) for $T$.

Then $c$ is a lower bound for $T \cup \left\{{\infty}\right\}$ as well since $c < \infty$.

Accordingly, $c$ is a lower bound for $S$ since $S$ is a subset of $T \cup \left\{{\infty}\right\}$.

Therefore:
 * the set of lower bounds for $T$ is a subset of the set of lower bounds for $S$

We have:
 * the set of lower bounds for $T$ is a subset of the set of lower bounds for $S$
 * the set of lower bounds for $S$ is a subset of the set of lower bounds for $T$

Therefore:
 * the set of lower bounds for $T$ equals the set of lower bounds for $S$ by definition

Next, we show that $S$ and $T$ have the same infima.

We have that $S$ and $T$ have the same set of lower bounds.

Therefore, $S$ and $T$ have the same greatest lower bound in $\overline \R$.

Accordingly, as a corollary, if one of the sets $S$ and $T$ admits an infimum (in $\R$), so does the other.

Furthermore, these infima are equal.

Lemma 10 (Infimum of Subset of Extended Real Numbers is Arbitrarily Close)
Let $A \subseteq \overline \R$ be a subset of the extended real numbers.

Let $b$ be an infimum (in $\R$) of $A$.

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

Then:
 * $\exists x \in \left({A \cap \R}\right): x - b < \epsilon$

Proof
We have that:
 * $A$ is a a set of extended real numbers
 * $A$ is bounded below (in $\R$) as the real number $b$ is a lower bound for $A$

From this follows by Lemma 9:
 * $\inf \left({A \cap \R}\right) \in \R$ as $\inf A \in \R$

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

Then:

Lemma 12 (Infimum of Set of Oscillations on Set)
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:
 * $S = \left\{{\omega_f \left({I}\right): I \in S_x}\right\}$

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 $S$ admits an infimum (in $\R$) if and only if $S$ contains a real number.

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 Lemma 7 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 Lemma 10 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 Lemma 7 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 Lemma 9 as $\inf SR \in \R$