TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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Real Stone–Weierstrass theorem for compact Hausdorff spaces

Statement

Let $X$ be a compact Hausdorff space. Every unital point-separating real function algebra $A \subseteq C (X, R)$ is uniformly dense in $C (X, R)$ .

Facts & Assumptions

Given: A compact Hausdorff space $X$ and a unital point-separating real function algebra $A \subseteq C (X, R)$ .

[L1]

The uniform closure of a real function algebra on a compact Hausdorff space is itself a real function algebra and a real vector sublattice of $C (X, R)$ (The uniform closure of a real function algebra is a vector lattice).

[L2]

On a compact Hausdorff space, a unital point-separating real vector sublattice of $C (X, R)$ contains, for every $f \in C (X, R)$ and every $ε > 0$ , a member within $ε$ of $f$ at every point; that is, it is uniformly dense (Lattice Stone–Weierstrass theorem on a compact Hausdorff space).

Proof

technique · direct

1.1given

If $X = \emptyset$ , then $C (X, R)$ contains only the empty function, which is a constant function and therefore belongs to the unital algebra $A$ ; thus $A = C (X, R)$ .

1.2L1given

Assume $X \neq \emptyset$ and let $B : = \overline{A}$ be the uniform closure. By [L1], $B$ is a real vector sublattice and a real function algebra; it is unital and point-separating because it contains $A$ .

2.1step 1.2L2∎

By [L2], the vector sublattice $B$ is dense in $C (X, R)$ , while by definition $B$ is closed; hence $B = C (X, R)$ , which says exactly that $A$ is uniformly dense.

Depends on

Used by

Dependency tree · two levels

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Sources

J. M. Erdman, A Companion to Real Analysis, Theorem 21.2.6 (standard reference, not scraped)
M. Xu, Math 205B notes from a course by R. Mazzeo (Stanford), Theorem 9.3 (standard reference, not scraped)
E. Carlen, Notes on Topology and the Stone-Weierstrass Theorem, Theorem 1.26 (standard reference, not scraped)