# 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 $\map {\omega_f} x$ be the oscillation of $f$ at $x$ based on $N_x$:

- $\map {\omega_f} x = \inf \set {\map {\omega_f} I: I \in N_x}$

where $\map {\omega_f} I$ is the oscillation of $f$ on $I$:

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

Let $\map {\omega^E_f} x$ be the oscillation of $f$ at $x$ based on $E_x$:

- $\map {\omega^E_f} x = \inf \set {\map {\omega_f} I: I \in E_x}$

Then:

- $\map {\omega_f} x \in \R$ if and only if $\map {\omega^E_f} x \in \R$

and, if $\map {\omega_f} x$ and $\map {\omega^E_f} x$ exist as real numbers:

- $\map {\omega_f} x = \map {\omega^E_f} x$

## Proof

### 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 $\map {\omega_f} x$ be the oscillation of $f$ at $x$ based on $N_x$:

- $\map {\omega_f} x = \inf \set {\map {\omega_f} I: I \in N_x}$

where $\map {\omega_f} I$ is the oscillation of $f$ on $I$:

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

Let $\map {\omega^E_f} x$ be the oscillation of $f$ at $x$ based on $E_x$:

- $\map {\omega^E_f} x = \inf \set {\map {\omega_f} I: I \in E_x}$

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

Let $\map {\omega^E_f} x \in \R$.

Then $\map {\omega_f} x = \map {\omega^E_f} x$.

### Proof

We have:

- $\map {\omega_f} x \in \R$
- $\map {\omega^E_f} x \in \R$

We need to prove that:

- $\map {\omega_f} x = \map {\omega^E_f} x$

Let:

- $N = \set {\map {\omega_f} I: I \in N_x}$
- $E = \set {\map {\omega_f} I: I \in E_x}$
- $\map {\omega_f} x = \inf N$
- $\map {\omega^E_f} x = \inf E$
- $\map {\omega_f} I = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$

We have:

- $\inf N \in \R$ as $\map {\omega_f} x \in \R$
- $\inf E \in \R$ as $\map {\omega^E_f} x \in \R$

Let:

- $NR = N \cap \R$

Let $I \in N_x$.

Therefore, $x \in I$.

Oscillation on Set is an Extended Real Number gives that $\map {\omega_f} I$ is an extended real number.

Therefore:

- $N$ is a set of extended real numbers as $N = \set {\map {\omega_f} I: I \in N_x}$.

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

This gives by Infimum of Real Subset:

- $\inf NR \in \R$ 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 Infimum of Real Subset:

- $\inf ER \in \R$ 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:

- $\map {\omega_f} I = s$ as $N = \set {\map {\omega_f} {I'}: I' \in N_x}$

We have that $\map {\omega_f} I \in \R$ 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 $\openint {x - h} {x + h}$ is a subset of $I$.

Observe that $\openint {x - h} {x + h} \in N_x$.

We have:

- $I \in N_x$
- $\map {\omega_f} I \in \R$
- $\openint {x - h} {x + h} \in N_x$
- $\openint {x - h} {x + h} \subset I$

This gives by Oscillation on Subset:

- $\map {\omega_f} {\openint {x - h} {x + h} } \in \R$
- $\map {\omega_f} {\openint {x - h} {x + h} } \le \map {\omega_f} I$

We have that $\openint {x - h} {x + h} \in E_x$ as $\openint {x - h} {x + h}$ is an $\epsilon$-neighborhood of $x$.

Therefore:

- $\map {\omega_f} {\openint {x - h} {x + h} } \in E$ as $E = \set {\map {\omega_f} {I'}: I' \in E_x}$

Let:

- $t = \map {\omega_f} {\openint {x - h} {x + h} }$

We have $t \in E$ as $\map {\omega_f} {\openint {x - h} {x + h} } \in E$.

Also, $t \in \R$ as $\map {\omega_f} {\openint {x - h} {x + h} } \in \R$.

Accordingly:

- $t \in ER$ as $ER = E \cap \R$

We have $t = \map {\omega_f} {\openint {x - h} {x + h} }$ and $s = \map {\omega_f} I$.

Therefore:

- $t \le s$ as $\map {\omega_f} {\openint {x - h} {x + h} } \le \map {\omega_f} I$

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

Therefore:

- $s \in NR$ as $NR = N \cap \R$

We have:

\(\displaystyle \forall s \in NR: \exists t \in ER: s\) | \(\ge\) | \(\displaystyle t\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \forall \epsilon \in \R_{>0}: \forall s \in NR: \exists t \in ER: s\) | \(\ge\) | \(\displaystyle t \land \epsilon > 0\) | as $\epsilon > 0$ is true since $\epsilon \in \R_{>0}$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \forall \epsilon \in \R_{>0}: \forall s \in NR: \exists t \in ER: s\) | \(\ge\) | \(\displaystyle t \land s + \epsilon > s\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \forall \epsilon \in \R_{>0}: \forall s \in NR: \exists t \in ER: s + \epsilon\) | \(>\) | \(\displaystyle s \ge t\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \forall \epsilon \in \R_{>0}: \forall s \in NR: \exists t \in ER: s + \epsilon\) | \(>\) | \(\displaystyle t\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \inf NR\) | \(=\) | \(\displaystyle \inf ER\) | by Condition for Infimum of Subset to equal Infimum of Set as $ER$ is a subset of $NR$, $\inf NR \in \R$, and $\inf ER \in \R$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \inf N\) | \(=\) | \(\displaystyle \inf ER\) | as $\inf N = \inf NR$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \inf N\) | \(=\) | \(\displaystyle \inf E\) | as $\inf E = \inf ER$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \map {\omega_f} x\) | \(=\) | \(\displaystyle \map {\omega^E_f} x\) | definitions |

$\Box$

### Necessary Condition

Let:

- $\map {\omega_f} x = \inf \set {\map {\omega_f} I: I \in N_x}$
- $\map {\omega^E_f} x = \inf \set {\map {\omega_f} I: I \in E_x}$
- $\map {\omega_f} I = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$

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

We need to prove:

- $\map {\omega^E_f} x \in \R$
- $\map {\omega^E_f} x = \map {\omega_f} x$

First, we intend to prove that $\map {\omega^E_f} x \in \R$.

We have $\map {\omega_f} x \in \R$.

Therefore, $\set {\map {\omega_f} I: I \in N_x}$ contains a real number by Infimum of Set of Oscillations on Set.

Accordingly, an $I \in N_x$ exists such that $\map {\omega_f} I$ 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 $\openint {x - h} {x + h}$ is a subset of $I$.

Observe that $\openint {x - h} {x + h} \in N_x$.

We have:

- $I \in N_x$
- $\map {\omega_f} I \in \R$
- $\openint {x - h} {x + h} \in N_x$
- $\openint {x - h} {x + h} \subset I$

This gives by Oscillation on Subset:

- $\map {\omega_f} {\openint {x - h} {x + h} } \in \R$

We observe that $\openint {x - h} {x + h}$ is an $\epsilon$-neighborhood of $x$.

Therefore, $\openint {x - h} {x + h} \in E_x$.

Accordingly, $\map {\omega_f} {\openint {x - h} {x + h} } \in \set {\map {\omega_f} I: I \in E_x}$.

Therefore, $\set {\map {\omega_f} I: I \in E_x}$ contains a real number as $\map {\omega_f} {\openint {x - h} {x + h} } \in \R$.

From this follows that $\map {\omega^E_f} x \in \R$ by Infimum of Set of Oscillations on Set.

This is the first statement that we intended to prove.

Next, we need to prove that $\map {\omega^E_f} x = \map {\omega_f} x$.

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

$\Box$

### Sufficient Condition

Let:

- $N = \set {\map {\omega_f} I: I \in N_x}$
- $E = \set {\map {\omega_f} I: I \in E_x}$
- $\map {\omega_f} x = \inf N$
- $\map {\omega^E_f} x = \inf E$
- $\map {\omega_f} I = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$

Let $\map {\omega^E_f} x \in \R$.

We need to prove:

- $\map {\omega_f} x \in \R$
- $\map {\omega_f} x = \map {\omega^E_f} x$

First, we intend to prove that $\map {\omega_f} x \in \R$.

We have $\map {\omega^E_f} x \in \R$.

Therefore, $E$ contains a real number by Infimum of Set of Oscillations on Set.

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:

\(\displaystyle \) | \(\) | \(\displaystyle E \text { contains a real number}\) | |||||||||||

\(\displaystyle \) | \(\leadsto\) | \(\displaystyle N \text { contains a real number}\) | as $E$ is a subset of $N$ | ||||||||||

\(\displaystyle \) | \(\leadsto\) | \(\displaystyle \map {\omega_f} x \in \R\) | Infimum of Set of Oscillations on Set |

This is the first statement that we intended to prove.

Next, we need to prove that $\map {\omega_f} x = \map {\omega^E_f} x$.

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

$\blacksquare$