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

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Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous

Statement

Let $(X,d_X)$ be a compact metric space (Open cover, subcover, compact metric space, and compact subset of a metric space), let $(Y,d_Y)$ be any metric space (Metric space: $d(x,y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric) and let $f : X \to Y$ be continuous (Continuity of a map between metric spaces, at a point and globally, in the $\varepsilon$ - $\delta$ form). Then $f$ is uniformly continuous (Uniform continuity of a map of metric spaces: one $\delta$ serving every point).

No choice principle is used: the cover built below is cut out by a property, and the Lebesgue number lemma it is fed to is itself choice free (Every open cover of a compact metric space has a Lebesgue number: a $\delta > 0$ such that every nonempty subset of diameter less than $\delta$ lies inside a single member of the cover).

Facts & Assumptions

Given: A compact metric space $(X,d_X)$ , a metric space $(Y,d_Y)$ and a continuous $f : X \to Y$ .

[L1]

$f$ is continuous at $a$ : for every real $\varepsilon > 0$ there is a real $\delta > 0$ with $f[B_X(a,\delta)] \subseteq B_Y(f(a),\varepsilon)$ (Continuity of a map between metric spaces, at a point and globally, in the $\varepsilon$ - $\delta$ form, Open ball, closed ball and sphere in a metric space).

[L2]

$f$ is uniformly continuous when for every real $\varepsilon > 0$ there is a real $\delta > 0$ such that $d_X(x,x') < \delta$ implies $d_Y(f(x),f(x')) < \varepsilon$ , for all $x,x' \in X$ (Uniform continuity of a map of metric spaces: one $\delta$ serving every point).

[L3]

Every open cover of a compact metric space has a Lebesgue number: a real $\delta > 0$ such that every nonempty subset of diameter less than $\delta$ lies in a single member of the cover (Every open cover of a compact metric space has a Lebesgue number: a $\delta > 0$ such that every nonempty subset of diameter less than $\delta$ lies inside a single member of the cover, Open cover, subcover, compact metric space, and compact subset of a metric space).

[L4]

Open balls are open (Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement).

[L5]

For nonempty bounded $A$ , $\operatorname{diam}(A) = \sup\{d_X(u,v) : u,v \in A\}$ ; in particular $\operatorname{diam}(\{x,x'\}) = d_X(x,x')$ , the set of distances being $\{0, d_X(x,x')\}$ and a metric being nonnegative (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, Nonnegativity of a metric is a consequence of the other axioms, not an axiom).

[L6]

A metric is symmetric and satisfies the triangle inequality (Metric space: $d(x,y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

Proof

technique · direct

1.1

If $X = \emptyset$ the condition of uniform continuity holds vacuously, so assume $X \ne \emptyset$ , and let $\varepsilon > 0$ be real.

2.1

Put $\mathcal{U} := \{\, U \subseteq X : U \text{ is open in } X \text{ and } f[U] \subseteq B_Y(z, \varepsilon/2) \text{ for some } z \in Y \,\}$ , a family cut out by a property and not by a selection.

L1step 1.1

3.1

$\mathcal{U}$ is an open cover of $X$ : given $x \in X$ , continuity at $x$ supplies a real $\delta_x > 0$ with $f[B_X(x,\delta_x)] \subseteq B_Y(f(x), \varepsilon/2)$ , and $B_X(x,\delta_x)$ is open and contains $x$ , so it belongs to $\mathcal{U}$ .

L1L4step 2.1

4.1

By the Lebesgue number lemma there is a real $\delta > 0$ such that every nonempty subset of $X$ of diameter less than $\delta$ is contained in a single member of $\mathcal{U}$ .

L3step 3.1

5.1

Let $x, x' \in X$ with $d_X(x,x') < \delta$ ; the set $\{x,x'\}$ is nonempty with diameter $d_X(x,x') < \delta$ , so $\{x,x'\} \subseteq U$ for some $U \in \mathcal{U}$ , and there is $z \in Y$ with $f[U] \subseteq B_Y(z,\varepsilon/2)$ .

L5step 2.1step 4.1

6.1

Then $d_Y(f(x),z) < \varepsilon/2$ and $d_Y(z,f(x')) < \varepsilon/2$ , so $d_Y(f(x),f(x')) \le d_Y(f(x),z) + d_Y(z,f(x')) < \varepsilon$ ; as $\varepsilon > 0$ was arbitrary, $f$ is uniformly continuous.

L2L6step 5.1∎

Remarks

The centre $z$ is not chosen, and that is why the proof is choice free. The family $\mathcal{U}$ is defined by the existence of a suitable $z$ , and the argument instantiates that existential once, at step 5.1, for the single member $U$ that the Lebesgue number produced. No function assigning a centre to every member of $\mathcal{U}$ is ever needed.

Compactness is not removable. The map $x \mapsto 1/x$ is continuous on the interval $(0,1)$ and is not uniformly continuous there ( $x \mapsto 1/x$ is continuous on $(0,1)$ and not uniformly continuous, so Heine-Cantor needs compactness of the domain ↗); $(0,1)$ is not compact.

The codomain is arbitrary. Nothing is assumed about $(Y,d_Y)$ — not completeness, not boundedness, not compactness. All the work is done on the domain side, which is where the finite subcover lives.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 79 results over 16 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Heine-Cantor theorem (Wikipedia) (standard reference, not scraped)
Lebesgue's number lemma (Wikipedia) (standard reference, not scraped)