Oscillation measures how much a real-valued function varies in small intervals around a point. It gives a precise criterion for continuity and shows that the discontinuity set of any real-valued function on $\mathbb{R}$ must be of type F-sigma. This provides a deep link between real functions and category theory. :::definition[Set of Type F-Sigma] A subset $S$ of $\mathbb{R}$ is of type $F-sigma$ if $S=\bigcup_{n=1}^{\infty}S_n$, where each $S_n$ is closed. ::: :::definition[Oscillation Over an Interval] For $f:\mathbb{R}\to\mathbb{R}$ and a bounded open interval $I$, the $oscillation$ of $f$ over $I$ is $\omega(f,I)=\sup_{x\in I}f(x)-\inf_{x\in I}f(x)$. ::: :::definition[Oscillation at a Point] For $a\in\mathbb{R}$, the $oscillation$ of $f$ at $a$ is $\omega(f,a)=\inf\omega(f,I)$, where $I$ runs over bounded open intervals containing $a$. ::: :::example[Rationals Are F-Sigma] Since $\mathbb{Q}$ is a countable union of singleton closed sets, $\mathbb{Q}$ is F-sigma. ::: :::example[Function Continuous Exactly at Irrationals] Thomae's function is continuous at irrational points and discontinuous at rational points. ::: :::theorem[Oscillation Criterion for Continuity] Let $f:\mathbb{R}\to\mathbb{R}$. Then $f$ is continuous at $a$ if and only if $\omega(f,a)=0$. ::: :::proof Given that $f:\mathbb{R}\to\mathbb{R}$ and $a\in\mathbb{R}$. To prove the criterion. If $f$ is continuous at $a$, then for every $\varepsilon>0$ there is $\delta>0$ such that $|f(x)-f(a)|<\varepsilon/2$ whenever $|x-a|<\delta$. On $I=(a-\delta,a+\delta)$, the oscillation is less than $\varepsilon$, so $\omega(f,a)=0$. Conversely, if $\omega(f,a)=0$, choose an interval $I$ containing $a$ with $\omega(f,I)<\varepsilon$. For a smaller interval around $a$ contained in $I$, every $x$ in that interval satisfies $|f(x)-f(a)|\leq\omega(f,I)<\varepsilon$. Hence $f$ is continuous at $a$. $\square$ ::: :::solved-problem Show that the discontinuity set of $f:\mathbb{R}\to\mathbb{R}$ is F-sigma. ::: :::solution Let $S_n=\{x:\omega(f,x)\geq1/n\}$. Each $S_n$ is closed: a limit point of $S_n$ has every neighbourhood with oscillation at least $1/n$. By the oscillation criterion, $f$ is discontinuous at $x$ exactly when $\omega(f,x)>0$, which is equivalent to $x\in S_n$ for some $n$. Hence the discontinuity set is $\bigcup_{n=1}^{\infty}S_n$. $\square$ ::: :::exercise [1] State the main definition from this lesson. [2] State the main theorem from this lesson. [3] Give one example illustrating the theorem. [4] Explain one common mistake related to this topic. ::: :::answer [1] The main definition is the first titled definition in the lesson. [2] The main theorem is the titled theorem proved in the lesson. [3] The examples in the lesson provide standard choices. [4] A common mistake is to ignore the metric or the ambient space when applying the definition. ::: :::faq[Frequently Asked Questions] Q: Why is the ambient metric space important? A: Openness, closedness, closure, and boundary are interpreted relative to the chosen metric space. Q: Can the same set behave differently in different metric spaces? A: Yes. Changing the metric or the ambient space may change topological behaviour. Q: How should students check a definition? A: They should return to balls, closures, and sequences, depending on the context. ::: :::call-to-action[Continue from Oscillation and Sets of Discontinuity] subtitle: Review the definitions and proofs before moving through the metric topology sequence. button: Next Lesson | /functional-analysis/metric-spaces/topological-structure-of-metric-spaces/boundary-and-exterior-of-sets button-ghost: Previous Lesson | /functional-analysis/metric-spaces/topological-structure-of-metric-spaces/baire-category-theorem-and-applications :::