Uniform Completeness, Completion, and the Samuel Compactification: Examples and Counterexamples

By [L1], the identity $K\to K$ is a Hausdorff completion: it is a uniform embedding, its image is dense, and its target is complete.

Apply [L2]: every Samuel completion is uniformly isomorphic over $K$ to the identity completion of step 1.1, and under the stated choice principles it is the Samuel compactification.

Thus no point is added; the singleton case is included.

Choose $n=m+1$, where $m$ is from [L2]; then $n\ge 2$ and $1\mathrm{/}n<\epsilon$. The map $j\mapsto (j+1)\mathrm{/}n$ is injective on the natural number $m$, so its image $F=\{(j+1)\mathrm{/}n:j\in m\}$ is finite and lies in $(0,1)$.

For $x\in (0,1)$, write $k=\lfloor nx\rfloor$. If $k=0$, then $(0+1)\mathrm{/}n\in F$ is within $1\mathrm{/}n$ of $x$. Otherwise $1\le k<n=m+1$, so $k-1\in m$, $k\mathrm{/}n\in F$, and $k\mathrm{/}n\le x<(k+1)\mathrm{/}n$. Thus $F$ is an $\epsilon$-net.

Hence $(0,1)$ is separated and totally bounded. The inclusion $i$ pulls back every metric entourage of $[0,1]$ to the same-radius metric entourage of $(0,1)$, and its image is dense because every interval about $0$, $1$, or an interior point meets $(0,1)$.

By [L4], $[0,1]$ is complete, and by [L1] its metric uniformity is separated; so step 3.1 verifies the completion conditions of A Hausdorff completion of a uniform space and its canonical dense map.

The identification in [L5] now gives the Samuel compactification $S((0,1))\cong [0,1]$.

The displayed $d$ satisfies the metric axioms: if $m\mathrm{\ne }n$, then for every $r$ at least one of $m\mathrm{\ne }r$ or $r\mathrm{\ne }n$ holds, which proves the triangle inequality; symmetry and separation are immediate.

The entourage $E_{1\mathrm{/}2}$ is the diagonal. Therefore every map from $\mathbb{N}$ to a uniform space is uniformly continuous, since every target entourage contains the diagonal; every map from the discrete topology is continuous.

The two extension properties in [L2] consequently quantify over the same maps from $\mathbb{N}$; using the Samuel compactification whose existence is given by [L4], they yield inverse maps between $S(\mathbb{N})$ and $\beta \mathbb{N}$ fixing $\mathbb{N}$.

The Stone--Cech object exists by [L3] and the Samuel compactification by [L4]; the two inverse maps give the asserted isomorphism over $\mathbb{N}$.

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

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

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

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.

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

If $f:X\to [0,1]$ is uniformly continuous and $x,y\in X$, then for every $\epsilon >0$ the sole source entourage forces $\mathrm{\mid }f(x)-f(y)\mathrm{\mid }<\epsilon$; hence $f(x)=f(y)$.

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

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

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