Oscillation at Point (Infimum) equals Oscillation at Point (Limit)

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 $\omega_f \left({x}\right)$ be the oscillation of $f$ at $x$:


 * $\omega_f \left({x}\right) = \displaystyle \inf \left\{{\omega_f \left({I}\right): I \in N_x}\right\}$

where $\omega_f \left({I}\right)$ is the oscillation of $f$ on a real set $I$:


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

Then:


 * $\omega_f \left({x}\right) \in \R$ $\displaystyle \lim_{h \to 0^+} \omega_f \left({\left({x - h \,.\,.\, x + h}\right)}\right) \in \R$

and, if $\omega_f \left({x}\right)$ and $\displaystyle \lim_{h \to 0^+} \omega_f \left({\left({x - h \,.\,.\, x + h}\right)}\right)$ exist as real numbers:


 * $\omega_f \left({x}\right) = \displaystyle \lim_{h \to 0^+} \omega_f \left({\left({x - h \,.\,.\, x + h}\right)}\right)$

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

We need to prove:


 * $\displaystyle \lim_{h \to 0^+} \omega_f \left({\left({x - h \,.\,.\, x + h}\right)}\right) \in \R$


 * $\displaystyle \lim_{h \to 0^+} \omega_f \left({\left({x - h \,.\,.\, x + h}\right)}\right) = \omega_f \left({x}\right)$

where:


 * $\omega_f \left({\left({x - h \,.\,.\, x + h}\right)}\right) = \displaystyle \sup \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in \left({x - h \,.\,.\, x + h}\right) \cap D}\right\}$


 * $\omega_f \left({x}\right) = \displaystyle \inf \left\{{\omega_f \left({I}\right): I \in N_x}\right\}$


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

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

Then an $I \in N_x$ exists such that:


 * $\omega_f \left({I}\right) - \omega_f \left({x}\right) < \epsilon$ by Lemma 2

Let $I$ be such an element of $N_x$.

We observe in particular that $\omega_f \left({I}\right) \in \R$.

A neighborhood in $N_x$ contains an open subset that contains the point $x$.

So, $I$ contains such an open subset as $I \in N_x$.

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

Let $h$ be a real number that satisfies: $0 < h < \delta$.

We observe that $\left({x - h \,.\,.\, x + h}\right) \subset I$.

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

We have:


 * $I \in N_x$


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


 * $\left({x - h \,.\,.\, x + h}\right) \subset I$


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

from which follows by Lemma 4:


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


 * $\omega_f \left({\left({x - h \,.\,.\, x + h}\right)}\right) \le \omega_f \left({I}\right)$

and by Lemma 5:


 * $\omega_f \left({x}\right) \le \omega_f \left({\left({x - h \,.\,.\, x + h}\right)}\right)$

We find:

which means that $\displaystyle \lim_{h \to 0^+} \omega_f \left({\left({x - h \,.\,.\, x + h}\right)}\right)$ exists and equals $\omega_f \left({x}\right)$ by the definition of limit.

Sufficient Condition
Let $\displaystyle \lim_{h \to 0^+} \omega_f \left({\left({x - h \,.\,.\, x + h}\right)}\right) \in \R$.

We need to prove:


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


 * $\omega_f \left({x}\right) = \displaystyle \lim_{h \to 0^+} \omega_f \left({\left({x - h \,.\,.\, x + h}\right)}\right)$

where:


 * $\omega_f \left({x}\right) = \displaystyle \inf \left\{{\omega_f \left({I}\right): I \in N_x}\right\}$


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

We have: $\displaystyle \lim_{h \to 0^+} \omega_f \left({\left({x - h \,.\,.\, x + h}\right)}\right) \in \R$.

Therefore, $\omega_f \left({\left({x - h \,.\,.\, x + h}\right)}\right) \in \R$ for a small enough $h$ in $\R_{>0}$ by the definition of limit.

Let $h$ be such a real number.

We observe that $\left({x - h \,.\,.\, x + h}\right)$ is a neighborhood in $N_x$.

We have:


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


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

Accordingly:


 * $\omega_f \left({x}\right) \in \R$ by Lemma 3

$\omega_f \left({x}\right) = \displaystyle \lim_{h \to 0^+} \omega_f \left({\left({x - h \,.\,.\, x + h}\right)}\right)$ follows by Lemma 1.

This finishes the proof of the theorem.

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 $\omega_f \left({x}\right)$ be the oscillation of $f$ at $x$:


 * $\omega_f \left({x}\right) = \displaystyle \inf \left\{{\omega_f \left({I}\right): I \in N_x}\right\}$

where $\omega_f \left({I}\right)$ is the oscillation of $f$ on a real set $I$:


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

Let $\displaystyle \lim_{h \to 0^+} \omega_f \left({\left({x - h \,.\,.\, x + h}\right)}\right) \in \R$.

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

Then $\displaystyle \lim_{h \to 0^+} \omega_f \left({\left({x - h \,.\,.\, x + h}\right)}\right) = \omega_f \left({x}\right)$.

Proof
Let $L = \displaystyle \lim_{h \to 0^+} \omega_f \left({\left({x - h \,.\,.\, x + h}\right)}\right)$.

We know that $L$ and $\omega_f \left({x}\right)$ are real numbers.

We need to prove that $\omega_f \left({x}\right) = L$.

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

First, we aim to prove that $\left\vert{\omega_f \left({\left({x - h \,.\,.\, x + h}\right)}\right) - \omega_f \left({x}\right)}\right\vert < \epsilon$ for a small enough $h \in R_{>0}$.

$L = \displaystyle \lim_{h \to 0^+} \omega_f \left({\left({x - h \,.\,.\, x + h}\right)}\right)$ means by the definition of limit that a strictly positive real number $h_1$ exists such that:


 * $\left\vert{\omega_f \left({\left({x - h \,.\,.\, x + h}\right)}\right) - L}\right\vert < \epsilon$

for every $h$ that satisfies: $0 < h < h_1$.

This means in particular that $\omega_f \left({\left({x - h \,.\,.\, x + h}\right)}\right) \in \R$ for every $h$ that satisfies: $0 < h < h_1$.

Let $h'$ be a real number that satisfies: $0 < h' < h_1$.

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

Therefore, $\omega_f \left({\left({x - h' \,.\,.\, x + h'}\right)}\right) \in \left\{{\omega_f \left({I}\right): I \in N_x}\right\}$.

By definition, $\omega_f \left({x}\right)$ is a lower bound for $\left\{{\omega_f \left({I}\right): I \in N_x}\right\}$.

Accordingly:


 * $\omega_f \left({x}\right) \le \omega_f \left({\left({x - h' \,.\,.\, x + h'}\right)}\right)$

The fact that $\omega_f \left({x}\right) \in \R$ implies that:


 * $\omega_f \left({I}\right) - \omega_f \left({x}\right) < \epsilon$ by Lemma 2

for an $I \in N_x$.

Let $I$ be such an element of $N_x$.

We observe in particular that $\omega_f \left({I}\right) \in \R$.

A neighborhood in $N_x$ contains an open subset that contains the point $x$.

So, $I$ contains such an open subset as $I \in N_x$.

Therefore, a number $h_2 \in \R_{>0}$ exists such that $\left({x - h_2 \,.\,.\, x + h_2}\right)$ is a subset of $I$.

Let $h$ be a real number that satisfies: $0 < h < h_2$.

We observe that $\left({x - h \,.\,.\, x + h}\right)$ is a subset of $I$.

We have:


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


 * $\left({x - h \,.\,.\, x + h}\right)$ is a subset of $I$

Therefore:


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

Putting all this together, we get for every $h$ that satisfies: $0 < h < \min \left({h_1, h_2}\right)$:

Thus, we achieved our first aim.

Next, we get for every $h$ that satisfies: $0 < h < \min \left({h_1, h_2}\right)$:

This holds for every $\epsilon \in \R_{>0}$.

Therefore, $\left\vert{\omega_f \left({x}\right) - L}\right\vert = 0$ as $\left\vert{\omega_f \left({x}\right) - L}\right\vert$ is independent of $\epsilon$.

Accordingly, $\omega_f \left({x}\right) = L$.

Lemma 2
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 $\omega_f \left({x}\right)$ be the oscillation of $f$ at $x$:


 * $\omega_f \left({x}\right) = \displaystyle \inf \left\{{\omega_f \left({I}\right): I \in N_x}\right\}$

where $\omega_f \left({I}\right)$ is the oscillation of $f$ on a real set $I$:


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

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

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

Then an $I \in N_x$ exists such that:
 * $\omega_f \left({I}\right) - \omega_f \left({x}\right) < \epsilon$

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

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

We need to prove that an $I \in N_x$ exists such that:
 * $\omega_f \left({I}\right) - \omega_f \left({x}\right) < \epsilon$

Observe that $\R \in N_x$.

Therefore, $N_x$ is nonempty.

We have:


 * $\omega_f \left({I}\right) \in \overline{\R}_{\ge 0}$ for every $I \in N_x$ by Lemma 6


 * $\overline{\R}_{\ge 0}$ is a subsets of $\overline{\R}$


 * $N_x$ is nonempty

Therefore:


 * $\left\{{\omega_f \left({I}\right): I \in N_x}\right\}$ is a nonempty subset of $\overline{\R}$

We have:


 * $\left\{{\omega_f \left({I}\right): I \in N_x}\right\}$ is a nonempty subset of $\overline{\R}$


 * $\displaystyle \inf \left\{{\omega_f \left({I}\right): I \in N_x}\right\} \in \R$ as $\displaystyle \inf \left\{{\omega_f \left({I}\right): I \in N_x}\right\} = \omega_f \left({x}\right)$

Therefore, an $I \in N_x$ exists such that:

Lemma 3
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 $\omega_f \left({x}\right)$ be the oscillation of $f$ at $x$:


 * $\omega_f \left({x}\right) = \displaystyle \inf \left\{{\omega_f \left({I}\right): I \in N_x}\right\}$

where $\omega_f \left({I}\right)$ is the oscillation of $f$ on a real set $I$:


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

Then:


 * $\omega_f \left({x}\right) \in \R$ $\omega_f \left({I}\right) \in \R$ for an $I$ in $N_x$

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

We need to prove that an $I \in N_x$ exists such that $\omega_f \left({I}\right) \in \R$.

Suppose that there is no $I \in N_x$ for which $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ is bounded above.

Note that $N_x$ is nonempty by observing that $\R \in N_x$.

We have:

This contradicts the fact that $\omega_f \left({x}\right) \in \R$.

Accordingly, an $I \in N_x$ exists such that $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ is bounded above.

For such an $I$, $\omega_f \left({I}\right) \in \R_{\ge 0}$ by Lemma 7.

Therefore, $\omega_f \left({I}\right) \in \R$ as $\R$ is a superset of $\R_{\ge 0}$.

Sufficient Condition
Let $\omega_f \left({I}\right) \in \R$ for an $I$ in $N_x$.

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

$\omega_f \left({I}\right) \in \overline{\R}_{\ge 0}$ for every $I \in N_x$ by Lemma 6.

So, $\left\{{\omega_f \left({I}\right): I \in N_x}\right\}$ is a subset of $\overline{\R}_{\ge 0}$.

$\left\{{\omega_f \left({I}\right): I \in N_x}\right\} \cap \R$ is nonempty as $\omega_f \left({I}\right) \in \R$ for an $I$ in $N_x$.

$\left\{{\omega_f \left({I}\right): I \in N_x}\right\} \cap \R$ is bounded below by zero as $\omega_f \left({I}\right) \in \overline{\R}_{\ge 0}$ for every $I \in N_x$.

It follows by the Continuum Property that $\displaystyle \inf \left({\left\{{\omega_f \left({I}\right): I \in N_x}\right\} \cap \R}\right)$ exists (in $\R$).

There are two cases.

Either:


 * $\left\{{\omega_f \left({I}\right): I \in N_x}\right\} = \left\{{\omega_f \left({I}\right): I \in N_x}\right\} \cap \R$

or:


 * $\left\{{\omega_f \left({I}\right): I \in N_x}\right\} = \left({\left\{{\omega_f \left({I}\right): I \in N_x}\right\} \cap \R}\right) \cup \left\{{\infty}\right\}$

Assume that $\left\{{\omega_f \left({I}\right): I \in N_x}\right\} = \left\{{\omega_f \left({I}\right): I \in N_x}\right\} \cap \R$.

Then $\displaystyle \inf \left\{{\omega_f \left({I}\right): I \in N_x}\right\} = \inf \left({\left\{{\omega_f \left({I}\right): I \in N_x}\right\} \cap \R}\right)$.

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

So, $\omega_f \left({x}\right) \in \R$ by the definition of $\omega_f \left({x}\right)$.

This finishes the proof for this case.

Now, assume that $\left\{{\omega_f \left({I}\right): I \in N_x}\right\} = \left({\left\{{\omega_f \left({I}\right): I \in N_x}\right\} \cap \R}\right) \cup \left\{{\infty}\right\}$.

$\infty$ is greater than $\displaystyle \inf \left({\left\{{\omega_f \left({I}\right): I \in N_x}\right\} \cap \R}\right)$ as $\displaystyle \inf \left({\left\{{\omega_f \left({I}\right): I \in N_x}\right\} \cap \R}\right) \in \R$.

In other words, $\infty$ is greater than the greatest lower bound of $\left\{{\omega_f \left({I}\right): I \in N_x}\right\} \cap \R$.

Therefore, $\infty$ is not a lower bound for $\left\{{\omega_f \left({I}\right): I \in N_x}\right\} \cap \R$.

Furthermore, $\infty$ is not a lower bound for $\left\{{\omega_f \left({I}\right): I \in N_x}\right\}$ either because then it would be a lower bound for $\left\{{\omega_f \left({I}\right): I \in N_x}\right\} \cap \R$ as well.

So, $\left\{{\omega_f \left({I}\right): I \in N_x}\right\}$ and $\left\{{\omega_f \left({I}\right): I \in N_x}\right\} \cap \R$ have the same set of lower bounds.

Therefore, they have the same infima.

In other words, $\displaystyle \inf \left\{{\omega_f \left({I}\right): I \in N_x}\right\} = \displaystyle \inf \left({\left\{{\omega_f \left({I}\right): I \in N_x}\right\} \cap \R}\right)$.

So, $\displaystyle \inf \left\{{\omega_f \left({I}\right): I \in N_x}\right\} \in \R$ as $\displaystyle \inf \left({\left\{{\omega_f \left({I}\right): I \in N_x}\right\} \cap \R}\right) \in \R$.

Therefore, $\omega_f \left({x}\right) \in \R$ by the definition of $\omega_f \left({x}\right)$.

Lemma 4
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 $\omega_f \left({I}\right)$ be the oscillation of $f$ on a real set $I$:


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

Let $I \in N_x$.

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

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

Then:


 * $\omega_f \left({J}\right) \in \R$

and:


 * $\omega_f \left({J}\right) \le \omega_f \left({I}\right)$

Proof
Let:


 * $I, J \in N_x$


 * $J \subset I$


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

where:


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

We need to prove:


 * $\omega_f \left({J}\right) \in \R$


 * $\omega_f \left({J}\right) \le \omega_f \left({I}\right)$

We observe that $\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert \in \R$ for every $y, z \in D$.

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

$\displaystyle \sup \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\} = \omega_f \left({I}\right)$ by the definition of $\omega_f \left({I}\right)$.

Therefore, $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ has a supremum as $\omega_f \left({I}\right) \in \R$.

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

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

Also, $x \in J \cap D$ since $x\in D$ and $x \in J$ as $J \in N_x$.

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

So, $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in J \cap D}\right\}$ is nonempty.

We have found:


 * $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ is a set of real numbers


 * $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ has a supremum


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

from which follows:


 * $\displaystyle \sup \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in J \cap D}\right\}$ exists ($\in \R$) by Supremum of Subset is Less Than or Equal to Supremum of Set


 * $\iff \omega_f \left({J}\right) \in \R$ by definition

and:

Lemma 5
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 $\omega_f \left({x}\right)$ be the oscillation of $f$ at $x$:


 * $\omega_f \left({x}\right) = \displaystyle \inf \left\{{\omega_f \left({I}\right): I \in N_x}\right\}$

where $\omega_f \left({I}\right)$ is the oscillation of $f$ on a real set $I$:


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

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

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

Then:


 * $\omega_f \left({x}\right) \le \omega_f \left({I}\right)$

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

Let $I$ be a set in $N_x$ such that $\omega_f \left({I}\right) \in \R$.

$\omega_f \left({x}\right)$ is a lower bound for $\left\{{\omega_f \left({I'}\right): I' \in N_x}\right\}$ according to the definition of $\omega_f \left({x}\right)$.

Therefore, $\omega_f \left({x}\right) \le \omega_f \left({I}\right)$ as $\omega_f \left({I}\right)$ is an element of $\left\{{\omega_f \left({I'}\right): I' \in N_x}\right\}$.

Lemma 6
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 $I \in N_x$.

Let $\omega_f \left({I}\right)$ be the oscillation of $f$ on $I$:


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

Then $\omega_f \left({I}\right) \in \overline{\R}_{\ge 0}$.

Proof
Let $I \in N_x$.

We observe that $\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert \in \R$ for every $y, z \in D$.

So, $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ is a real set.

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.

if $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ is bounded above (in $\R$), $\omega_f \left({I}\right) \in \R_{\ge 0}$ by Lemma 7.

Otherwise, $\omega_f \left({I}\right) = \infty$ by Lemma 7.

In either case, $\omega_f \left({I}\right) \in \overline{\R}_{\ge 0}$.

Lemma 7
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 $I \in N_x$.

Let $\omega_f \left({I}\right)$ be the oscillation of $f$ on $I$:


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

Then:


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


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


 * otherwise:


 * $\omega_f \left({I}\right) = \infty$

Proof
Let $I \in N_x$.

We observe that $\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert \in \R$ for every $y, z \in D$.

Also, $\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert = 0$ for $y = z = x$.

We have that $x \in I \cap D$ since $x \in I$ as $I \in N_x$, and $x \in D$.

So, $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ is a nonempty real set.

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 $\overline{\R}$.

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.

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 $\R$).

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

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

Therefore, $\displaystyle \sup \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\} \ge 0$ as $\displaystyle \sup \left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ is greater than or equal to every element of $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$.

So, $\omega_f \left({I}\right) \in \R_{\ge 0}$ by the definition of $\omega_f \left({I}\right)$.

Now, 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 $\R$).

Then $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ has $\infty$ as an upper bound as $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ is bounded above in $\overline{\R}$.

Also, $\infty$ is its least upper bound because otherwise it would be bounded above in $\R$.

In other words, the supremum in $\overline{\R}$ of $\left\{{\left\vert{f \left({y}\right) - f \left({z}\right)}\right\vert: y, z \in I \cap D}\right\}$ equals $\infty$.

So, $\omega_f \left({I}\right)$ equals $\infty$ by the definition of $\omega_f \left({I}\right)$.

Lemma 8
Let $S$ be a nonempty subset of $\overline{\R}$.

Let $\inf S$ exist in $\R$.

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

Then:


 * $\exists x \in S: x − \inf S < \epsilon$

Proof
$S$ does not contain the number $-\infty$ because then $\inf S$ would equal $-\infty$.

$S$ does not contain only $\infty$ because then $\inf S$ would equal $\infty$.

Accordingly, $S$ contains a real number.

Let $T = S \cap\R$.

$\inf T = \inf S$ by Lemma 9.

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

We observe:


 * $T$ is a nonempty subset of the real numbers


 * $\inf T \in \R$

Therefore, an $x \in T$ exists such that:

Lemma 9
Let $S$ be a subset of $\overline{\R}$.

Let $S$ contain at least one real number.

Let $\inf S$ exist in $\R$.

Let $T = S \cap \R$.

Then $\inf T = \inf S$.

Proof
$S$ does not contain the number $-\infty$ because otherwise $\inf S$ would equal $-\infty$.

So, the only possible difference between $S$ and $T$ is that $S$ may contain the number $\infty$.

Either $S$ and $T$ are equal or $S = T \cup \left\{{\infty}\right\}$.

Assume that $S$ and $T$ are equal.

Then $\inf T = \inf S$.

This finishes the proof for this case.

Now, assume that $S = T \cup \left\{{\infty}\right\}$.

We get by observing $T = S \cap \R$:


 * $T$ is a real set


 * $T$ is nonempty as $S$ contains a real number

So, $T$ is a nonempty real set.

$\inf S$ is a lower bound for $T$ as $T$ is a subset of $S$.

Therefore, $\inf T$ exists (in $\R$) by the Continuum Property.

$\inf T$ is the greatest lower bound of $T$.

Therefore, $\inf T \ge \inf S$ as $\inf S$ is a lower bound for $T$.

We have:


 * $\inf T < \infty$ as $\inf T \in \R$


 * $\inf T$ is a lower bound for $T$

Therefore:
 * $\inf T$ is a lower bound for $S$ as $S = T \cup \left\{{\infty}\right\}$

and:
 * $\inf T \le \inf S$ as $\inf S$ is the greatest lower bound of $S$

From this follows:


 * $\inf T = \inf S$ as $\inf T \ge \inf S$