TheoremStatement: Literature-sourcedProof: AI-adaptedverified 2026-09-26 (gpt-6-sol)

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Heine-Cantor in $R$ : a continuous real function on a compact subset of $R$ is uniformly continuous

Statement

Let $K \subseteq R$ be compact (Open cover, subcover, compact subset of $R$ (every open cover has a finite subcover), and sequentially compact subset) and let $f : K \to R$ be continuous on $K$ (Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point). Then $f$ is uniformly continuous on $K$ (Uniform continuity of $f : A \to R$ : one $δ$ serving every pair of points of $A$ ).

This theorem is stated twice in this library, on purpose. Its metric-space twin is Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous, proved there from the cover machinery of metric spaces; the proof below uses the open-cover definition of compactness in $R$ directly. That the two statements are the same statement in two vocabularies is Dictionary: for $A \subseteq R$ with the metric $d (x, y) = ∣ x - y ∣$ , continuity and uniform continuity of $f : A \to R$ agree with the metric-space notions, the Lipschitz and Hölder conditions are the metric ones instantiated, and a subset of $R$ is compact in the open-cover sense of $R$ exactly when it is a compact metric subspace, clauses 1, 2 and 5, immediately above.

Choice-free proof. The argument below uses the open-cover definition of compactness directly. It considers every continuity neighbourhood at once and extracts only a finite subcover, so no countable or global choice function is selected.

Facts & Assumptions

Given: A compact set $K \subseteq R$ and a continuous function $f : K \to R$ .

[L1]

Continuity at $x \in K$ means that for every $η > 0$ there is $r > 0$ such that $∣ f (y) - f (x) ∣ < η$ whenever $y \in K$ and $∣ y - x ∣ < r$ (Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point).

[L2]

Every open cover of compact $K$ has a finite subcover (Open cover, subcover, compact subset of $R$ (every open cover has a finite subcover), and sequentially compact subset).

[L3]

Uniform continuity means that for every $ε > 0$ one $δ > 0$ works for every pair of points of $K$ at distance less than $δ$ (Uniform continuity of $f : A \to R$ : one $δ$ serving every pair of points of $A$ ).

[L4]

The real absolute value satisfies the triangle inequality (The triangle inequality, Basic properties of the absolute value).

Proof

technique · direct

1.1givenL1

If $K = \emptyset$ , the assertion is immediate. Otherwise fix $ε > 0$ . For each pair $(x, r)$ with $x \in K$ , $r > 0$ , and $∣ f (y) - f (x) ∣ < ε / 2$ for every $y \in K$ satisfying $∣ y - x ∣ < 2 r$ , put $U_{x, r} : = (x - r, x + r)$ . The family of all such open intervals is an open cover of $K$ : for each $x$ , [L1] gives the existence of at least one admissible $r$ , and $x \in U_{x, r}$ . This defines the full family by a property, without selecting one radius for each $x$ .

2.1step 1.1L2

By [L2], finitely many members $U_{x_{1}, r_{1}}, \dots, U_{x_{m}, r_{m}}$ cover $K$ . Since $K$ is nonempty, $m \geq 1$ . Set $δ = \min_{1 \leq i \leq m} r_{i} > 0$ .

3.1step 1.1step 2.1L3L4∎

Let $y, z \in K$ with $∣ y - z ∣ < δ$ , and choose one index $i$ with $y \in U_{x_{i}, r_{i}}$ . Then $∣ y - x_{i} ∣ < r_{i} < 2 r_{i}$ and, by [L4], $∣ z - x_{i} ∣ \leq ∣ z - y ∣ + ∣ y - x_{i} ∣ < δ + r_{i} \leq 2 r_{i}$ . Admissibility of $(x_{i}, r_{i})$ gives $∣ f (y) - f (x_{i}) ∣ < ε / 2$ and $∣ f (z) - f (x_{i}) ∣ < ε / 2$ . Thus $∣ f (y) - f (z) ∣ < ε$ . The same $δ$ works for every such pair, proving uniform continuity by [L3].

Remarks

Where compactness is used. Continuity supplies all admissible local balls in step 1.1; compactness extracts finitely many in step 2.1. Taking the least of their radii gives the uniform bound. No simultaneous selection over the points of $K$ is made.
The converse is sharp. For every noncompact $E \subseteq R$ that is bounded there is a continuous function on $E$ that is not uniformly continuous, and for every noncompact $E$ there is an unbounded continuous function and a bounded continuous one with no greatest value. That is Rudin 4.20, the sharp converse: on a noncompact $E \subseteq R$ there is an unbounded continuous function and a bounded continuous function with no greatest value, and if $E$ is bounded there is a continuous function on $E$ that is not uniformly continuous, later on this page, and together with this theorem it says that compactness is exactly the hypothesis these results need.

Depends on

Used by

Dependency tree · two levels

37 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Heine-Cantor theorem (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 4 (Thm 4.19) (standard reference, not scraped)
J. Lebl, Basic Analysis I, §3.4 (standard reference, not scraped)
W. Trench, Introduction to Real Analysis, Ch. 8: Metric Spaces (standard reference, not scraped)
J. Lebl, Basic Analysis I, §3.3: Uniform continuity (standard reference, not scraped)

Heine-Cantor in RR: a continuous real function on a compact subset of RR is uniformly continuous