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 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 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 a 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:
\(\ds \forall s \in NR: \exists t \in ER: s\) | \(\ge\) | \(\ds t\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall \epsilon \in \R_{>0}: \forall s \in NR: \exists t \in ER: s\) | \(\ge\) | \(\ds t \land \epsilon > 0\) | as $\epsilon > 0$ is true since $\epsilon \in \R_{>0}$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall \epsilon \in \R_{>0}: \forall s \in NR: \exists t \in ER: s\) | \(\ge\) | \(\ds t \land s + \epsilon > s\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall \epsilon \in \R_{>0}: \forall s \in NR: \exists t \in ER: s + \epsilon\) | \(>\) | \(\ds s \ge t\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall \epsilon \in \R_{>0}: \forall s \in NR: \exists t \in ER: s + \epsilon\) | \(>\) | \(\ds t\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \inf NR\) | \(=\) | \(\ds \inf ER\) | by Condition for Infimum of Subset to equal Infimum of Set as $ER$ is a subset of $NR$, $\inf NR \in \R$, and $\inf ER \in \R$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \inf N\) | \(=\) | \(\ds \inf ER\) | as $\inf N = \inf NR$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \inf N\) | \(=\) | \(\ds \inf E\) | as $\inf E = \inf ER$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map {\omega_f} x\) | \(=\) | \(\ds \map {\omega^E_f} x\) | 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:
\(\ds \) | \(\) | \(\ds E \text { contains a real number}\) | ||||||||||||
\(\ds \) | \(\leadsto\) | \(\ds N \text { contains a real number}\) | as $E$ is a subset of $N$ | |||||||||||
\(\ds \) | \(\leadsto\) | \(\ds \map {\omega_f} x \in \R\) | 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.
$\blacksquare$