Statement and proof of Heine's theorem

Statement of the theorem

Heine's theorem. Every function that is continuous on a closed bounded interval $[a \,;\, b]$ is uniformly continuous on it.

The words closed bounded interval are essential: they denote an interval of $\mathbb{R}$ that is both closed and bounded, that is, a compact set. In its general form, due to Heine in 1872, the statement says that a continuous map from a compact metric space to a metric space is uniformly continuous.

The theorem is purely existential: it guarantees that a uniform $\delta$ exists, without ever providing it.

Proof by contradiction

Suppose $f$ is continuous on $[a \,;\, b]$ but not uniformly continuous. The negation of the definition reads:

$$\exists \varepsilon_0 > 0,\; \forall \delta > 0,\; \exists (x, y) \in [a \,;\, b]^2,\quad \lvert x - y \rvert \leq \delta \;\text{ and }\; \lvert f(x) - f(y) \rvert > \varepsilon_0$$

Apply it successively to $\delta = \dfrac{1}{n}$ for each integer $n \geq 1$. We obtain two sequences $(x_n)$ and $(y_n)$ in $[a \,;\, b]$ such that

$$\lvert x_n - y_n \rvert \leq \frac{1}{n} \qquad \text{and} \qquad \lvert f(x_n) - f(y_n) \rvert > \varepsilon_0$$

The sequence $(x_n)$ lives in $[a \,;\, b]$, so it is bounded: the Bolzano–Weierstrass theorem provides a strictly increasing extraction map $\varphi$ and a real number $c$ such that $x_{\varphi(n)} \to c$. Since $[a \,;\, b]$ is closed, the limit $c$ still belongs to $[a \,;\, b]$.

By the squeeze theorem, $y_{\varphi(n)} \to c$ as well, since

$$\lvert y_{\varphi(n)} - c \rvert \leq \lvert y_{\varphi(n)} - x_{\varphi(n)} \rvert + \lvert x_{\varphi(n)} - c \rvert \leq \frac{1}{\varphi(n)} + \lvert x_{\varphi(n)} - c \rvert \xrightarrow[n \to +\infty]{} 0$$

The continuity of $f$ at the single point $c$ then gives $f(x_{\varphi(n)}) \to f(c)$ and $f(y_{\varphi(n)}) \to f(c)$, hence

$$\lvert f(x_{\varphi(n)}) - f(y_{\varphi(n)}) \rvert \xrightarrow[n \to +\infty]{} 0$$

which contradicts the strict lower bound $\varepsilon_0 > 0$. The initial assumption is therefore absurd, and $f$ is uniformly continuous on $[a \,;\, b]$.

Variant using a covering

A second proof uses the Heine–Borel theorem. For $\varepsilon > 0$, continuity at each point $x$ gives a $\delta_x > 0$ such that $\lvert f(y) - f(x) \rvert \leq \tfrac{\varepsilon}{2}$ whenever $\lvert y - x \rvert \leq \delta_x$. The open intervals centered at $x$ with radius $\tfrac{\delta_x}{2}$ cover $[a \,;\, b]$; we extract a finite subcover indexed by $x_1, \dots, x_m$, and set

$$\delta = \min\left(\frac{\delta_{x_1}}{2}, \dots, \frac{\delta_{x_m}}{2}\right) > 0$$

This minimum is strictly positive because the family is finite: that is exactly where compactness comes in. The triangle inequality then lets us conclude.

Classic pitfall

The theorem fails as soon as either of the two hypotheses on the interval is dropped.

Interval	Example	Uniformly continuous?
$[0 \,;\, 1]$, closed and bounded	$x \mapsto x^2$	yes, by Heine
$\left]0 \,;\, 1\right]$, not closed	$x \mapsto \dfrac{1}{x}$	no
$\left[0 \,;\, +\infty\right[$, not bounded	$x \mapsto x^2$	no

Another frequent confusion: Heine says nothing about differentiability. The function $x \mapsto \lvert x \rvert$ on $[-1 \,;\, 1]$ is not differentiable at $0$ and is still uniformly continuous.