TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02

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Under dependent choice and the ultrafilter lemma, uniformly continuous maps to compact Hausdorff spaces extend uniquely over the Samuel compactification

Statement

Assume dependent choice and the ultrafilter lemma. Let $X$ be a separated uniform space, let $η : X \to S (X)$ be a Samuel compactification, and let $f : X \to K$ be uniformly continuous, where $K$ is compact Hausdorff with its unique compatible uniformity. Then there is a unique uniformly continuous $\overset{ˉ}{f} : S (X) \to K$ such that $\overset{ˉ}{f} η = f$ .

Facts & Assumptions

Given: The stated choice principles, a separated uniform space $X$ , a uniformly continuous $f : X \to K$ , and compact Hausdorff $K$ .

[L1]

A compact Hausdorff topology has exactly one compatible uniformity; that uniformity is separated, the topology is Tychonoff under dependent choice, and its continuous maps to uniform spaces are uniformly continuous (A nonempty compact Hausdorff space carries exactly one compatible uniformity, A uniformity is separated if and only if its induced topology is Hausdorff, Under dependent choice a compact Hausdorff space is Tychonoff, and its disjoint closed sets are separated by continuous functions, Every continuous map from a nonempty compact Hausdorff space to a uniform space is uniformly continuous).

[L2]

The gauge $q_{h} (u, v) = ∣ h (u) - h (v) ∣$ over continuous $h : K \to [0, 1]$ induces the topology of a completely regular space (The topology of a nonempty completely regular space is induced by the gauge of its continuous $[0, 1]$ -valued pseudometrics).

[L3]

A compact uniform space is complete, and a uniformly continuous map from a uniform space to a complete separated uniform space extends uniquely over its Hausdorff completion (Every compact uniform space is complete, Every uniformly continuous map into a complete Hausdorff uniform space extends uniquely across the Hausdorff completion; consequently completions are unique up to a unique uniform isomorphism).

[L4]

Under the two choice principles, a Samuel completion of separated $X$ is a Samuel compactification (Under the ultrafilter lemma the Samuel completion is compact, and under dependent choice plus the ultrafilter lemma it compactifies every separated uniform space).

Proof

technique · direct

1.1

By [L1] and [L2], the gauge of the continuous maps $h : K \to [0, 1]$ is a compatible uniformity on $K$ , hence it is the unique compatible uniformity of $K$ .

L1L2

2.1

For such $h$ , [L1] makes $h$ uniformly continuous, so $h \circ f$ is a Samuel coordinate; therefore every finite basic entourage of the gauge in step 1.1 has Samuel-entourage preimage under $f$ .

L1step 1.1

3.1

Thus $f : (X, U_{S}) \to K$ is uniformly continuous, while $K$ is complete and separated by [L1] and [L3].

L1L3step 2.1

4.1

Apply [L3] to the Samuel completion to obtain the unique uniformly continuous extension $\overset{ˉ}{f} : S (X) \to K$ .

L3step 3.1

5.1

By [L4] this completion is the Samuel compactification claimed in the statement, and its dense canonical image also gives uniqueness among continuous extensions.

L4step 4.1∎

Depends on

Used by

Dependency tree · two levels

51 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

A. Zucker, Big Ramsey Degrees and Topological Dynamics (standard reference, not scraped)
Garrido and Meroño, The Samuel realcompactification (standard reference, not scraped)