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ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02
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The Samuel reflection of a nonempty indiscrete uniform space is a singleton

Example

Let XX\ne\varnothing carry the indiscrete uniformity {X×X}\{X\times X\}. Its Samuel completion, equivalently its Hausdorff Samuel reflection, is a singleton. This is not called a compactification unless XX itself is a singleton.

Facts & Assumptions

Verification

technique · direct
1.1

If f:X[0,1]f:X\to[0,1] is uniformly continuous and x,yXx,y\in X, then for every ε>0\varepsilon>0 the sole source entourage forces f(x)f(y)<ε|f(x)-f(y)|<\varepsilon; hence f(x)=f(y)f(x)=f(y).

L1
2.1

Every Samuel coordinate is constant by step 1.1, so every Samuel pseudometric vanishes and the Samuel uniformity is again indiscrete.

L1step 1.1
3.1

Let η:XS(X)\eta:X\to S(X) be a Hausdorff completion of the Samuel uniformity. If η(x)η(y)\eta(x)\ne\eta(y), separatedness in [L2] gives a target entourage excluding that pair, while uniform continuity pulls it back to the sole source entourage X×XX\times X, a contradiction. Thus η[X]\eta[X] is a singleton.

L2step 2.1
4.1

The nonempty singleton η[X]\eta[X] is dense by [L2] and closed by [L2], so it is all of S(X)S(X). Thus the Samuel reflection is a singleton; if XX has at least two points its canonical map is not injective.

L2step 3.1

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

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