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The Samuel compactification map need not be a uniform embedding for the original uniformity

Statement refuted

Refuted claim: for every separated uniform space, the Samuel compactification map is a uniform embedding for the original uniformity.

Let $\mathbb N$ carry the zero-one discrete metric. Its Samuel compactification map is not a uniform embedding when its domain is read with that original discrete uniformity. Under dependent choice and the ultrafilter lemma it is nevertheless a topological embedding.

Facts & Assumptions

Given: The zero-one metric $d$ on $\mathbb N$ , its original metric uniformity, and its Samuel uniformity.

[L1]

A metric must satisfy separation, symmetry, and the triangle inequality; its radius- $1/2$ entourage is the diagonal for the zero-one metric, and every metric uniformity is separated (Metric space: $d(x,y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric, A metric on a nonempty set generates an entourage uniformity whose induced topology and uniformly continuous maps are the usual metric notions, and this uniformity is separated).

[L2]

A totally bounded uniform space has a finite centre set for every entourage, while $\mathbb N$ is not finite (Totally bounded uniform space, Finite, countably infinite, countable, uncountable, The pigeonhole principle on $\mathbb{N}$ ).

[L3]

The Samuel uniformity is totally bounded, and a Hausdorff completion pulls its target uniformity back exactly to its source uniformity (The Samuel uniformity is totally bounded, A Hausdorff completion of a uniform space and its canonical dense map).

[L4]

A uniform embedding identifies its source uniformity with the subspace uniformity on its image (Uniform embedding and uniform isomorphism).

[L5]

Under dependent choice and the ultrafilter lemma, the Samuel completion map is a topological embedding for a separated original uniform space (Under the ultrafilter lemma the Samuel completion is compact, and under dependent choice plus the ultrafilter lemma it compactifies every separated uniform space).

Counterexample

technique · direct

1.1

The zero-one function is a metric: if $x\ne z$ , at least one of $x\ne y$ or $y\ne z$ holds, so $d(x,z)=1\le d(x,y)+d(y,z)$ ; its radius- $1/2$ balls are singletons.

1.2

If the original discrete uniformity were totally bounded, finitely many radius- $1/2$ singleton balls would cover $\mathbb N$ , making $\mathbb N$ finite, contrary to [L2].

L1L2

1.3

By [L3], the Samuel uniformity on $\mathbb N$ is totally bounded and the Samuel completion map pulls back exactly that uniformity.

2.1

If that map were a uniform embedding for the original discrete uniformity, [L4] would identify that uniformity with its pullback uniformity; steps 1.2 and 1.3 would then give the contradiction that the original uniformity is totally bounded.

L4step 1.2step 1.3

3.1

Under the choice hypotheses of [L5], the map is still a topological embedding, which isolates the failure as uniform rather than topological.

L1L5step 2.1∎

Depends on

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Sources

M. Megrelishvili, Samuel and Smirnov compactifications (standard reference, not scraped)