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

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Every continuous $[0,1]$ -valued function extends uniquely over the closure of the full evaluation image

Statement

Let $e:X\to[0,1]^{C(X,[0,1])}$ be the full evaluation map and let $B=\overline{e[X]}$ . Every continuous $g:X\to[0,1]$ has a unique continuous $\bar g:B\to[0,1]$ with $\bar g\circ e=g$ .

Facts & Assumptions

Given: The full evaluation map $e$ , its closure $B$ , and a continuous $g:X\to[0,1]$ .

[L1]

Coordinate projections from a product 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]

Two continuous maps to a Hausdorff target that agree on a dense subset agree everywhere (Two continuous maps into a Hausdorff space that agree on a dense subset are equal).

Proof

technique · direct

1.1

The coordinate projection $\pi_g$ is continuous by [L1]. Its restriction $\bar g=\pi_g|_B$ is continuous and satisfies $\bar g(e(x))=e(x)(g)=g(x)$ .

2.1

If $h:B\to[0,1]$ is another such extension, then $h$ and $\bar g$ agree on $e[X]$ , which is dense in $B$ . Since $[0,1]$ is Hausdorff, [L2] gives $h=\bar g$ .

L2step 1.1∎

Depends on

Used by

Under the ultrafilter lemma and dependent choice, the closure of the full evaluation image is the Stone–Čech compactification Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 40 results over 15 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)

Every continuous [0,1][0,1][0,1]-valued function extends uniquely over the closure of the full evaluation image