=== 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: {{begin-eqn}} {{eqn | l = \forall s \in NR: \exists t \in ER: s       | o = \ge       | r = t }} {{eqn | ll= \leadsto       | l = \forall \epsilon \in \R_{>0}: \forall s \in NR: \exists t \in ER: s       | o = \ge       | r = t \land \epsilon > 0       | c = as $\epsilon > 0$ is true since $\epsilon \in \R_{>0}$ }} {{eqn | ll= \leadsto       | l = \forall \epsilon \in \R_{>0}: \forall s \in NR: \exists t \in ER: s       | o = \ge       | r = t \land s + \epsilon > s }} {{eqn | ll= \leadsto       | l = \forall \epsilon \in \R_{>0}: \forall s \in NR: \exists t \in ER: s + \epsilon       | o = >       | r = s \ge t }} {{eqn | ll= \leadsto       | l = \forall \epsilon \in \R_{>0}: \forall s \in NR: \exists t \in ER: s + \epsilon       | o = >       | r = t }} {{eqn | ll= \leadsto       | l = \inf NR       | r = \inf ER       | c = by Condition for Infimum of Subset to equal Infimum of Set as $ER$ is a subset of $NR$, $\inf NR \in \R$, and $\inf ER \in \R$ }} {{eqn | ll= \leadsto       | l = \inf N       | r = \inf ER       | c = as $\inf N = \inf NR$ }} {{eqn | ll= \leadsto       | l = \inf N       | r = \inf E       | c = as $\inf E = \inf ER$ }} {{eqn | ll= \leadsto       | l = \map {\omega_f} x       | r = \map {\omega^E_f} x       | c = definitions }} {{end-eqn}} {{qed|lemma}} === 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. {{qed|lemma}} === 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: {{begin-eqn}} {{eqn | o =        | r = E \text { contains a real number} }} {{eqn | o = \leadsto       | r = N \text { contains a real number}       | c = as $E$ is a subset of $N$ }} {{eqn | o = \leadsto       | r = \map {\omega_f} x \in \R       | c = Infimum of Set of Oscillations on Set }} {{end-eqn}} 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. {{qed}} Category:Real Analysis n8wxkt68gjl4mpocw12jqmyoddgztw7