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Lattice Stone–Weierstrass theorem on a compact Hausdorff space

Statement

Let X be a compact Hausdorff space and let LC(X,R) be a unital point-separating real vector sublattice. Then for every fC(X,R) and every ε>0 there is gL with g(x)f(x)<ε for every xX; that is, L is uniformly dense in C(X,R). When X is nonempty this is exactly density for the topology of uniform convergence, which Uniform convergence, and the topology of uniform convergence: the metric topology of the uniform metric on YX and on C(X,Y) defines only on a nonempty domain.

Facts & Assumptions

Given: A compact Hausdorff space X, a unital point-separating real vector sublattice LC(X,R), a target fC(X,R), and a real ε>0.

[L1]

For distinct x,yX and arbitrary α,βR, a unital separating real function lattice contains h with h(x)=α and h(y)=β (A unital separating real function lattice interpolates arbitrary values at two distinct points).

[L2]

On a nonempty compact space, a family closed under pointwise maxima and minima and having the two-point duplication property relative to f contains, for every positive error, a member within that error of f at every point (A function lattice with the two-point duplication property uniformly approximates its target).

[L3]

On nonempty X, the topology of uniform convergence on C(X,R) is the metric topology of the restricted uniform metric ρˉ(f,g)=supxXmin{f(x)g(x),1} (Uniform convergence, and the topology of uniform convergence: the metric topology of the uniform metric on YX and on C(X,Y)).

Proof

technique · direct
1.1

If X=, then C(X,R) contains only the empty function, which is a constant function and hence belongs to the unital lattice L; the displayed approximation condition holds vacuously, there being no x to test. The topological reading is not asserted here, because [L3] supplies the uniform metric only on a nonempty domain.

L3given
1.2

Assume X. For distinct x,y, apply [L1] with α=f(x) and β=f(y); for x=y, the constant function with value f(x) belongs to L. Thus L has the two-point duplication property relative to f.

L1given
2.1

Apply [L2] with the positive error min{ε,1}/2 to obtain gL satisfying g(x)f(x)<min{ε,1}/2<ε for every xX.

step 1.2L2
3.1

Suppose further that X, which is where [L3] defines the uniform metric. The approximant of step 2.1 then satisfies ρˉ(f,g)min{ε,1}/2, so every uniform-metric neighbourhood of every f meets L; hence L is dense in the topology of uniform convergence.

step 2.1L3

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Dependency tree · next 3 levels

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