TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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Heine-Cantor in $\mathbb{R}$ : a continuous real function on a compact subset of $\mathbb{R}$ is uniformly continuous, proved $\mathbb{R}$ -natively from sequential compactness

Statement

Let $K \subseteq \mathbb{R}$ be compact (Open cover, subcover, compact subset of $\mathbb{R}$ (every open cover has a finite subcover), and sequentially compact subset) and let $f : K \to \mathbb{R}$ be continuous on $K$ (Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ : the $\varepsilon$ - $\delta$ 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 $f : A \to \mathbb{R}$ : one $\delta$ 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 is $\mathbb{R}$ -native and runs through A subset of $\mathbb{R}$ is compact iff it is sequentially compact, which is order-based. That the two statements are the same statement in two vocabularies is Dictionary: for $A \subseteq \mathbb{R}$ with the metric $d(x,y) = |x-y|$ , continuity and uniform continuity of $f : A \to \mathbb{R}$ agree with the metric-space notions, the Lipschitz and Hölder conditions are the metric ones instantiated, and a subset of $\mathbb{R}$ is compact in the open-cover sense of $\mathbb{R}$ exactly when it is a compact metric subspace, clauses 1, 2 and 5, immediately above.

The choice cost, named. The proof invokes the axiom of countable choice (The Axiom of Countable Choice ( $\mathrm{AC}_\omega$ )) exactly once, at step 3.1, to select one bad pair of points from each of countably many nonempty sets. The backward implication of A subset of $\mathbb{R}$ is compact iff it is sequentially compact also spends countable choice, and that item names its own uses; the forward implication used here, from compact to sequentially compact, does not. No claim is made that the axiom is necessary for either.

Facts & Assumptions

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

[L1]

Uniform continuity on $K$ : for every real $\varepsilon > 0$ there is a real $\delta > 0$ such that all $x, x' \in K$ with $|x - x'| < \delta$ satisfy $|f(x) - f(x')| < \varepsilon$ . Its negation: there is a real $\varepsilon_0 > 0$ such that for every real $\delta > 0$ some pair $x, x' \in K$ has $|x - x'| < \delta$ and $|f(x) - f(x')| \ge \varepsilon_0$ (Uniform continuity of $f : A \to \mathbb{R}$ : one $\delta$ serving every pair of points of $A$ , Ordered field).

[L2]

A compact subset of $\mathbb{R}$ is sequentially compact: every sequence with all terms in $K$ has a subsequence converging to a point of $K$ (A subset of $\mathbb{R}$ is compact iff it is sequentially compact, Open cover, subcover, compact subset of $\mathbb{R}$ (every open cover has a finite subcover), and sequentially compact subset, Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals).

[L3]

Countable choice: for a family $(P_k)_{k \in \mathbb{N}}$ of nonempty sets there is a function on $\mathbb{N}$ picking an element of each (The Axiom of Countable Choice ( $\mathrm{AC}_\omega$ )).

[L4]

A strictly increasing index map satisfies $n_j \ge j$ (A strictly increasing index map satisfies $n_k \ge k$ , Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L5]

Archimedean property in reciprocal form: for every real $\eta > 0$ there is a natural $n \ge 1$ with $1/n < \eta$ ; and $0 < s \le t$ implies $1/t \le 1/s$ (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ , Every complete ordered field is Archimedean, Inverses of positives are positive, and reciprocation reverses order).

[L6]

Sequential criterion, the choice-free direction: if $f$ is continuous at $p \in K$ and $(z_j)$ has terms in $K$ with $z_j \to p$ , then $f(z_j) \to f(p)$ ( $f$ is continuous at $c \in A$ if and only if $f(x_k) \to f(c)$ for every sequence in $A$ converging to $c$ , the converse direction costing countable choice, Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ : the $\varepsilon$ - $\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point).

[L7]

Triangle inequality and absolute value: $|u + v| \le |u| + |v|$ , $|{-u}| = |u|$ , $|u| \ge 0$ (The triangle inequality, Basic properties of the absolute value).

[L8]

Convergence of real sequences is tested at rational $\varepsilon > 0$ , and below every positive real lies a positive rational, so the test may equally be run at every real $\varepsilon > 0$ (Limits and Cauchy sequences of reals, The rationals embed densely in the reals).

Proof

technique · contradiction

1.1

Suppose $f$ is not uniformly continuous on $K$ . By [L1] fix a real $\varepsilon_0 > 0$ such that for every real $\delta > 0$ there are $x, x' \in K$ with $|x - x'| < \delta$ and $|f(x) - f(x')| \ge \varepsilon_0$ .

L1assume-contra

2.1

For $k \in \mathbb{N}$ put $P_k := \{\, (x,x') \in K \times K \ : \ |x - x'| < 1/(k+1) \text{ and } |f(x) - f(x')| \ge \varepsilon_0 \,\}$ . Since $1/(k+1) > 0$ , step 1.1 makes every $P_k$ nonempty.

step 1.1L5

3.1

By [L3] applied to the family $(P_k)_{k \in \mathbb{N}}$ fix a function $k \mapsto (x_k, x'_k)$ with $(x_k, x'_k) \in P_k$ for every $k$ . This is the single use of countable choice in this proof.

step 2.1L3choose

4.1

$(x_k)$ is a sequence of reals with all terms in $K$ , so by [L2] there are a strictly increasing $n : \mathbb{N} \to \mathbb{N}$ and $p \in K$ with $x_{n_j} \to p$ .

step 3.1L2choose

5.1

The second sequence converges to $p$ as well. Let a rational $\varepsilon > 0$ be given. By [L5] and [L8] fix $J_1$ with $1/(j+1) < \varepsilon/2$ for every $j \ge J_1$ , and by step 4.1 fix $J_2$ with $|x_{n_j} - p| < \varepsilon/2$ for every $j \ge J_2$ . For $j \ge \max\{J_1, J_2\}$ , using $(x_{n_j}, x'_{n_j}) \in P_{n_j}$ and $n_j \ge j$ from [L4], we get $|x'_{n_j} - x_{n_j}| < 1/(n_j+1) \le 1/(j+1) < \varepsilon/2$ , hence $|x'_{n_j} - p| \le |x'_{n_j} - x_{n_j}| + |x_{n_j} - p| < \varepsilon$ by [L7]. So $x'_{n_j} \to p$ .

step 3.1step 4.1L4L5L7L8

6.1

The point $p$ lies in $K$ and $f$ is continuous at $p$ , so [L6] applied to the two sequences of steps 4.1 and 5.1, both with terms in $K$ , gives $f(x_{n_j}) \to f(p)$ and $f(x'_{n_j}) \to f(p)$ .

step 4.1step 5.1L6

7.1

By [L8] fix a rational $\varepsilon$ with $0 < \varepsilon < \varepsilon_0/2$ , and by step 6.1 fix $J$ with $|f(x_{n_j}) - f(p)| < \varepsilon$ and $|f(x'_{n_j}) - f(p)| < \varepsilon$ for every $j \ge J$ . For such $j$ , [L7] gives $|f(x_{n_j}) - f(x'_{n_j})| \le |f(x_{n_j}) - f(p)| + |f(p) - f(x'_{n_j})| < 2\varepsilon < \varepsilon_0$ .

step 6.1L7L8choose

8.1

But $(x_{n_j}, x'_{n_j}) \in P_{n_j}$ gives $|f(x_{n_j}) - f(x'_{n_j})| \ge \varepsilon_0$ for every $j$ , which contradicts step 7.1. The assumption of step 1.1 is therefore false, and $f$ is uniformly continuous on $K$ .

step 3.1step 7.1discharge-contradiction∎

Remarks

Where compactness is used, and where continuity is used. Compactness is used once, in step 4.1, to extract a convergent subsequence; continuity is used once, in step 6.1, at the single point $p$ that the extraction produces. Neither can be weakened: $x \mapsto 1/x$ on $(0,1)$ is continuous on a bounded non-closed set and not uniformly continuous ( $x \mapsto 1/x$ is continuous on $(0,1)$ and not uniformly continuous there, the pairs $1/(k+2)$ and $1/(k+3)$ defeating every $\delta$ ↗), and $x \mapsto x^{2}$ on $\mathbb{R}$ is continuous on a closed unbounded set and not uniformly continuous ( $x \mapsto x^2$ is continuous on $\mathbb{R}$ and not uniformly continuous, the pairs $k+1$ and $k+1+1/(k+1)$ defeating every $\delta$ ↗).
The converse is sharp. For every noncompact $E \subseteq \mathbb{R}$ 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 $E \subseteq \mathbb{R}$ 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.
The pairs, not the points, are what is chosen. A common presentation selects two sequences separately and then extracts twice. Selecting the pair once, as above, keeps the count of choice applications at one and makes the second sequence's convergence a consequence rather than a second extraction.

Depends on

Used by

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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 R\mathbb{R}R: a continuous real function on a compact subset of R\mathbb{R}R is uniformly continuous, proved R\mathbb{R}R-natively from sequential compactness