Uniform Completeness, Completion, and the Samuel Compactification

The diagonal and symmetry axioms for $p_f$ follow from [L1], while $p_f(x,z)\le p_f(x,y)+p_f(y,z)$ follows by applying [L2] to $f(x)-f(z)=(f(x)-f(y))+(f(y)-f(z))$; thus $p_f$ is a pseudometric.

For every $\varepsilon>0$, uniform continuity of $f$ and [L3] give an entourage $U$ of $\mathcal U$ with $(x,y)\in U$ implying $p_f(x,y)<\varepsilon$.

If $g:X\to\mathbb R$ is uniformly continuous with $|g|\le M$, then for $M>0$ the map $h=(g+M)/(2M)$ is $[0,1]$-valued and uniformly continuous: for a target tolerance $\delta>0$, use uniform continuity of $g$ with tolerance $2M\delta$. Moreover $p_g=2M p_h$; if $M=0$, $g$ is constant. Thus the two gauges have the same basic entourages.

For a basic Samuel entourage, step 1.2 gives an original entourage inside each of its finitely many coordinate balls; their finite intersection is therefore an original entourage contained in the basic Samuel entourage. Hence every basic Samuel entourage belongs to $\mathcal U$, and so does the generated filter.

Steps 1.1, 2.1, and 1.3 prove all assertions.

For each $f\in F$, [L2] supplies a finite set $A_f\subseteq[0,1]$ such that every value of $f$ is within $\varepsilon/3$ of some member of $A_f$.

The product $A:=\prod_{f\in F}A_f$ is finite, and for $a\in A$ let $C_a$ be the set of $x\in X$ with $|f(x)-a_f|<\varepsilon/3$ for every $f\in F$.

The index set $A':=\{a\in A:C_a\ne\varnothing\}$ of nonempty cells is a finite subset of $A$. Choose a natural $n$ and a bijection $e:n\to A'$, form the explicitly $n$-indexed family $i\mapsto C_{e(i)}$, and use [L3] to choose $c_{e(i)}\in C_{e(i)}$; let $C$ be the set of chosen points.

If $x\in X$, choose $a\in A$ with $x\in C_a$ using step 1.1; then $a\in A'$ and $|f(x)-f(c_a)|<2\varepsilon/3<\varepsilon$ for every $f\in F$, so $x\in E(F,\varepsilon)[c_a]$.

Thus $C$ is a finite net for each basic Samuel entourage. Every Samuel entourage contains one of these basic entourages, so the same finite centres cover it; when $F=\varnothing$, use the empty centre set if $X=\varnothing$ and any singleton centre otherwise. Hence $\mathcal U_S$ is totally bounded.

Since $\mathcal U_S\subseteq\mathcal U$, every Samuel-open set is original-open.

Choose $U\in\mathcal U$ with $U[x]\subseteq O$, take the sequence of [L3], and take the pseudometric $p$ of [L4]; then $\{y:p(x,y)\le1/4\}\subseteq O$.

Put $f(y)=\min\{1,4p(x,y)\}$. The reverse triangle inequality for a pseudometric and the uniform continuity of $p$ make $f$ uniformly continuous, so $f\in\mathcal F_{\mathcal U}$ and $f(x)=0$.

The Samuel neighbourhood $\{y:|f(y)-f(x)|<1\}$ lies in $\{y:p(x,y)<1/4\}\subseteq O$, so every original-open set is Samuel-open.

The two inclusions in steps 1.1 and 2.1 give equality of the topologies; for $X=\varnothing$ both are the empty topology.

Choose $V$ and $U$ as in [L1] and [L2], and choose the finite $U$-net $F$ from [L3].

For $y\in Y$, density gives $x\in X$ with $i(x)\in V[y]$; choose $a\in F$ with $x\in U[a]$.

Step 1.2 gives $(i(a),i(x))\in V$ and, by symmetry, $(i(x),y)\in V$, hence $(i(a),y)\in V\circ V\subseteq E$.

The finite set $i[F]$ has $E$-balls covering $Y$, proving total boundedness; if $Y=\varnothing$, the empty set is the required finite net.

The completion map has dense image and is uniformly continuous, so [L1] and [L3] make $S(X)$ totally bounded.

The space $S(X)$ is complete by [L2], so [L4] makes it compact under the ultrafilter lemma.

Under dependent choice, [L5] identifies the original and Samuel topologies; if $\mathcal U$ is separated, the Samuel uniformity is separated as well, so $\eta$ is a uniform embedding for $\mathcal U_S$ and a topological embedding for the original topology.

The image is dense by [L2], the source topology is Tychonoff by [L5], and step 1.2 gives compact Hausdorff target; hence the pair is a compactification and therefore a Samuel compactification.

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 $K$.

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 $f$.

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

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

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

Apply [L1] to $\eta_2:X\to S_2$ and to $\eta_1:X\to S_1$, obtaining uniformly continuous maps $F:S_1\to S_2$ and $G:S_2\to S_1$ with $F\eta_1=\eta_2$ and $G\eta_2=\eta_1$.

The maps $GF$ and $\operatorname{id}_{S_1}$ agree on the dense set $\eta_1[X]$, while $FG$ and $\operatorname{id}_{S_2}$ agree on $\eta_2[X]$, so [L2] makes both composites identities.

Therefore $F$ and $G$ are inverse uniform isomorphisms, and uniqueness of $F$ is the uniqueness clause in [L1].

By [L1], it is enough to show that every original entourage contains a Samuel entourage.

Take $p$ as in [L2] and a finite $p$-net $F$ as in [L3]; for $z\in F$ put $f_z(y)=\min\{1,p(z,y)\}$.

Each $f_z$ is $[0,1]$-valued and uniformly continuous: the pseudometric triangle inequality gives $|p(z,x)-p(z,y)|\le p(x,y)$, and truncation at $1$ does not increase this difference. Thus every $f_z$ is a Samuel coordinate.

If $|f_z(x)-f_z(y)|<1/8$ for every $z\in F$, choose $z\in F$ with $p(z,x)<1/16$. Then $f_z(x)=p(z,x)<1/16$ and $f_z(y)<3/16<1$, so $p(z,y)=f_z(y)$; hence $p(x,y)\le p(x,z)+|f_z(x)-f_z(y)|+p(z,x)<1/16+1/8+1/16=1/4$.

The finite-coordinate Samuel entourage in step 3.1 lies in $\{p<1/4\}\subseteq U$, so step 1.1 proves $\mathcal U=\mathcal U_S$; the empty space is immediate.

By [L1], a Samuel completion is a Hausdorff completion of the original uniformity.

The uniqueness theorem [L2] identifies it with every other Hausdorff completion by the unique uniform isomorphism fixing $X$.

Its dense canonical image and [L3] make it totally bounded, while a Hausdorff completion is complete; hence [L3] makes it compact under the ultrafilter lemma.

This proves the DC identification and the separately qualified compactness assertion.

The map $\eta:X\to S(X)$ is continuous by [L2], so [L1] and [L3] give a continuous $q:\beta X\to S(X)$ with $qj=\eta$.

By [L1], $\beta X$ is compact. Thus the image $q[\beta X]$ is compact and therefore closed in the Hausdorff space $S(X)$ by [L4].

Since $q[\beta X]$ contains $qj[X]=\eta[X]$, it contains a dense subset of $S(X)$; its closedness from step 2.1 gives $q[\beta X]=S(X)$.

Hence $q$ is the asserted continuous surjection.