Endpoints in T_0-quasi-metric spaces

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Topology and its Applications 168 (2014) 82–93

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Topology and its Applications

www.elsevier.com/locate/topol

Endpoints in T0-quasi-metric spaces ✩

Collins Amburo Agyingi, Paulus Haihambo, Hans-Peter A. Künzi ∗

Department of Mathematics and Applied Mathematics, University of Cape Town, Rondebosch 7701,South Africa

a r t i c l e i n f o a b s t r a c t

MSC:54D3554E1554E3554E5554E50

Keywords:HyperconvexityT0-quasi-metric spaceJoincompactEndpointStartpointInjective hullq-hyperconvexTight span

In his well-known paper dealing with the construction of the injective hull of ametric space Isbell introduced the concept of an endpoint of a (compact) metricspace.In the present article we introduce similarly the notion of an endpoint in a(joincompact) T0-quasi-metric space. It turns out that in a (joincompact) T0-quasi-metric space there is a dual concept which we shall call a startpoint. With the helpof these concepts we are able to generalize some of the classical results on endpointsin metric spaces to the quasi-metric setting.

© 2014 Elsevier B.V. All rights reserved.

1. Introduction

In [9] Isbell showed that every metric space X has a hyperconvex hull TX . It turns out to be compactprovided that X is totally bounded (see for instance [5, (1.7)]). It is known that a metric space is hyperconvexif and only if it is injective in the category of metric spaces and nonexpansive maps (compare also [5,6,11]).

In [10, Propositions 7 and 8] Kemajou et al. similarly proved that each T0-quasi-metric space X has aq-hyperconvex hull QX , which is joincompact provided that X is totally bounded. They also showed that aT0-quasi-metric space is q-hyperconvex if and only if it is injective in the category of T0-quasi-metric spacesand nonexpansive maps [10, Theorem 1].

✩ The authors would like to thank the National Research Foundation of South Africa for partial financial support. This researchwas also supported by a Marie Curie International Research Staff Exchange Scheme Fellowship within the 7th European CommunityFramework Programme.* Corresponding author.

E-mail addresses: [email protected] (C.A. Agyingi), [email protected] (P. Haihambo),[email protected] (H.-P.A. Künzi).

http://dx.doi.org/10.1016/j.topol.2014.02.0100166-8641/© 2014 Elsevier B.V. All rights reserved.


C.A. Agyingi et al. / Topology and its Applications 168 (2014) 82–93 83

In this paper we intend to generalize further results due to Isbell [9] and Dress [5] to the category ofT0-quasi-metric spaces and nonexpansive maps. During his investigations on the hyperconvex hull of a metricspace Isbell introduced the concept of an endpoint of a metric space and proved among other things thatthe hyperconvex hull of a compact metric space is equal to the hyperconvex hull of the subspace consistingof its endpoints (compare also [5,8]).

It turns out that in the quasi-metric context it is natural to consider also the dual concept of an endpoint,which we shall call a startpoint in this paper. In this way we succeed in generalizing several results onendpoints of metric spaces to the quasi-metric setting. For instance we show that the q-hyperconvex hull ofa joincompact T0-quasi-metric space X can be identified with the q-hyperconvex hull of the subspace B ofX which consists of all the startpoints and endpoints of X.

2. Preliminaries

In this section we recall some of the basic definitions from asymmetric topology needed to read this paper.

Definition 1. Let X be a set and d : X × X → [0,∞) be a function mapping into the set [0,∞) of thenonnegative reals. Then d is a quasi-pseudometric on X if

(a) d(x, x) = 0 whenever x ∈ X, and(b) d(x, z) � d(x, y) + d(y, z) whenever x, y, z ∈ X.

We shall say that (X, d) is a T0-quasi-metric space provided that d also satisfies the following condition:For each x, y ∈ X, d(x, y) = 0 = d(y, x) implies that x = y.

Let d be a quasi-pseudometric on a set X. Then d−1 : X × X → [0,∞) defined by d−1(x, y) = d(y, x)whenever x, y ∈ X is also a quasi-pseudometric, called the conjugate quasi-pseudometric of d. Observe thatif d is a T0-quasi-metric on X, then ds = max{d, d−1} = d ∨ d−1 is a metric on X.

Let (X, d) be a quasi-pseudometric space. For each x ∈ X and ε > 0, Bd(x, ε) = {y ∈ X: d(x, y) < ε}denotes the open ε-ball at x. The collection of all “open” balls yields a base for a topology τ(d). It is calledthe topology induced by d on X. Similarly we set Cd(x, ε) = {y ∈ X: d(x, y) � ε} whenever x ∈ X andε � 0. Observe that Cd(x, ε) is τ(d−1)-closed.

A map f : (X, d) → (Y, e) between quasi-pseudometric spaces (X, d) and (Y, e) is called isometric providedthat d(x, y) = e(f(x), f(y)) whenever x, y ∈ X. Note that each isometric map with a T0-quasi-metric domainis a one-to-one map.

Furthermore a map f : (X, d) → (Y, e) between quasi-pseudometric spaces is called nonexpansive providedthat e(f(x), f(y)) � d(x, y) whenever x, y ∈ X.

Given two real numbers a and b we shall write a − b for max{a − b, 0}, which we shall also denote by(a − b) ∨ 0. Note that u(x, y) = x − y with x, y ∈ R defines a T0-quasi-metric on the set R of the reals.Observe that x �→ −x defines a bijective isometric map from (R, u) to (R, u−1).

For further basic concepts used from the theory of asymmetric topology we refer the reader to [7] and[12]. The reader can find some recent work about quasi-pseudometric spaces in [1,4,13–15]. Many basic factsabout hyperconvexity in metric spaces can be found in [6,11].

Remark added during revision: The authors continued their investigations in [3]. The results of the presentpaper were used by Otafudu in [16].

3. Collinearity in quasi-pseudometric spaces

The following definition is crucial for our paper.


84 C.A. Agyingi et al. / Topology and its Applications 168 (2014) 82–93

Definition 2. (Compare [8].) Let (X, d) be a quasi-pseudometric space.

(1) A finite sequence (x1, x2, . . . , xn) in X is called collinear in (X, d) provided that i < j < k � n impliesthat d(xi, xk) = d(xi, xj) + d(xj , xk).

(2) An element x ∈ X is called an endpoint of (X, d) provided that there exists an element y in (X, d) suchthat d(y, x) > 0 and for any z ∈ X collinearity of (y, x, z) in (X, d) implies that x = z. We shall saythat y witnesses that x is an endpoint.The set of endpoints of a T0-quasi-metric space (X, d) will be denoted by Ed.

(3) An element x ∈ X is called a startpoint of (X, d) if it is an endpoint of (X, d−1). By Ad we shall denotethe set of startpoints of (X, d).

Note that a quasi-pseudometric space possessing only one point cannot have an endpoint resp. a start-point. Observe also that we obtain the standard definition of an endpoint in the case that (X, d) is a metricspace (compare [9, p. 73]).

Remark 1. Observe that the sequence (x1, x2, x3) is collinear in the quasi-pseudometric space (X, d) if andonly if the sequence (x3, x2, x1) is collinear in (X, d−1). �Example 1. Consider the four point set X = {1, 2, 3, 4}. Let the T0-quasi-metric ρ be defined by the distancematrix

M =

⎛⎜⎜⎝0 1 2 11 0 1 22 1 0 12 1 1 0

⎞⎟⎟⎠ ,

that is, mi,j = ρ(i, j) whenever i, j ∈ X. One easily checks that indeed ρ is a T0-quasi-metric on X.Furthermore the sequences (1, 2, 3), (3, 2, 1) and (4, 2, 1) are readily checked to be collinear in (X, ρ).

They show that 2 is not an endpoint of (X, ρ). On the other hand neither of the sequences (1, 2, 4) nor(3, 2, 4) nor (4, 2, 4) are collinear in (X, ρ), which shows that 4 witnesses that 2 is indeed a startpoint of(X, ρ).

As an illustration, observe also that both (1, 2, 3) and (2, 3, 4) are collinear, but (1, 2, 3, 4) is not collinear,since (1, 3, 4) is not collinear. �Lemma 1. ([8, p. 184]) If (x1, x2, x3) and (x1, x3, x4) are collinear sequences in a quasi-pseudometric space(X, d), then so are (x1, x2, x4) and (x2, x3, x4). (Hence (x1, x2, x3, x4) is collinear.)

Proof. As in [8] we have that d(x1, x4) � d(x1, x2)+d(x2, x4) � d(x1, x2)+d(x2, x3)+d(x3, x4) = d(x1, x3)+d(x3, x4) = d(x1, x4) imply d(x1, x4) = d(x1, x2) + d(x2, x4) and d(x2, x4) = d(x2, x3) + d(x3, x4). �Proposition 1. Let (X, d) be a T0-quasi-metric space. Fix y ∈ X. Set a1 �y a2 if (y, a1, a2) is collinear in(X, d). Then �y is a partial order on X.

Proof. For any a ∈ X we have that (y, a, a) is collinear in (X, d), since d(y, a) + d(a, a) = d(y, a). So �y isreflexive.

Let a1, a2 ∈ X be such that a1 �y a2 and a2 �y a1. Then d(y, a1) + d(a1, a2) = d(y, a2) and d(y, a2) +d(a2, a1) = d(y, a1). Thus d(y, a1) + d(a1, a2) + d(a2, a1) = d(y, a2) + d(a2, a1) = d(y, a1). Then d(a1, a2) +d(a2, a1) = 0. By the T0-property therefore a1 = a2. So �y is antisymmetric.

Suppose that (y, a1, a2) and (y, a2, a3) are collinear in (X, d). We need to show that (y, a1, a3) is collinearin (X, d). But this follows from Lemma 1. So �y is transitive. �



Let (X, d) be a quasi-pseudometric space. Then d is called joincompact provided that the topology τ(ds)is compact.

Example 2. Let (X, d) be a T0-quasi-metric space such that for some a, b ∈ X, d(a, b) = sup{d(x, y): x, y ∈X} > 0. (Note that such a and b exist for instance if (X, d) is joincompact and X has at least two points.)Then a is a startpoint of (X, d) if {a} is τ(d)-closed and b is an endpoint of (X, d) if {b} is τ(d−1)-closed.

Proof. Consider any x ∈ X such that (x, a, b) is collinear in (X, d). Hence d(x, a)+d(a, b) = d(x, b). Thereforeby our assumption d(x, a) = 0 and thus x = a, since {a} is τ(d)-closed. Thus a is a startpoint in (X, d), sinced(a, b) > 0. Consider any x ∈ X such that (a, b, x) is collinear in (X, d). Hence d(a, b) + d(b, x) = d(a, x).Therefore d(b, x) = 0 and b = x, since {b} is τ(d−1)-closed. Thus b is an endpoint in (X, d), witnessedby a. �4. q-hyperconvex hulls of T0-quasi-metric spaces

A T0-quasi-metric space (X, d) is said to be q-hyperconvex if for each family (xi)i∈I of points of X andfamilies (ri)i∈I and (si)i∈I of nonnegative real numbers the following conditions hold: If d(xi, xj) � ri + sjwhenever i, j ∈ I, then

⋂i∈I(Cd(xi, ri) ∩ Cd−1(xi, si)) �= ∅ (see [10, Definition 2]).

Example 3. (Compare [6, discussion after Theorem 3.1].) Each T0-quasi-metric space X can be isomet-rically embedded into a q-hyperconvex T0-quasi-metric space, as the following well-known analogue of aconstruction for metric spaces shows.

Let (X, d) be a T0-quasi-metric space and a ∈ X fixed. Furthermore let B(X, d) be the set of real-valuedfunctions which are bounded on (X, d).

Given f, g ∈ B(X, d), set N(f, g) = supx∈X(f(x) − g(x)). Obviously (B(X, d), N) is a T0-quasi-metricspace. We next show that (B(X, d), N) is q-hyperconvex.

Indeed suppose that (fi)i∈I is a family of points in B(X, d) and (ri)i∈I and (si)i∈I are families ofnonnegative real numbers such that N(fi, fj) � ri + sj whenever i, j ∈ I.

Then for each x ∈ X we have u(fi(x), fj(x)) � ri + sj . Hence by q-hyperconvexity of (R, u) (see [10,Example 1]) we have that for each x ∈ X there is h(x) ∈

⋂i∈I(Cu(fi(x), ri) ∩ Cu−1(fi(x), si)). Hence

h ∈⋂

i∈I(CN (fi, ri) ∩ CN−1(fi, si)), since obviously h ∈ B(X, d).Note that by the triangle inequality, for any b ∈ X the function eb(x) = d(b, x)− d(a, x) whenever x ∈ X

is bounded by ds(b, a).Then the map e : X → B(X, d) defined by b �→ eb yields an isometric embedding of (X, d) into B(X, d),

since for any b, b′ ∈ X, we see that

N(eb, eb′) = supx∈X

[(d(b, x) − d(a, x)

)−(d(b′, x

)− d(a, x)

)]= sup

x∈X

(d(b, x) − d

(b′, x

))= d

(b, b′

).

Recall that a T0-quasi-metric space (X, d) is called bicomplete provided that the metric space (X, ds)is complete. Each q-hyperconvex T0-quasi-metric space is known to be bicomplete (see [10, Corollary 3]).Hence the τ(Ns)-closure of e(X) in B(X, d) yields the (quasi-)metric bicompletion of (X, d) (compare [12,Example 2.7.1]). In general however it is a nontrivial task to identify a subspace of (B(X, d), N) that isisometric to the q-hyperconvex hull QX of X, although such minimal q-hyperconvex extensions of e(X) mustexist in (B(X, d), N) (compare [10, Proposition 7]): Indeed using injectivity of B(X, d), by [10, Theorem 1]we can extend the isometric embedding e : X → B(X, d) to a nonexpansive map e : QX → B(X, d), whichmust be an (injective) isometric map by the fact that QX is a T0-quasi-metric tight extension of X (see [2,Remark 4]).



If we apply the function space construction above to (X, d−1) and note that B(X, d) = B(X, d−1), thenwe see that (X, d) also embeds isometrically into (B(X, d), N−1) via b �→ d(x, b) − d(x, a) (x ∈ X).

Next we recall various facts from the theory of the construction of the q-hyperconvex hull of a T0-quasi-metric space [10].

Let (X, d) be a T0-quasi-metric space. We shall say that a function pair f = (f1, f2) on (X, d) wherefi : X → [0,∞) (i = 1, 2) is ample if d(x, y) � f2(x) + f1(y) whenever x, y ∈ X.

Let PX denote the set of all ample function pairs on (X, d). (In such situations we may also writeP(X,d) in cases where d is not obvious.) For each f, g ∈ PX we set D(f, g) = supx∈X(f1(x) − g1(x)) ∨supx∈X(g2(x) − f2(x)). Then D is an extended1 T0-quasi-metric on PX .

A function pair f with fi : X → [0,∞) (i = 1, 2) is called minimal on (X, d) (among the ample functionpairs on (X, d)) if it is ample and whenever g is ample on X and for each x ∈ X we have g1(x) � f1(x) andg2(x) � f2(x),2 then g = f . It is well known that Zorn’s Lemma implies that below each ample function pairthere is a minimal ample pair (for a more constructive and a global approach, essentially due to Dress [5],see [2, Proposition 1]). By QX we shall denote the set of all minimal ample pairs on (X, d) equipped withthe restriction of D to QX × QX , which for convenience we shall also denote by D. It is known that D isindeed a (real-valued) T0-quasi-metric on QX ×QX [10, Remark 6].

Furthermore f ∈ PX belongs to QX if and only if

f1(x) = sup{d(y, x) − f2(y): y ∈ X

}and

f2(x) = sup{d(x, y) − f1(y): y ∈ X

}whenever x ∈ X (see [13, Remark 2]).

Note (compare [5, (1.2)]) that for any f ∈ PX there exists a unique maximal subset Y ⊆ X with f |Y ∈ QY

(which may be empty), since f ∈ PX , f |Yα∈ QYα

for a family {Yα ⊆ X: α ∈ A} of subsets of X andY =

⋃α∈A Yα imply that for y ∈ Yα and α ∈ A,

f2(y) � sup{d(y, z) − f1(z): z ∈ X

}� sup

{d(y, z) − f1(z): z ∈ Y

}� sup

{d(y, z) − f1(z): z ∈ Yα

}= f2(y)

and similarly

f1(y) � sup{d(z, y) − f2(z): z ∈ X

}� sup

{d(z, y) − f2(z): z ∈ Y

}� sup

{d(z, y) − f2(z): z ∈ Yα

}= f1(y).

Hence the assertion is verified.It is an important fact (see [10, Lemma 3]) that f ∈ QX implies that f1(x) − f1(y) � d(y, x) and

f2(x) − f2(y) � d(x, y) whenever x, y ∈ X.Moreover the definition of the distance between pairs belonging to QX is simplified by the fact that

supx∈X(f1(x) − g1(x)) = supx∈X(g2(x) − f2(x)) whenever f, g ∈ QX (compare [10, Lemma 7]).

1 If we replace in the definition of a quasi-pseudometric [0,∞) by [0,∞], we obtain the definition of an extended quasi-pseudometric. As usual, the triangle inequality for extended quasi-pseudometrics is interpreted in the obvious way.2 In this case we shall write g � f .



For each x ∈ X we can define the minimal function pair

fx(y) =(d(x, y), d(y, x)

)(whenever y ∈ X) on (X, d). The map e determined by x �→ fx whenever x ∈ X defines an isometricembedding of (X, d) into (QX , D) (see [10, Lemma 1]). The T0-quasi-metric space (QX , D) is called theq-hyperconvex hull of (X, d). Of course, it is q-hyperconvex. In fact a T0-quasi-metric space X is known tobe q-hyperconvex if and only if f ∈ QX implies that there is an x ∈ X such that f = fx (compare [10,Corollary 4]).

We finally observe that D(f, fx) = f1(x) and D(fx, f) = f2(x) whenever x ∈ X and f ∈ QX [10,Lemma 8].

The following illustrating example generalizes [10, Examples 7 and 8] by treating the case of an arbitraryT0-quasi-metric space possessing two points.

Remark added during revision: Independently the example was also discussed in recent work of Willerton[17].

Example 4. Let X = {0, 1} be equipped with a T0-quasi-metric d defined as follows: Set d(0, 1) = b andd(1, 0) = a, where a and b are nonnegative reals such that a+ b �= 0. We shall denote this space by (Xa,b, d)in the following.

We can identify the q-hyperconvex hull QXa,bof (Xa,b, d) with the rectangle [0, a] × [0, b] equipped with

the T0-quasi-metric

d((x, y),

(x′, y′

))= max

{x −x′, y − y′

}whenever (x, y), (x′, y′) ∈ [0, a] × [0, b].

Indeed it is readily checked that we obtain exactly the following minimal function pairs (f1, f2) on(Xa,b, d): Namely (f1(0), f1(1)) = (x, y) and (f2(0), f2(1)) = (b− y, a− x) where (x, y) ∈ [0, a] × [0, b].

Hence we can identify the points of QXa,bwith the points of [0, a] × [0, b].

Obviously in this way via the isometric embedding e : (Xa,b, d) → (QXa,b, d) the point 0 of (Xa,b, d) is

identified with the function (f0)1 = (0, b) on Xa,b and the point 1 of (Xa,b, d) is identified with the function(f1)1 = (a, 0) on Xa,b.

We also note that for any (α, β) ∈ [0, a] × [0, b] the sequences

((a, 0), (α, β), (0, b)

)and

((0, b), (α, β), (a, 0)

)are collinear in QXa,b

.Suppose for the following that a, b > 0. Let us first show that (0, 0) is not an endpoint of QXa,b

. Assumethat (y1, y2) ∈ QXa,b

and (y1, y2) �= (0, 0) witnesses that (0, 0) is an endpoint. Note that d((y1, y2), (0, 0)) > 0.Let y = max{y1, y2}. Then ((y1, y2), (0, 0), (0, y)) is collinear if y = y1 and ((y1, y2), (0, 0), (y, 0)) is collinearif y = y2. Thus no element of QXa,b

can witness that (0, 0) is an endpoint. Observe that d((0, 0), (y1, y2)) = 0for all (y1, y2) ∈ QXa,b

. So there is no point (y1, y2) in QXa,bwhich can witness that (0, 0) is a startpoint,

too.



Example 5. Note that the diagonal Δ of (R2, u × u−1) is isometric to (R, us) where we set E = u × u−1,that is,

E((x1, x2), (y1, y2)

)= (x1 − y1) ∨ (y2 −x2)

whenever (x1, x2), (y1, y2) ∈ R2.As a product of q-hyperconvex factor spaces (compare [10, Proposition 2]) (R2, u×u−1) is readily seen to

be q-hyperconvex. We want to show that the q-hyperconvex hull of the diagonal Δ is equal to (R2, u×u−1).To this end we show that each point of this product needs to be added to the points of the diagonal in

order to obtain a q-hyperconvex extension of the diagonal contained in the product (R2, u× u−1).Given (x1, x2) ∈ R2 and ε, ε′ � 0 we have that

CE

((x1, x2), ε

)∩ CE−1

((x1, x2), ε′

)=([x1 − ε,∞

)×(−∞, x2 + ε]

)∩((−∞, x1 + ε′] × [x2 − ε′,∞

))=[x1 − ε, x1 + ε′

]×[x2 − ε′, x2 + ε

].

Let (a, b) ∈ R2. Choose x, y ∈ R such that y < a < x and y < b < x.Hence we have that

CE

((x, x), x− a

)∩ CE−1

((x, x), x− b

)∩ CE

((y, y), b− y

)∩ CE−1

((y, y), a− y

)={(a, b)

}.

Since indeed we have E((x, x), (y, y)) = E((y, y), (x, x)) = us(x, y) = x−y = (x−a)+(a−y) = (b−y)+(x−b),by the definition of q-hyperconvexity we conclude that (a, b) must belong to any q-hyperconvex extensionof Δ in (R2, u× u−1). Hence we see that (R2, u× u−1) is the q-hyperconvex hull of its subspace Δ and thestatement is verified.

5. A crucial lemma related to collinearity

Obviously the following result is related to the concept of collinearity that we have studied before (compareExample 4).

Lemma 2. (Compare [9, 2.6].) Let (X, d) be a joincompact T0-quasi-metric space and f ∈ QX . Given x ∈ X

such that f2(x) > 0 there is y ∈ X such that d(x, y) = f2(x) + f1(y). (Note that necessarily x �= y.)Similarly for each x ∈ X such that f1(x) > 0 there is y′ ∈ X such that d(y′, x) = f2(y′) + f1(x).

Proof. Assume that m ∈ N is such that 1m < f2(x). Consider any n ∈ N such that n � m.

Suppose that for all y ∈ X we have d(x, y) + 1n � f2(x) + f1(y).

Then set h2(z) = f2(x)− 1n if z = x, and h2(z) = f2(z) whenever z ∈ X \ {x}. We note that

(0, 0) � (f1, h2) < (f1, f2), but (f1, h2) is an ample function pair on (X, d), so that (f1, f2) is not minimal—a contradiction.

Therefore for each n ∈ N such that n � m there is yn ∈ X such that f2(x) + f1(yn) < d(x, yn) + 1n .

By joincompactness of (X, d) there are y ∈ X and a subsequence (ynk)k∈N of (yn)n�m such that

ds(ynk, y) → 0. We get that f2(x)+f1(y) � d(x, y) � f2(x)+f1(y), because we have that |f1(ynk

)−f1(y)| �ds(ynk

, y) and |d(x, ynk) − d(x, y)| � ds(ynk

, y) whenever k ∈ N, because f is minimal ample. The secondstatement is proved analogously. �

We note that the first part of the preceding proof yields the following corollary.

Corollary 1. (Compare [9, 2.6].) Let (X, d) be a T0-quasi-metric space, f ∈ QX , ε > 0 and x ∈ X. Supposethat f2(x) > 0. Then there is y ∈ X such that d(x, y) + ε > f2(x) + f1(y). Similarly suppose that f1(x) > 0.Then there is y′ ∈ X such that d(y′, x) + ε > f2(y′) + f1(x). �



Lemma 3. (Compare [9, 2.6].) Let (X, d) be a T0-quasi-metric space, (f1, f2) be an ample function pairon (X, d) and given arbitrary ε > 0. Suppose for each x ∈ X with f2(x) > 0 there is y ∈ X such thatd(x, y) + ε > f2(x) + f1(y); and similarly for each x ∈ X with f1(x) > 0 there is y′ ∈ X such thatd(y′, x) + ε > f2(y′) + f1(x). Then (f1, f2) is minimal on (X, d).

Proof. Suppose that (h1, h2) is a function pair on (X, d) such that

(0, 0) � (h1, h2) < (f1, f2),

say that there are x0 ∈ X and δ > 0 such that h2(x0) + δ < f2(x0). Then by our assumption on f there isy ∈ X such that d(x0, y) > f2(x0) − δ + f1(y) > h2(x0) + h1(y). Hence (h1, h2) is not ample. The secondcase is dealt with analogously. Therefore (f1, f2) is minimal ample on (X, d). �

Let us next discuss the positivity hypothesis in Lemma 2.

Example 6. Consider the subspace Y = {(0, 0), (a, 0), (0, b)} of (QXa,b, d) with a, b > 0 (compare Example 4).

Let us first show that for any (α, β) ∈ [0, a]× [0, b] the restriction of f(α,β) to Y (which we shall also denoteby f(α,β)) belongs to QY .

We shall check that for each z ∈ Y the two equations

(f(α,β))2(z) = supy∈Y

(d(z, y) −(f(α,β))1(y)

)and

(f(α,β))1(z) = supy∈Y

(d(y, z) −(f(α,β))2(y)

)are satisfied.

This follows however from the fact that QXa,bis a T0-quasi-metric tight extension of Xa,b (compare [2]),

that is, according to [2, Proposition 5(c)] we have that for any y1, y2 ∈ QXa,b,

d(y1, y2) = sup{(fy2)2(x) −(fy1)2(x): x ∈ Xa,b

}and

d(y1, y2) = sup{(fy1)1(x) −(fy2)1(x): x ∈ Xa,b

}.

Indeed for instance the first equation immediately yields

(f(α,β))1(z) = d((α, β), z

)= sup

{(fz)2(x) −(f(α,β))2(x): x ∈ Xa,b

}whenever z ∈ QXa,b

. Similarly the second equation yields

(f(α,β))2(z) = d(z, (α, β)

)= sup

{(fz)1(x) −(f(α,β))1(x): x ∈ Xa,b

}whenever z ∈ QXa,b

. Hence the stated two equations obviously hold, since Xa,b ⊆ Y (for a less explicitapproach compare also [2, Proposition 4]).

Observe now that we cannot find y ∈ Y such that

d((0, 0), y

)= (f(a,b))2

((0, 0)

)+ (f(a,b))1(y),



since

supy∈Y

(d((0, 0), y

)− (f(a,b))1(y)

)= −(a ∨ b) ∨ (−a) ∨ (−b) < 0 = (f(a,b))2

((0, 0)

).

On the other hand however given z = (0, 0) and f(a/2,0) ∈ QY , although (f(a/2,0))2(z) = 0, we can findy = (a, 0) ∈ Y such that d(z, y) − (f(a/2,0))1(y) = (f(a/2,0))2(z).

So we see that Lemma 2 may, but need not hold, under the assumption that f2(x) = 0.Finally set

(f(a,b))2∗((0, 0)

):= sup

y∈Y

(d((0, 0), y

)− (f(a,b))1(y)

).

Then (f(a,b))2∗((0, 0)) < 0, as established above.Observe that the (real-valued!) function pair h(a,b) on Y defined as follows satisfies the inequality of

ampleness, but is no longer nonnegative and so is not a function pair on Y in our sense: (h(a,b))1 = (f(a,b))1,and (h(a,b))2(z) = (f(a,b))2(z) if z ∈ Y \ {(0, 0)} and (h(a,b))2((0, 0)) = (f(a,b))2∗((0, 0)).

We next generalize a metric result due to Isbell to our asymmetric setting.

Proposition 2. (Compare [9, Remark 3.2].) A bicomplete T0-quasi-metric space (X, d) is q-hyperconvex iffor every ε > 0 there is a q-hyperconvex subspace S of X such that for every point x ∈ X we find s ∈ S

such that ds(x, s) < ε.

Proof. Let f be a minimal ample pair of functions on (X, d). By assumption for each n ∈ N we can finda subspace Sn of X such that Sn is q-hyperconvex and given any y ∈ X there is sn ∈ Sn such thatds(y, sn) < 2−n.

Consider the restriction f |Snof f to Sn. That restriction is ample on Sn, hence has a minimal function

pair on Sn below it. Thus there is pn ∈ Sn with (d(pn, ·), d(·, pn)) � f |Sn, since Sn is q-hyperconvex. We

shall show that (pn)n∈N is a Cauchy sequence in (X, ds).Let ε > 0 and let m ∈ N be such that 2−m < ε. Suppose now first that f2(pm) > 0.Then by Corollary 1 there is zm ∈ X such that d(pm, zm) + ε > f2(pm) + f1(zm), since f is minimal on

(X, d).Furthermore there is am ∈ Sm such that ds(zm, am) < ε.Hence f2(pm) < ε+(d(pm, zm)−d(pm, am))+(d(pm, am)−f1(am))+(f1(am)−f1(zm)) � ε+ds(zm, am)+

0 + ds(am, zm) � 3ε.We conclude that in any case f2(pm) < 3ε whenever m ∈ N such that 2−m < ε.Similarly one shows that f1(pm) < 3ε whenever m ∈ N such that 2−m < ε.Hence for any m,n ∈ N such that max{2−n, 2−m} < ε we have

d(pm, pn) � f2(pm) + f1(pn) < 3(ε + ε).

Therefore (pn)n∈N is a Cauchy sequence in (X, ds). By bicompleteness of (X, d) there is p ∈ X such thatds(pn, p) → 0.

We are going to show that f = fp on (X, d). Consider any ε > 0 and y ∈ X. There is m ∈ N such that2−m < ε, and also there is k ∈ N such that k � m and ds(p, pk) < ε.

Furthermore there is b ∈ Sk such that ds(y, b) < ε.Therefore d(p, y)−f1(y) � (d(p, y)−d(pk, b))+(d(pk, b)−f1(b))+(f1(b)−f1(y)) � (ds(p, pk)+ds(y, b))+

0 + ds(y, b) � ε + ε + 0 + ε = 3ε.



Since ε > 0 was arbitrary, we have d(p, y) � f1(y) whenever y ∈ X. Analogously one shows that for anyy ∈ X we have d(y, p) � f2(y). By minimality of f we conclude that f1(y) = d(p, y) and f2(y) = d(y, p)whenever y ∈ X. We have shown that (X, d) is q-hyperconvex. �6. Endpoints in q-hyperconvex hulls of T0-quasi-metric spaces

The following result shows in particular that each joincompact T0-quasi-metric space with at least twopoints has an endpoint resp. a startpoint.

Proposition 3. Let (X, d) be a joincompact T0-quasi-metric space. Furthermore let two points y, a ∈ X withd(y, a) > 0 be given. Then there is an endpoint e of (X, d) such that (y, a, e) is collinear. Moreover there isa startpoint s in (X, d) such that (s, y, a) is collinear in (X, d).

Proof. Consider My,a = {a′ ∈ X: (y, a, a′) is collinear in (X, d)}. It is nonempty, since a ∈ My,a. ThenMy,a equipped with the restriction of the partial order �y on X is a partially ordered set. Let K ⊆ My,a

be a nonempty chain.We consider the net xk = k where k ∈ K, which is directed by the linear order of the chain K. Since

(X, τ(ds)) is compact, we know that there is a subnet (xke)e∈E of (xk)k∈K converging to some point x in

(X, ds).We next show that x is an upper bound of K in My,a: Indeed for each e ∈ E we have that d(y, a) +

d(a, xke) = d(y, xke

) whenever e ∈ E, since xke∈ My,a whenever e ∈ E.

Taking limits in R where R is equipped with its usual topology, we have d(y, a) + d(a, x) = d(y, x), since|d(a, xke

) − d(a, x)| � ds(xke, x) and |d(y, xke

) − d(y, x)| � ds(xke, x) whenever e ∈ E.

Thus (y, a, x) is collinear in (X, d) and x ∈ My,a. Since (xke)e∈E is a subnet of (xk)k∈K, given k ∈ K we

see that k �y ke eventually. By definition of �y, if k �y ke then we have d(y, xk) + d(xk, xke) = d(y, xke

).Taking as above the limit, we get for each k ∈ K that d(y, xk)+d(xk, x) = d(y, x). Consequently for each

k ∈ K, (y, xk, x) is collinear in (X, d). Thus x is an upper bound of K in My,a.Hence by Zorn’s Lemma, My,a has a maximal element m. We show that the maximal element m of My,a

is an endpoint of (X, d), witnessed by y. Observe first that d(y,m) � d(y, a) > 0. Suppose now that forsome x ∈ X we have that (y,m, x) is collinear in (X, d). Since (y, a,m) is collinear in (X, d), we know thata �y m �y x. Thus (y, a, x) is collinear in (X, d) and therefore x ∈ My,a and m = x by maximality of m inMy,a. Hence m is an endpoint of (X, d). Applying the analogous argument to the space (X, d−1) we obtainthe part of the statement dealing with startpoints. �

Observe that the last part of the preceding proof yields the following result.

Corollary 2. Let (X, d) be a T0-quasi-metric space and let y, a ∈ X with d(y, a) > 0 be given. Equip theset My,a = {a′ ∈ X: (y, a, a′) is collinear in (X, d)} with the restriction of the partial order �y (seeProposition 1). Then each maximal element of My,a is an endpoint of (X, d). �Corollary 3. Let (X, d) be a joincompact T0-quasi-metric space. Moreover let two distinct points y1, y2 ∈ X

with d(y1, y2) > 0 be given. Then there are an endpoint e and a startpoint s in (X, d) such that (s, y1, y2, e)is collinear in (X, d).

Proof. By Proposition 3 there is a startpoint s ∈ X such that (s, y1, y2) is collinear. In particular d(s, y2) > 0.Similarly by Proposition 3 there is an endpoint e ∈ X such that (s, y2, e) is collinear. Then (s, y1, y2, e) iscollinear by Lemma 1. �



Proposition 4. (Compare [9, Theorem 2.12].) Let (X, d) be a joincompact T0-quasi-metric space. Each end-point of (X, d) is an endpoint of its q-hyperconvex hull (Q(X,d), D).

Proof. Suppose that x is an endpoint of (X, d), that is, there is y ∈ X such that d(y, x) > 0 and suchthat for any a ∈ X collinearity of (y, x, a) in (X, d) implies that x = a. Assume that for some w ∈ QX

we have that (y, x, w) is collinear in (QX , D). (Of course, here we consider X a subspace of QX .) Then0 < D(y, x) � D(y, w). We are going to show that x = w and then x is an endpoint in (QX , D). Giveny ∈ X, according to Lemma 2 by joincompactness of (X, d) there is u ∈ X such that (y, w, u) is collinearin (QX , D). Because of transitivity of �y we have that (y, x, u) is collinear in (QX , D). So x = u, by ourassumption. Hence x �y w �y u = x. Consequently w = x. It follows that x is an endpoint of (QX , D). �Corollary 4. Let (X, d) be a joincompact T0-quasi-metric space. Each startpoint of (X, d) is a startpoint ofits q-hyperconvex hull (Q(X,d), D).

Proof. By the previous result each endpoint of (X, d−1) is an endpoint of (Q(X,d−1), D). Thus each startpointof (X, d) is a startpoint of (Q(X,d−1), D

−1). By [10, Proposition 4] t : (Q(X,d), D) → (Q(X,d−1), D−1) is an

isometric bijection where t(f1, f2) = (f2, f1) whenever (f1, f2) ∈ Q(X,d). �Example 7. Consider the space (Xa,b, d) of Example 4 with a, b > 0. Then neither (0, 1, 0) nor (1, 0, 1) arecollinear, since (Xa,b, d) satisfies the T0-condition. By our assumption we conclude that both 0 and 1 areendpoints and startpoints of (Xa,b, d).

By Proposition 4 and Corollary 4 (0, b) and (a, 0) are both startpoints and endpoints of (QXa,b, d).

Proposition 5. ([9, p. 73] or [8, Theorem 4]) Let (X, d) be a joincompact T0-quasi-metric space. Then thereis a bijective isometric map QX → QB where B = Ad ∪ Ed.

Proof. Indeed we want to show that X is a T0-quasi-metric tight extension of B in the sense of [2] (compare[10, Remark 7]). Let (y1, y2) ∈ X2. Suppose first that d(y1, y2) = 0. By the triangle inequality for anyb1, b2 ∈ B we have that d(b1, b2) � d(b1, y1)+d(y1, y2)+d(y2, b2). Therefore d(y1, y2) = 0 = sup{(d(b1, b2)−d(b1, y1) − d(y2, b2)) ∨ 0: b1, b2 ∈ B}. Consider now the case that d(y1, y2) > 0. By Corollary 3 there are astartpoint s and an endpoint e in (X, d) such that (s, y1, y2, e) is collinear in (X, d).

Thus d(s, y2)+ d(y2, e) = d(s, e) and therefore d(s, y1)+ d(y1, y2)+ d(y2, e) = d(s, y2)+ d(y2, e) = d(s, e).Consequently d(y1, y2) = d(s, e) − d(s, y1) − d(y2, e).Hence the pair (B,X) satisfies the condition (b) of [2, Proposition 5]. Thus X is a T0-quasi-metric tight

extension of B. Hence by [2, Proposition 4] there exists a bijective isometric map from QX to QB , namelyf �→ f |B where f ∈ QX . �Proposition 6. (Compare [9, p. 73].) Let (X, d) be a joincompact T0-quasi-metric space with a nonemptyτ(ds)-closed set K not containing some point of B = Ad ∪ Ed. Then X is not a T0-quasi-metric tightextension of K.

Proof. Suppose that K is a nonempty τ(ds)-closed subset of X such that there is x ∈ B \K. Assume firstthat x is an endpoint of (X, d). Then there is y ∈ X which witnesses that x is an endpoint of (X, d). Inparticular d(y, x) > 0.

Then H = K∪{y} is τ(ds)-closed and thus joincompact. In the light of condition (c) of [2, Proposition 5]it suffices to show that fx|H is not minimal. Then X is not a T0-quasi-metric tight extension of H, socertainly not a T0-quasi-metric tight extension of the smaller subspace K of X (compare condition (b) of[2, Proposition 5]).



If fx|H ∈ QH , then by joincompactness of H and Lemma 2 there is k ∈ K such that d(y, k) = d(y, x) +d(x, k). Since y witnesses that x is an endpoint of (X, d), then x = k ∈ K—a contradiction. We concludethat fx|H /∈ QH .

The case that x is a startpoint of (X, d) is treated analogously. �References

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