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

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

Statement

Assume the ultrafilter lemma and dependent choice. If $X$ is Tychonoff, $e:X\to[0,1]^{C(X,[0,1])}$ is its full evaluation map, and $B=\overline{e[X]}$ , then $(B,e)$ is a Stone–Čech compactification of $X$ .

Facts & Assumptions

Given: The two stated choice principles, a Tychonoff space $X$ , its full evaluation closure $B$ , a compact Hausdorff space $K$ , and a continuous map $f:X\to K$ .

[L1]

Under the ultrafilter lemma, the evaluation closure gives a Hausdorff compactification (Assuming the ultrafilter lemma, every Tychonoff space has a Hausdorff compactification).

[L2]

Under dependent choice, $K$ has an embedding $j:K\to[0,1]^J$ (Under dependent choice, every compact Hausdorff space embeds in a unit cube).

[L3]

A compact subset of a Hausdorff space is closed, and continuous maps agreeing on a dense subset with Hausdorff target agree everywhere (In a Hausdorff space a point and a disjoint compact set, and two disjoint compact sets, have disjoint open neighbourhoods; hence every compact subset is closed, and in a compact Hausdorff space the compact subsets are exactly the closed ones, Two continuous maps into a Hausdorff space that agree on a dense subset are equal).

[L4]

A family of continuous component maps assembles uniquely to a continuous map into the product (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).

[L5]

Every continuous unit-interval-valued map extends uniquely over the full evaluation closure (Every continuous $[0,1]$ -valued function extends uniquely over the closure of the full evaluation image).

[L6]

A Stone–Čech compactification is a Hausdorff compactification with the stated unique compact-Hausdorff extension property (The Stone–Čech compactification by its compact-Hausdorff extension property).

Proof

technique · direct

1.1

By [L1], $B$ is compact Hausdorff and $e[X]$ is dense in it.

1.2

Use [L2] to fix an embedding $j:K\to[0,1]^J$ . For each $a\in J$ , the map $\pi_a\circ j\circ f:X\to[0,1]$ extends uniquely to a continuous $h_a:B\to[0,1]$ by [L5].

L2L5

2.1

The family $(h_a)_{a\in J}$ assembles to a continuous map $h:B\to[0,1]^J$ by [L4], and $h\circ e=j\circ f$ coordinatewise.

step 1.2L4

3.1

The subset $j[K]$ is compact, hence closed in the Hausdorff cube by [L3]. It contains $h[e[X]]$ , so it contains $h[B]$ : the inverse image $h^{-1}[j[K]]$ is closed in $B$ and contains the dense subset $e[X]$ .

L3step 1.1step 2.1

4.1

Thus $\bar f=j^{-1}\circ h:B\to K$ is continuous and satisfies $\bar f\circ e=f$ . If $q:B\to K$ is another extension, then $j\circ q$ and $h$ agree on dense $e[X]$ , so [L3] gives equality and injectivity of $j$ gives $q=\bar f$ .

L3step 2.1step 3.1

5.1

Step 1.1 gives a Hausdorff compactification and step 4.1 gives the required unique extension for every compact Hausdorff target. This is exactly [L6].

step 1.1step 4.1L6∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 95 results over 16 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

Stacks Project, Stone–Čech compactification (standard reference, not scraped)