Heine-Cantor Theorem¶

Theorem¶

Let $a < b$ be real numbers, and consider a function $f\colon [a, b] \to \mathbb{R}$ that is continuous on $[a, b]\text{.}$ Then, $f$ is uniformly continuous on $[a, b]\text{.}$

Interpretation¶

Continuity of $f$ at each point gives, for every $\varepsilon>0$ $0$ , a $\delta$ that may depend on the point. The Heine-Cantor theorem says that on a closed bounded interval this dependence can be removed: one $\delta$ works at every point of $[a,b]$ at once. The function cannot become steeper and steeper towards an edge, because the endpoints belong to the interval and $f$ is continuous there as well.

The graph of x squared over 2 on the interval from a to b with three equal-sized rectangles of width 2 delta and height 2 epsilon centred on the graph, each containing the graph over its whole width.

The figure takes $f(x)=\tfrac{1}{2}x^2$ with $a=0$ and $b=2$ $2$ . Each rectangle has width $2\delta$ and height $2\varepsilon$ and is centred on a point of the graph, and the graph stays inside every rectangle over its whole width. Since $|f'(x)|=|x|\le2$ on $[0,2]$ , the mean value theorem gives $|f(x_1)-f(x_2)|\le2|x_1-x_2|$ $x_{2} ∣$ , so $\delta=\varepsilon/2$ works everywhere.

Both properties of the domain matter. On the open interval $(0,1)$ the continuous function $1/x$ is not uniformly continuous: it becomes arbitrarily steep near the missing endpoint $0$ . On the closed but unbounded interval $[0,\infty)$ the function $x^2$ fails too: the points $x$ and $x+\tfrac{\delta}{2}$ are closer than $\delta$ , yet their values differ by $x\delta+\tfrac{\delta^2}{4}$ $\frac{δ ^{2}}{4}$ , which exceeds any $\varepsilon$ for large $x$ . The theorem gives a sufficient condition only, as the Examples show.

The theorem is the usual step in showing that continuous functions on $[a,b]$ are Riemann integrable: one $\delta$ controls the variation of $f$ on every subinterval of a partition at once.

"Uniform" does not mean that $\delta$ is independent of $\varepsilon$ . It still depends on $\varepsilon$ , only no longer on the point.

The theorem returns later with $[a,b]$ replaced by more general closed and bounded sets, the compact sets.

Key Results¶

This theorem has a generalized version if $[a,b]$ is replaced by a compact set.
This theorem does not imply that uniform continuity holds only on closed intervals.

Examples¶

The function $f(x)=\dfrac{x^2+\sin x}{\mathrm{e}^x}$ is uniformly continuous on any closed interval since we know it is continuous everywhere.
The function $g(x)=x$ is uniformly continuous on $(0,1)$ though $(0,1)$ is not a closed interval. This does not contradict the theorem since the "closed interval" condition is sufficient but not necessary.

Prerequisites¶

Sources¶

Math III: Analysis · Lecture 1

Symbols: delta, epsilon, infinity, real numbers

Last updated: 20 October 2026 03:02