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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

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.

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

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

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

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

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