Real Function is Continuous at Point iff Oscillation is Zero

From ProofWiki
Jump to navigation Jump to search

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:

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


Then $\omega_f \left({x}\right) = 0$ if and only if $f$ is continuous at $x$.


Proof

Necessary Condition

Let $\omega_f \left({x}\right) = 0$.

Let $\epsilon > 0$.

Suppose that $\forall I: I \in N_x: \omega_f \left({I} \right) \ge \epsilon$.

Then by definition, $\omega_f \left({x}\right) \ge \epsilon$.

This contradicts $\omega_f \left({x}\right) = 0$.

From this contradiction we deduce that:

$\exists I: I \in N_x: \omega_f \left({I}\right) < \epsilon$

For this particular $I$, there is an open set $O \subset I$ by the definition of open subset neighborhood.

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

So for our specific $x$, if $y$ satisfies:

$\left \vert {x - y} \right \vert < \delta$

then:

$y \in I$

and, if $y \in D$:

$\left\vert {f \left({x}\right) - f \left({y}\right)} \right\vert \le \omega_f \left({I}\right)$

Since $\omega_f \left({I}\right) < \epsilon$ it follows by the definition of continuity that $f$ is continuous at $x$.

$\Box$


Sufficient Condition

Let $f$ be continuous at $x$.

Then $\forall \epsilon > 0: \exists \delta \in \R_{>0}$ such that, if $y \in D$:

$\left\vert{x-y}\right\vert < \delta \implies \left \vert{f \left({x}\right) - f \left({y}\right)}\right\vert < \epsilon$


Let the interval $I_\delta$ be defined as:

$I_\delta := \left({x - \delta \,.\,.\, x + \delta}\right)$

$I_\delta$ is an element of $N_x$ as $I_\delta$ is a neighborhood of $x$.

We have:

\(\displaystyle \omega_f \left({I_\delta}\right)\) \(=\) \(\displaystyle \sup \left\{ {\left \vert{f \left({y}\right) - f \left({z}\right)}\right \vert : y, z \in I_\delta \cap D}\right\}\) Definition of $\omega_f \left({I_\delta}\right)$
\(\displaystyle \) \(=\) \(\displaystyle \sup \left\{ {\left \vert{f \left({y}\right) - f \left({x}\right) + f \left({x}\right) - f \left({z}\right)}\right \vert : y, z \in I_\delta \cap D}\right\}\)
\(\displaystyle \) \(\le\) \(\displaystyle \sup \left\{ {\left \vert{f \left({y}\right) - f \left({x}\right)}\right \vert + \left \vert{f \left({x}\right) - f \left({z}\right)}\right \vert : y, z \in I_\delta \cap D}\right\}\) Triangle Inequality and Supremum of Function is less than Supremum of Greater Function
\(\displaystyle \) \(=\) \(\displaystyle \sup \left\{ {\left \vert{f \left({y}\right) - f \left({x}\right)}\right \vert : y \in I_\delta \cap D}\right\} + \sup \left\{ {\left \vert{f \left({x}\right) - f \left({z}\right)}\right \vert : z \in I_\delta \cap D}\right\}\) Supremum of Sum equals Sum of Suprema
\(\displaystyle \) \(=\) \(\displaystyle \sup \left\{ {\left \vert{f \left({x}\right) - f \left({y}\right)}\right \vert : y \in I_\delta \cap D}\right\} + \sup \left\{ {\left \vert{f \left({x}\right) - f \left({y}\right)}\right \vert : y \in I_\delta \cap D}\right\}\)
\(\displaystyle \) \(=\) \(\displaystyle 2 \sup \left\{ {\left \vert{f \left({x}\right) - f \left({y}\right)}\right \vert : y \in I_\delta \cap D} \right\}\)
\(\displaystyle \) \(<\) \(\displaystyle 2 \epsilon\) as $\left \vert{f \left({x}\right) - f \left({y}\right)}\right \vert < \epsilon$ whenever $y \in I_\delta \cap D$

This gives:

\(\displaystyle \omega_f \left({x}\right)\) \(=\) \(\displaystyle \inf \left\{ {\omega_f \left({I}\right): I \in N_x}\right\}\) Definition of $\omega_f \left({x}\right)$
\(\displaystyle \) \(\le\) \(\displaystyle \omega_f \left({I_{\delta} }\right)\) Definition of Infimum and that $I_\delta \in N_x$
\(\displaystyle \) \(<\) \(\displaystyle 2 \epsilon\)


This holds true for any value of $\epsilon$.

Thus $\omega_f \left({x}\right)$ must be $0$.

Hence the result.

$\blacksquare$