# Continuity

Some inclusions on the real line:

• Differentiable with a bounded derivative $$\subset$$ Lipschitz continuous $$\subset$$ absolutely continuous $$\subset$$ uniformly continuous $$\subset$$ continuous

Proofs:

• Mean Value Theorem,
• Triangle inequality,
• Definition of absolute continuity specialized to one interval,
• Definition of uniform continuity

There is a function discontinuous precisely on $${\mathbf{Q}}$$.

$$f(x) = \frac 1 n$$ if $$x = r_n \in {\mathbf{Q}}$$ is an enumeration of the rationals, and zero otherwise. The limit at every point is 0.

There do not exist functions that are discontinuous precisely on $${\mathbf{R}}\setminus {\mathbf{Q}}$$.

$$D_f$$ is always an $$F_\sigma$$ set, which follows by considering the oscillation $$\omega_f$$. Use that $$\omega_f(x) = 0 \iff f$$ is continuous at $$x$$, and $$D_f = \cup_n A_{\frac 1 n}$$ where $$A_\varepsilon = \left\{{\omega_f \geq \varepsilon}\right\}$$ is closed.

An alternative characterization of uniform continuity: \begin{align*} \left\|\tau_{y} f-f\right\|_{u} \rightarrow 0 \text { as } y \rightarrow 0 \end{align*}

If $$f$$ is Lipschitz on $$X$$, then $$f$$ is uniformly continuous on $$X$$.

Supposing that \begin{align*} {\left\lVert {f(x) - f(y)} \right\rVert} \leq C {\left\lVert {x-y} \right\rVert} ,\end{align*} for a fixed $${\varepsilon}$$ take $$\delta({\varepsilon}) \coloneqq{\varepsilon}/C$$, then \begin{align*} {\left\lVert {f(x) - f(y)} \right\rVert} &\leq C {\left\lVert {x-y} \right\rVert} \\ &\leq C \delta \\ &= C \qty{{\varepsilon}/C} \\ &= {\varepsilon} .\end{align*}

Every continuous function $$f:X\to Y$$ where $$X$$ is a compact metric space is uniformly continuous. As a result, if $$f:U\to {\mathbf{R}}$$ is continuous, then $$f$$ is uniformly continuous on any $$K \subseteq U$$ compact.

Fix $${\varepsilon}>0$$, we’ll find a $$\delta$$ that works for all $$x\in X$$ uniformly. For every $$x\in X$$, pick a $$\delta_x$$ neighborhood satisfying the conditions for (assumed) continuity. Take an open cover by $$\delta_x/2$$ balls, extract a finite subcover, take $$\delta$$ the minimal radius.

If $$\mathcal F \subset C(X)$$ is a family of continuous functions on $$X$$, then $$\mathcal F$$ equicontinuous at $$x$$ iff

\begin{align*} \forall \varepsilon > 0 ~~\exists U \ni x \text{ such that } y\in U \implies {\left\lvert {f(y) - f(x)} \right\rvert} < \varepsilon \quad \forall f\in \mathcal{F} .\end{align*}