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

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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.

$\blacksquare$