ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-16

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The lattice generated by the constants and the distance functions is dense on every compact metric space

Example

Let $(X, d)$ be a compact metric space. Let $L$ be the smallest real vector sublattice of $C (X, R)$ containing every constant function and every distance function $d_{a} : X ⟶ R, d_{a} (x) : = d (a, x) (a \in X) .$ Then $L$ is uniformly dense in $C (X, R)$ : for every $f \in C (X, R)$ and every $ε > 0$ there is $g \in L$ with $∣ g (x) - f (x) ∣ < ε$ for every $x \in X$ . When $X$ is nonempty this is density for the topology of uniform convergence.

Facts & Assumptions

Given: A compact metric space $(X, d)$ and the real vector sublattice $L$ generated by constants and the distance functions $d_{a}$ .

[L1]

On a compact Hausdorff space, a unital point-separating real vector sublattice contains, for every $f \in C (X, R)$ and every $ε > 0$ , a member within $ε$ of $f$ at every point; for nonempty $X$ this is density for the topology of uniform convergence (Lattice Stone–Weierstrass theorem on a compact Hausdorff space).

[L2]

A unital real vector sublattice contains all constants and separates points when every distinct pair is distinguished by one member (Unital point-separating real vector sublattices of $C (X, R)$ ).

[L3]

A metric satisfies $d (x, y) = 0$ exactly when $x = y$ , symmetry, and $d (x, z) \leq d (x, y) + d (y, z)$ (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[L4]

A metric space is compact if and only if it is compact as a topological space in its metric topology (For a metric space with its metric topology, compactness in the topological sense is compactness in the metric sense, and the two notions of compact subset coincide, clause 1).

[L5]

Every metric space is Hausdorff: distinct points are separated by disjoint open balls (Distinct points of a metric space have disjoint balls around them).

Verification

technique · direct

1.1L4L5

By [L4] and [L5], the metric topology makes $X$ a compact Hausdorff space.

1.2L3algebra

For $a, x, y \in X$ , the triangle inequality and symmetry in [L3] give $d (a, x) \leq d (a, y) + d (x, y)$ and $d (a, y) \leq d (a, x) + d (x, y)$ ; hence $∣ d_{a} (x) - d_{a} (y) ∣ \leq d (x, y)$ , so every $d_{a}$ is continuous.

2.1step 1.2L2L3

The generated lattice $L$ is unital by construction. If $x \neq y$ , then $d_{x} (x) = 0$ while $d_{x} (y) = d (x, y) \neq 0$ by [L3], so $d_{x} \in L$ separates $x$ and $y$ ; thus $L$ is point-separating in the sense of [L2].

3.1step 1.1step 2.1L1∎

Apply [L1] to the unital point-separating real vector sublattice $L$ on the compact Hausdorff space of step 1.1.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

33 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

E. Carlen, Notes on Topology and the Stone-Weierstrass Theorem, Lemma 1.27 (standard reference, not scraped)