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$$0$, a $\delta$ that may depend on the point. The Heine-Cantor theorem says that on a closed bounded interval this dependence can be removed: one $\delta$ works at every point of $[a,b]$ at once, including near the endpoints. This does not require bounded slopes: $\sqrt{x}$ is uniformly continuous on $[0,1]$, although its derivative $1/(2\sqrt{x})$ is unbounded as $x\downarrow0$$0$.

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: for $n\ge2$ the points $1/n$ and $1/(n+1)$ lie in $(0,1)$ and get arbitrarily close as $n$ grows, yet their values $n$ and $n+1$ always differ by $1$. 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{{\delta }^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 bounded interval $[a,b]$ 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¶

• Interval
• Continuity

Related Concepts¶

• Uniform Continuity between Metric Spaces
• Uniform Continuity
• Compact Set
• Extreme Value Theorem on Compact Sets

Sources¶

• Math III: Analysis · Lecture 1

Symbols: delta, epsilon, infinity, real numbers, square root

Last updated: 27 October 2026 11:52