LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-02

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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The uniform closure of a unital real function algebra is closed under absolute value, maximum, and minimum

Statement

If $A \subseteq C (K, R)$ is a unital real function algebra and $\overline{A}$ is its uniform closure, then $u \in \overline{A}$ implies $∣ u ∣ \in \overline{A}$ ; consequently $u \lor v$ and $u \land v$ lie in $\overline{A}$ whenever $u, v$ do.

Facts & Assumptions

Given: A unital real function algebra $A$ and $u, v \in \overline{A}$ .

[L1]

The algebra operations and constants are those in A unital point-separating real subalgebra of $C (K, R)$ .

[L2]

Polynomials uniformly approximate $∣ t ∣$ on every bounded closed interval (Polynomials are uniformly dense in $C ([a, b], R)$ for every closed interval).

Proof

technique · direct

1.1

Choose $a_{n} \in A$ converging uniformly to $u$ . Their ranges, together with that of $u$ , lie in one bounded interval.

givenchoose

1.2

By [L2], choose polynomials $p_{j}$ with $p_{j} (0) = 0$ converging uniformly to $∣ t ∣$ on that interval. Then $p_{j} (a_{n}) \in A$ by [L1].

L1L2choose

2.1

A diagonal choice of $j, n$ makes $p_{j} (a_{n})$ uniformly converge to $∣ u ∣$ , so $∣ u ∣ \in \overline{A}$ .

step 1.1step 1.2algebra

3.1

The identities $u \lor v = (u + v + ∣ u - v ∣) / 2$ and $u \land v = (u + v - ∣ u - v ∣) / 2$ and [L1] give the remaining closure.

step 2.1L1algebra∎

Depends on

Used by

Dependency tree · two levels

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Sources

The Stone--Weierstrass Theorem and its Applications (standard reference, not scraped)