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TheoremStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)
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C(K,R) is complete in the supremum metric for every nonempty compact metric space K

Statement

Let (K,d) be a nonempty compact metric space. Every member of C(K,R) is bounded, so the supremum metric

d∞(f,g):=sup⁡x∈K∣f(x)−g(x)∣

is defined on C(K,R). With this metric, C(K,R) is complete.

Facts & Assumptions

Given: A nonempty compact metric space (K,d) and the set C(K,R) of continuous real-valued functions on it.

[L1]

Every continuous real-valued function on a nonempty compact metric space has a bounded range (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value).

[L2]

If S is nonempty, the formula d∞(f,g)=sup⁡x∈S∣f(x)−g(x)∣ defines a metric on the set of bounded functions S→R (The supremum metric d∞(f,g)=sup⁡x∣f(x)−g(x)∣ is a metric on the bounded real-valued functions on a nonempty set).

[L3]

A sequence is Cauchy in a metric d when, for every positive error, all pairwise distances sufficiently far out are below that error; it converges to p when its distances to p tend to zero (Cauchy sequence in a metric space, Convergence of a sequence in a metric space: xk→x iff d(xk,x)→0 in R).

[L4]

A sequence of real-valued functions converges uniformly if and only if it is uniformly Cauchy (A sequence of real-valued functions converges uniformly if and only if it is uniformly Cauchy).

[L5]

A uniform limit of continuous real-valued functions on a metric space is continuous (The uniform limit of continuous real-valued functions on a metric space is continuous).

[L6]

A metric space is complete when every Cauchy sequence in it converges to one of its points (Complete metric space: every Cauchy sequence converges in the space).

Proof

technique · direct
1.1

By [L1], every f∈C(K,R) is bounded. Thus C(K,R) is a subset of the bounded functions on K, and the restriction of the metric in [L2] is a metric on C(K,R).

L1L2
1.2

Let (fj) be a Cauchy sequence in this supremum metric.

givenL3
2.1

Given a real ε>0, choose J such that d∞(fm,fn)<ε for all m,n≥J. Then ∣fm(x)−fn(x)∣≤d∞(fm,fn)<ε for all such m,n and every x∈K, so (fj) is uniformly Cauchy.

step 1.2L2L3
3.1

By [L4] there is a function f:K→R such that fj→f uniformly on K.

step 2.1L4
4.1

The function f is continuous by [L5], hence belongs to C(K,R) and is bounded by [L1].

step 3.1L1L5
5.1

Let ε>0. Uniform convergence gives J such that ∣fj(x)−f(x)∣<ε/2 for every j≥J and x∈K; hence d∞(fj,f)≤ε/2<ε, so fj→f in the supremum metric.

step 3.1step 4.1L2L3
6.1

Every Cauchy sequence in C(K,R) therefore converges in the supremum metric to a member of C(K,R), so the metric space is complete.

step 1.1step 5.1L6∎

Depends on

Used by

Dependency tree · two levels

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Sources