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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 $π_{a} \circ j \circ f : X \to [0, 1]$ 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 $\overset{ˉ}{f} = j^{- 1} \circ h : B \to K$ is continuous and satisfies $\overset{ˉ}{f} \circ e = f$ . 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 $q = \overset{ˉ}{f}$ .

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

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Sources

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