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Real Stone--Weierstrass theorem for compact metric spaces

Statement

Let K be a nonempty compact metric space and let A⊆C(K,R) be a unital subalgebra which separates points. Then A is dense in C(K,R) for the supremum metric.

Facts & Assumptions

Given: f∈C(K,R) and ε>0.

[L1]

The uniform closure A‾ is closed under pointwise maximum and minimum (The uniform closure of a unital real function algebra is closed under absolute value, maximum, and minimum).

[L2]

For distinct x,y∈K and prescribed real values at x,y, A contains a function taking those two values (A unital separating real function algebra interpolates arbitrary values at two distinct points).

[L3]

Every open cover of the compact metric space K has a finite subcover (Open cover, subcover, compact metric space, and compact subset of a metric space).

Proof

technique · constructive
1.1givenL2algebra

Put δ=ε/3. For a fixed x∈K, let Ax={a∈A:a(x)=f(x)}. The open sets Ua={z:a(z)>f(z)−δ} indexed by all a∈Ax cover K: for each y≠x, [L2] gives one a∈Ax with a(y)=f(y), and the constant f(x)1 handles y=x. This asserts existence separately at each point without selecting an interpolant for every pair.

2.1L1L3step 1.1construct

By [L3] this cover has a finite subcover. Choose representatives a1,…,ar∈Ax for its finitely many members; r≥1 since K is nonempty. Their maximum g=max⁡jaj lies in A‾ by [L1], satisfies g(x)=f(x), and is greater than f−δ throughout K. This proves existence of such a finite maximum for every fixed x; it does not select a family (gx)x∈K.

3.1L1step 2.1construct

Let G be the set of all finite nonempty maxima of members of A which satisfy g>f−δ everywhere on K. Every member of G belongs to A‾ and is continuous by [L1]. The open sets Vg={z:g(z)<f(z)+δ} indexed by all g∈G cover K: for each x, the existential conclusion of step 2.1 supplies a member with g(x)=f(x). Thus this second cover also requires no choice indexed by K.

4.1L1L3step 3.1algebra

By [L3] take a finite subcover and representatives g1,…,gs∈G, where s≥1. Then h=min⁡igi∈A‾ by [L1]. Every gi>f−δ globally, and the subcover supplies at each point an index with gi<f+δ. Hence f−δ<h<f+δ pointwise and ∥h−f∥∞≤δ. Only finitely many representatives were selected in either compactness step.

5.1givenstep 4.1discharge-construct∎

Since h∈A‾, choose one a∈A with ∥a−h∥∞<δ. Then ∥a−f∥∞<2δ<ε. For every f and positive target error this gives an approximant in A itself, proving density without any countable or arbitrary-index choice assumption.

Depends on

Used by

Dependency tree · two levels

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Sources