LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02

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The evaluation map of a point–closed-set separating family is a topological embedding

Statement

If a family $\mathcal F$ separates points from closed sets (A family of continuous unit-interval-valued functions that separates points from closed sets), then its evaluation map $e_{\mathcal F}$ (The evaluation map from a space into the unit cube indexed by a family of continuous functions) is a homeomorphism of $X$ onto the subspace $e_{\mathcal F}[X]$ . In particular it is a topological embedding.

Facts & Assumptions

Given: A space $X$ , a point–closed-set separating family $\mathcal F$ , and its evaluation map $e=e_{\mathcal F}$ .

[L1]

A map into a product is continuous exactly when all of its coordinate maps are continuous (A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice).

[L2]

A homeomorphism is a bijection whose map and inverse are continuous (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological).

Proof

technique · direct

1.1

Each coordinate $\pi_f\circ e$ equals $f$ and is continuous, so $e$ is continuous by [L1].

1.2

If $x\ne y$ , point separation supplies $f\in\mathcal F$ with $f(x)\ne f(y)$ , and then $e(x)(f)\ne e(y)(f)$ . Thus $e$ is injective.

given

1.3

Let $U$ be open in $X$ and $x\in U$ . The complement $C=X\setminus U$ is closed, so choose $f\in\mathcal F$ with $f(x)=1$ and $f[C]=\{0\}$ . Then $e(x)$ belongs to $e[X]\cap\pi_f^{-1}((1/2,1])$ , and this subspace-open set is contained in $e[U]$ .

givenconstruct

2.1

Step 1.3 shows that $e[U]$ is open in $e[X]$ for every open $U$ , so $e^{-1}:e[X]\to X$ is continuous. Together with step 1.1 and injectivity from step 1.2, [L2] proves the assertion.

step 1.1step 1.2step 1.3L2∎

Depends on

Used by

A space is Tychonoff if and only if it embeds in a cube [0,1]^J Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 40 results over 10 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

E. Moorhouse, The Stone–Čech Compactification (standard reference, not scraped)