Heine-Cantor Theorem

From Closed Intervals to Compact Domains

The Heine-Cantor Theorem for intervals established that a function continuous at every point of a closed bounded interval is uniformly continuous there. The key feature is not that the domain fills an interval. The same conclusion holds on any compact subset of the real line, including sets with gaps or isolated points.

Recall that a set is compact if every open cover of the set has a finite subcover. In \(\mathbb{R}\), the Heine-Borel Theorem says that a set is compact if and only if it is closed and bounded. Thus compact domains include closed intervals, finite unions of closed bounded intervals, and closed bounded sets that may have a more irregular shape.

The conclusion is stronger than pointwise continuity. At each point \(a\in E\), continuity gives a radius that may depend on \(a\). Uniform continuity requires one choice of radius for each \(\varepsilon\) that works for every pair of points in the domain. Compactness makes it possible to pass from local control to a single global choice. In a proof, a finite subcover reduces the local information to finitely many neighborhoods; an additional argument then ensures that sufficiently close points are controlled within one of them.

What the Theorem Allows

The theorem does not require the domain to be an interval, nor does it require the function to be given by one formula on an interval containing the domain. It requires continuity on the domain itself and compactness of that domain. Continuity on \(E\) is understood relative to \(E\): at \(a\in E\), only inputs \(x\in E\) need to be considered.

In this example the isolated points impose no continuity restriction beyond the values at those points: each has a neighborhood containing no other points of \(E\). The point \(0\), where the domain accumulates, is where continuity needs to be checked by a limit estimate. Once continuity holds everywhere on this compact domain, the theorem gives the global conclusion.

A Useful Consequence: Continuous Functions on Compact Sets Are Bounded

The same finite-subcover idea gives a related result. It uses the local boundedness theorem from “Continuity at a Point”: a function continuous at \(a\) is bounded on some neighborhood of \(a\), relative to its domain.

Proof. If \(E=\varnothing\), it is bounded, for example with bound \(M=0\). Now suppose \(E\ne\varnothing\). For each \(a\in E\), continuity at \(a\) and the local boundedness theorem give a radius \(r_a>0\) and a finite bound \(M_a\geq0\) such that \(|f(x)|\leq M_a\) whenever \(x\in E\) and \(|x-a|<r_a\). The open intervals \((a-r_a,a+r_a)\), as \(a\) ranges over \(E\), cover \(E\). Compactness gives a finite subcover, say the intervals centered at \(a_1,\ldots,a_k\). Set \(M=\max\{M_{a_1},\ldots,M_{a_k}\}\). Every \(x\in E\) lies in at least one of these finitely many intervals, so \(|f(x)|\leq M_{a_j}\leq M\) for some \(j\). Therefore \(f\) is bounded on \(E\). \(\square\)

The finite subcover is essential to this argument: it reduces potentially many local bounds to a finite list, whose maximum is still finite. This boundedness conclusion and the Heine-Cantor Theorem are related consequences of compactness, but boundedness alone does not establish uniform continuity. For that, the proof must control the output difference between two nearby inputs, not just the size of the function values.

Proof. If \(E\) is empty, the defining condition for uniform continuity holds vacuously. If \(E\) has one point, choose \(\delta=1\) for any \(\varepsilon>0\); every pair \(x,y\in E\) is equal, so \(|f(x)-f(y)|=0<\varepsilon\). Now suppose \(E\) contains at least two points. There are finitely many positive distances \(|x-y|\) with \(x,y\in E\) and \(x\ne y\), so their minimum \(d\) exists and is positive. For any \(\varepsilon>0\), choose \(\delta=d\). If \(x,y\in E\) satisfy \(|x-y|<\delta=d\), they cannot be distinct, because distinct points have distance at least \(d\). Thus \(x=y\), and \(|f(x)-f(y)|=0<\varepsilon\). This proves uniform continuity. \(\square\)

Why Boundedness Alone Is Not Enough

The Heine-Cantor Theorem depends on compactness, not merely boundedness. In \(\mathbb{R}\), boundedness without closedness does not guarantee compactness. A continuous function on a bounded, noncompact domain can fail to be uniformly continuous.

This example identifies the role of compactness: continuity alone gives control near each fixed point, but near the missing endpoint \(0\), the reciprocal function changes rapidly. A compact domain cannot have such a missing accumulation point. More generally, the theorem guarantees that no location in a compact domain can require an arbitrarily smaller input tolerance than every other location.

How to Apply the Heine-Cantor Theorem

When using the theorem, check the hypotheses separately. First establish that the domain is compact; for subsets of \(\mathbb{R}\), it is enough to verify that the set is closed and bounded, by Heine-Borel. Then establish continuity at every point of that domain. Only after both conditions are verified can the theorem be applied.

The conclusion is an existence statement: for each \(\varepsilon>0\), some \(\delta>0\) works uniformly. The theorem may not give a convenient formula for \(\delta\). If a quantitative estimate is available, such as a Lipschitz bound, it can provide an explicit choice. Otherwise, compactness still guarantees that a suitable choice exists.

A common error is to argue that a continuous function must be uniformly continuous simply because it is continuous at every point. The example \(1/x\) on \((0,1)\) shows why this is false. Another is to assume that the domain must be an interval: compactness permits arbitrary closed bounded subsets of the real line, including finite sets and disconnected sets. The Heine-Cantor Theorem is precisely the principle that continuity becomes uniform when the domain has this compactness property.