# Ultra-Quasi-Metrically Tight Extensions of Ultra-Quasi-Metric Spaces

Authors: Collins Amburo Agyingi
Journal: Chinese Journal of Mathematics
 (2015)

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## Abstract

The concept of the tight extension of a metric space was introduced and studied by Dress. It is known that Dress theory is equivalent to the theory of the injective hull of a metric space independently discussed by Isbell some years earlier. Dress showed in particular that for a metric spaceX the tight extension TX is maximal among the tight extensions of X. In a previous work with P. Haihambo and H.-P. Künzi, we constructed the tight extension of a T0-quasi-metric space. In this paper, we continue these investigations by presenting a similar construction in the category of UQP-metric spaces and nonexpansive maps.


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## Body

## 1. Introduction

In [1] a concept of tight extension that is appropriate in the category of T0-quasi-metric spaces and nonexpansive maps was studied. In particular such an extension was constructed and it was shown that this extension is maximal among the tight extensions.In this paper we will show how the studies of [1] can be modified in order to obtain a theory that is appropriate for UQP-metric spaces.  By UQP-metric space in the following, we mean T0-ultra-quasi-metric spaces.  Even though our studies follow essentially [1, 2], we found it imperative to work out every detail of this theory in this paper.We will show that everyUQP-metric space X has a uq-tight extension which is maximal amongst the uq-tight extensions of X. This agrees with the result we have for T0-quasi-metric spaces (check [1]).


## 2. Preliminaries

We mention that the ultra-quasi-pseudometric spaces should not be confused with the quasi-ultrametrics as they are discussed in the theory of dissimilarities (check, e.g., [3]).Definition 1 (compare [4, page 2]).
LetX be a set and let d:X×X→[0,∞) be a mapping into the set [0,∞) of nonnegative reals. Then d is an ultra-quasi-pseudometric or, for short, a uqp-metric on on X if(a)
d(x,x)=0 for all x∈X,(b)
d(x,z)≤max⁡{d(x,y),d(y,z)} whenever x,y,z∈X.
We remark here that the conjugatedt of d where dt(x,y)=d(y,x) whenever x,y∈X is also uqp-metric on X.
Ifd also satisfies the condition,(c)
for anyx,y∈X, d(x,y)=0=d(y,x) implies that x=y,then d is called UQP-metric on X.
Notice thatds=max⁡{d,dt}=d∨dt is ultrametric on X.Example 2 (compare [4, Example 2]).
Note that, forX=[0,∞), the pair (X,n) is UQP-metric space, where n is such that n(x,y)=x if x,y∈X and x>y, and n(x,y)=0 if x,y∈X and x≤y. We show the strong triangle inequality n(x,z)≤max⁡{n(x,y),n(y,z)} whenever x,y,z∈X since the other conditions are obvious. For n(x,y)=x, the result is trivial, since n(x,z)≤n(x,y). Similarly the case that n(x,y)=0 and n(y,z)=y is obvious, since x≤y and n(x,z)≤n(y,z). In the remaining case that n(x,y)=0=n(y,z), we have by transitivity of ≤ that x≤z, and thus n(x,z)=0. It is obvious that n satisfies the T0-condition.
Notice also that, forx,y∈[0,∞), we have ns(x,y)=max⁡{x,y} if x≠y and ns(x,y)=0 if x=y. The ultrametric ns is complete on [0,∞) since n and n-1 are bicomplete on [0,∞). Recall that a UQP-metric space (X,d) is said to bebicomplete if the ultrametric space (X,ds) is complete.
Furthermore0 is the only nonisolated point of τ(ns). Indeed A={0}∪{1/n:n∈N} is a compact subspace of ([0,∞),ns).
In some cases we will replace[0,∞) with [0,∞] and, in this case, we will speak of an extended uqp-metric.Lemma 3 (compare [5, Proposition 2.1]).
Letα,β,γ∈[0,∞). Then the following are equivalent:(a)
n(α,β)≤γ,(b)
α≤max⁡{β,γ}.Proof.
(a)⇒(b). To reach a contradiction, suppose that α>max⁡{β,γ}. Since α>β, we have n(α,β)=α≤γ by part (a) and the way n was defined. Thus, we would get α≤max⁡{β,γ}<α, a contradiction.
(b)⇒(a). Suppose that, on the contrary, n(α,β)>γ would hold. Then α>β and, hence, α>γ would hold which would imply α>max⁡{β,γ} in contradiction to our assumption α≤max⁡{β,γ}.The following corollaries are immediate. Their proofs rely on Lemma3.Corollary 4 (see [4]).
Let(X,d) be uqp-metric space. Consider a map f:X→[0,∞) and let x,y∈X. Then the following are equivalent:(a)
n(f(x),f(y))≤d(x,y),(b)
f(x)≤max⁡{f(y),d(x,y)}.Definition 5.
A mapf:(X,dX)→(Y,dY) between two uqp-metric spaces (X,dX) and (Y,dY) is called nonexpansive if dY(f(x),f(y))≤dX(x,y) holds for any x,y∈X.Corollary 6 (see [4]).
Let(X,d) be uqp-metric space. Then(a)
f:(X,d)→([0,∞),n) is a nonexpansive map if and only if f2(x)≤max⁡{f1(y),d(x,y)} holds for all x,y∈X,(b)
f:(X,d)→([0,∞),n-1) is a nonexpansive map if and only if f1(x)≤max⁡{f2(y),d(y,x)} holds for all x,y∈X.Definition 7.
A mapf:(X,dX)→(Y,dY) between two uqp-metric spaces (X,dX) and (Y,dY) is said to be an isometry provided that dY(f(x),f(y))=dX(x,y) whenever x,y∈X. Two uqp-metric spaces (X,dX) and (Y,dY) are said to be isometric provided that there exists a bijective isometry between them. Note that if (X,dX) is a UQP-metric space, then f is injective.


## 3. Ultra-Ample Function Pairs onUQP-Metric Space

We will recall some results from the theory of hyperconvex hulls ofUQP-metric spaces due to [4].Definition 8 (compare [4, Definition 1, page 4]).
Let(X,d) be UQP-metric space. One will say that a function pair f=(f1,f2) on (X,d) where fi:X→[0,∞)(i=1,2) isultra-ample if, for all x,y∈X, one has d(x,y)≤max⁡{f2(x),f1(y)}.
Let us denote byuPX the set of all ultra-ample function pairs on UQP-metric space (X,d). For each f,g∈uPX, define(1)Nf,g=max⁡supx∈X⁡nf1x,g1x,supx∈X⁡ng2x,f2x,where the UQP-metric (of course we can use UQP-metric n here since the function pairs take values in [0,∞))n is as defined in Example 2. Then N is an extended UQP-metric on uPX.Lemma 9.
Let(X,d) be UQP-metric space and let a∈X. Then one has that fa(x)≔(d(a,x),d(x,a)) whenever x∈X is an ultra-ample function pair belonging to uPX.
We say that a function pairf is uq-minimal among the ultra-ample function pairs on (X,d) if it is an ultra-ample function pair and if g=(g1,g2) is an ultra-ample function pair on (X,d) and for each x∈Xg1(x)≤f1(x) and g2(x)≤f2(x), which implies f=g. We will also call uq-minimal ultra-ample function pair uq-extremal (ultra-ample) function pair. By uQX we will denote the set of all uq-extremal function pairs on (X,d) equipped with the restriction of N to uQX, which we will still denote by N. Of course N is actually UQP-metric on uQX (compare Corollary 5 of [4]). We will call (uQX,N) the ultra-quasi-metrically injective hull of (X,d).Lemma 10.
Let(X,d) be UQP-metric space and let f∈uQX. For all x,y∈X,(f1(x)>f1(y)) implies that f1(x)≤d(y,x) and (f2(x)>f2(y)) implies that f2(x)≤d(x,y).Proof.
See the proof of Lemma3 of [4].As a corollary, we have the following.Corollary 11.
Let(X,d) be UQP-metric space. If f is a minimal ultra-ample function pair on (X,d), then f1(x)≤max⁡{f1(y),d(y,x)} and f2(x)≤max⁡{f2(y),d(x,y)} whenever x,y∈X. Thus the maps f1:(X,d)→([0,∞),n-1) and f2:(X,d)→([0,∞),n) are contracting maps (check, e.g., Corollary 6).Lemma 12.
Letf be a minimal ultra-ample function pair on a T0-ultra-quasi-metric space (X,d). Then (2)f1x=sup⁡dy,x:y∈X,dy,x>f2y=sup⁡fx2y:y∈X,fx2y>f2y,f2x=sup⁡dx,y:y∈X,dx,y>f1y=sup⁡fx1y:y∈X,fx1y>f1ywhenever x∈X.Proof.
See the proof of Lemma4 of [4].The following lemma and its proof are found in [4].Lemma 13.
Iff and g are minimal ultra-ample function pairs on a UQP-metric space (X,d), then(3)Nf,g=supx∈X⁡nf1x,g1x=supx∈X⁡ng2x,f2x.
As a consequence of Lemmas12 and 13 we have the following corollary.Corollary 14.
Any minimal ultra-ample function pairf=(f1,f2) on UQP-metric space (X,d) satisfies the following:(4)f1x=sup⁡ndy,x,f2y:y∈X,f2x=sup⁡ndx,y,f1y:y∈Xwhenever x∈X.


## 4.UQP-Metric Tight Extensions

In this paper we will studyuq-tight extensions as defined below in Definition 19. Moreover, by uq-tight extensions, we will mean tight extensions of UQP-metric spaces.Proposition 15 (compare [2, Section 1.3]).
Let(X,d) be UQP-metric space. Then uQX consists of all functions pairs which are “minimal” in uPX.Proof.
To prove this proposition, we prove that there is nog∈uPX with g<f but g≠f. This is so since, on the one hand, g≤f∈uQX and g∈UPX imply (5)f2x=sup⁡dx,y:y∈X,dx,y>f1y≥g1y=sup⁡dx,y:y∈X,dx,y>gy≤g2x.Thus(6)f2x≤g2x.
Using (6) and the condition that g2≤f2, we have that f2=g2.
In the same manner we can show thatf1=g1 so as to conclude that f=g.
On the other hand, suppose that, for somex∈X and f∈uPX, we have that f2(x)>sup⁡{d(x,y):y∈Xandd(x,y)>f1(y)}.
For eachx∈X and f∈uPX set (px(f))1(z)=f1(z) if z∈X∖{x} and (px(f))1(x)=sup⁡{d(y,x):y∈Xandd(y,x)>f2(y)}.
Similarly for eachx∈X and f∈uPX set (px(f))2(z)=f2(z) if z∈X∖{x} and (px(f))2(x)=sup⁡{d(x,y):y∈Xandd(x,y)>f1(y)}.
We show first thatpx(f) is ultra-ample. We will consider the following cases.
Case 1. If z=x and y=x, then the result holds since d(x,x)=0.
Case 2. If z≠x and y≠x, then (px(f))2(z)=f2(z), so that(7)max⁡pxf2z,pxf1y=max⁡f2z,f1y≥dz,y.
Case 3. Consider z=x and y≠x. In this case (px(f))1(y)=f1(y) and (px(f))2(z)=sup⁡{d(z,y):y∈Xandd(z,y)>f1(y)}, so that (8)max⁡pxf2z,pxf1y=max⁡sup⁡dz,y:y∈X,dz,y>f1y,f1y≥dz,y.
Case 4. In a manner similar to case 3, the result can be shown.
Thuspx(f) is ultra-ample and also satisfies px(f)≤f by the way it was constructed.
Thus by takingg=px(f), we can conclude that, for any f∈uPX, g≤f.Lemma 16 (compare [4, Lemma 8]).
Let(X,d) be UQP-metric space. Then, for any f,g∈uQX, we have that(9)Nf,g=sup⁡dx1,x2:x1,x2∈X,dx1,x2>f2x1,dx1,x2>g1x2.Proposition 17 (compare [1, Proposition 1]).
Let(X,d) be UQP-metric space. There exists a retraction map p:uPX→uQX, that is, a map that satisfies the following conditions:(a)
N(p(f),p(g))≤N(f,g) whenever f,g∈uPX.(b)
p(f)≤f whenever f∈uPX.(In particular we have thatp(f)=f whenever f∈uQX.)Proof.
We will prove Proposition17 by the use of Zorn’s lemma.
Indeed, let(X,d) be UQP-metric space and let P be the set of all maps from uPX to uPX satisfying conditions (a) and (b) in Proposition 17.
OrderP by(10)p⪯q⟺pf≤qf,⟺Npf,pg≤Nqf,qgfor all f,g∈uPX and p,q∈P. Then P≠∅ since the identity map belongs to P.
We have to check now that⪯ is actually a partial order.
Reflexivity is obvious since every map is equal to itself.
Let nowp,q∈P such that p⪯q and q⪯p. Consider (11)p⪯q⟹pf1≤qf1,⟹pf2≤qf2,⟹Npf,pg≤Nqf,qgq⪯p⟹qf1≤pf1,⟹qf2≤pf2,⟹Nqf,qg≤Npf,pg.(p(f))1≤(q(f))1 and (q(f))1≤(p(f))1 imply that (p(f))1=(q(f))1. In a similar manner, we have that (p(f))2=(q(f))2 so that we can conclude that p=q.
AlsoN(p(f),p(g))≤N(q(f),q(g)) and N(q(f),q(g))≤N(p(f),p(g)) imply that p=q. This shows that ⪯ is antisymmetric.
Suppose now thatp,q,s∈P such that p⪯q and q⪯s: (12)p⪯q⟹pf1≤qf1,⟹pf2≤qf2,⟹Npf,pg≤Nqf,qgq⪯s⟹qf1≤sf1,⟹qf2≤sf2,⟹Nqf,qg≤Nsf,sg.(p(f))1≤(q(f))1 and (q(f))1≤(s(f))1 imply that (p(f))1≤(s(f))1 by transitivity of [0,∞) as a subset of R with the usual ordering ≤. Similarly, we can show that (p(f))2≤(s(f))2.
AlsoN(p(f),p(g))≤N(q(f),q(g)) and N(q(f),q(g))≤N(s(f),s(g)) imply that N(p(f),p(g))≤N(s(f),s(g)). Thus p⪯s. This proves that ⪯ is transitive. Therefore (P,⪯) is a partially ordered set.
Next we show that every chain inP has a lower bound.
Let∅≠K⊆P be a chain and define s:uPX→uPX by(13)sfx≔infk∈K⁡kf1x,infk∈K⁡kf2xwhenever x∈X. Since k(f)∈P, we have that s(f)∈P.
Indeed observe thats(f)≤k(f)≤f, for all f∈uPX. Thus s(f)≤f and condition (a) is satisfied.
To check condition(b), we check that N(s(f),s(g))≤N(k(f),k(g))≤N(f,g).
Indeed(14)Nf,g≥Nkf,kg,sincek∈P≥supx∈X⁡nkf1x,kg1x∨supx∈X⁡nkg2x,kf2x≥supx∈X⁡infk∈K⁡nkf1x,kg1x∨supx∈X⁡infk∈Knkg2x,kf2x≥supx∈X⁡ninfk∈K⁡kf1x,infk∈Kkg1x∨supx∈X⁡ninfk∈K⁡kg2x,infk∈K⁡kf2x=Nsf,sg.
Thus we have that condition (b) is satisfied and sinces is a map from uPX to uPX, we conclude that s∈P and s is a lower bound of the chain K by construction. We therefore appeal to Zorn’s lemma to conclude that P has a minimal element, say m, with respect to the partial order ⪯.
To complete the proof, we show thatm(f)∈uQX whenever f∈uPX.
For eachx∈X, we have that px∘m∈P and px∘m⪯m (where px is as defined in the proof of Proposition 15). Thus by minimality of m, we have px∘m=m. It therefore follows that, for each x∈X, px(m(f))=m(f) whenever f∈uPX. Thus by the way elements in uQX are defined, we conclude that m(f)∈uQX whenever f∈uPX.Proposition 18 (compare [1, Proposition 3]).
Let(Y,d) be UQP-metric space and let ∅≠X be a subspace of (Y,d). Then there exists an isometric embedding τ:uQX→uQY such that τfX=f whenever f∈uQX.Proof.
Fixx0∈X and let p:uPY→uQY be a retraction satisfying the conditions of Proposition 17. Also let s:uQX→uPY be such that s(f)=f′, where f1′(y)=f1(y) whenever y∈X, and f1′(y)=max⁡{f1(x0),d(x0,y)} whenever y∈Y∖X. The coordinate f2′ of pair f′ is defined similarly.
We will consider the following cases to prove thatf′ belongs to uPY.
Case 1. Consider that x∈X and y∈X.
Thenmax⁡{f2′(x),f1′(y)}=max⁡{f2(x),f1(y)}≥d(x,y).
Case 2. One has that x∈Y∖X and y∈Y∖X.
Then(15)max⁡f2′x,f1′y=max⁡f2x0,f1x0,dx,x0,dx0,y≥max⁡dx,x0,dx0,y≥dx,y.
Case 3. Consider that x∈X and y∈Y∖X.
Thenmax⁡{f2′(x),f1′(y)}=max⁡{f2(x),f1(x0),d(x0,y)}≥max⁡{d(x,x0),d(x0,y)}≥d(x,y).
Case 4. Consider that x∈Y∖X and y∈X.
Thenmax⁡{f2′(x),f1′(y)}=max⁡{f2(x0),f1(y),d(x,x0)}≥max⁡{d(x,x0),d(x0,y)}≥d(x,y).
Thusf′∈uPY.
Define mapτ=p∘s. Then τfX=pf′X=f whenever f∈uQX since p(f′)≤f′. Thus pf′X≤f′X=f, and f is minimal on X.
Moreover for anyf,g∈uQX, we have (16)Nf,g=NτfX,τgX≤Nτf,τg=Npf′,pg′≤Nf′,g′=Nf,g.The last equality follows from the definition of f′ and g′. Hence we have that N(f,g)≤N(τ(f),τ(g))≤N(f,g) which implies that N(τ(f),τ(g))=N(f,g) and hence τ is an isometric map.Definition 19.
LetX be a subspace of UQP-metric space (Y,dY). Then (Y,dY) is called uq-tight extension of X if for any uqp-metric ρ on Y that satisfies ρ≤dY and agrees with dY on X×X, we have that ρ=dY.Remark 20.
For anyUQP-metric uq-tight extension Y1 of X, any UQP-metric extension (Y2,d) of X and any nonexpansive map φ:Y1→Y2 satisfying φ(x)=x whenever x∈X must necessarily be an isometric map.
Indeed if that is not the case, then theuqp-metric ρ:Y1×Y1→[0,∞) defined by (x,y)↦ρ(x,y)=d(φ(x),φ(y)) would contradict the fact that the span Y1 of X is ultra-ample.
It was shown in [4] that the map eX:(X,d)→(uQX,N) from UQP-metric space (X,d) to its ultra-quasi-metrically injective hull (uQX,N) defined by eX(a)=fa whenever a∈X is an isometric embedding. We will proceed now with the help of Lemma 16 to show that uQX is uq-tight extension of eX(X).Proposition 21.
Let(X,d) be UQP-metric space and let eX:X→uQX be as defined above. Then uQX is uq-tight extension of eX(X).Proof.
Letρ be uqp-metric on uQX such that ρ≤N and ρ(fx,fy)=N(fx,fy) whenever x,y∈X. By Lemma 16 and the fact that ρ≤N, for any f,g∈uQX, we have (17)Nf,g=supx1,x2∈X⁡Nfx1,fx2:Nfx1,fx2>Nfx1,f,Ng,fx2≤supx1,x2∈X⁡ρfx1,fx2:ρfx1,fx2>ρfx1,f,ρg,fx2≤ρf,g,sinceρfx1,fx2≤max⁡ρfx1,f,ρf,g,ρg,fx2.Thus ρ=N.Proposition 22.
Let(Y,d) be UQP-metric uq-tight extension of X. Then the restriction map defined by f↦fX whenever f∈uQY is a bijective isometric map uQY→uQX.Proof.
Letp:uPX→uQX be a retraction map that satisfies the conditions of Proposition 17 and let φ:uQY→uQX:f↦p(fX) denote the composition of the retraction map p with the restriction map. Then one sees immediately that φ is nonexpansive. Thus by Lemma 3, φ must be an isometry, because uQY is uq-tight extension of X (this is so since uQY is uq-tight extension of Y and Y is uq-tight extension of X).
Chooseτ:uQX→uQY an isometric embedding such that τfX=f for every f∈uQX (compare Proposition 18). We therefore have(18)φτf=pτfX=pf=ffor  everyf∈uQX.This implies that φ is onto. The fact that φ is injective is clear since (uQX,N) is UQP-metric space (see, e.g., the last line of Definition 7). Thus φ is bijective. In this case, the inverse of φ has to be the inverse of τ and hence for any f∈uQY, we have f|X=τ(φ(f))|X=φ(f)∈uQX, which is the map(19)uQY⟶uPX:f⟼fXthat maps uQY onto uQX, without it being composed of p. Hence for any UQP-metric uq-tight extension Y of X, the map(20)uQY⟶uQX:f⟼fXis a bijective isometry between uQX and uQY.Theorem 23.
LetX be a subspace of the UQP-metric space (Y,d). Then the following are equivalent:(a)
Y is UQP-metric uq-tight extension of X.(b)
d(y1,y2)=sup⁡{d(x1,x2):x1,x2∈X,d(x1,x2)>d(x1,y1),d(x1,x2)>d(y2,x2)} whenever y1,y2∈Y.(c)
fy|X(x)=(d(y,x),d(x,y)), x∈X, is minimal on X whenever y∈Y and the map (Y,d)→(uQX,N) defined by y↦fy|X is an isometric embedding.Proof.
(a)⇒(b). Let Y be UQP-metric uq-tight extension of X. By Proposition 22, the restriction map uQY→uQX is a bijective isometry between uQY and uQX. Thus the extension Y⊆uQY satisfies condition (b), since uQX satisfies it by Lemma 16.
(b)⇒(c). Let x1,x2∈X and y1∈Y. Then we have that d(x1,x2)≤max⁡{d(x1,y1),d(y1,x2)}. Thus by condition (b) we have that d(x1,x2)≤d(y1,x2). Also(21)dx1,x2≤ndy1,x2,dy2,x2≤dy1,y2.Hence for y1,y2∈Y we have by condition (b) that (22)dy1,y2=sup⁡dx1,x2:x1,x2∈X,dx1,x2>dx1,y1,dx1,x2>dy2,x2≤sup⁡dy1,x2:x2∈X,dy1,x2>dy2,x2≤dy1,y2.Similarly we have that d(x1,x2)≤max⁡{d(x1,y2),d(y2,x2)} whenever x1,x2∈X and y2∈Y so that by condition (b) we get d(x1,x2)≤d(x1,y2). It therefore follows that, for each x1,x2∈X and y1,y2∈Y, d(x1,x2)≤d(x1,y2). Thus for y1,y2∈Y we see by (b) that (23)dy1,y2=sup⁡dx1,x2:x1,x2∈X,dx1,x2>dx1,y1,dx1,x2>dy2,x2≤sup⁡dx1,y2:x1∈X,dx1,y2>dx1,y1≤dy1,y2.Thus we conclude that d(y1,y2)=N(fy1X,fy2X).
As we have above, for anyy1,y2∈Y we have that(24)dy1,y2=sup⁡dy1,x2:x2∈X,dy1,x2>dy2,x2,(25)dy1,y2=sup⁡dx1,y2:x1∈X,dx1,y2>dx1,y1.Observe that if we substitute x1∈X for y1 in (24) and x2∈X for y2 in (25) we obtain the following equations:(26)fy22x1=dx1,y2=sup⁡dx1,x2:x2∈X,dx1,x2>dy2,x2whenever y2∈Y and x1∈X and(27)fy11x2=dy1,x2=sup⁡dx1,x2:x1∈X,dx1,x2>dx2,y2whenever y1∈Y and x2∈X. We have therefore that restriction fyX is minimal on X whenever y∈Y (compare Lemma 12).
(c)⇒(a). Let ρ be uqp-metric on Y such that ρ(y1,y2)≤d(y1,y2) whenever y1,y2∈Y and ρ(x1,x2)=d(x1,x2) whenever x1,x2∈X. Then by part (c) and the fact that fyX is minimal whenever y∈X, we have (28)dy1,y2=Nfy1X,fy2X=sup⁡dy1,x:x∈X,dy1,x>dy2,xby24=supdx,y2:x∈X,dx,y2>dx,y1by25.By substituting(29)dx1,y2=sup⁡dx1,x2:x2∈X,dx1,x2>dy2,x2into formula(30)dy1,y2=sup⁡dx1,y2:x1∈X,dx1,y2>dx1,y1we obtain (31)dy1,y2=supdx1,y2:x1∈X,dx1,y2>dx1,y1=sup⁡dx1,x2:x1,x2∈X,dx1,y2>dx1,y1,dx1,x2>dy2,x2=sup⁡ρx1,x2:x1,x2∈X,ρx1,y2>ρx1,y1,ρx1,x2>ρy2,x2≤ρy1,y2whenever y1,y2∈Y. The last inequality holds by the light of the inequality(32)ρx1,x2≤max⁡ρx1,y1,ρy1,y2,ρy2,x2and the fact that ρ(x1,x2)>ρ(x1,y1) and ρ(x1,x2)>ρ(y2,x2). Thus we have that ρ(y1,y2)=d(y1,y2) whenever y1,y2∈Y and hence (a) follows.Remark 24.
We see from Theorem23 that there is only one isometric embedding φ:Y→uQX satisfying φ(x)=fx whenever x∈X, since for such an embedding we have(33)fy2Xx=dx,y=Nφx,φy=Nfx,φy=φy2x;therefore (fy)2|X=(φ(y))2. Similarly, one can show that (fy)1|X=(φ(y))1 whenever y∈Y.
Thus we see that theuq-tight extension Y of X can be understood as a subspace of extension uQX of X. Hence uQX is maximal among the UQP-metric uq-tight extensions of X.



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