Uniformly Continuous iff Compact and Continuous

Theorem

A continuous function on a compact set is uniformly continuous.

The idea here is that in our example in Lecture 26 - Uniform Continuity#Back to Uniform Continuity the function $f$ would be still continuous if $A = [0, 1]$ instead (same definition of $f$ , just include the endpoints on the interval). In contrast, $g$ couldn't get that same extension.

Proof

Let $f : K \to R$ be continuous, with $K$ being compact.

To show $f$ is uniformly continuous, follow the definition. Let $ε > 0$ . Since $f$ is continuous, then $δ > 0$ where $\forall x \in K$ then picking $\forall x^{'} \in K$ has $0 < | x - x^{'} | < δ \to | f (x) - f (x^{'}) | < \frac{ε}{2}$ .

The problem here is that $δ$ may change with each $x$ , so first notice that if I consider the following collection (we denote the $δ$ we choose for each $x$ as $δ (x)$ ):

{V_{\frac{δ (x)}{2}} (x) | x \in K}

This is an Open Cover of $K$ . By the Heine-Borel Theorem (Compact Equivalences) then there is a finite subcover $\exists x_{1}, x_{2}, \dots, x_{n}$ such that:

K \subseteq V_{\frac{δ (x)}{2}} (x_{1}) \cup \dots \cup V_{\frac{δ (x_{n})}{2}} (x_{n})

Now choose $δ = min {\frac{δ (x_{1})}{2}, \dots, \frac{δ (x_{n})}{2}}$ . Let $x, y \in K$ where $| x - y | < δ$ . Consider $x \in K$ then by our construction then $\exists x_{k}$ such that $x \in V_{\frac{δ (x_{k})}{2}} (x_{k})$ . Therefore, using the given for being continuous then:

\Rightarrow | x - x_{k} | < \frac{δ (x_{k})}{2} < δ (x_{k}) \Rightarrow | f (x) - f (x_{k}) | < \frac{ε}{2}

Now notice:

\begin{aligned} | y - x_{k} | & \leq | y - x | + | x - x_{k} | \\ < δ + \frac{δ (x_{k})}{2} \\ \leq \frac{δ (x_{k})}{2} + \frac{δ (x_{k})}{2} \\ = δ (x_{k}) \\ \Rightarrow | f (y) - f (x_{k}) | < \frac{ε}{2} & Def’n Continuity \end{aligned}

Now using that we got $| f (x) - f (x_{k}) | < \frac{ε}{2}$ then:

\begin{aligned} | f (x) - f (y) | & \leq | f (x) - f (x_{k}) | + | f (x_{k}) - f (y) | \\ < \frac{ε}{2} + \frac{ε}{2} \\ = ε \end{aligned}

☐