Gluing Constructions and Local-To-Global Results for Hyperconvex Metric Spaces

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Doctoral Thesis

Gluing Constructions and Local-to-Global Results for Hyperconvex Metric Spaces

Author(s): Miesch, Benjamin

Publication Date: 2017

Permanent Link: https://doi.org/10.3929/ethz-a-010867059

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ETH Library Diss. ETH No. 23997

Gluing Constructions and Local-to-Global Results for Hyperconvex Metric Spaces

A thesis submitted to attain the degree of

DOCTOR OF SCIENCES of ETH ZURICH¨ (Dr. sc. ETH Z¨urich)

presented by Benjamin Raphael Miesch

MSc ETH Mathematik, ETH Z¨urich born May 14, 1987 citizen of Titterten BL

accepted on the recommendation of Prof. Dr. Urs Lang, examiner Prof. Dr. Rafael Esp´ınola,co-examiner

2017

Soli Deo Gloria.

Zusammenfassung

Hyperkonvexität beschreibt eine Schnitteigenschaft von Bällen in metrischen Räumen und wurde 1956 von N. Aronszajn and P. Panitchpakdi fur¨ das Studi- um von Erweiterungen gleichmässig stetiger Abbildungen eingefuhrt.¨ In dieser Arbeit studieren wir nun verschiedene Eigenschaften von hyperkonvexen metrischen Räumen. Zuerst untersuchen wir Verklebungen von solchen Räumen entlang isometrischer Teilmengen. Fur¨ schwach extern hyperkonvexe Klebmengen können wir eine notwendige und hinreichende Bedingung angeben, so dass der resultierende Raum wieder hyperkonvex ist. Als Konsequenz erhal- ten wir, dass das Verkleben von metrischen Räumen entlang extern hyperkonvexer oder stark konvexer Teilmengen die Hyperkonvexität erhaltet. Wir geben auch eine Klassifizierung fur¨ Verklebungen von hyperkonvexen Vektorräumen entlang eines linearen Unterraumes. In einem zweiten Teil analysieren wir, wann ein lokal hyperkonvexer metrischer Raum hyperkonvex ist. Zu diesem Zweck beweisen wir einen Cartan- Hadamard-Satz fur¨ metrische Räume, in denen wir lokal Geodäten mit einer Konvexitätseigenschaft wählen können. Es folgt, dass ein vollständiger, ein- fach-zusammenhängender, lokal kompakter und lokal hyperkonvexer metrischer Raum mit einer Längenmetrik, der lokal endliche kombinatorische Dimension hat, ein hyperkonvexer metrischer Raum ist. Des Weiteren beschäftigen wir uns mit schwächeren Formen der Hyper- konvexität. Wir zeigen, dass fur¨ n ≥ 3 jeder vollständige, fast n-hyperkonvexe metrische Raum n-hyperkonvex ist, und beweisen dann, dass jeder vollständige, 4-hyperkonvexe metrische Raum bereits n-hyperkonvex fur¨ alle n ∈ N ist. Anschliessend untersuchen wir Konvexitätseigenschaften von schwach extern hyperkonvexen Teilmengen und verbinden diese mit Lokal-global-Resulta- ten fur¨ diese Mengen. Dies fuhrt¨ unter anderem zu einem Helly-artigen Satz fur¨ n schwach extern hyperkonvexe Teilmengen von l∞. Zum Schluss wenden wir uns noch kurz den Medianräumen und Wurfel-¨ komplexen zu. Wir beweisen, dass hyperkonvex metrisierte Wurfelkomplexe¨ mit der entsprechenden Metrik CAT(0)-Räume sind. Zudem konstruieren wir eine bi-Lipschitz-äquivalente, hyperkonvexe Metrik auf geodätischen Medianräumen und zeigen, dass ein lokaler Medianraum mit einem geodätischen Bicombing ein Medianraum ist.

iii

Abstract

Hyperconvexity is an intersection property of balls in metric spaces and was introduced in 1956 by N. Aronszajn and P. Panitchpakdi to study extensions of uniformly continuous transformations. In this thesis, we study different properties of hyperconvex metric spaces. First, we investigate gluings of such spaces along isometric subsets. For weakly externally hyperconvex gluing sets, we can give necessary and sufficient conditions, so that the resulting space is hyperconvex as well. As a consequence, we get that the gluing of metric spaces along externally hyperconvex or strongly convex subsets preserves hyperconvexity. We then give a classification for gluings of hyperconvex vector spaces along linear subspaces. In the second part, we analyze under which conditions a locally hyperconvex metric space is hyperconvex. For this purpose, we prove a Cartan-Hadamard Theorem for metric spaces with a local geodesic bicombing. It follows that a complete, simply-connected, locally compact and locally hyperconvex metric space with finite combinatorial dimension, endowed with the length metric, is hyperconvex. Furthermore, we consider some relaxed notions of hyperconvexity. We show that, for n ≥ 3, every complete, almost n-hyperconvex metric space is n-hyperconvex and prove that every complete, 4-hyperconvex metric space is n-hyperconvex for every n ∈ N. Afterwards, we investigate convexity of weakly externally hyperconvex subsets and connect them with local-to-global results for these sets. This leads to n a Helly-type theorem for weakly externally hyperconvex subsets of l∞. Finally, we turn our attention to median metric spaces and cube complexes. We prove that cube complexes which are hyperconvex with respect to some metric also possess a metric such that they become CAT(0)-spaces. Moreover, we construct a bi-Lipschitz equivalent hyperconvex metric on geodesic median metric spaces and show that a locally median metric space with a geodesic bicombing is a median metric space.

Acknowledgments

First of all, I would like to thank Prof. Dr. Urs Lang for his encouragement and guidance during the past years. He was always ready to take time for my questions and could give me the advise needed. I am also indebted to Prof. Dr. Rafael Esp´ınolafor agreeing to act as co-examiner. I joyfully remember my visit to Sevilla. Moreover, I really appreciated the companionship with my colleagues from Assistant Groups 1 & 4 of the Department of Mathematics at ETH Zürich. Especially, I would like to mention my office mate Christian, who always had a helping hand. With Nicolas I could also conduct projects beyond mathematics. Maëland Giuliano were excellent coworkers and travel companions. Above all, I thank my parents for their unconditional love and support.

I gratefully acknowledge ﬁnancial support from the Swiss National Science Foundation.

vii

CONTENTS

Contents

Abstract v

I Introduction 1

II Hyperconvex Metric Spaces 7 II.1 Basic Properties ...... 7 II.2 An Intersection Property for Externally Hyperconvex Subsets 9 II.3 Weakly Externally Hyperconvex Subsets ...... 13 II.4 Retracts ...... 21

III Gluing Hyperconvex Metric Spaces 23 III.1 Gluing along Weakly Externally Hyperconvex Subsets . . . . 23 III.2 Gluing Isometric Copies ...... 29 III.3 Gluing Hyperconvex Linear Spaces ...... 31

IV Local to Global 43 IV.1 The Cartan-Hadamard Theorem for Metric Spaces with Local Geodesic Bicombings ...... 43 IV.2 Locally Hyperconvex Metric Spaces ...... 49 IV.3 Absolute 1-Lipschitz Neighborhood Retracts ...... 51

V Finite Hyperconvexity 55 V.1 Basic Properties ...... 55 V.2 Almost n-Hyperconvex Metric Spaces ...... 56 V.3 4-Hyperconvex Metric Spaces ...... 61

VI Convexity of Weakly Externally Hyperconvex Subsets 65 VI.1 Main Results ...... 65 VI.2 σ-Convexity ...... 66 VI.3 Locally Weakly Externally Hyperconvex Subsets ...... 69

VII Hyperconvex Metrics on Median Metric Spaces 75 VII.1 Median Metric Spaces ...... 75 VII.2 Cube Complexes ...... 77 VII.3 Hyperconvex Metrics on Median Metric Spaces ...... 83 VII.4 Locally Median Metric Spaces ...... 86

ix CONTENTS

A Some Results on Geodesic Bicombings 87 A.1 Reversible Geodesic Bicombings ...... 87 A.2 A Non-Consistent Convex Geodesic Bicombing ...... 89

Bibliography 97

Index 101

x Chapter I

Introduction

Hyperconvex metric spaces appear in various contexts. They were introduced by N. Aronszajn and P. Panitchpakdi and later studied by J. Lindenstrauss to receive extension results for linear operators between Banach spaces [AP56, Lin64]. Furthermore, they occur as the injective hull or tight span of a metric space [Isb64, Dre84], which has applications to geometric group theory [Lan13] and theoretical computer science [CL91, CL94]. Hyperconvex metric spaces also play an important role in metric ﬁxed point theory [EK01]. A metric space (X, d) is called hyperconvex if for every family {B(xi, ri)}i∈I of closed balls with d(xi, xj) ≤ ri + rj, we have \ B(xi, ri) 6= ∅. i∈I

In other words, (X, d) is a metrically convex metric space whose closed balls have the binary intersection property; compare [BL00, Definition 1.3]. Already N. Aronszajn and P. Panitchpakdi showed that hyperconvex metric spaces are the same as injective metric spaces and absolute 1-Lipschitz retracts [AP56]. From the construction of the injective hull by J. Isbell [Isb64] it follows that hyperconvex metric spaces possess a geodesic bicombing which is invariant under isometries [Lan13, Proposition 3.8]. This means that we can select geodesics in such a way that the distance between them fulfills a weak convexity property. Therefore, we can look at hyperconvex metric spaces as some kind of weakly negatively curved spaces. It turns out that many results for CAT(0) and Buse- mann spaces transfer to metric spaces with geodesic bicombings [DL16, Des16] and hence also to hyperconvex metric spaces. In this thesis, we continue in this spirit by providing gluing results for hyperconvex metric spaces and proving a local-to-global theorem for hyperconvex metric spaces which is based on a Cartan-Hadamard Theorem for metric spaces with geodesic bicombings. Furthermore, we investigate finite hyperconvexity and study σ-convexity of weakly externally hyperconvex subsets. Then a section on median metric spaces follows, where we construct a bi-Lipschitz equivalent hyperconvex metric. In the appendix we finally include some further results on geodesic bicombings.

1 CHAPTER I. INTRODUCTION

Gluing Hyperconvex Metric Spaces. It is a classical question if we can glue two metric spaces along a common subset such that the properties of the spaces are preserved. It is well known that gluing along a single point preserves hyperconvexity [JLPS02, Lemma 2.1]. Recently, B. Piatek proved that if we ‘ glue two hyperconvex metric spaces along a unique interval, then the resulting metric space is hyperconvex as well [Pia14]. This result can be generalized by studying externally hyperconvex and weakly externally hyperconvex subsets. A subset A of a metric space (X, d) is externally hyperconvex in X if for every family of closed balls {B(xi, ri)}i∈I in X with d(xi, xj) ≤ ri + rj and d(xi,A) ≤ ri, we have \ B(xi, ri) ∩ A 6= ∅. i∈I Furthermore, a subset A of a metric space (X, d) is weakly externally hyperconvex in X if for every x ∈ X the set A is externally hyperconvex in A ∪ {x}. As a ﬁrst result we establish the binary intersection property for bounded externally hyperconvex subsets. This will turn out to have great impact.

Proposition II.2.1. Let (X, d) be a hyperconvex metric space and {Ai}i∈I a family of pairwise intersecting, externally hyperconvex subsets such that at least T one of them is bounded. Then we have i∈I Ai 6= ∅. Eventually, we get the following characterization of gluings along weakly externally hyperconvex subsets. Theorem III.1.8. Let (X, d) be the metric space obtained by gluing a family of hyperconvex metric spaces (Xλ, dλ)λ∈Λ along some set A such that for each λ ∈ Λ, A is weakly externally hyperconvex in Xλ. Then X is hyperconvex if and only if for all λ ∈ Λ and all x ∈ X \ Xλ, the set B(x, d(x, A)) ∩ A is externally hyperconvex in Xλ. Moreover, if X is hyperconvex, the subspaces Xλ are weakly externally hyperconvex in X. As a consequence, we obtain that gluing along strongly convex and gluing along externally hyperconvex subsets preserves hyperconvexity. A subset A of a metric space X is called strongly convex if for all x, y ∈ A, we have I(x, y) := {z ∈ X : d(x, z) + d(z, y) = d(x, y)} ⊂ A. We call I(x, y) the metric interval between x and y. Corollary III.1.9. Let (X, d) be the metric space obtained by gluing a collection (Xλ, dλ)λ∈Λ of hyperconvex metric spaces along some space A such that A is closed and strongly convex in Xλ for each λ ∈ Λ. Then X is hyperconvex as well. Corollary III.1.10. Let (X, d) be the metric space obtained by gluing a family of hyperconvex metric spaces (Xλ, dλ)λ∈Λ along some set A such that A is externally hyperconvex in each Xλ. Then X is hyperconvex. Moreover, A is externally hyperconvex in X.

2 Furthermore, we apply this theorem to give a full characterization for gluings of ﬁnite dimensional hyperconvex vector spaces along linear subspaces. In the n n following, l∞ denotes the vector space R endowed with the maximum norm. n m Theorem III.3.13. Let X1 = l∞ and X2 = l∞. Moreover, let V be a linear subspace of both X1 and X2 such that V 6= X1,X2. Then X = X1tV X2 is k 0 k 0 hyperconvex if and only if there is some k such that X1 = l∞×X1, X2 = l∞×X2, k 0 0 0 n−k 0 m−k V = l∞ × V , and V is strongly convex in both X1 = l∞ and X2 = l∞ . With the exception of Section II.4, the results so far have been published in two papers, one of them joint with M. Pav´on[Mie15, MP17]. Section II.4 is part of the preprint [MP16].

Local to Global. The classical Cartan-Hadamard Theorem was generalized by W. Ballmann [Bal90] for metric spaces with non-positive curvature, and by S. Alexander and R. Bishop [AB90] for locally convex metric spaces. We now prove the Cartan-Hadamard Theorem in a more general setting, namely for spaces which are not uniquely geodesic but locally possess a suitable selection of geodesics. In a metric space (X, d), a geodesic bicombing is a selection of a geodesic between each pair of points. This is a map σ : X × X × [0, 1] → X such that for all x, y ∈ X, the path σxy := σ(x, y, ·) is a geodesic from x to y. Moreover, we say that the geodesic bicombing σ is consistent if

σσxy(s1)σxy(s2)(t) = σxy((1 − t)s1 + ts2) for all 0 ≤ s1 ≤ s2 ≤ 1 and t ∈ [0, 1]. A geodesic bicombing σ is called convex if the function t 7→ d(σxy(t), σx¯y¯(t)) is convex for all x, y, x,¯ y¯ ∈ X. Furthermore, we say that σ is reversible if σyx(t) = σxy(1 − t) for all x, y ∈ X and t ∈ [0, 1]. Theorem IV.1.2. Let X be a complete, simply-connected metric space with a convex local geodesic bicombing σ. Then the induced length metric on X admits a unique consistent, convex geodesic bicombing σ˜ which is consistent with σ. As a consequence, X is contractible. Moreover, if the local geodesic bicombing σ is reversible, then σ˜ is reversible as well.

This especially applies to certain locally hyperconvex metric spaces and results in the following local-to-global theorem for hyperconvex metric spaces and the analogue for absolute 1-Lipschitz retracts.

Theorem IV.2.1. Let X be a complete, locally compact, simply-connected, locally hyperconvex length space with locally ﬁnite combinatorial dimension. Then X is a hyperconvex metric space.

Theorem IV.3.1. Let X be a locally compact absolute 1-Lipschitz uniform neighborhood retract with locally ﬁnite combinatorial dimension. Then X is an absolute 1-Lipschitz retract.

Chapter IV is contained in a paper in preparation [Mie16].

3 CHAPTER I. INTRODUCTION

Finite Hyperconvexity. For the study of extensions of uniformly continuous functions and compact linear operators a weaker form of hyperconvexity is considered. In [Lin64], J. Lindenstrauss characterizes all Banach spaces B with the property that any compact linear operator with target B possesses an ”almost” norm preserving extension in pure metric terms, namely as the Banach spaces which are n-hyperconvex for every n ∈ N. A counterpart for uniformly continuous maps between metric spaces was later proven by R. Esp´ınolaand G. López, see [EL02]. This motivates a closer look on results concerning n- hyperconvexity in general metric spaces. Note that the following definitions are slightly different from the ones given in [AP56]. Let A be a subset of a metric space (X, d). The subset A is . . .

n • n-hyperconvex if for every family {B(xi, ri)}i=1 of n closed balls with Tn centers xi ∈ A and d(xi, xj) ≤ ri + rj, we have i=1 B(xi, ri) ∩ A 6= ∅;

n • almost n-hyperconvex if for every family {B(xi, ri)}i=1 of n closed balls Tn with xi ∈ A and d(xi, xj) ≤ ri + rj, we have i=1 B(xi, ri + ) ∩ A 6= ∅, for every > 0;

n • externally n-hyperconvex in X if for every family {B(xi, ri)}i=1 of n closed balls with xi ∈ X, d(xi,A) ≤ ri and d(xi, xj) ≤ ri + rj, we have that the Tn intersection i=1 B(xi, ri) ∩ A 6= ∅ is non-empty;

• weakly externally n-hyperconvex in X if for every x ∈ X, the set A is externally n-hyperconvex in A ∪ {x}.

The following two theorems supplement results proven for Banach spaces by J. Lindenstrauss, see [Lin64, Lemma 4.2] and [BL00, Lemma 2.13], and hence completely answer Problem 1 and Problem 4 raised by N. Aronszajn and P. Panitchpakdi in [AP56].

Theorem V.2.1. Let X be a complete, almost n-hyperconvex metric space for n ≥ 3. Then X is n-hyperconvex.

This implies for instance that the metric completion of an n-hyperconvex metric space is n-hyperconvex as well. Note that there are complete metric spaces which are almost 2-hyperconvex but not 2-hyperconvex, see [AP56].

Theorem V.3.1. Let X be a complete metric space and let A ⊂ X be an arbitrarily chosen non-empty subset. Then, the following hold:

(i) X is 4-hyperconvex if and only if X is n-hyperconvex for every n ∈ N.

(ii) A is externally 4-hyperconvex in X if and only if A is externally n-hyperconvex in X for every n ∈ N.

(iii) A is weakly externally 4-hyperconvex in X if and only if A is weakly externally n-hyperconvex in X for every n ∈ N.

4 Observe that this is the best we can hope for, since there are metric spaces 3 which are 3-hyperconvex but not 4-hyperconvex, e.g. l1, and there is a subset A of l∞(N) which is externally n-hyperconvex for every n ∈ N, but fails to be hyperconvex, see Example V.3.5. σ-Convexity. Given a geodesic bicombing σ on a metric space, we can investigate σ-convex subsets. A subset A of a metric space X, endowed with a geodesic bicombing σ, is σ-convex if for all x, y ∈ A, it holds that σxy([0, 1]) ⊂ A. For suﬃciently strong assumptions on the geodesic bicombing, we show that externally and weakly externally hyperconvex subsets are σ-convex. Moreover, we prove that σ-convex subsets which are uniformly locally (weakly) externally hyperconvex are (weakly) externally hyperconvex.

Theorem VI.1.1. Let X be a hyperconvex metric space, let A ⊂ X be any subset and let σ denote a convex geodesic bicombing on X.

(I) The following are equivalent:

(i) A is externally hyperconvex in X. (ii) A is σ-convex and uniformly locally externally hyperconvex.

(II) If straight curves in X are unique, the following are equivalent:

(i) A is weakly externally hyperconvex and possesses a consistent, convex geodesic bicombing. (ii) A is σ-convex and uniformly locally weakly externally hyperconvex.

Chapter V and Chapter VI are a publication in preparation joint with M. Pav´on[MP16].

Hyperconvex Metrics on Median Metric Spaces. Median metric spaces are another class of metric spaces which where studied intensively. They are strongly related with CAT(0) cube complexes and B. Bowditch showed that complete connected median metric spaces of ﬁnite rank admit a bi-Lipschitz equivalent CAT(0) metric [Bow16a]. Adopting his methods, we also construct a hyperconvex metric on certain median metric spaces.

Theorem VII.3.1. Let (M, ρ) be a proper, geodesic median metric space of ﬁnite rank, then M possesses a bi-Lipschitz equivalent metric d such that (M, d) is hyperconvex.

Independently, B. Bowditch recently gave another construction for a hyperconvex metric on median metric spaces, see [Bow16b]. Finally, we use the methods from Chapter IV to prove the following local- to-global result for median metric spaces.

Theorem VII.4.1. Let (X, d) be a metric space with a reversible, conical geodesic bicombing σ, such that for all x ∈ X and some r > 0, the neighborhood B(x, r) is a median metric space, then X is a median metric space.

5 CHAPTER I. INTRODUCTION

Geodesic Bicombings. A geodesic bicombing σ : X × X × [0, 1] → X is called reversible if we have

σyx(t) = σxy(1 − t) for all x, y ∈ X and t ∈ [0, 1]. We prove that a result of D. Descombes [Des16, Proposition 1.2] on the existence of reversible geodesic bicombings holds in general metric spaces.

Proposition A.1.1. Let (X, d) be a complete metric space with a conical geodesic bicombing σ. Then X also admits a reversible, conical geodesic bicombing.

Moreover, we give a ﬁrst example of a convex geodesic bicombing which is not consistent. These last results are part of a publication in preparation, co-authored with G. Basso [BM16].

6 Chapter II

Hyperconvex Metric Spaces

II.1 Basic Properties

First, we ﬁx some notation and then prove some basic facts. Let (X, d) be a metric space. We denote by

B(x0, r) := {x ∈ X : d(x, x0) ≤ r} the closed ball of radius r with center in x0. For any subset A ⊂ X, let

B(A, r) := {x ∈ X : d(x, A) := inf d(x, y) ≤ r} y∈A be the closed r-neighborhood of A. A metric space (X, d) is injective if for every isometric embedding ι: A,→ Y of metric spaces and every 1-Lipschitz map f : A → X, there is some 1-Lipschitz map f¯: Y → X such that f = f¯ ◦ ι. Similarly, a metric space (X, d) is an absolute 1-Lipschitz retract if for every isometric embedding ι: X,→ Y of X into a metric spaces Y , there is a 1-Lipschitz retraction r : Y → X. It is a classical result due to N. Aronszajn and P. Panitchpakdi that these deﬁnitions coincide with the one of hyperconvex metric spaces [AP56].

Proposition 1.1. Let (X, d) be a metric space. The following statements are equivalent:

(i) X is hyperconvex.

(ii) X is injective.

(iii) X is an absolute 1-Lipschitz retract.

Proof. First we show (i) ⇒ (ii). Let A ⊂ Y and let f : A → X be a 1-Lipschitz map. Consider the following set

F := {(Z, g): A ⊂ Z ⊂ Y, f : Z → X 1-Lipschitz and g|A = f}

0 0 0 0 with partial order (Z, g) (Z , g ) if and only if Z ⊂ Z and g |Z = g.

7 CHAPTER II. HYPERCONVEX METRIC SPACES

By Zorn’s Lemma, there is a maximal element (Z,¯ g¯) ∈ F. Assume that there is some y ∈ Y \ Z¯. For z ∈ Z, we deﬁne rz := d(z, y) and consider the collection {B(¯g(z), rz)}z∈Z of closed balls. We have

0 0 d(g(z), g(z )) ≤ d(z, z ) ≤ rz + rz0 and therefore there is some \ x ∈ B(¯g(z), rz). z∈Z ¯ ¯ ¯ ¯ But then (Z ∪ {y}, f) ∈ F with f : Z ∪ {y} → X, f Z =g ¯ and f(y) = x is strictly bigger than (Z,¯ g¯), a contradiction. Hence, we have Z¯ = Y and g¯: Y → X is a 1-Lipschitz extension of f. The implication (ii) ⇒ (iii) is immediate. Note that any 1-Lipschitz extension f¯: Y → X of the identity map id: X → X is a 1-Lipschitz retraction. Finally, to prove (iii) ⇒ (i), recall that l∞(X) is hyperconvex and X embeds via the Kuratowski embedding k : X,→ l∞(X), x 7→ dx−dx0 for some ﬁxed x0 ∈ X, where dx denotes the map z 7→ d(x, z). Hence, X is a 1-Lipschitz retract of the hyperconvex space l∞(X) and therefore hyperconvex itself. Indeed, ﬁx a retraction r : l∞(X) → X and consider a family {B(xi, ri)}i∈I of closed balls in X with d(xi, xj) ≤ ri + rj. Then there is some \ z ∈ B(k(xi), ri) ⊂ l∞(X) i∈I T and thus r(z) ∈ i∈I B(xi, ri) ∩ X 6= ∅.

We call a non-empty subset of a metric space admissible if it can be written T as an intersection of closed balls A = i∈I B(xi, ri). Furthermore, we denote by A(X), E(X), W(X), and H(X) the collection of all admissible, externally hyperconvex, weakly externally hyperconvex, and hyperconvex subsets of X. We always have E(X) ⊂ W(X) ⊂ H(X). Moreover, it holds that A(X) ⊂ E(X) if and only if X is hyperconvex.

Lemma 1.2. If A ∈ A(X) and E ∈ E(X) such that A ∩ E 6= ∅, then we have A ∩ E ∈ E(X). Especially, if X is hyperconvex, we have A(X) ⊂ E(X).

Proof. Since A is admissible, there is a collection of balls {B(xi, ri)}i∈I such T 0 0 that A = i∈I B(xi, ri). Now, given a family of closed balls {B(xj, rj)}j∈J 0 0 0 0 0 0 0 0 with d(xj, xk) ≤ rj + rk and d(xj,A ∩ E) ≤ rj, we have d(xi, xj) ≤ ri + rj and d(xi,E) ≤ ri, and therefore

\ 0 0 \ \ 0 0 A ∩ E ∩ B(xj, rj) = E ∩ B(xi, ri) ∩ B(xj, rj) 6= ∅ j∈J i∈I j∈J since E is externally hyperconvex. If X is hyperconvex, then X ∈ E(X) and therefore A(X) ⊂ E(X).

8 II.1. INTERSECTION PROPERTY

Recall the following well known facts about admissible and externally hyperconvex subsets; cf. [Sin89, KKM00]. T Lemma 1.3. Let X be a hyperconvex metric space, A = i∈I B(xi, ri) ∈ A(X) and s ≥ 0. Then one has \ B(A, s) = B(xi, ri + s) ∈ A(X). i∈I T Proof. Clearly, B(A, s) is contained in i∈I B(xi, ri + s). Conversely, for any T z ∈ i∈I B(xi, ri + s) we have d(xi, z) ≤ ri + s. Hence, since X is hyperconvex, there is some \ y ∈ B(z, s) ∩ B(xi, ri) = B(z, s) ∩ A i∈I and thus z ∈ B(y, s) ⊂ B(A, s).

Lemma 1.4. Let X be a hyperconvex metric space and A ∈ E(X). Then also B(A, r) ∈ E(X).

Proof. Let {B(xi, ri)}i∈I be a collection of closed balls with d(xi, xj) ≤ ri + rj and d(xi,B(A, r)) ≤ ri. Then we also have d(xi,A) ≤ ri + r. Since A is T externally hyperconvex, there is some y ∈ i∈I B(xi, ri + r) ∩ A. Especially, we have d(xi, y) ≤ ri + r and therefore, since X is hyperconvex, we get \ \ ∅= 6 B(xi, ri) ∩ B(y, r) ⊂ B(xi, ri) ∩ B(A, r) i∈I i∈I as desired.

II.2 An Intersection Property for Externally Hyperconvex Subsets

In this section, we establish the following important intersection property of externally hyperconvex subsets.

Proposition 2.1. Let (X, d) be a hyperconvex metric space and {Ai}i∈I a family of pairwise intersecting, externally hyperconvex subsets such that at least one T of them is bounded. Then we have i∈I Ai 6= ∅. Observe that externally hyperconvex subsets are closed. The following tech- nical lemma turns out to be the initial step in proving Proposition 2.1.

Lemma 2.2. Let X be a hyperconvex metric space. Let A, A0 ∈ E(X) with y ∈ A ∩ A0 6= ∅ and x ∈ X with d(x, A), d(x, A0) ≤ r. Denote d := d(x, y) and s := d − r. Then we have

A ∩ A0 ∩ B(x, r) ∩ B(y, s) 6= ∅, given s ≥ 0. In any case, the intersection A ∩ A0 ∩ B(x, r) is non-empty.

9 CHAPTER II. HYPERCONVEX METRIC SPACES

Proof. For s ≤ 0, we have y ∈ A ∩ A0 ∩ B(x, r). Therefore, let us assume s > 0. Claim. For each 0 < l ≤ s, there are a ∈ A, a0 ∈ A0 such that d(a, a0) ≤ l and a, a0 ∈ B(x, r) ∩ B(y, s). We start choosing

a1 ∈ B(y, l) ∩ B(x, d − l) ∩ A and 0 0 a1 ∈ B(y, l) ∩ B(x, d − l) ∩ B(a1, l) ∩ A . Then, we inductively take

0 an ∈ B(y, nl) ∩ B(x, d − nl) ∩ B(an−1, l) ∩ A and 0 0 an ∈ B(y, nl) ∩ B(x, d − nl) ∩ B(an, l) ∩ A s as long as n ≤ b l c =: n0. Finally, there are

0 a ∈ B(y, s) ∩ B(x, r) ∩ B(an0 , l) ∩ A and a0 ∈ B(y, s) ∩ B(x, r) ∩ B(a, l) ∩ A0 as desired.

We now construct recursively two converging sequences (an)n∈N ⊂ A and 0 0 0 (an)n∈N ⊂ A such that an, an ∈ B(x, r) ∩ B(y, s) with 1 1 d(a , a0 ) ≤ and d(a , a ), d(a0 , a0 ) ≤ . n n 2n+1 n−1 n n−1 n 2n

0 0 1 First, choose a0, a0 ∈ B(x, r) ∩ B(y, s) with d(a0, a0) ≤ 2 according to the 0 0 1 claim. Given an−1, an−1 with d(an−1, an−1) ≤ 2n , by hyperconvexity, there is 1 0 1 1 some xn ∈ B(an−1, 2n+1 ) ∩ B(an−1, 2n+1 ) ∩ B(x, r − 2n+1 ). Now applying the claim to xn and y, we ﬁnd

0 1 an, an ∈ B(y, s) ∩ B(xn, 2n+1 ) ⊂ B(y, s) ∩ B(x, r)

0 1 with d(an, an) ≤ 2n+1 . Moreover, we have 1 1 1 d(a , a ) ≤ d(a , x ) + d(x , a ) ≤ + = . n−1 n n−1 n n n 2n+1 2n+1 2n For m ≥ n, we get

m ∞ X X 1 1 d(a , a ) ≤ d(a , a ) ≤ = , n m k−1 k 2k 2n k=n+1 k=n+1

0 0 and similarly for an. Hence, the two sequences converge and, since d(an, an) → 0, they have a common limit point a ∈ B(y, s) ∩ B(x, r) ∩ A ∩ A0.

10 II.2. INTERSECTION PROPERTY

Lemma 2.3. Let X be a hyperconvex metric space and let A0,A1,A2 ∈ E(X) be pairwise intersecting, externally hyperconvex subsets. Then we get

A0 ∩ A1 ∩ A2 6= ∅.

Proof. Choose some point x0 ∈ A1 ∩ A2 and let r := d(x0,A0). By Lemma 2.2, 0 there is y0 ∈ A0 ∩ A1 ∩ B(x0, r). Deﬁne A0 := A0 ∩ B(y0, r) ∈ E(X). Using 0 again the lemma, we have A0 ∩A2 = A0 ∩A2 ∩B(y0, r) 6= ∅ and therefore, there is some

0 z0 ∈ A0 ∩ A2 ∩ B(x0, r) = A0 ∩ A2 ∩ B(x0, r) ∩ B(y0, r).

Then, since A0 is externally hyperconvex, there is some

r r x¯0 ∈ B(x0, r) ∩ B(y0, 2 ) ∩ B(z0, 2 ) ∩ A0 and using again Lemma 2.2, we ﬁnd

r r x1 ∈ A1 ∩ A2 ∩ B(¯x0, 2 ) ∩ B(x0, 2 ). r Proceeding this way, we get some sequence (xn)n ⊂ A1 ∩A2 with d(xn,A0) ≤ 2n r Pm r r and d(xn−1, xn) ≤ 2n . Hence, d(xn, xm) ≤ k=n+1 2k ≤ 2n and therefore, (xn)n converges to x ∈ A0 ∩ A1 ∩ A2 6= ∅.

Lemma 2.4. Let X be a hyperconvex metric space. If A0,A1 ∈ E(X) and A0 ∩ A1 6= ∅, then it holds A0 ∩ A1 ∈ E(X).

Proof. Let {B(xi, ri)}i∈I be a collection of closed balls with d(xi, xj) ≤ ri + rj T and d(xi,A0 ∩ A1) ≤ ri. Deﬁne A := i∈I B(xi, ri). Since, for k = 0, 1, the set Ak is externally hyperconvex, we have \ A ∩ Ak = B(xi, ri) ∩ Ak 6= ∅, i∈I and since admissible sets are externally hyperconvex, we have \ (A0 ∩ A1) ∩ B(xi, ri) = A0 ∩ A1 ∩ A 6= ∅ i∈I by Lemma 2.3. By induction, we therefore get the following proposition.

Proposition 2.5. Let X be a hyperconvex metric space and A0,...,An ∈ E(X) Tn with Ai ∩ Aj 6= ∅. Then we have ∅= 6 k=0 Ak ∈ E(X). As a consequence of Baillon’s theorem on the intersection of hyperconvex spaces [Bai88], the following theorem was proven by Esp´ınolaand Khamsi in [EK01].

Theorem 2.6. [EK01, Theorem 5.4]. Let {Ai}i∈I be a descending chain of non-empty externally hyperconvex subsets of a bounded hyperconvex metric space T X. Then i∈I Ai is non-empty and externally hyperconvex in X.

11 CHAPTER II. HYPERCONVEX METRIC SPACES

Similarly to Corollary 8 in [Bai88], we can deduce the following corollary which implies Proposition 2.1.

Corollary 2.7. Let {Ai}i∈I be a family of pairwise intersecting externally hy- T perconvex subsets of a bounded hyperconvex metric space X. Then i∈I Ai is non-empty and externally hyperconvex in X.

Proof. Consider the set

( ) \ F := J ⊂ I : ∀F ⊂ I ﬁnite, Ai 6= ∅ is externally hyperconvex . i∈J∪F

By Proposition 2.5, clearly ∅ ∈ F. Considering a chain (Jk)k∈N ∈ F and some finite set F ⊂ I, the sets A := T A build a descending chain Jk i∈Jk∪F i of non-empty externally hyperconvex subsets. Define J := S J . We have k∈N k that A := T A = T A is non-empty and externally hyperconvex i∈J∪F i k∈N Jk by Theorem 2.6. Therefore, J ∈ F is an upper bound for (Jk)k∈N. Hence, F satisfies the hypothesis of Zorn’s Lemma and therefore there is some maximal element J0 ∈ F. But for i ∈ I, we have J0 ∪ {i} ∈ F and by maximality of J0, we conclude that I = J0 ∈ F.

A ﬁrst application of Proposition 2.1 yields the following.

Proposition 2.8. Let Y be an externally hyperconvex subset of the metric space X. Moreover, let A be externally hyperconvex in Y . Then A is also externally hyperconvex in X.

Proof. Let {B(xi, ri)}i∈I be a collection of closed balls with d(xi, xj) ≤ ri + rj and d(xi,A) ≤ ri. Then the sets Ai := B(xi, ri) ∩ Y are externally hyperconvex subsets of X and therefore also of Y . We have

\ 1 Ai ∩ A = B(xi, ri + n ) ∩ A 6= ∅ n∈N and, since Y is externally hyperconvex, we also get

Ai ∩ Aj = B(xi, ri) ∩ B(xj, rj) ∩ Y 6= ∅.

Therefore, this is a collection of pairwise intersecting externally hyperconvex subsets of Y and hence, by Proposition 2.1, it follows that

\ \ A ∩ B(xi, ri) = A ∩ Ai 6= ∅. i∈I i∈I

12 II.3. WEAKLY EXTERNALLY HYPERCONVEX SUBSETS

II.3 Weakly Externally Hyperconvex Subsets

Let us now have a look at weakly externally hyperconvex subsets . We can characterize them by the following properties.

Lemma 3.1. Let A be a subset of the hyperconvex metric space X. Then A is weakly externally hyperconvex in X if and only if for every x ∈ X and s := d(x, A), the following hold:

(i) The intersection B(x, s) ∩ A is externally hyperconvex in A, and

(ii) for every y ∈ A, there is some a ∈ B(x, s) ∩ A such that

d(x, y) = d(x, a) + d(a, y).

Proof. If A is weakly externally hyperconvex, (i) clearly holds. Moreover, for y ∈ A, it also follows by weak external hyperconvexity of A that there is some

a ∈ A ∩ B(x, s) ∩ B(y, d(x, y) − s) 6= ∅ and therefore

d(x, y) ≤ d(x, a) + d(a, y) ≤ s + d(x, y) − s = d(x, y).

For the converse, ﬁrst observe that A must be hyperconvex. Indeed, for T 0 x ∈ i∈I B(xi, ri) with xi ∈ A, let A := B(x, d(x, A)) ∩ A ∈ E(A). Then, we 0 have d(xi,A ) ≤ ri by (ii) and therefore, there is also

0 0 \ \ x ∈ A ∩ B(xi, ri) ⊂ A ∩ B(xi, ri) i∈I i∈I by property (i). Now ﬁx x ∈ X, r ≥ s = d(x, A) and let {B(xi, ri)}i∈I be a family of closed balls with d(xi, xj) ≤ ri + rj and d(x, xi) ≤ r + ri. Then, by (ii), we have

B(x, r) ∩ A = BA(B(x, s) ∩ A, r − s) and therefore B(x, r) ∩ A is externally hyperconvex in A by (i) and Lemma 1.4. Moreover, by (ii), for xi ∈ A with d(x, xi) ≤ r + ri, we have

d(xi,B(x, r) ∩ A) ≤ ri and therefore \ B(x, r) ∩ B(xi, ri) ∩ A 6= ∅ i∈I since B(x, r) ∩ A is externally hyperconvex in A.

Lemma 3.2. Let X be a metric space, A ∈ W(X) and s ≥ 0. Then there is an s-constant retraction ρ: B(A, s) → A, i.e. d(ρ(x), ρ(y)) ≤ d(x, y) and d(x, ρ(x)) ≤ s for all x, y ∈ B(A, s).

13 CHAPTER II. HYPERCONVEX METRIC SPACES

Proof. Consider the partially ordered set F := {(B, ρ): B ⊂ B(A, s) and ρ: B → A is an s-constant retraction}.

By Zorn’s Lemma, there is some maximal element (B,˜ ρ˜) ∈ F. Assume that there is some x ∈ B(A, s) \ B˜. For all y ∈ B˜, define ry = d(x, y). Then, we have d(x, ρ(y)) ≤ d(x, y) + d(y, ρ(y)) ≤ ry + s and therefore, since A is weakly externally hyperconvex, there is some \ z ∈ B(x, s) ∩ B(ρ(y), ry) ∩ A. y∈B˜ But then, definingρ ˜(x) := z, we can extendρ ˜ to B˜ ∪ {x} contradicting maximality of (B,˜ ρ˜). Hence, we conclude B˜ = B(A, s). This and the first part of the following result can also be found in [EKL00, Esp05], where a complete characterization of weakly externally hyperconvex subsets in terms of retractions is given. Lemma 3.3. Let A be a weakly externally hyperconvex subset of a hyperconvex metric space X. Then, for any s ≥ 0, the closed neighborhood B(A, s) is weakly externally hyperconvex in X. Moreover, if for all x ∈ X we have B(x, d(x, A)) ∩ A ∈ E(X), then this also holds for B(A, s), i.e. B(x, d(x, B(A, s))) ∩ B(A, s) ∈ E(X).

Proof. Let x ∈ X and {xi}i∈I ⊂ B(A, s) with d(xi, xj) ≤ ri+rj, d(x, xi) ≤ r+ri and d(x, B(A, s)) ≤ r. By Lemma 3.2, there is a retraction ρ: B(A, s) → A such that d(y, ρ(y)) ≤ s. Then, we have d(ρ(xi), x) ≤ d(ρ(xi), xi)+d(xi, x) ≤ s+r+ri and therefore, there is some \ y ∈ B(x, s + r) ∩ B(ρ(xi), ri) ∩ A. i∈I

As d(x, y) ≤ r + s and d(xi, y) ≤ d(xi, ρ(xi)) + d(ρ(xi), y) ≤ s + ri, by hyperconvexity of X, there is some \ \ z ∈ B(x, r) ∩ B(xi, ri) ∩ B(y, s) ⊂ B(x, r) ∩ B(xi, ri) ∩ B(A, s) i∈I i∈I as required. Now, deﬁne r := d(x, B(A, s)). We claim that B(x, r) ∩ B(A, s) = B(x, r) ∩ B(B(x, d(x, A)) ∩ A, s) and therefore, if B(x, d(x, A)) ∩ A ∈ E(X), we also have B(x, r) ∩ B(A, s) ∈ E(X). Indeed, if y ∈ B(x, r)∩B(A, s), there is some a ∈ A such that d(y, a) = s and hence d(x, a) ≤ d(x, y)+d(y, a) ≤ r+s = d(x, A), i.e. a ∈ B(x, d(x, A))∩A.

14 II.3. WEAKLY EXTERNALLY HYPERCONVEX SUBSETS

Example 3.4. In general, it is not true that the neighborhood of a hyperconvex 3 subset is hyperconvex as well. Consider the isometric embedding ι: R → l∞ of the real line given by  (t, t, −t), t ≤ 0,  ι(t) = (t, t, t), 0 ≤ t ≤ 10, (20 − t, t, t) t ≥ 10,

3 and deﬁne A := ι(R) ∈ H(l∞). Then the two points x = (−2, 0, 2), y = (8, 10, 12) are contained in B(A, 1) since d(x, ι(−1)) = d(y, ι(11)) = 1. Now look at the intersection B(x, 5) ∩ B(y, 5) = {z = (3, 5, 7)}. But d(z, A) = d(z, ι(5)) = 2 and therefore B(x, 5) ∩ B(y, 5) ∩ B(A, 1) = ∅, i.e. B(A, 1) is not hyperconvex, even not geodesic. Lemma 3.5. Let X be a hyperconvex metric space and let A = I(x, y) be a metric interval. Then A ∈ W(X) and, for all z ∈ X, we have B(z, d(z, A))∩A ∈ E(X). Proof. Fix some z ∈ X. First, observe that

B(z, d(z, A)) ∩ A = B(x, rx) ∩ B(y, ry) ∩ B(z, rz) for rx := (y|z)x, ry := (x|z)y and rz := (x|y)z, where

1 (y|z)x := 2 (d(x, y) + d(x, z) − d(y, z)) denotes the Gromov product. Hence, B(z, d(z, A)) ∩ A ∈ E(X) since it is admissible. Now, let p ∈ A and t := d(p, z) − rz. Then we have d(x, p) = d(x, y) − d(y, p) ≤ d(x, y) − d(y, z) + d(p, z)

= rx + ry − ry − rz + rz + t = rx + t and similarly d(y, p) ≤ ry + t. Therefore, since X is hyperconvex, there is some

p¯ ∈ B(x, rx) ∩ B(y, ry) ∩ B(z, rz) ∩ B(p, t) 6= ∅, that isp ¯ ∈ B(z, d(z, A))∩A with d(p, z) = d(p, p¯)+d(¯p, z). Thus, by Lemma 3.1, we get A ∈ W(X). Although we cannot expect a similar result for weakly externally hyperconvex subsets as in Proposition 2.1, it is possible to say more about the intersection of weakly externally hyperconvex subsets. Example 3.6. The intersection property as we stated for externally hyperconvex subsets in Proposition 2.1 does not hold for weakly externally hyperconvex subsets. Indeed, consider the points z := (1, 1), w := (−1, −1) and the half- 2 2 space H := {y ∈ l∞ : y1 − y2 ≥ 2} in l∞. Clearly, B(z, 1), B(w, 1) and H are 2 all three elements of W(l∞) and they are pairwise intersecting. However, B(z, 1) ∩ B(w, 1) ∩ H = ∅.

15 CHAPTER II. HYPERCONVEX METRIC SPACES

Lemma 3.7. Let X be a metric space, Y ∈ W(X) and A ∈ E(Y ). Then we have A ∈ W(X).

Proof. Let x ∈ X, r ≥ d(x, A) and let {B(xi, ri)}i∈I be a collection of closed balls with xi ∈ A, d(xi, xj) ≤ ri + rj and d(xi, x) ≤ ri + r. Then the sets B := B(x, r + ) ∩ Y , Ai := B(xi, ri) ∩ Y and A are pairwise intersecting, externally hyperconvex subsets of Y and therefore, by Proposition 2.1, we have \ \ \ B(x, r) ∩ A ∩ B(xi, ri) = B ∩ A ∩ Ai 6= ∅. i∈I >0 i∈I

Lemma 3.8. Let X be a metric space and A ∈ E(X), Y ∈ W(X) such that A ∩ Y 6= ∅. Let {xi}i∈I ⊂ Y be a collection of points with d(xi, xj) ≤ ri + rj T and d(xi,A) ≤ ri. Then, for any s > 0, there are a ∈ A ∩ i∈I B(xi, ri) and T y ∈ Y ∩ i∈I B(xi, ri) with d(a, y) ≤ s. T Proof. Let y0 ∈ A∩Y and d := d(y0, i∈I B(xi, ri)). Without loss of generality, we may assume that s ≤ d. Then, since A ∈ E(X) and Y ∈ W(X), there are \ a1 ∈ B(xi, ri + d − s) ∩ B(y0, s) ∩ A, i∈I \ y1 ∈ B(xi, ri + d − s) ∩ B(a1, s) ∩ Y. i∈I

Proceeding this way, we can choose inductively \ an ∈ B(xi, ri + d − ns) ∩ B(yn−1, s) ∩ A, i∈I \ yn ∈ B(xi, ri + d − ns) ∩ B(an, s) ∩ Y, i∈I

d for n ≤ b s c =: n0 and ﬁnally, there are \ a ∈ B(xi, ri) ∩ B(yn0 , s) ∩ A, i∈I \ y ∈ B(xi, ri) ∩ B(a, s) ∩ Y, i∈I as desired.

Proposition 3.9. Let X be a metric space and let A ∈ E(X), Y ∈ W(X) such that A ∩ Y 6= ∅. Then we have A ∩ Y ∈ E(Y ), and therefore A ∩ Y ∈ W(X).

Proof. Let {xi}i∈I ⊂ Y with d(xi, xj) ≤ ri + rj and d(xi,A) ≤ ri. We now T construct inductively two converging sequences (an)n∈N ⊂ i∈I B(xi, ri) ∩ A T 1 and (yn)n∈N ⊂ i∈I B(xi, ri) ∩ Y with d(an, yn) ≤ 2n+1 as follows:

16 II.3. WEAKLY EXTERNALLY HYPERCONVEX SUBSETS

By Lemma 3.8, we may choose \ a0 ∈ B(xi, ri) ∩ A, i∈I \ y0 ∈ B(xi, ri) ∩ Y i∈I

1 with d(a0, y0) ≤ 2 . Assume now, that we have \ an ∈ B(xi, ri) ∩ A, i∈I \ yn ∈ B(xi, ri) ∩ Y i∈I

1 with d(an, yn) ≤ 2n+1 . Then, applying Lemma 3.8 for the collection of balls 1 {B(xi, ri)}i∈I ∪ {B(yn, 2n+1 )}, we ﬁnd

\ 1 an+1 ∈ B(xi, ri) ∩ B(yn, 2n+1 ) ∩ A, i∈I \ 1 yn+1 ∈ B(xi, ri) ∩ B(yn, 2n+1 ) ∩ Y i∈I

1 1 with d(an+1, yn+1) ≤ 2n+2 . Especially, we have d(yn+1, yn) ≤ 2n+1 and

1 1 1 d(an+1, an) ≤ d(an+1, yn) + d(yn, an) ≤ 2n+1 + 2n+1 = 2n . Hence the two sequences are Cauchy and therefore converge to some common T limit point z ∈ A ∩ Y ∩ i∈I B(xi, ri) 6= ∅. Finally, we have A ∩ Y ∈ W(X) by Lemma 3.7.

Looking at the proof carefully, we see that only d(xi,A) ≤ ri is assumed. Therefore, we may deduce the following corollary:

Corollary 3.10. Let X be a metric space and A ∈ E(X), Y ∈ W(X) such that A ∩ Y 6= ∅. Then, for all x ∈ Y , we have d(x, A) = d(x, A ∩ Y ).

Corollary 3.11. Let X be a metric space and A ∈ W(Y ) for Y ∈ W(X). Then we have A ∈ W(X).

Proof. Let x ∈ X with d(x, A) ≤ r and {xi}i∈I ⊂ A with d(xi, xj) ≤ ri + rj, d(xi, x) ≤ ri + r. Then we have

B(x, r) ∩ Y ∈ E(Y ) and B(x, r) ∩ A = (B(x, r) ∩ Y ) ∩ A ∈ E(A) by Proposition 3.9. Moreover, by applying Corollary 3.10 twice, we have

d(xi, (B(x, r) ∩ Y ) ∩ A) = d(xi,B(x, r) ∩ Y ) = d(xi,B(x, r)) ≤ ri T and hence i∈I B(xi, ri) ∩ B(x, r) ∩ A 6= ∅.

17 CHAPTER II. HYPERCONVEX METRIC SPACES

Corollary 3.12. Let X be a hyperconvex metric space and let A ∈ E(X), Y ∈ W(X). Then there are a ∈ A, y ∈ Y with d(a, y) = d(A, Y ).

1 Proof. Let s := d(A, Y ). For any n ∈ N, we have B(A, s + 2n ) ∈ E(X) and 1 An := B(A, s+ 2n )∩Y ∈ E(Y ) by Proposition 3.9. Clearly, we have An∩Am 6= ∅ and therefore, there is some y ∈ Y ∩ B(A, s) = T A by Proposition 2.1. n∈N n Now, since A is proximinal, there is some a ∈ A with d(a, y) ≤ s = d(A, Y ) as required.

The following proposition answers an open question on the intersection of weakly externally hyperconvex sets stated in [EK01] for proper metric spaces, i.e. for spaces where all closed balls are compact.

Proposition 3.13. Let X be a proper hyperconvex metric space and let Y and Y 0 be two weakly externally hyperconvex subsets with non-empty intersection. Then we have Y ∩ Y 0 ∈ W(X).

Proof. By Corollary 3.11, it is enough to show that Y ∩ Y 0 ∈ W(Y ). Therefore, 0 let {B(xi, ri)}i∈I be a collection of balls with xi ∈ Y ∩Y and d(xi, xj) ≤ ri +rj 0 and x ∈ Y with d(x, Y ∩ Y ) ≤ r and d(xi, x) ≤ ri + r. Furthermore, let 0 0 s > 0. Since d(x, Y ∩ Y ) ≤ r, there is some y0 ∈ Y ∩ Y ∩ B(x, r + s). Deﬁne T d := d(y0, i∈I B(xi, ri)). Then there is some

0 0 \ y0 ∈ B(y0, s) ∩ Y ∩ B(x, r) ∩ B(xi, ri + d). i∈I

d Now, for n ≤ b s c =: n0, we can choose inductively

0 \ yn ∈ B(yn−1, s) ∩ Y ∩ B(xi, ri + d − ns), i∈I 0 0 \ yn ∈ B(yn, s) ∩ Y ∩ B(x, r) ∩ B(xi, ri + d − ns). i∈I

Finally, there are

0 \ y ∈ B(yn0 , s) ∩ Y ∩ B(xi, ri), i∈I 0 0 \ y ∈ B(y, s) ∩ Y ∩ B(x, r) ∩ B(xi, ri) i∈I

T 0 T 0 and hence d(Y ∩ i B(xi, ri),Y ∩ i B(xi, ri) ∩ B(x, r)) ≤ d(y, y ) ≤ s, i.e. T 0 T we get d(Y ∩ i∈I B(xi, ri),Y ∩ i∈I B(xi, ri) ∩ B(x, r)) = 0, and since X is proper, both sets are compact and therefore their intersection

0 \ Y ∩ Y ∩ B(xi, ri) ∩ B(x, r) i∈I is non-empty.

18 II.3. WEAKLY EXTERNALLY HYPERCONVEX SUBSETS

n Note that not even linearly convex hyperconvex subsets of l∞ are stable under non-empty intersections, as observed in [Pav16, Example 1.4].

Proposition 3.14. Let X be a metric space and let (Yn)n∈N ⊂ W(X) be an increasing sequence such that Y := S Y is proper. Then we have Y ∈ n∈N n W(X).

Proof. Consider a family {B(xi, ri)}i∈I of closed balls with xi ∈ Y such that d(xi, xj) ≤ ri+rj. Moreover, let B(x, r) be a closed ball with x ∈ X, d(x, Y ) ≤ r and d(x, xi) ≤ r + ri. There is a decreasing sequence sn ↓ 0 such that d(x, Yn) ≤ r + sn. Now ﬁx := 1 > 0. Then, for every i ∈ I, there is some y ∈ B(x , ) ∩ S Y . For m i i n∈N n n ∈ N, let In := {i ∈ I : yi ∈ Yn} and since Yn is weakly externally hyperconvex, there is some \ zn ∈ Yn ∩ B(x, r + sn) ∩ B(yi, ri + ).

i∈In

Since Y is proper and (zn)n∈N ⊂ Y ∩ B(x, r + s0), it follows that there is a m T 1 convergent subsequence znk → z ∈ Y ∩B(x, r)∩ i∈I B(xi, ri + m ). Moreover, ml T since Y is proper, there is a subsequence z → z ∈ Y ∩B(x, r)∩ i∈I B(xi, ri). This proves that Y is weakly externally hyperconvex in X.

It turns out that a subset of a hyperconvex metric space is already (weakly) externally hyperconvex if it is (weakly) externally hyperconvex in a uniform neighborhood. We say that a subset A of a metric space X is proximinal if, for any x ∈ X, we have B(x, d(x, A)) ∩ A 6= ∅. Note that weakly externally hyperconvex subsets are proximinal.

Lemma 3.15. Let X be a hyperconvex metric space and A ∈ W(B(A, r)). Then A is proximinal in B(A, 2r).

Proof. Let x ∈ B(A, 2r) with d(x, A) = r + t. Then for every > 0, there is some x ∈ B(A, r) with d(x, x) ≤ t + . Since A ∈ W(B(A, r)), there is a proximinal non-expansive retraction ρ: B(A, r) → A by Proposition 4.1. We have d(x, ρ(x)) ≤ r + d(x, x) and hence, there is some \ y ∈ B(x, r) ∩ B(ρ(x), d(x, x)). >0

In particular, we have d(y, A) ≤ d(y, ρ(x)) ≤ t + for all > 0 and therefore d(y, A) = t ≤ r. Hence, since A is proximinal in B(A, r), there is some z ∈ A with d(z, y) = t. It follows that

d(x, z) ≤ d(x, y) + d(y, z) ≤ r + t = d(x, A) as desired.

19 CHAPTER II. HYPERCONVEX METRIC SPACES

Lemma 3.16. Let X be a hyperconvex metric space and A ⊂ X. Assume that there is some s > 0 such that A ∈ W(B(A, s)). Then we have A ∈ W(X).

Proof. We show that A ∈ W(B(A, r)) implies A ∈ W(B(A, 2r)). It then follows that A ∈ W(B(A, r)) for any r > 0 and thus A ∈ W(X). By Corollary 2.1 in [EKL00], it is enough to show that there is a proximinal non-expansive retraction R: A ∪ {x0} → A for any x0 ∈ B(x, 2r). For some given x0 ∈ B(x, 2r) \ B(A, r), let t := d(x0,A) − r ≤ r. By Lemma 3.15, A is proximinal in B(A, 2r) and hence, there is somex ¯0 ∈ B(A, r) with d(x0, x¯0) = t. Then, for all x ∈ A, deﬁne rx := d(x, x0). We have

d(x, x¯0) ≤ d(x, x0) + d(x0, x¯0) ≤ rx + r.

Hence, since A ∈ W(B(A, r)), there is some \ y ∈ B(x, rx) ∩ B(¯x0, r) ∩ A. x∈A

Especially, we get d(x0, y) ≤ d(x0, x¯0)+d(¯x0, y) ≤ t+r = d(x0,A) and therefore, we can deﬁne a proximinal non-expansive retraction R: A ∪ {x0} → A by R(x0) := y and R(x) := x for x ∈ A. Lemma 3.17. Let X be a hyperconvex metric space and A ⊂ X. Assume that there is some s > 0 such that A ∈ E(B(A, s)). Then we have A ∈ E(X).

Proof. By Lemma 3.16, we already have A ∈ W(X). Let us ﬁrst show that A ∈ E(B(A, r)) for any r ≥ 0. By assumption, this holds for r = s > 0. Therefore, it is enough to prove A ∈ E(B(A, r)) ⇒ A ∈ E(B(A, 2r)). Let {B(xi, ri)}i∈I be a family of closed balls with xi ∈ B(A, 2r), such that d(xi, xj) ≤ ri + rj and d(xi,A) ≤ ri. Deﬁne

Ai := B(xi, ri) ∩ B(A, r) ∈ E(B(A, r)).

Clearly A∩Ai = A∩B(xi, ri) 6= ∅. By Lemma 3.3, B(A, r) ∈ W(X) and hence, by Lemma 3.2, there is a retraction ρ: B(A, 2r) → B(A, r) with d(ρ(x), x) ≤ r. T Set yi := ρ(xi). Since A ∈ E(B(A, r)), there is some z ∈ A ∩ i∈I B(yi, ri) with d(xi, z) ≤ ri + r. Therefore, since X is hyperconvex, we get \ ∅= 6 B(z, r) ∩ B(xi, ri) ⊂ B(A, r). i∈I Especially, Ai ∩ Aj = B(xi, ri) ∩ B(xj, rj) ∩ B(A, r) 6= ∅ and hence, \ \ A ∩ B(xi, ri) = A ∩ Ai 6= ∅ i∈I i∈I by Proposition 2.1. To deduce that A ∈ E(X), let now {B(xi, ri)}i∈I be a family of closed balls in X with d(xi, xj) ≤ ri + rj and d(xi,A) ≤ ri. Deﬁne Ai := B(xi, ri) ∩ A ∈ E(A).

20 II.4. RETRACTS

For ﬁxed i, j ∈ I, we have xi, xj ∈ B(A, r) for some r ≥ 0 and hence, by the ﬁrst step, we have

Ai ∩ Aj = A ∩ B(xi, ri) ∩ B(xj, rj) 6= ∅.

Therefore, we get \ \ A ∩ B(xi, ri) = Ai 6= ∅ i∈I i∈I by Proposition 2.1 as before.

To conclude this section, we give some properties for products of hyperconvex metric spaces.

1 2 Lemma 3.18. Let X = X ×∞ X be the product of two metric spaces with λ λ 1 2 d(x, y) = maxλ=1,2 dλ(x , y ). Moreover, let A = A × A be a subset of X. Then, the following properties hold:

(i) B(x, r) = B1(x1, r) × B2(x2, r).

(ii) A ∈ E(X) if and only if Aλ ∈ E(Xλ) for each λ ∈ {1, 2}.

(iii) A ∈ W(X) if and only if Aλ ∈ W(Xλ) for each λ ∈ {1, 2}.

(iv) X is hyperconvex if and only if Xλ is hyperconvex for each λ ∈ {1, 2}.

Proof. First, property (i) follows from the fact that d(x, y) ≤ r if and only if λ λ dλ(x , y ) ≤ r for λ = 1, 2. For (ii), let first {xi}i∈I ⊂ X be any collection of points with the property λ λ d(xi, xj) ≤ ri+rj and d(xi,A) ≤ ri. Then we have d(xi , xj ) ≤ d(xi, xj) ≤ ri+rj λ λ λ λ and d(xi ,A ) ≤ d(xi,A) ≤ ri and therefore, if A ∈ E(X ), there is some λ λ T λ λ 1 2 T y ∈ A ∩ i∈I B (xi , ri) and hence y = (y , y ) ∈ A ∩ i∈I B(xi, ri). 1 1 1 1 2 2 For the converse, if d(xi , xj ) ≤ ri + rj and d(xi ,A ) ≤ ri, fix some x ∈ A . 1 2 Then the points xi = (xi , x ) fulfill d(xi, xj) ≤ ri + rj and d(xi,A) ≤ ri, 1 2 T 1 1 i.e. there is some y = (y , y ) ∈ A ∩ i∈I B(xi, ri) and hence y ∈ A ∩ T 1 1 i∈I B (xi , ri) 6= ∅. The proof of (iii) is similar and (iv) follows from (ii) by setting Aλ = Xλ.

II.4 Retracts

Weakly externally hyperconvex subsets were recognized as the proximinal 1-Lip- schitz retracts by Esp´ınolain [Esp05].

Proposition 4.1. [Esp05, Theorem 3.6] Let X be a hyperconvex metric space and let A ⊂ X be non-empty. Then A is a proximinal 1-Lipschitz retract of X if and only if A is a weakly externally hyperconvex subset of X.

Similarly, we can characterize externally hyperconvex subsets as 1-Lipschitz retracts with some further properties.

21 CHAPTER II. HYPERCONVEX METRIC SPACES

Proposition 4.2. Let X be a hyperconvex metric space and A ⊂ X a subset. Then A is externally hyperconvex in X if and only if there is a proximinal 1-Lipschitz retraction ρ: X → A with d(ρ(x), y) ≤ max{d(x, y), d(A, y)} for all x, y ∈ X.

Proof. First, assume that there is a proximinal 1-Lipschitz retraction ρ: X → A, with d(ρ(x), y) ≤ max{d(x, y), d(A, y)} for all x, y ∈ X. Let {B(xi, ri)}i∈I be a family of closed balls in X with d(xi, xj) ≤ ri +rj and T d(xi,A) ≤ ri. Then, by hyperconvexity of X, there is some z ∈ i∈I B(xi, ri). We get d(ρ(z), xi) ≤ max{d(z, xi), d(A, xi)} ≤ ri and hence \ ρ(z) ∈ A ∩ B(xi, ri) 6= ∅. i∈I For the converse, consider the set

F := {(Y, ρ): Y ⊂ X, ρ: Y → A a good retraction} , i.e. ρ: Y → A is a retraction with d(ρ(x), y) ≤ max{d(x, y), d(A, y)} for all 0 0 x, y ∈ X. We endow F with the usual order relation (Y, ρ) 4 (Y , ρ ) if and 0 0 only if Y ⊂ Y and ρ |Y = ρ. Clearly, F is non-empty, since (A, id) ∈ F, and for any chain (Yn, ρn)n∈N, the element (Y, ρ) with Y := S Y and ρ(x) := ρ (x) for x ∈ Y is an upper n∈N n n n bound. Hence, there is some maximal element (Y,¯ ρ¯) ∈ F. Assume that there is some x0 ∈ X \ Y¯ . Then for all x ∈ X, let us deﬁne rx := max{d(x0, x), d(A, x)} and for all y ∈ Y¯ , deﬁne sy := d(x0, y). We have

0 0 d(x, x ) ≤ d(x0, x) + d(x0, x ) ≤ rx + rx0 ,

d(x, A) ≤ rx, 0 0 d(¯ρ(y), ρ¯(y )) ≤ d(y, y ) ≤ sy + sy0 ,

d(¯ρ(y), x) ≤ max{d(y, x), d(A, x)} ≤ sy + rx.

Hence, since A is externally hyperconvex, there is some \ \ z ∈ A ∩ B(x, rx) ∩ B(¯ρ(y), sy). x∈X y∈Y¯

0 0 0 But then (Y¯ ∪ {x0}, ρ ) with ρ (y) :=ρ ¯(y) for y ∈ Y¯ and ρ (x0) := z is a strictly bigger element in F, contradicting maximality of (Y,¯ ρ¯). Therefore, Y¯ = X andρ ¯: X → A is the desired retraction.

22 Chapter III

Gluing Hyperconvex Metric Spaces

III.1 Gluing along Weakly Externally Hyperconvex Subsets

Definition 1.1. Let (Xλ, dλ)λ∈Λ be a family of metric spaces with closed subsets Aλ ⊂ Xλ. Suppose that all Aλ are isometric to some metric space A. For every λ ∈ Λ, fix an isometry ϕλ : A → Aλ. We define an equivalence relation F on the disjoint union λ∈Λ Xλ generated by ϕλ(a) ∼ ϕλ0 (a) for a ∈ A. The F resulting space X := ( λ∈Λ Xλ)/ ∼ is called the gluing of the Xλ along A.

X admits a natural metric. For x ∈ Xλ and y ∈ Xλ0 it is given by ( d (x, y), if λ = λ0, d(x, y) = λ (1.1) 0 infa∈A{dλ(x, ϕλ(a)) + dλ0 (ϕλ0 (a), y)}, if λ 6= λ .

For more details see for instance [BH99, Lemma I.5.24]. In the following, if there is no ambiguity, indices for dλ are dropped and the sets Aλ = ϕλ(A) ⊂ Xλ are identiﬁed with A. λ Balls inside the subset Xλ are denoted by B (x, r).

Lemma 1.2. Let X be a hyperconvex metric space obtained by gluing a family of hyperconvex metric spaces (Xλ, dλ)λ∈Λ along some set A. Then A is hyperconvex.

Proof. Let {xi}i∈I ⊂ A such that d(xi, xj) ≤ ri + rj. Then for each λ, since Xλ T T is hyperconvex, there is some yλ ∈ i∈I B(xi, ri)∩Xλ. Moreover, i∈I B(xi, ri) is path-connected and a path from yλ to yλ0 must intersect A, i.e. \ B(xi, ri) ∩ A 6= ∅. i∈I

23 CHAPTER III. GLUING HYPERCONVEX METRIC SPACES

In general, we cannot say more about necessary conditions on A such that the gluing along A is hyperconvex. For instance, gluing a hyperconvex space X and any hyperconvex subset A ∈ H(X) along A, the resulting space is isometric to X and therefore hyperconvex. But there are also plenty of nontrivial examples.

Example 1.3. Let f : R → R be any 1-Lipschitz function. Consider its graph 2 2 A = {(x, y) ∈ l∞ : y = f(x)} and the two sets X1 = {(x, y) ∈ l∞ : y ≤ f(x)}, 2 2 X2 = {(x, y) ∈ l∞ : y ≥ f(x)}. Then l∞ = X1 tA X2 is hyperconvex and occurs as the gluing of two hyperconvex spaces X1,X2. But if we assume that the gluing set is weakly externally hyperconvex, we can do better.

Lemma 1.4. Let (X, d) be the metric space obtained by gluing a family of hyperconvex metric spaces (Xλ, dλ)λ∈Λ along some set A, such that A is weakly 0 externally hyperconvex in Xλ for each λ ∈ Λ. For x ∈ Xλ and x ∈ Xλ0 with λ 6= λ0, there are then points a ∈ B(x, d(x, A)) ∩ A and a0 ∈ B(x0, d(x0,A)) ∩ A such that d(x, x0) = d(x, a) + d(a, a0) + d(a0, x0).

Proof. As A is weakly externally hyperconvex in each Xλ, by Lemma II.3.1, for every y ∈ A there are points a ∈ B(x, d(x, A)) ∩ A and a0 ∈ B(x0, d(x0,A)) ∩ A such that both d(x, y) = d(x, a) + d(a, y) and d(y, x0) = d(y, a0) + d(a0, x0) hold. Hence

d(x, x0) = d(x, A) + d(B(x, d(x, A)) ∩ A, B(x0, d(x0,A)) ∩ A) + d(x0,A).

But the sets B(x, d(x, A))∩A and B(x0, d(x0,A))∩A are externally hyperconvex in A and therefore, by Corollary II.3.12, there are a ∈ B(x, d(x, A)) ∩ A and a0 ∈ B(x0, d(x0,A)) ∩ A with

d(a, a0) = d(B(x, d(x, A)) ∩ A, B(x0, d(x0,A)) ∩ A).

Lemma 1.5. Let (X, d) be the metric space obtained by gluing a family of hyperconvex metric spaces (Xλ, dλ)λ∈Λ along some set A, such that A is weakly 0 externally hyperconvex in Xλ for each λ ∈ Λ. Then for λ 6= λ , x ∈ Xλ and r ≥ s := d(x, A), one has

λ0 λ B(x, r) ∩ Xλ0 = B (B (x, s) ∩ A, r − s).

λ Therefore, if B (x, s) ∩ A ∈ E(Xλ0 ), then we also have B(x, r) ∩ Xλ0 ∈ E(Xλ0 ).

0 λ Proof. Let x ∈ B(x, r) ∩ Xλ0 . By Lemma 1.4, there is some a ∈ B (x, s) ∩ A with d(x, x0) = d(x, a) + d(a, x0) . We have d(a, x0) ≤ r − s and hence

0 x0 ∈ Bλ (Bλ(x, s) ∩ A, r − s).

24 III.1. WEAKLY EXTERNALLY HYPERCONVEX

Proposition 1.6. Let (X, d) be the metric space obtained by gluing a family of hyperconvex metric spaces (Xλ, dλ)λ∈Λ along some set A, such that A is weakly externally hyperconvex in Xλ for each λ ∈ Λ. If X is hyperconvex, then for all λ ∈ Λ and all x ∈ X \ Xλ, the set B(x, d(x, A)) ∩ A is externally hyperconvex in Xλ.

Proof. Set s := d(x, A) and let {xi}i∈I be a collection of point in Xλ and {ri}i∈I such that d(xi, xj) ≤ ri+rj and d(xi,B(x, s)∩A) ≤ ri. Then, by hyperconvexity of X, there is some \ y ∈ B(x, s) ∩ B(xi, ri). i∈I 0 Since y ∈ B(x, s), we have y ∈ Xλ0 for some λ 6= λ. Therefore, by Lem- ma 1.4, for each i ∈ I, there is some yi ∈ B(xi, d(xi,A)) ∩ A with d(y, xi) = 0 d(y, yi) + d(yi, xi). Deﬁne ri := ri − d(yi, xi). We have 0 0 d(yi, yj) ≤ d(yi, y) + d(y, yj) ≤ ri + rj

0 and d(x, yi) ≤ s + ri. Hence, since A is weakly externally hyperconvex in Xλ0 , there is some

\ 0 \ z ∈ B(yi, ri) ∩ B(x, s) ∩ A ⊂ B(xi, ri) ∩ B(x, s) ∩ A 6= ∅. i∈I i∈I

Proposition 1.7. Let (X, d) be the metric space obtained by gluing a family of hyperconvex metric spaces (Xλ, dλ)λ∈Λ along some set A, such that, for each λ ∈ Λ, the set A is weakly externally hyperconvex in Xλ and, for all x ∈ X \Xλ, the intersection B(x, d(x, A)) ∩ A is externally hyperconvex in Xλ. Then X is hyperconvex and Xλ ∈ W(X) for every λ ∈ Λ.

Proof. Let {B(xi, ri)}i∈I be a family of balls in X with d(xi, xj) ≤ ri + rj. We divide the proof into two cases.

Case 1. If, for every i, j ∈ I, one has

B(xi, ri) ∩ B(xj, rj) ∩ A 6= ∅, setting Ci := A ∩ B(xi, ri), we obtain that the family {Ci}i∈I is pairwise intersecting. Moreover, {Ci}i∈I is contained in E(A), since A is weakly externally T hyperconvex. By Proposition II.2.1, we obtain that i∈I Ci 6= ∅, and hence T i∈I B(xi, ri) 6= ∅.

Case 2. Otherwise, there are i0, j0 ∈ I with xi0 , xj0 ∈ Xλ0 such that

B(xi0 , ri0 ) ∩ B(xj0 , rj0 ) ∩ A = ∅.

Indeed, either there is some i0 ∈ I such that d(xi0 ,A) > ri0 and we may take i0 = j0, or if

B(xi0 , ri0 ) ∩ B(xj0 , rj0 ) ∩ A = ∅

25 CHAPTER III. GLUING HYPERCONVEX METRIC SPACES

with d(xi0 ,A) ≤ ri0 and d(xj0 ,A) ≤ rj0 , we get xi0 , xj0 ∈ Xλ0 by Lemma 1.4.

Observe that, in both cases, we may assume that, for xi ∈ Xλ 6= Xλ0 , we have d(xi,A) ≤ ri. λ0 Deﬁne Ai := B(xi, ri) ∩ Xλ0 . The goal is now to show the following claim: λ0 λ0 Claim. For every i, j ∈ I, one has Ai ∩ Aj 6= ∅.

λ0 Then, by Lemma 1.5, we have Ai ∈ E(Xλ0 ) and, by Proposition II.2.1, we get \ \ λ0 B(xi, ri) ∩ Xλ0 = Ai 6= ∅. To prove the claim, consider ﬁrst the following two easy cases.

• If xi, xj ∈ Xλ0 , then we are done by hyperconvexity of Xλ0 .

• If xi ∈ Xλ 6= Xλ0 3 xj, we have B(xi, ri)∩B(xj, rj)∩A 6= ∅ by Lemma 1.4 and we are done.

The remaining case is when xi, xj ∈ Xλ 6= Xλ0 . We do this in two steps.

Step I. Set

0 λ0 λ0 A := B(xi0 , ri0 ) ∩ B(xj0 , rj0 ) = B (xi0 , ri0 ) ∩ B (xj0 , rj0 ) and s := d(A, A0). By Corollary II.3.12, we have B(A0, s)∩A 6= ∅. Furthermore, we get 0 B(A , s) = B(xi0 , ri0 + s) ∩ B(xj0 , rj0 + s) and hence B(xi0 , ri0 + s) ∩ B(xj0 , rj0 + s) ⊂ Xλ0 . To see this, observe ﬁrst that by Lemma II.1.3, we have

0 λ0 λ0 B(A , s) = B (xi0 , ri0 + s) ∩ B (xj0 , rj0 + s) ⊂ Xλ0 and therefore

0 (B(xi0 , ri0 + s) ∩ B(xj0 , rj0 + s)) \ B(A , s) ⊂ X \ Xλ0 .

Thus, assume that there is some

y ∈ B(xi0 , ri0 + s) ∩ B(xj0 , rj0 + s) ∩ (Xλ \ Xλ0 ) .

Now since

B(xi0 , ri0 + s) ∩ B(xj0 , rj0 + s) ∩ A 6= ∅ and

B(xi0 , ri0 + s) ∩ B(xj0 , rj0 + s) ∩ Xλ is externally hyperconvex in Xλ and thus path-connected, there is some

0 y ∈ B(xi0 , ri0 + s) ∩ B(xj0 , rj0 + s) ∩ Xλ \ Xλ0

26 III.1. WEAKLY EXTERNALLY HYPERCONVEX with d(y0,A) ≤ s. But then by Lemma 1.4 we have

0 B(xi0 , ri0 ) ∩ B(y , s) ∩ Xλ0 6= ∅, 0 B(xj0 , rj0 ) ∩ B(y , s) ∩ Xλ0 6= ∅ and therefore

0 B(xi0 , ri0 ) ∩ B(xj0 , rj0 ) ∩ B(y , s) ∩ Xλ0 6= ∅,

0 0 0 i.e. y ∈ B(A , s) contradicting y ∈/ Xλ0 .

Step II. We now show that the family

λ λ F := {B(xi0 , ri0 + s) ∩ Xλ,B(xj0 , rj0 + s) ∩ Xλ,B (xi, ri),B (xj, rj)} is pairwise intersecting. We already observed that

(B(xi0 , ri0 + s) ∩ Xλ) ∩ (B(xj0 , rj0 + s) ∩ Xλ) 6= ∅.

Further, since xi0 ∈ Xλ0 6= Xλ 3 xi, by Lemma 1.4 one has

λ (B(xi0 , ri0 + s) ∩ Xλ) ∩ B (xi, ri) 6= ∅ and similarly for (i0, i) replaced by (i0, j) as well as by (j0, i) and (j0, j). Finally,

λ λ B (xi, ri) ∩ B (xj, rj) ∩ Xλ 6= ∅ by hyperconvexity of Xλ. Hence, we have shown that F is pairwise intersecting. Since F ⊂ E(Xλ), it follows by Proposition II.2.1 that

λ λ λ λ C := B (xi0 , ri0 + s) ∩ B (xj0 , rj0 + s) ∩ B (xi, ri) ∩ B (xi, ri) 6= ∅.

Since B(xi0 , ri0 + s) ∩ B(xj0 , rj0 + s) ∩ Xλ ⊂ A, we have in particular C ⊂ A. Hence λ λ B (xi, ri) ∩ B (xj, rj) ∩ A ⊃ C ∩ A = C 6= ∅, and this is the desired result.

To see that Xλ is weakly externally hyperconvex in X, use that for x ∈ X, r ≥ d(x, Xλ) and {xi}i∈I ∈ Xλ with d(x, xi) ≤ r + ri, d(xi, xj) ≤ ri + rj, we have B(x, r) ∩ B(xi, ri) ∩ Xλ 6= ∅ by Lemma 1.4 and therefore the set

λ {B(x, r) ∩ Xλ} ∪ {B (xi, ri)}i∈I is a family of pairwise intersecting externally hyperconvex subsets of Xλ and hence \ B(xi, ri) ∩ B(x, r) ∩ Xλ 6= ∅ i∈I by Proposition II.2.1.

27 CHAPTER III. GLUING HYPERCONVEX METRIC SPACES

Combining Proposition 1.6 and Proposition 1.7, we get the following.

Theorem 1.8. Let (X, d) be the metric space obtained by gluing a family of hyperconvex metric spaces (Xλ, dλ)λ∈Λ along some set A, such that, for each λ ∈ Λ, A is weakly externally hyperconvex in Xλ. Then X is hyperconvex if and only if for all λ ∈ Λ and all x ∈ X \ Xλ, the set B(x, d(x, A)) ∩ A is externally hyperconvex in Xλ. Moreover, if X is hyperconvex, the subspaces Xλ are weakly externally hyperconvex in X.

As a consequence, we get the following results for gluing along strongly convex and along externally hyperconvex subsets.

Corollary 1.9. Let (X, d) be the metric space obtained by gluing a collection (Xλ, dλ)λ∈Λ of hyperconvex metric spaces along some space A, such that A is closed and strongly convex in Xλ for each λ ∈ Λ. Then X is hyperconvex as well.

Corollary 1.10. Let (X, d) be the metric space obtained by gluing a family of hyperconvex metric spaces (Xλ, dλ)λ∈Λ along some set A, such that A is externally hyperconvex in each Xλ. Then X is hyperconvex. Moreover, A is externally hyperconvex in X.

In the second case, we clearly have that the gluing set is weakly externally hyperconvex in each Xλ. Furthermore, it holds that B(x, d(x, A)) ∩ A ∈ E(A) and therefore, by Proposition II.2.8, we get B(x, d(x, A)) ∩ A ∈ E(Xλ). The reason that Corollary 1.9 holds is that a closed, strongly convex subsets A of a hyperconvex metric space X is gated, that is that for each x ∈ X there is somex ¯ ∈ A such that for all y ∈ A, we havex ¯ ∈ I(x, y). Clearly, if such an x¯ exists, it is unique and we call it the gate of x in A.

Lemma 1.11. Let A be a subset of a hyperconvex metric space X. Then A is strongly convex and closed if and only if it is gated.

Proof. First assume that A is strongly convex and closed. Fix x ∈ X. Let xn 1 be a sequence of points in A with d(x, xn) ≤ d(x, A) + n . Note that, for all x, y, z ∈ X, we have I(x, y) ∩ I(y, z) ∩ I(z, x) 6= ∅, see Proposition V.1.1. Hence, for n, k ∈ N, we can choose mn,k ∈ I(x, xn) ∩ I(x, xk) ∩ I(xn, xk). By strong convexity, we get mn,k ∈ A and hence 1 1 d(x , m ) = d(x, x ) − d(x , m ) ≤ d(x, A) + − d(x, A) = . n n,k n n n,k n n

1 By interchanging xn and xk, we also get d(xk, mn,k) ≤ k and therefore 1 1 d(x , x ) = d(x , m ) + d(m , x ) ≤ + . n k n n,k n,k k n k

Thus (xn)n∈N is a Cauchy sequence and since A is closed, it converges to some x¯ ∈ A. Moreover, we have d(x, x¯) = d(x, A). We claim thatx ¯ is a gate for x in A. Let y ∈ A. Then, there is some z ∈ I(x, x¯) ∩ I(¯x, y) ∩ I(y, x). By strong

28 III.2. GLUING ISOMETRIC COPIES convexity, we have z ∈ A and therefore d(x, z) ≥ d(x, A) = d(x, x¯). Since z ∈ I(x, x¯), this implies z =x ¯ and hencex ¯ ∈ I(x, y) as desired. On the other hand, if A is gated, for all points x, y ∈ A and z ∈ I(x, y) we have z =z ¯ and hence I(x, y) ⊂ A. Moreover, for all x ∈ X, we have d(x, A) = d(x, x¯) and thereforex ¯ ∈ B(x, d(x, A)) ∩ A, i.e. A is proximinal and therefore closed.

Therefore, if A is closed and strongly convex in each Xλ, the intersection B(x, d(x, A)) ∩ A is a point (the gate) and thus externally hyperconvex in X. Moreover, A is weakly externally hyperconvex in Xλ by Lemma II.3.1. We can apply Theorem 1.8 to the situation where we glue several hyperconvex metric spaces onto a given space and get the following result:

Theorem 1.12. Let X0 be a hyperconvex metric space and {Xλ}λ∈Λ a family of hyperconvex metric spaces with subsets Aλ ∈ W(Xλ), such that, for every 0 λ ∈ Λ, there is an isometric copy Aλ ∈ W(X0) and Aλ ∩ Aλ0 = ∅ for λ 6= λ . If for every xλ ∈ Xλ and every x ∈ X0, we have B(xλ, d(xλ,Aλ)) ∩ Aλ ∈ E(X0) and B(x, d(x, A )) ∩ A ∈ E(X ), then X = X F X is hyperconvex. λ λ λ 0 {Aλ:λ∈Λ} λ

Proof. First, by Theorem 1.8, we get that Yλ = X0 tAλ Xλ is hyperconvex and X0 ∈ W(Yλ). Observe that X can be obtained by gluing the spaces Yλ along X , i.e. X = F Y . Therefore, it remains to prove that for λ 6= λ0 and x ∈ Y 0 X0 λ λ the intersection B := B(x, d(x, X0)) ∩ X0 ∈ E(Yλ0 ). By Corollary II.3.11, we already have B ∈ W(Yλ0 ) and without loss of generality, we may assume that x∈ / X0. Then we have d(x, X0) = d(x, Aλ) and therefore B = B(x, d(x, Aλ)) ∩ Aλ ∈ E(X0), especially B ⊂ Aλ. Hence, by Corollary II.3.12, we get d(B,Aλ0 ) > 0, i.e. there is some s > 0, such λ0 λ0 that B (B, s) ⊂ X0. Thus B ∈ E(B (B, s)) and, by Lemma II.3.17, we get B ∈ E(Yλ0 ) as desired.

III.2 Gluing Isometric Copies

We now turn our attention to the case where we glue two copies of the same hyperconvex space. Here we get that the gluing set must be weakly externally hyperconvex.

Proposition 2.1. Let X be a metric space and A ⊂ X such that X tA X is hyperconvex. Then the following hold:

(i) A is weakly externally hyperconvex in X.

(ii) For every x ∈ X, the intersection B(x, d(x, A)) ∩ A is externally hyperconvex in X.

Proof. Let us denote the second copy of X by X0 and for any y ∈ X, let y0 denote its corresponding copy in X0. Pick x ∈ X and r ≥ 0 such that d(x, A) ≤ r and let {B(xi, ri)}i∈I be a family of closed balls with xi ∈ A,

29 CHAPTER III. GLUING HYPERCONVEX METRIC SPACES

0 d(x, xi) ≤ r + ri and d(xi, xj) ≤ ri + rj. It follows that d(x, x ) ≤ 2r and, since X tA X is hyperconvex, we have

\ 0 B := B(xi, ri) ∩ B(x, r) ∩ B(x , r) 6= ∅. i∈I

By symmetry, there are y, y0 ∈ B with y ∈ X and y0 ∈ X0. Then, since intersections of balls are hyperconvex, there is some geodesic [y, y0] ⊂ B, which must intersect A. Therefore, we get \ B(xi, ri) ∩ B(x, r) ∩ A 6= ∅ i∈I and hence, A is weakly externally hyperconvex. For (ii), observe that

B(x, d(x, A)) ∩ A = B(x, d(x, A)) ∩ B(x0, d(x, A)) is admissible in X tA X and therefore externally hyperconvex in X.

Condition (i) is not enough, as the following example shows.

3 Example 2.2. Let X1 and X2 be two copies of l∞. Consider now the gluing X := X1 tV X2 where

3 V := {x ∈ l∞ : x1 = x2 and x3 = 0}. and where the gluing maps are given by the inclusion maps for V . To see that X is not hyperconvex, consider p1 := (0, 0, 1) in X1, as well as p2 := (2, 0, 0) and 0 p2 := (0, −2, 0) both in X2. Note that B(p1, 1) ∩ X2 = {(t, t, 0) : t ∈ [−1, 1]} 0 and hence B(p1, 1) ∩ B(p2, 1) = {(1, 1, 0)} ⊂ V , as well as B(p1, 1) ∩ B(p2, 1) = 0 {(−1, −1, 0)} ⊂ V . Moreover, B(p2, 1) ∩ B(p2, 1) = {1} × {−1} × [−1, 1] ⊂ X2. Hence, 0 B(p1, 1) ∩ B(p2, 1) ∩ B(p2, 1) = ∅.

As a consequence of Proposition 1.7 and Proposition 2.1, we get a necessary and suﬃcient condition for gluings of isometric copies.

Theorem 2.3. Let X be a hyperconvex metric space and let A be a subset. Then X tA X is hyperconvex if and only if A is weakly externally hyperconvex in X and for every x ∈ X, the intersection B(x, d(x, A)) ∩ A is externally hyperconvex in X.

Example 2.4. Let X be a hyperconvex metric space, A ⊂ X and r ≥ 0. If either A is strongly convex or A = I(x, y) is a metric interval, then B(A, r) is weakly externally hyperconvex and, for all x ∈ X, the set B(x, d(x, B(A, r))) ∩ B(A, r) is externally hyperconvex in X by Lemma II.3.3 and Lemma II.3.5. Hence, gluing two copies of X along B(A, r) preserves hyperconvexity.

30 III.3. GLUING HYPERCONVEX LINEAR SPACES

III.3 Gluing Hyperconvex Linear Spaces

We start this section with a classiﬁcation of externally hyperconvex, strongly n convex and weakly externally hyperconvex subsets of l∞. We use coordinates n x = (x1, . . . , xn) ∈ l∞. n Qn A cuboid in l∞ is the product i=1 Ii of (possibly unbounded) closed, non- empty intervals Ii ⊂ R. Note that these are exactly the sets that can be described by inequalities of the form σxi ≤ C with σ ∈ {±1} and C ∈ R.

n Proposition 3.1. A subset A of l∞ is externally hyperconvex if and only if it is a cuboid.

Proof. On the one hand, cuboids are externally hyperconvex by Lemma II.3.18. On the other hand, if A is externally hyperconvex, it is closed. Hence it is n enough to show that for any points x, y ∈ A and z ∈ l∞ with zi ∈ I(xi, yi) for each i ∈ {1, . . . , n}, it follows that z ∈ A. Without loss of generality, we may assume that xi ≤ zi ≤ yi. Let

r := max {zi − xi, yi − zi}. i∈{1,...,n}

For each i ∈ {1, . . . , n}, deﬁne

i p := (z1, . . . zi−1, xi − r, zi+1, . . . , zn) and ri := zi − xi + r, i q := (z1, . . . zi−1, yi + r, zi+1, . . . , zn) and si := yi − zi + r.

Then we have n n \ i \ i B(p , ri) ∩ B(q , si) = {z}. i=1 i=1

i i i i Moreover, it holds d(p ,A) ≤ d(p , x) ≤ ri as well as d(q ,A) ≤ d(q , y) ≤ si and therefore, since A is externally hyperconvex,

n n \ i \ i ∅= 6 B(p , ri) ∩ B(q , si) ∩ A ⊂ {z}. i=1 i=1

It follows that z ∈ A and this concludes the proof.

Theorem 3.2. Let I 6= ∅ be any index set. Suppose that Q is a non-empty subset of l∞(I), given by an arbitrary system of inequalities of the form σxi ≤ C or σxi + τxj ≤ C with σ, τ ∈ {±1} and C ∈ R. Then Q is weakly externally hyperconvex in l∞(I).

Before giving a proof of Theorem 3.2, we show that a set Q given by a system of inequalities as in the theorem is proximinal. Recall that l∞(I) is the dual space of l1(I) and observe that Q is a weak*-closed subset of l∞(I). In general, the following holds.

31 CHAPTER III. GLUING HYPERCONVEX METRIC SPACES

Lemma 3.3. Let V be a normed vector space and V ∗ its dual space. Then every non-empty weak*-closed subset of V ∗ is proximinal. Especially, if Q is a non-empty subset of l∞(I), given by an arbitrary system of inequalities of the form σxi ≤ C or σxi + τxj ≤ C with σ, τ ∈ {±1} and C ∈ R, then Q is proximinal. Proof. Let A be a non-empty weak*-closed subset of V ∗, let x be any point in V ∗ and set r := d(x, A). It follows from the theorem of Banach-Alaoglu that the set A0 := B(x, r + 1) is weak*-compact. Now, the sets

B x, r + 1 ∩ A n n∈N form a decreasing sequence of non-empty weak*-closed subsets of A0. By the closed set criterion for compact sets, it follows that

\ 1 B(x, r) ∩ A = B(x, r + n ) ∩ A 6= ∅. n∈N This shows that A is proximinal. To see that Q is weak*-closed, note that the maps from l∞(I) to R given by ϕσei : f 7→ σfi and ϕσei+τej : f 7→ σfi + τfj are continuous in the weak*- topology on l∞(I) since σei and σei + τej are both in l1(I). Since, by assumption, \ \ Q = ϕ−1 ((−∞,C ]) ∩ ϕ−1 ((−∞,C ]), σei (i,σ) σei+τej (i,j,σ,τ) (i,σ) (i,j,σ,τ) we deduce that Q is weak*-closed and therefore proximinal.

Now, the proof of Theorem 3.2 is a direct adaptation of the proof of [Lan13, Proposition 2.4].

Proof of Theorem 3.2. For i ∈ I, denote by Ri the reﬂection of l∞(I) that interchanges xi with −xi. Let A ∪ {b} be a metric space with ∅= 6 A and let y ∈ l∞(I) \ Q. For any f ∈ Lip1(A, Q ∪ {y}) satisfying that d(y, Q) ≤ d(a, b) if f(a) = y, we show that there is an extension f ∈ Lip1(A ∪ {b},Q ∪ {y}) such that f(b) ∈ Q. It is easy to see that this implies that Q ∈ W(l∞(I)), it is similar to showing that injectivity implies hyperconvexity. Let A0 := A \ f −1({y}). By proximinality of Q, we have Q ∩ B(y, d(y, Q)) 6= ∅. We can thus assume that \ 0 ∈ Q ∩ B(y, d(y, Q)) ⊂ Q ∩ B(f(a0), d(a0, b)), (3.1) a0∈f −1({y}) so that all constants on the right sides of the inequalities describing Q are non-negative. First, for a real valued function f ∈ Lip1(A, R), we combine the smallest and largest 1-Lipschitz extensions and deﬁne f : A ∪ {b} → R by n o n o f(b) := sup 0, sup(f(a) − d(a, b)) + inf 0, inf (f(a0) + d(a0, b)) . a∈A a0∈A

32 III.3. HYPERCONVEX LINEAR SPACES

Note that at most one of the two summands is nonzero since

f(a) − d(a, b) ≤ f(a0) + d(a, a0) − d(a, b) ≤ f(a0) + d(a0, b).

It is not diﬃcult to check that f is a 1-Lipschitz extension of f and that

R ◦ f = R ◦ f for the reﬂection R: x 7→ −x of R. Moreover, by (3.1), it follows that

sup (f(a) − d(a, b)) ≤ 0 a∈f −1({y}) and inf (f(a) + d(a, b)) ≥ 0. a∈f −1({y})

It follows that f can be defined by taking suprema and infima on A0 instead of A without changing f(b). We define the extension operator

φ: Lip1(A, l∞(I)) → Lip1(A ∪ {b}, l∞(I)) such that φ(f) satisﬁes φ(f)i = fi for every i. Clearly φ(f) ∈ Lip1(A ∪ {b}, l∞(I)) and φ(Ri ◦ f) = Ri ◦ φ(f) (3.2) for every i. To see that φ(f)(b) ∈ Q, it thus suﬃces to show that the components of φ(f) satisfy φ(f)i + φ(f)j ≤ C (3.3) whenever fi + fj ≤ C for some pair of possibly equal indices i, j and some constant C ≥ 0. Suppose that fi(a) + fj(a) ≤ C for some indices i, j, some constant C ≥ 0 0 −1 and every a ∈ A := A \ f ({y}). Assume that φ(f)i(b) ≥ φ(f)j(b). If φ(f)j(b) > 0, then

0 0 φ(f)i(b) + φ(f)j(b) = sup (fi(a) + fj(a ) − d(a, b) − d(a , b)) a,a0∈A0 0 0 ≤ sup (fi(a) + fj(a ) − d(a, a )) a,a0∈A0

≤ sup(fi(a) + fj(a)) ≤ C. a∈A0

If φ(f)i(b) > 0 ≥ φ(f)j(b), then

0 0 φ(f)i(b) + φ(f)j(b) ≤ sup(fi(a) − d(a, b)) + inf (fj(a ) + d(a , b)) a∈A0 a0∈A0

≤ sup(fi(a) + fj(a)) ≤ C. a∈A0

Finally, if φ(f)i(b) ≤ 0, then φ(f)i(b) + φ(f)j(b) ≤ 0 ≤ C.

33 CHAPTER III. GLUING HYPERCONVEX METRIC SPACES

Remark 3.4. By Theorem 3.2, and borrowing the notation and terminology from [Lan13], for any metric space X and for any set A ∈ A (X), note that the set P (A) := ∆(X) ∩ H(A) is isometric to a weakly externally hyperconvex subset of l∞(X). X X Indeed, for any point x0 ∈ X, let us denote by τ : R → R the translation f 7→ f − dx0 , where dx0 : x 7→ d(x0, x). One has

τ(P (A)) ⊂ τ(E(X)) ⊂ l∞(X), see [Lan13, Sections 3 and 4]. Recall that

X ∆(X) := {f ∈ R : f(x) + f(y) ≥ d(x, y) for all x, y ∈ X} and X H(A) := {g ∈ R : g(x) + g(y) = d(x, y) for all {x, y} ∈ A}.

By Theorem 3.2, it follows that τ(P (A)) is in W(l∞(X)). In particular, it follows that τ(P (A)) is weakly externally hyperconvex in τ(E(X)) and since τ is an isometry, P (A) ∈ W(E(X)). Finally, recall that under suitable assumptions on X, the family of cells

{P (A)}A∈A (X) endows E(X) with a canonical polyhedral structure. Thereby for any A ∈ A (X), P (A) is a cell of E(X) isometric to a convex polytope in n l∞.

In the following, we use the notation In := {1, . . . , n}. To characterize n strongly convex and weakly externally hyperconvex linear subspaces of l∞, it is helpful to recall the characterization for hyperconvex linear subspaces.

n Theorem 3.5. [Pav16, Theorem 2.1] Let ∅= 6 V ⊂ l∞ be a linear subspace and let k := dim(V ). Then the following are equivalent:

(i) V is hyperconvex.

(ii) There is a subset J ⊂ In with |J| = k such that, for all i ∈ In \ J, there P exist real numbers {ci,j}j∈J such that j∈J |ci,j| ≤ 1 and ( ) n X V = (x1, . . . , xn) ∈ l∞ : for all i ∈ In \ J, one has xi = ci,jxj . j∈J

n Proposition 3.6. Let n ≥ 1 and let V denote any linear subspace of l∞. Then n V is strongly convex in l∞ if and only if one of the following three cases occurs:

(i) V = {0},

n (ii) V = Re where e denotes a vertex of the hypercube [−1, 1] , or

n (iii) V = l∞.

34 III.3. HYPERCONVEX LINEAR SPACES

Proof. Let k = dim V . Since strongly convex subspaces are hyperconvex, by Theorem 3.5, we may assume that V is of the form

( k ) n X V = (x1, . . . , xn) ∈ l∞ : xl = cl,ixi for l = k + 1, . . . n i=1

Pk with i=1 |cl,i| ≤ 1. Fix some 1 ≤ i ≤ k and consider

x = (0,... 0, 1, 0,... 0, ck+1,i, . . . , cn,i) ∈ V.

We have cl,i 6= 0 since otherwise

1 ck+1,i cl−1,i 1 cl+1,i cn,i y = (0,... 0, 2 , 0,... 0, 2 ,..., 2 , 2 , 2 ,..., 2 ) ∈ I(0, x) \ V.

Now take l such that |cl,i| is minimal. Then

|cl,i| |cl,i| y = (0,... 0, |cl,i|, 0,... 0, ck+1,i,..., cn,i) ∈ I(0, x) ⊂ V |ck+1,i| |cn,i| and therefore cl,i = cl,i · |cl,i|, i.e. |cl,i| = 1. Hence, one has cl,i ∈ {±1} for Pk all l = k + 1, . . . , n and i = 1, . . . , k. Since i=1 |cl,i| ≤ 1, it thus follows that either k = 0, which corresponds to (i), k = 1, which corresponds to (ii) or k = n, which corresponds to (iii).

In the following, relint(S) denotes the relative interior of S and Fi denotes n n n n the facet [−1, 1] ∩ {x ∈ l∞ : xi = 1} of the unit ball [−1, 1] in l∞. Half-spaces n are denoted by Hv := {x ∈ R : x · v ≥ 0} (with · denoting the standard scalar product) and ∂Hv is the boundary of Hv.

n Lemma 3.7. Let V denote any hyperconvex linear subspace of l∞, such that V is not contained in any hyperplane of the form ∂Hσei+τej . Then the following hold:

(i) There is (i, σ) ∈ In × {±1} such that V ∩ relint(σFi) 6= ∅.

n n (ii) If V 6= l∞, there is i ∈ In and ν ∈ R \{0} such that V ⊂ ∂Hν and

∂Hν ∩ (Fi ∪ (−Fi)) = ∅.

Proof. By Theorem 3.5, V can, without loss of generality, be written as

 k k   X X  V := (x1, . . . , xk, ck+1,jxj,..., cn,jxj): x1, . . . , xk ∈ R ,  j=1 j=1 

Pk where, for every m ∈ {k + 1, . . . , n}, one has j=1 |cm,j| ≤ 1. It follows that V ⊂ ∂Hνm , where

m ν := (cm,1, . . . , cm,k, 0,..., 0, −1, 0,..., 0)

35 CHAPTER III. GLUING HYPERCONVEX METRIC SPACES

(i.e. −1 in the m-th entry) and

k m X m kν k1 = 1 + |cm,j| ≤ 2 = 2 kν k∞ . j=1

Since we assume that V 6⊂ ∂Hσei+τej , it follows that, for every m ∈ {k + 1, . . . , n} and every i ∈ {1, . . . , k}, we have |cm,i| < 1. In particular, (i) follows from the fact that

(1, 0,..., 0, ck+1,1, . . . , cn,1) ∈ V ∩ relint(F1).

Now, (ii) follows from the fact that V ⊂ ∂Hν, where

ν := (ck+1,1, . . . , ck+1,k, −1, 0,..., 0) ∈ relint(−Fk+1) and ∂Hν ∩ (Fk+1 ∪ (−Fk+1)) = ∅. We now proceed to the characterization of weakly externally hyperconvex n polyhedra in l∞. n Theorem 3.8. Let n ≥ 1 and let V denote any linear subspace of l∞. Then n V is weakly externally hyperconvex in l∞ if and only if V can be written as the intersection of hyperplanes of the form ∂Hσei or ∂Hσei+τej , where σ, τ ∈ {±1} n and i, j ∈ {1, ··· , n}. Equivalently, V is weakly externally hyperconvex in l∞ if and only if k m V = {0} × l∞ × Sp1 × · · · × Spq , pj where Spj denotes a one-dimensional strongly convex linear subspace of l∞ for Pq each j ∈ {1, . . . , q}. Note that then n = k + m + j=1 pj and dim(V ) = m + q. Proof. First, it is easy to see that both representations for V given in the statement of Theorem 3.8 are equivalent. To see this, note that, if V has the desired product representation, then using Proposition 3.6, it is easy to express V as an intersection of hyperplanes as desired. Conversely, if V can be written as such an intersection, then V is given by a system of linear equations which one can decompose into a set of minimal linear subsystems. Applying Proposi- tion 3.6, one obtains the desired product representation for V . n Note that the hyperplanes of l∞ of the form ∂Hσei or ∂Hσei+τej are exactly the ones to which Theorem 3.2 applies. Hence, whenever V can be written as an intersection of such hyperplanes, it follows that V is weakly externally hyperconvex. This proves one implication in Theorem 3.8. n We now assume that V ∈ W(l∞) and we show that V can be written as \ \ V = ∂Hσei ∩ ∂Hσei+τej . (i,σ) (i,σ),(j,τ)

If V ⊂ W , then V ∈ W(W ). Now assume that W = ∂Hσei or W = ∂Hσei+τej . Then there is a canonical bijective linear isometry

n−1 π : W → l∞ , (x1, . . . , xn) 7→ (x1, . . . , xi−1, xi+1, . . . xn)

36 III.3. HYPERCONVEX LINEAR SPACES

n−1 with π(V ) ∈ W(l∞ ) and we can iterate this process. We can therefore, without loss of generality, assume that V is not contained in any hyperplane of the form ∂Hσei or ∂Hσei+τej . It follows by (i) in Lem- ma 3.7 and hyperconvexity of V that there is (i, σ) ∈ In × {±1} such that V ∩ relint(σFi) 6= ∅. n By (ii) in Lemma 3.7, there is a facet τFj of [−1, 1] such that τFj ∩ V = ∅. 0 0 It follows that the facet τFj := τFj ∩ σFi of σFi satisﬁes τFj ∩ V = ∅. But 0 0 (0) τFj can be written as τFj = ∩i∈{1,2,3}Bi where we pick v ∈ V ∩ relint(σFi) (0) so that [0, ∞)v ∩ ∂Hτej 6= ∅ and with

B1 = B(0, 1), (0) (0) B2 = B(Rv , kRv k∞) and

B3 = B(σei + τej + Rτee j, kσei + τej + Rτee j − (σei + τej)k∞).

Indeed, for Re big enough, one has B3 ∩ V 6= ∅ since V 6⊂ ∂Hej and by the choice of v(0). For Re chosen even bigger (if needed) and for R big enough, one 0 has σFi = B1 ∩ B2 and τFj = B1 ∩ B2 ∩ B3. It follows that the balls pairwise intersect and 0 V ∩ B1 ∩ B2 ∩ B3 = V ∩ τFj = ∅. n This implies that V 6⊂ W(l∞), which is a contradiction. Let V be a ﬁnite dimensional real vector space and let Q be any convex non-empty subset of V . The tangent cone to Q at p ∈ Q is deﬁned to be the set T Q := S (p + n(Q − p)) where Q − p := {q − p : q ∈ Q} and p n∈N nQ := {nq : q ∈ Q}.

n Proposition 3.9. Suppose that Q is a subset of l∞. Then the following are equivalent:

(i) Q is given by a finite system of inequalities of the form σxi + τxj ≤ C or σxi ≤ C with |σ|, |τ| = 1 and C ∈ R, n (ii) Q ∈ W(l∞). Proof. The fact that (i) implies (ii) is a direct consequence of Theorem 3.2. On the other hand, the fact that (ii) implies (i) follows by considering the tangent cones TpQ for each smooth point p in the boundary of Q. By Theorem VI.1.1, the set Q is convex and as it is for instance observed at the beginning of the proof of [BEL05, Theorem 3.1], by virtue of Mazur density theorem, cf. [Hol75], the set of smooth points of the boundary of Q is dense in ∂Q. It follows that Q can be written as an intersection of half-spaces which correspond to tangent cones of Q at smooth points on the boundary. These tangent cones are weakly externally hyperconvex by Proposition II.3.14 and therefore they are of the n n form {x ∈ l∞ : σxi ≤ C} or {x ∈ l∞ : σxi + τxj ≤ C} by Theorem 3.8. Since there are only finitely many such cones up to translation, Q is an intersection of finitely many half-spaces and is thus a convex polyhedron of the desired form.

37 CHAPTER III. GLUING HYPERCONVEX METRIC SPACES

We now go back to the study of gluings.

n m Lemma 3.10. Let X1 = l∞ and X2 = l∞, and let V be a linear subspace of both X1 and X2. Consider X = X1 tV X2. Then, for x ∈ X1 and r ≥ 0, we have [ 2 B(x, r) ∩ X2 = B (y, r − d(x, y)). y∈B1(x,r)∩V

Moreover, B(x, r) ∩ X2 is connected.

0 Proof. Since V is proper, for any x ∈ X1 and x ∈ X2, there is some y ∈ V with d(x, x0) = d(x, y) + d(y, x0). Since B1(x, r) ∩ V is connected, this is also true for B(x, r) ∩ X2. If V is any finite dimensional real vector space and Q is any convex non- empty subset of V , we say that e ∈ Q is an extreme point of Q, and we write e ∈ ext(Q), if for any c, c0 ∈ Q and t ∈ [0, 1] such that e = (1 − t)c + tc0, it follows that t ∈ {0, 1}. Finally, by a convex polytope in V , we mean a bounded intersection of finitely many closed half-spaces or equivalently the convex hull of finitely many points. A compact convex set in V is a convex polytope if and only if it has finitely many extreme points, cf. [Gru02, Section 3.1].

n m Lemma 3.11. Let X1 = l∞ and X2 = l∞. Moreover, let V be a linear subspace of both X1 and X2 such that V 6= X1,X2 and

X = X1 tV X2 is hyperconvex. Then, for x ∈ X1 and r > d(x, V ), the set B(x, r) ∩ X2 is a cuboid.

Proof. We ﬁrst show that Q := B(x, r) ∩ X2 is convex. Let p, q ∈ Q. Then there arep, ¯ q¯ ∈ V such that

d(x, p¯) + d(¯p, p) = d(x, p) and d(x, q¯) + d(¯q, q) = d(x, q). (3.4)

Consider now any point z := p + t(q − p) ∈ p + [0, 1](q − p) ⊂ X2 and let z¯ :=p ¯ + t(¯q − p¯) ∈ V . On the one hand,

d(z, z¯) ≤ (1 − t)d(p, p¯) + td(q, q¯) and on the other hand,

d(x, z¯) ≤ (1 − t)d(x, p¯) + td(x, q¯).

Summing the above two inequalities, applying the triangle inequality and using (3.4), we obtain d(x, z) ≤ (1 − t)d(x, p) + td(x, q) ≤ r. It follows that z ∈ Q and shows that Q is convex. Since X is hyperconvex, for every y ∈ X2 with d(x, y) ≤ r + d(y, V ), the set

B(y, d(y, V )) ∩ B(x, r) ∩ X2 = B(y, d(y, V )) ∩ B(x, r)

38 III.3. HYPERCONVEX LINEAR SPACES

is externally hyperconvex in X2 and is therefore a cuboid by Proposition 3.1. Hence B(x, r) ∩ X2 \ V is locally a cuboid. To see that Q is a convex polytope, we show that Q has only finitely many extreme points. Observe that Q has only finitely many extreme points inside V since Q ∩ V = B1(x, r) ∩ V and the right-hand side is a convex polytope contained in V . Now, note that, since Q is locally a cuboid outside of V , Q has only finitely many different tangent cones at points outside of V , and this up to translation (of such cones). It is easy to see that Q cannot have both a cone and a nontrivial translate of this cone as tangent cones. It follows that Q can only have finitely many different tangent cones at points outside V . In particular, Q has only finitely many extreme points and it is thus a convex polytope. Furthermore, V meets the interior of Q and V has codimension at least one in X2, hence V cannot contain any facet of Q. It follows that Q can be written as the intersection of tangent cones to Q at points in Q \ V . Once again, since Q is locally a cuboid outside of V , it follows that Q is a cuboid.

n m Lemma 3.12. Let X1 = l∞ and X2 = l∞. Moreover, let V be a linear subspace of both X1 and X2 such that V 6= X1,X2 and

X = X1 tV X2 is hyperconvex. Then V must be weakly externally hyperconvex in X.

Proof. Note that it is enough to show that V is weakly externally hyperconvex in X1. First we introduce coordinates:

( k ) m X V = x ∈ l∞ : for every l ∈ {k + 1, . . . , n}, one has xl = cl,ixi . i=1

Pk As V must be hyperconvex by Lemma 1.2, we may assume that i=1 |cl,i| ≤ 1 for all l by Theorem 3.5. For x ∈ X1 and r > d(x, V ), we can write

m Y −1 1 A := B(x, r) ∩ X2 = [ai , ai ] i=1 by Lemma 3.11. Let ν be a vertex of [−1, 1]m, that is ν ∈ {−1, 1}m, and ν ν1 νm a = (a1 , . . . , am ) the corresponding vertex of A. By Lemma 3.10, for every aν there must be somea ¯ν ∈ V such that d(aν, a¯ν) = r − d(x, a¯ν). Since aν is a ν ν vertex of the cuboid A,a ¯ must lie on the diagonal a + Rν or more precisely

ν ν a¯ = a − tνν (3.5)

ν ν for tν = d(a , a¯ ) ≥ 0. Hence, for every l ∈ {k + 1, . . . , m}, one has

k k k ν X νi X νi X a¯l = cl,ia¯i = cl,iai − tν cl,iνi. (3.6) i=1 i=1 i=1

39 CHAPTER III. GLUING HYPERCONVEX METRIC SPACES

Fix now some l ∈ {k + 1, . . . , m}. Then, for every ν ∈ {−1, 1}m, consider − + m + − + − ν , ν ∈ {−1, 1} with νl = 1, νl = −1 and νi = νi = νi for i 6= l. Clearly, ν coincides with either ν+ or ν−. From (3.5), we get

+ − + − ν ν νl + νl − 1 −1 a¯l − a¯l = (al − tν+ νl ) − (al − tν− νl ) = al − al − tν+ − tν− .

Observe that, for ν+, ν−, on the right-hand side of (3.6) all parameters except Pk tν+ , tν− are equal. Hence, setting c := i=1 cl,iνi ∈ [−1, 1], we get

ν+ ν− a¯l − a¯l = c(tν+ − tν− ) ≤ |tν+ − tν− |.

Let us assume that tν+ ≥ tν− (the other case is analogous). We then get

+1 −1 al − al − tν+ − tν− ≤ tν+ − tν− ,

+1 −1 or equivalently 2tν+ ≥ al − al . But this inequality must be an equality since 2 ν+ ν+ ν− 2 ν+ B (¯a , tν+ ) ⊂ A. Moreover, we have a , a ∈ B (¯a , tν+ ). We conclude thata ¯ν can always be chosen such that 1 d(aν, a¯ν) = t = a+1 − a−1 . ν 2 l l

ν 1 +1 −1 Especially, we havea ¯l = 2 al + al . Hence, all vertices of A and therefore, by convexity, all points a ∈ A are contained in a ball B2(¯a, t) witha ¯ ∈ B1(x, r)∩ V and maximal radius 1 t = r − d(x, a¯) = a+1 − a−1 . 2 l l Note that, for y ∈ B1(x, r) ∩ V , the radius r − d(x, y) is maximal if and only if d(x, y) = d(x, V ) and thereforea ¯ ∈ B(x, d(x, V )) ∩ V . We now use Lemma II.3.1 to conclude that V is weakly externally hyperconvex in X1. First, we have

\ 1 B(x, d(x, V )) ∩ V = B(x, d(x, V ) + n ) ∩ X2 ∈ E(X2) n∈N using Lemma 3.11 and therefore B(x, d(x, V )) ∩ V ∈ E(V ). Furthermore, for y ∈ V , assume that r = d(x, y) > d(x, V ). Then, by the above, there is some a ∈ V ∩ B(x, d(x, V )) such that

r = d(x, y) ≤ d(x, a) + d(a, y) ≤ d(x, V ) + r − d(x, V ) = r, i.e. d(x, y) = d(x, a) + d(a, y).

n m Theorem 3.13. Let X1 = l∞ and X2 = l∞. Moreover, let V be a linear subspace of both X1 and X2 such that V 6= X1,X2. Then X = X1tV X2 is k 0 k 0 hyperconvex if and only if there is some k such that X1 = l∞×X1, X2 = l∞×X2, k 0 0 0 n−k 0 m−k V = l∞ × V , and V is strongly convex in both X1 = l∞ and X2 = l∞ .

40 III.3. HYPERCONVEX LINEAR SPACES

Proof. If X is hyperconvex, by Lemma 3.12, the subspace V is weakly externally hyperconvex in X1 and in X2, we can thus apply Theorem 3.8 and write

k1 k1 µ0 µ1 µp V = l∞ × S0 × S1 × ... × Sp ⊂ l∞ × l∞ × l∞ × ... × l∞ = X1,

µi µi where for all i ∈ {0, . . . , p}, the set Si ( l∞ is strongly convex in l∞ and for i 6= 0, Si is one dimensional (for i = 0 we also might have S0 = {0}) and similarly

k2 k2 ν0 ν1 νq V = l∞ × T0 × T1 × ... × Tq ⊂ l∞ × l∞ × l∞ × ... × l∞ = X2.

Let {e1, . . . , en} and {f1, . . . , fm} be the standard basis for X1 and X2, respectively. First, observe that, if Rei ⊂ V , then there is some fj with Rei = Rfj ⊂ V . Indeed, if Rei ⊂ V for every r ≥ 0, we have \ [−r, r]ei = B(re, r) ∈ E(X). i−1 n−i e∈{−1,1} ×{0}×{−1,1} ⊂X1

Note that, as [−r, r]ei ∈ E(X2), it is a cuboid in X2 and therefore we get [−r, r]ei = [−r, r]fj. k Hence, we have k = k1 = k2 and can split oﬀ the maximal component l∞ k 0 k 0 from X1, X2 and V , that is, we can write V = l∞ × V ⊂ l∞ × Xi = Xi with

0 µ0 µ1 µp 0 V = S0 × S1 × ... × Sp ⊂ l∞ × l∞ × ... × l∞ = X1 as well as

0 ν0 ν1 νq 0 V = T0 × T1 × ... × Tq ⊂ l∞ × l∞ × ... × l∞ = X2.

Assume now that there are at least two non-trivial factors in the ﬁrst de- µi µi composition, i.e. p ≥ 1. Since, for every factor l∞, there is some xi ∈ l∞ with 0 d(xi,Si) = 1, there is some x = (x0, x1, . . . , xp−1, 0) ∈ X1 with

0 0 0 p Sp B(x, d(x, V )) ∩ V = B(x, 1) ∩ V = {0} × Sp ∩ B (0, 1) ,

Sp where B (0, 1) denotes the unit ball inside Sp. Sp 0 0 But {0} × (Sp ∩ B (0, 1)) ⊂ X2 is not externally hyperconvex in X2 since 0 0 0 it is not a cuboid. It follows that B(x, d(x, V )) ∩ V ∈/ E(X2) which is a contradiction to Proposition 1.6. Therefore, V must consist of only one strongly convex factor. k 0 0 For the other direction, note that X = l∞ × (X1 tV 0 X2). The second factor is hyperconvex by Corollary 1.9 since V 0 is strongly convex. Hence also the product is hyperconvex by Lemma II.3.18.

Note that Theorem 3.13 further strengthens the weak optimality of the assumption in Theorem 1.8 that the gluing set is weakly externally hyperconvex.

Chapter IV

Local to Global

IV.1 The Cartan-Hadamard Theorem for Metric Spaces with Local Geodesic Bicombings

Let us ﬁrst ﬁx some notation. In a metric space X, we denote by

U(x, r) := {y ∈ X : d(x, y) < r} the open ball of radius r around x ∈ X and by

B(x, r) := {y ∈ X : d(x, y) ≤ r} the closed one. Let X be a metric space and γ : [0, 1] → X a continuous curve. The length of γ is given by

( n ) X L(γ) := sup d(γ(tk−1), γ(tk)) : 0 = t0 < . . . < tn = 1 . k=1 Then d¯(x, y) := inf {L(γ): γ : [0, 1] → X, γ(0) = x, γ(1) = y} deﬁnes a metric on X, called the induced length metric. If we have d = d¯, we say that (X, d) is a length space. For a metric space X, let G(X) := {c: [0, 1] → X} be the set of all geodesics in X, i.e. continuous maps c: [0, 1] → X with

d(c(s), c(t)) = |s − t| · d(c(0), c(1)) for all s, t ∈ [0, 1]. We equip G(X) with the metric

D(c, c0) := sup d(c(t), c0(t)). t∈[0,1]

Let c ∈ G(X) and 0 ≤ a ≤ b ≤ 1, then we denote by c[a,b] the reparametrized geodesic given by c[a,b] : [0, 1] → X with

c[a,b](t) := c((1 − t)a + tb).

43 CHAPTER IV. LOCAL TO GLOBAL

For a metric space X, a geodesic bicombing is a map σ : X × X × [0, 1] → X that selects, for every pair of points (x, y) ∈ X × X, a geodesic σxy := σ(x, y, ·) from x to y. If this selection satisﬁes the inequality

0 0 d(σxy(t), σx0y0 (t)) ≤ (1 − t)d(x, x ) + td(y, y ) for all x, y, x0y0 ∈ X and all t ∈ [0, 1], we call it a conical geodesic bicombing. Note that this does not imply that the map t 7→ d(σxy(t), σx0y0 (t)) is convex. If this is the case, we say that the geodesic bicombing is convex. A suﬃcient (but not necessary) condition for a conical geodesic bicombing to be convex is that the selection of geodesics is consistent. This is

σσxy(s1)σxy(s2)(t) = σxy((1 − t)s1 + ts2) for x, y ∈ X, 0 ≤ s1 ≤ s2 ≤ 1 and t ∈ [0, 1]. Finally, we call a geodesic bicombing reversible if σyx(t) = σxy(1 − t) for all x, y ∈ X and t ∈ [0, 1]. Deﬁnition 1.1. A local geodesic bicombing on a metric space X is a local selection of geodesics, i.e. a map σ : U × [0, 1] → X,(y, z, t) 7→ σyz(t), with the following properties:

(i) For all x ∈ X, there is some rx > 0 such that, for all y, z ∈ U(x, rx), there is a geodesic σyz : [0, 1] → U(x, rx) from y to z, and

U = {(y, z) ∈ X × X : y, z ∈ U(x, rx) for some x}.

(ii) The selection is consistent with taking subsegments of geodesics, i.e.

σσyz(s1)σyz(s2)(t) = σyz((1 − t)s1 + ts2)

for (y, z) ∈ U, 0 ≤ s1 ≤ s2 ≤ 1 and t ∈ [0, 1]. We call a local geodesic bicombing σ convex if it is locally convex, i.e. for 0 0 y, z, y , z ∈ U(x, rx), it holds that

t 7→ d(σyz(t), σy0z0 (t)) is a convex function. Furthermore, σ is reversible if

σzy(t) = σyz(1 − t) for all (y, z) ∈ U and t ∈ [0, 1]. Remark. Observe that, by consistency, a local geodesic bicombing is convex if and only if 0 0 d(σyz(t), σy0z0 (t)) ≤ (1 − t)d(y, y ) + td(z, z ) 0 0 for all y, z, y , z ∈ U(x, rx) and t ∈ [0, 1]. A (local) geodesic c: [0, 1] → X is consistent with the local geodesic bicombing σ if

c[s1,s2](t) = σc(s1)c(s2)(t) for all 0 ≤ s1 ≤ s2 ≤ 1 with (c(s1), c(s2)) ∈ U.

44 IV.1. CARTAN-HADAMARD THEOREM

The goal of this section is now to prove the following generalization of the Cartan-Hadamard Theorem for metric spaces with a local geodesic bicombing.

Theorem 1.2. Let X be a complete, simply-connected metric space with a convex local geodesic bicombing σ. Then the induced length metric on X admits a unique consistent, convex geodesic bicombing σ˜ which is consistent with σ. As a consequence, X is contractible. Moreover, if the local geodesic bicombing σ is reversible, then σ˜ is reversible as well.

To prove this, we roughly follow the structure of Chapter II.4 in [BH99]. Adapting the methods of S. Alexander and R. Bishop [AB90], we can prove the following key lemma.

Lemma 1.3. Let X be a complete metric space with a convex local geodesic bicombing σ and let c be a local geodesic from x to y which is consistent with σ. Then there is some > 0 such that, for all x,¯ y¯ ∈ X with d(x, x¯), d(y, y¯) < , there is a unique local geodesic c¯ from x¯ to y¯ with D(c, c¯) < which is consistent with σ. Moreover, we have

L(¯c) ≤ L(c) + d(x, x¯) + d(y, y¯) and if c˜ is another consistent geodesic from x˜ to y˜ with D(c, c˜) < , then

t 7→ d(˜c(t), c¯(t)) is convex.

Proof. Let > 0 be such that σ U(c(t),2)×U(c(t),2)×[0,1] is a convex geodesic bicombing for all t ∈ [0, 1]. Now, let P(A) be the following statement:

P(A): For all a, b ∈ [0, 1] with 0 ≤ b − a ≤ A and for all p, q ∈ X with d(c(a), p), d(c(b), q) < , there is a unique local geodesicc ¯pq : [0, 1] → X from p to q with D(c[a,b], c¯pq) < which is consistent with σ. Moreover, for all such local geodesics the map t 7→ d(¯cpq(t), c¯p0q0 (t)) is convex.

3A By our assumption, P( l(c) ) holds. Therefore, let us show P(A) ⇒ P( 2 ). 3A 2 1 Given a, b ∈ [0, 1] with 0 ≤ b − a ≤ 2 , we deﬁne p0 := c( 3 a + 3 b) and 1 2 q0 := c( 3 a + 3 b). Then, by P(A), there are consistent local geodesics c1 from 0 1 0 1 p to q0 and c1 from p0 to q. Inductively, we set pn := cn( 2 ) and qn := cn( 2 ), 0 where cn is a consistent local geodesic from p to qn−1 and cn from pn−1 to q. 0 Observe that, by convexity of the cn, cn, we have d(pn−1, pn), d(qn−1, qn) < 2n and hence the sequences (pn)n∈N and (qn)n∈N converge to some p∞ and q∞, respectively, and we have d(p∞, p0), d(q∞, q0) < . Furthermore, by convexity, 0 0 the cn, cn converge to the consistent local geodesics c∞ from p to q∞ and c∞ 1 0 1 from p∞ to q, which coincide between p∞ = c∞( 2 ) and q∞ = c∞( 2 ). Hence, they deﬁne a new local geodesic cpq from p to q which is consistent with σ and 1 2 p∞ = cpq( 3 ), q∞ = cpq( 3 ).

45 CHAPTER IV. LOCAL TO GLOBAL

Now, given two local geodesics cpq and cp0q0 with D(c[a,b], cpq) < and 0 0 0 1 1 D(c[a,b], cp0q0 ) < , set s := d(p, p ), t := d(q, q ), s := d(cpq( 3 ), cp0q0 ( 3 )) and 0 2 2 0 s+t0 0 s0+t t := d(cpq( 3 ), cp0q0 ( 3 )). Then we have s ≤ 2 , t ≤ 2 and therefore, we get

0 s s0 t 0 2s+t s ≤ 2 + 4 + 4 ⇒ s ≤ 3 ,

0 s+2t and similarly, t ≤ 3 follows. Hence, we get convexity of t 7→ d(cpq(t), cp0q0 (t)) and therefore also uniqueness follows. It remains to prove that L(¯c) ≤ L(c) + d(x, x¯) + d(y, y¯). Letc ˜ be the unique consistent local geodesic from x toy ¯ with D(c, c˜) < . For t small enough, we have

tL(˜c) = d(˜c(0), c˜(t)) = d(c(0), c˜(t)) ≤ d(c(0), c(t)) + d(c(t), c˜(t)) ≤ tL(c) + td(c(1), c˜(1)), i.e. L(˜c) ≤ L(c) + d(y, y¯) and similarly L(¯c) ≤ L(˜c) + d(x, x¯).

Definition 1.4. Let X be some metric space with a local geodesic bicombing σ. For some fixed x0 ∈ X, we define ˜ Xx0 := {c: [0, 1] → X local geodesic with c(0) = x0, consistent with σ}. ˜ We equip Xx0 with the metric

D(c, c0) = sup d(c(t), c0(t)) t∈[0,1] and deﬁne the map ˜ exp: Xx0 → X, c 7→ c(1). If X is complete, this map has the following properties.

Lemma 1.5. Let X be a complete metric space with a convex local geodesic bicombing σ. Then the following holds: ˜ (i) The map exp: Xx0 → X is locally an isometry. Hence, σ naturally induces ˜ a convex local geodesic bicombing σ˜ on Xx0 . ˜ (ii) Xx0 is contractible. ˜ (iii) For each x˜ ∈ Xx0 , there is a unique local geodesic from x˜0 to x˜ which is consistent with σ˜, where x˜0 is the constant path x˜0(t) = x0. ˜ (iv) Xx0 is complete. ˜ Proof. By Lemma 1.3, for every c ∈ Xx0 , there is some > 0 such that the map

exp U(c,) : U(c, ) → U(c(1), ) is an isometry. Hence, σ naturally induces a convex local geodesic bicombingσ ˜ ˜ on Xx0 .

46 IV.1. CARTAN-HADAMARD THEOREM

˜ ˜ Consider the map r : Xx0 × [0, 1] → Xx0 ,(c, s) 7→ (rs(c): t 7→ c(st)). This ˜ deﬁnes a deformation retraction of Xx0 ontox ˜0. ˜ ˜ A continuous pathc ˜: [0, 1] → Xx0 is a local geodesic in Xx0 which is consistent withσ ˜ if and only if exp ◦c˜ is a local geodesic in X which is consistent with ˜ σ. Therefore, for any c ∈ Xx0 , the map s 7→ rs(c) is the unique local geodesic fromx ˜0 to c. ˜ Finally, if (cn)n∈N is a Cauchy sequence in X, by completeness of X, for every t ∈ [0, 1], the sequences (cn(t))n∈N converge in X, to c(t) say. Locally, i.e. inside U(c(t), rc(t)), the subsegment c|[t−,t+] is the limit of the consistent geodesics (cn|[t−,t+])n∈N and hence c is consistent with σ by the convexity of the local geodesic bicombing.

The following criterion will ensure that exp is a covering map.

Lemma 1.6. Let p: X˜ → X be a map of length spaces such that

(i) X is connected,

(ii) p is a local homeomorphism,

(iii) for all rectiﬁable curves c˜: [0, 1] → X˜, we have L(˜c) ≤ L(p ◦ c˜),

(iv) X has a convex local bicombing σ, and

(v) X˜ is complete.

Then p is a covering map.

Proof. The proof of Proposition I.3.28 in [BH99] also works in our setting. In the second step, take U = U(x, rx) and deﬁne the maps sx˜ : U(x, rx) → X˜ by sx˜(y) =σ ˜xy(1), whereσ ˜xy is the unique lift of σxy withσ ˜xy(0) =x ˜.

Remark. For a local isometry p, conditions (ii) and (iii) are satisﬁed.

Corollary 1.7. Let (X, d) be a complete, connected metric space with a convex ˜ local geodesic bicombing σ. Then exp: Xx0 → X is a universal covering map. ¯ ¯ ˜ Proof. Consider the induced length metrics d and D on X and Xx0 . Since (X, d) locally is a length space, the metrics d and D locally coincide with d¯ and D¯, respectively. Hence p still is a local isometry with respect to the length metrics and σ is a convex local bicombing. Thus Lemma 1.6 applies.

Proof of Theorem 1.2. First, we show that for all x, y ∈ X, there is a unique consistent local geodesic from x to y. Since X is simply-connected, the covering map exp: X˜x → X is a homeomorphism which is a local isometry and, by Lem- ma 1.5, there is a unique consistent local geodesicσ ˜xy from x to y. Next, we prove thatσ ˜xy is a geodesic. To do so, it is enough to show that, for every curve γ : [0, 1] → X and every t ∈ [0, 1], we have

L(˜σγ(0)γ(t)) ≤ L(γ|[0,t]).

47 CHAPTER IV. LOCAL TO GLOBAL

Let n o A := s ∈ [0, 1] : ∀t ∈ [0, s] we have L(˜σγ(0)γ(t)) ≤ L(γ|[0,t]) .

Clearly, A is non-empty and closed. To prove that A is open, consider s ∈ A. For δ > 0 small enough, by Lemma 1.3, we have

L(˜σγ(0)γ(s+δ)) ≤ L(˜σγ(0)γ(s)) + d(γ(s), γ(s + δ))

≤ L(γ|[0,s]) + L(γ|[s,s+δ]) = L(γ|[0,s+δ]).

Hence, A = [0, 1] as desired.

Finally, we show that t 7→ d(˜σxy(t), σ˜x¯y¯(t)) is convex. By Lemma 1.3, there is a sequence 0 = t1 < . . . < tn = 1 and k > 0 such that

• the balls U(˜σxx¯(t1), 1),...,U(˜σxx¯(tn), n) coverσ ˜xx¯,

• the balls U(˜σyy¯(t1), 1),...,U(˜σyy¯(tn), n) coverσ ˜yy¯, and

• for all p, p¯ ∈ U(˜σxx¯(tk), k) and for all q, q¯ ∈ U(˜σyy¯(tk), k) the map

t 7→ d(˜σpq(t), σ˜p¯q¯(t))

is convex.

Consider now a sequence 0 = s0 < s1 < . . . < sn = 1 with

σ˜xx¯(sk) ∈ U(˜σxx¯(tk), k) ∩ U(˜σxx¯(tk+1), k+1),

σ˜yy¯(sk) ∈ U(˜σyy¯(tk), k) ∩ U(˜σyy¯(tk+1), k+1), for k = 1, . . . , n − 1. Then we get d(˜σxy(t), σ˜x¯y¯(t)) n X ≤ d(˜σσ˜xx¯(sk−1)˜σyy¯(sk−1)(t), σ˜σ˜xx¯(sk)˜σyy¯(sk)(t)) k=1 n ! n ! X X ≤ (1 − t) d(˜σxx¯(sk−1), σ˜xx¯(sk)) + t d(˜σyy¯(sk−1), σ˜yy¯(sk)) k=1 k=1 = (1 − t)d(x, x¯) + td(y, y¯).

Hence,σ ˜ is a convex geodesic bicombing on X. ∗ If σ is reversible, thenσ ˜xy(t) :=σ ˜yx(1 − t) also deﬁnes a convex geodesic bicombing on X which is consistent with σ. Therefore, by uniqueness,σ ˜∗ and σ˜ coincide, i.e.σ ˜ is reversible.

48 IV.2. LOCALLY HYPERCONVEX METRIC SPACES

IV.2 Locally Hyperconvex Metric Spaces

Hyperconvex metric spaces are an important example of metric spaces with a convex geodesic bicombing. In fact, D. Descombes and U. Lang showed that every proper, hyperconvex metric space of ﬁnite combinatorial dimension admits a unique convex geodesic bicombing, which is consistent and reversible, see [DL15]. Such spaces occur, for instance, as injective hulls of hyperbolic groups [Lan13, Theorem 1.4] and therefore, every hyperbolic group acts prop- erly and cocompactly by isometries on a space with a reversible, consistent, convex geodesic bicombing [DL15, Theorem 1.3]. Recall that every hyperconvex metric space is complete, geodesic and contractible. Now, knowing that under the above conditions hyperconvex metric spaces possess a consistent, convex geodesic bicombing, we deduce the following local-to-global theorem for hyperconvex metric spaces.

Theorem 2.1. Let X be a complete, locally compact, simply-connected, locally hyperconvex length space with locally ﬁnite combinatorial dimension. Then X is a hyperconvex metric space.

A metric space X is locally hyperconvex if, for every x ∈ X, there is some rx > 0 such that B(x, rx) is hyperconvex. If we can choose rx = r for all x ∈ X, we call X uniformly locally hyperconvex.

Lemma 2.2. Let X be a metric space with the property that every closed ball B(x, r) is hyperconvex, then X is itself hyperconvex.

Proof. Let {B(xi, ri)}i∈I be a family of closed balls with d(xi, xj) ≤ ri + rj.

Fix some i0 ∈ I and set Ai := B(xi, ri) ∩ B(xi0 , ri0 ). Since, for r big enough, we have xi, xj ∈ B(xi0 , r), we get that the Ai’s are externally hyperconvex in

Ai0 and Ai ∩ Aj 6= ∅ for all i, j ∈ I. Hence, it follows \ \ B(xi, ri) = Ai 6= ∅ i∈I i∈I by Proposition II.2.1.

Proposition 2.3. Let X be a uniformly locally hyperconvex metric space with a reversible, conical geodesic bicombing σ. Then X is hyperconvex.

Proof. Consider the following property:

P(R): For every family {B(xi, ri)}i∈I with d(xi, xj) ≤ ri + rj and ri ≤ R, there T is some x ∈ i∈I B(xi, ri).

Since X is uniformly locally hyperconvex, this clearly holds for some R0 > 0. Next, we show P(R) ⇒ P(2R) and therefore P(R) holds for any R ≥ 0. Let {B(xi, ri)}i∈I be a family of closed balls with d(xi, xj) ≤ ri + rj and 1 ri ≤ 2R. For i, j ∈ I, deﬁne yij := σxixj ( 2 ). By convexity of σ, we have

1 1 1 rj rk d(yij, yik) = d(σxixj ( 2 ), σxixk ( 2 )) ≤ 2 d(xj, xk) ≤ 2 + 2 .

49 CHAPTER IV. LOCAL TO GLOBAL

Hence, for every i ∈ I, there is some

\ rj zi ∈ B(yij, 2 ). j∈I

Now, observe that

rj ri ri rj d(zi, zj) ≤ d(zi, yij) + d(yij, zj) ≤ 2 + 2 = 2 + 2 and therefore, we ﬁnd

\ ri \ x ∈ B(zi, 2 ) ⊂ B(xi, ri). i∈I i∈I

Since all balls with center in B(x, r) and radius larger than 2r contain B(x, r), P(R) for R = 2r implies that B(x, r) is hyperconvex. Hence, by Lem- ma 2.2, the metric space X is hyperconvex.

Since compact, locally hyperconvex metric spaces are always uniformly locally hyperconvex, we conclude the following.

Corollary 2.4. Let X be a compact, locally hyperconvex metric space with a reversible, conical geodesic bicombing σ. Then X is hyperconvex.

Corollary 2.5. Let X be a proper, locally hyperconvex metric space with a reversible, conical geodesic bicombing σ. Then X is hyperconvex.

Proof. Let {B(xi, ri)}i∈I be a family of balls with d(xi, xj) ≤ ri + rj. Fix some i0 ∈ I and deﬁne In := {i ∈ I : d(xi, xi0 ) ≤ n} for n ∈ N. Since B(xi0 , n) is compact, by the previous corollary, there is some \ yn ∈ B(xi, ri).

i∈In

Especially, (yn)n∈N ⊂ B(xi0 , ri0 ) and hence, there is some converging subsequence \ ynk → y ∈ B(xi, ri). i∈I

Remark. In [Lan13], U. Lang proves that every hyperconvex metric space admits a reversible, conical geodesic bicombing (Proposition 3.8). Therefore, we get the following equivalence statement: A metric space is hyperconvex if and only if it is uniformly locally hyperconvex and admits a reversible, conical geodesic bicombing. If a hyperconvex metric space X is proper, it also admits a convex geodesic bicombing [DL15, Theorem 1.1] and if X has ﬁnite combinatorial dimension, this convex geodesic bicombing is consistent, reversible and unique [DL15, The- orem 1.2]. The combinatorial dimension of a metric space X was introduced by

50 IV.3. ABSOLUTE 1-LIPSCHITZ NEIGHBORHOOD RETRACTS

A. Dress [Dre84] and is given by the supremum of the dimensions of the polyhedral complexes E(Y ) that arise as the tight span of finite subsets Y ⊂ X. For instance, every finite dimensional hyperconvex metric space also has finite combinatorial dimension as E(Y ) embeds isometrically into X. Proposition 2.6. Every proper, hyperconvex metric space with finite combinatorial dimension admits a unique reversible, convex geodesic bicombing. Recall that, by the Hopf-Rinow Theorem, any complete, locally compact length space is proper. Corollary 2.7. Let X be a locally compact, locally hyperconvex metric space with locally finite combinatorial dimension. Then X admits a reversible, convex local geodesic bicombing.

Proof. For every x ∈ X, there is some rx > 0 such that B(x, 3rx) is compact, hyperconvex and has ﬁnite combinatorial dimension. This also holds for x B(x, rx) and therefore, there is a reversible, convex geodesic bicombing σ on B(x, rx), i.e. x σ : B(x, rx) × B(x, rx) × [0, 1] → B(x, rx).

We will check that for B(x, rx) and B(y, ry) with B(x, rx)∩B(y, ry) 6= ∅ the two geodesic bicombings σx, σy coincide on the intersection. Assume without loss of generality that rx ≥ ry and hence B(x, rx),B(y, ry) ⊂ B(x, 3rx). Then the convex geodesic bicombing τ on B(x, 3rx) restricts to both B(x, rx) and B(y, ry) since, for p, q ∈ B(z, rz), we have

d(z, τpq(t)) ≤ (1 − t)d(z, p) + td(z, q) ≤ rz. Hence, by uniqueness, the geodesic bicombings σx, σy are both restrictions of τ and thus coincide on B(x, rx) ∩ B(y, ry). Therefore σ, deﬁned by

x σ|B(x,rx)×B(x,rx) := σ |B(x,rx)×B(x,rx), is a reversible, convex local geodesic bicombing on X. Proof of Theorem 2.1. Let X be a complete, locally compact, simply-connected, locally hyperconvex length space with locally ﬁnite combinatorial dimension. By Corollary 2.7, X has a reversible, convex local geodesic bicombing, which induces a reversible, convex geodesic bicombing by Theorem 1.2. Hence, we can apply Corollary 2.5 and deduce that X is hyperconvex.

IV.3 Absolute 1-Lipschitz Neighborhood Retracts

It is well known that hyperconvex metric spaces are the same as absolute 1-Lip- schitz retracts. For Lipschitz retracts, the weaker notion of absolute Lipschitz uniform neighborhood retracts is common; e.g. see [HJ14]. Absolute 1-Lipschitz uniform neighborhood retracts are locally hyperconvex but the converse is not true as we can see in Example 3.3. In fact, it turns out that the following holds.

51 CHAPTER IV. LOCAL TO GLOBAL

Theorem 3.1. Let X be a locally compact absolute 1-Lipschitz uniform neighborhood retract with locally ﬁnite combinatorial dimension. Then X is an absolute 1-Lipschitz retract.

A metric space X is an absolute 1-Lipschitz neighborhood retract if, for every metric space Y with X ⊂ Y , there is some neighborhood U of X in Y and a 1-Lipschitz retraction ρ: U → X. Furthermore, if we can take U = U(X, ) for some > 0, we call X an absolute 1-Lipschitz uniform neighborhood retract. In this case, can be chosen independent of Y ; see [HJ14, Proposition 7.78].

Lemma 3.2. Let X be an absolute 1-Lipschitz (uniform) neighborhood retract. Then X is (uniformly) locally hyperconvex.

Proof. Consider X ⊂ l∞(X). Since X is an absolute 1-Lipschitz neighborhood retract, there is some neighborhood U of X and a 1-Lipschitz retraction ρ: U → X. For x ∈ X, there is some rx > 0 such that B(x, rx) ⊂ U. Let now {B(xi, ri)}i∈I be a family of closed balls with xi ∈ B(x, rx) ∩ X and d(xi, xj) ≤ ri + rj. Then, since l∞(X) is hyperconvex, there is some \ y ∈ B(x, rx) ∩ B(xi, ri) ⊂ U. i∈I

Hence, we have \ ρ(y) ∈ B(x, rx) ∩ B(xi, ri) ∩ X 6= ∅ i∈I and therefore B(x, rx) ∩ X is hyperconvex. If X is an absolute 1-Lipschitz uniform neighborhood retract, we have that U = U(X, ) for some > 0 and therefore, we can choose rx = 2 for all x ∈ X.

The converse is not true, as the following example shows.

Example 3.3. Consider the unit sphere S1 endowed with the inner metric. 1 π Since, for every x ∈ S and ∈ (0, 2 ], the ball B(x, ) is isometric to the interval [−, ], the unit sphere S1 is uniformly locally hyperconvex. But S1 is not an absolute 1-Lipschitz neighborhood retract. Fix some in- 1 1 1 clusion S ⊂ l∞(S ). We choose three points x, y, z ∈ S with

2π r := d(x, y) = d(x, z) = d(y, z) = 3 .

1 1 r Let U be a neighborhood of S in l∞(S ). As U is open, there is some ∈ (0, 2 ) 1 such that B(x, ) ⊂ U. By hyperconvexity of l∞(S ), there is some

p ∈ B(x, ) ∩ B(y, r − ) ∩ B(z, r − ) ⊂ U.

But since B(x, ) ∩ B(y, r − ) ∩ B(z, r − ) ∩ S1 = ∅, there is no 1-Lipschitz retraction ρ: S1 ∪ {p} → S1.

52 IV.3. ABSOLUTE 1-LIPSCHITZ NEIGHBORHOOD RETRACTS

In fact, the notion of an absolute 1-Lipschitz uniform neighborhood retract is quite restrictive.

Lemma 3.4. Let X be an absolute 1-Lipschitz uniform neighborhood retract. Then, X is

(i) complete,

(ii) geodesic, especially a length space, and

(iii) simply-connected.

Proof. Fix some inclusion X ⊂ l∞(X) and r = 2 > 0 such that there is a 1-Lipschitz retraction ρ: U(X, ) → X.

First, if (xn)n∈N is a Cauchy sequence in X it converges to some x ∈ U(X, ). It follows that x = ρ(x) ∈ X. Next, assume that there is a geodesic in X between points at distance less than d. By Lemma 3.2, this is clearly true for d = r. Consider two points x, y ∈ X with d(x, y) ≤ d + r. Now, since l∞(X) is geodesic, there is some z ∈ l∞(X) with d(x, y) = d(x, z)+d(z, y), d(x, z) ≤ r and d(z, y) ≤ d. But then we have ρ(z) ∈ X with d(x, y) = d(x, ρ(z)) + d(ρ(z), y) and, by our hypothesis, there are geodesics from x to ρ(z) and from ρ(z) to y, which combine to a geodesic from x to y. Finally, since X is locally simply-connected, every curve is homotopic to a curve of finite length and hence it is enough to consider loops of finite length. We show that every such loop in X is contractible. Let γ be a loop in X of length L(γ) = 2πR with R > r. We denote by 2 3 SR := {x ∈ R : |x| = R} the sphere of radius R endowed with the inner metric 2 r and let A := {x ∈ SR : 0 ≤ x3 ≤ R sin( R )} be the region bounded by the 2 0 2 r two circles c := {x ∈ SR : x3 = 0} and c := {x ∈ SR : x3 = R sin( R )}. Let f : c → X be a parametrization of γ by arclength and let f¯: A → l∞(X) be a 1-Lipschitz extension. Then γ0 := ρ ◦ f¯(c0) is a loop of length L(γ0) ≤ L(c0) = 0 0 r r r 2πR with R := R cos( R ), which is homotopic to γ. Since cos( R0 ) ≤ cos( R ), r n we find inductively a loop γn with L(γn) ≤ 2πR cos( R ) , which is homotopic to γ. If L(γ) = 2πR with R ≤ r, we can use the same argument with A replaced by the upper halfsphere of radius R to show that γ is contractible.

We conclude that an absolute 1-Lipschitz uniform neighborhood retract is a complete, simply-connected, locally hyperconvex length space and therefore Theorem 3.1 follows directly from Theorem 2.1.

Chapter V

Finite Hyperconvexity

V.1 Basic Properties

We denote the collection of all n-hyperconvex, externally n-hyperconvex and weakly externally n-hyperconvex subsets of X by Hn(X), En(X) and Wn(X), respectively. It holds that En(X) ⊂ Wn(X) ⊂ Hn(X) and Hn+1(X) ⊂ Hn(X). Remark. Obviously, every metric space is 1-hyperconvex. A subset A of X is (weakly) externally 1-hyperconvex if and only if it is proximinal, i.e. for every x ∈ X, there is some a ∈ A with d(x, a) = d(x, A). Recall that proximinal subsets are closed. Furthermore, a metric space X is 2-hyperconvex if and only if it is metrically convex, i.e. for every x0, x1 ∈ X and every t ∈ [0, 1], there is some xt ∈ X with d(x0, xt) = td(x0, x1) and d(xt, x1) = (1 − t)d(x0, x1), compare [BL00, Deﬁnition 1.3]. A metric space (X, d) is called modular if, for all x, y, z ∈ X, the median set M(x, y, z) := I(x, y) ∩ I(y, z) ∩ I(z, x) is non-empty, where I(x, y) := {z ∈ X : d(x, y) = d(x, z) + d(z, y)} denotes the metric interval between x and y. Proposition 1.1. A metric space X is 3-hyperconvex if and only if it is metrically convex and modular. Proof. First, assume that X is 3-hyperconvex. Recall the Gromov product 1 (y|z)x := 2 (d(x, y) + d(x, z) − d(y, z)) for x, y, z ∈ X. Let x1, x2, x3 ∈ X and deﬁne ri := (xj|xk)i for {i, j, k} = {1, 2, 3}. Then we have

3 \ M(x1, x2, x3) = B(xi, ri) 6= ∅. i=1

Conversely, let X be metrically convex and modular. Let x1, x2, x3 ∈ X and T3 r1, r2, r3 ∈ R with d(xi, xj) ≤ ri + rj. If M(x1, x2, x3) ∩ i=1 B(xi, ri) = ∅, we might assume without loss of generality that r3 < (x1|x2)x3 . We may then take m ∈ M(x1, x2, x3) and deﬁne rm := min{r1 − d(x1, m), r2 − d(x2, m)}. Since X T3 is metrically convex, we get ∅= 6 B(m, rm) ∩ B(x3, r3) ⊂ i=1 B(xi, ri).

55 CHAPTER V. FINITE HYPERCONVEXITY

V.2 Almost n-Hyperconvex Metric Spaces

We now turn our attention to almost n-hyperconvex metric spaces and eventually prove the following.

Theorem 2.1. Let X be a complete, almost n-hyperconvex metric space for n ≥ 3. Then X is n-hyperconvex.

A subset A of a metric space X is called almost externally n-hyperconvex in n X if, for every family {B(xi, ri)}i=1 of closed balls with d(xi, xj) ≤ ri + rj and d(xi,A) ≤ ri, we have n \ B(xi, ri + ) ∩ A 6= ∅ i=1 for every > 0. We denote the set of all closed, almost externally n-hyperconvex subsets of X by E˜n(X). Similarly, we say that a subset A of a metric space X is almost weakly externally n-hyperconvex in X if it is almost externally n-hyperconvex in A∪{x} for every x ∈ X. We denote all such subsets which are closed by W˜ n(X). N. Aronszajn and P. Panitchpakdi already showed that every complete, almost (n+1)-hyperconvex metric space is n-hyperconvex [AP56, Theorem 3.4]. This result easily extends to almost (weakly) externally hyperconvex subsets.

Lemma 2.2. Let X be a complete metric space. Then we have

(i) A ∈ E˜n+1(X) ⇒ A ∈ En(X),

(ii) A ∈ W˜ n+1(X) ⇒ A ∈ Wn(X), and

(iii) if X is almost (n + 1)-hyperconvex, then X is n-hyperconvex.

Proof. We will prove (i). The other statements then follow since

• A ∈ Wn(X) ⇔ A ∈ En(A ∪ {x}) for all x ∈ X, and

• A is n-hyperconvex ⇔ A ∈ En(A).

n Let {B(xi, ri)}i=1 be a family of closed balls with d(xi, xj) ≤ ri + rj and Tn 1 d(xi,A) ≤ ri. Starting with y1 ∈ i=1 B(xi, ri+ 2 )∩A, we construct inductively a sequence (yk)k by choosing

n 1 1 \ 1 yk+1 ∈ B(yk, 2k + 2k+1 ) ∩ B(xi, ri + 2k+1 ) ∩ A. i=1

1 1 Since d(yk, yk+1) ≤ 2k + 2k+1 , this sequence is Cauchy and converges to some y ∈ A with 1 d(xi, y) = lim d(xi, yk) ≤ lim ri + k = ri. k→∞ k→∞ 2 Tn That is y ∈ i=1 B(xi, ri) ∩ A 6= ∅.

56 V.2. ALMOST n-HYPERCONVEX METRIC SPACES

Lemma 2.3. Let X be a complete metric space, A ∈ E˜n(X) and k ≤ n − 2. k Then, for every family {B(xi, ri)}i=1 with d(xi, xj) ≤ ri + rj and d(xi,A) ≤ ri, 0 Tk ˜ we have A := A ∩ i=1 B(xi, ri) ∈ En−k(X).

n Proof. Let {B(xi, ri)}i=k+1 be a family of closed balls with d(xi, xj) ≤ ri + rj 0 Tn and d(xi,A ) ≤ ri. Fix some > 0. Then, there is some y ∈ A ∩ i=1 B(xi, ri + 2 ) and by Lemma 2.2, we get

k n \ 0 \ ∅= 6 A ∩ B(xi, ri) ∩ B(y, 2 ) ⊂ A ∩ B(xi, ri + ). i=1 i=k+1

Corollary 2.4. Let X be a complete, almost n-hyperconvex metric space and k let k ≤ n − 2. Then for every family {B(xi, ri)}i=1 with d(xi, xj) ≤ ri + rj, we Tk ˜ have B := i=1 B(xi, ri) ∈ En−k(X).

By investigating the proof of Lemma II.2.2, we see that the requirements there can be weakened as follows.

Lemma 2.5. Let X be a complete, almost 3-hyperconvex metric space. Let 0 0 0 A, A ∈ E˜2(X) with y ∈ A ∩ A 6= ∅ and x ∈ X with d(x, A), d(x, A ) ≤ r. Denote d := d(x, y) and s := d − r. Then, for every > 0 and every δ > 0, we have A ∩ A0 ∩ B(x, r + δ) ∩ B(y, s + ) 6= ∅, given s ≥ 0. In any case, the intersection A ∩ A0 ∩ B(x, r + δ) is non-empty.

Proof. We will split the proof into three steps.

Step I. For all > 0 and δ > 0, there are a ∈ Cδ := B(x, r + δ) ∩ A and 0 0 0 0 a ∈ Cδ := B(x, r + δ) ∩ A such that d(a, a ) ≤ and d(y, a) ≤ s + .

Let 0 ≤ ˜ ≤ , n := s and 0 < δ ≤ ˜ . We start by choosing 2 0 ˜ n0+1

δ δ a1 ∈ B(y, ˜+ 2 ) ∩ B(x, d − ˜+ 2 ) ∩ A, δ δ δ δ 0 a2 ∈ B(a1, ˜+ 2 + 4 ) ∩ B(x, d − 2˜ + 2 + 4 ) ∩ A .

Then, as long as n ≤ n0, we can inductively pick

an ∈ B(an−1, ˜+ ∆n) ∩ B(x, d − n˜+ ∆n) ∩ A if n is odd, and

0 an ∈ B(an−1, ˜+ ∆n) ∩ B(x, d − n˜+ ∆n) ∩ A

Pn δ if n is even, where ∆n := i=1 2i ≤ δ. Observe that we have d(y, an) ≤ n(˜+δ).

57 CHAPTER V. FINITE HYPERCONVEXITY

0 Finally, assuming without loss of generality an0 ∈ A (otherwise interchange A and A0), there are

a ∈ B(an0 , ˜+ ∆n0+1) ∩ B(x, r + ∆n0+1) ∩ A ⊂ Cδ and 0 0 0 a ∈ B(a, ˜+ ∆n0+2) ∩ B(x, r + ∆n0+2) ∩ A ⊂ Cδ. 0 We have d(a, a ) ≤ ˜+δ ≤ and d(a, y) ≤ (n0+1)(˜+δ) ≤ s+˜+(n0+1)δ ≤ s+.

0 0 Step II. For all > 0, δ > 0 and all n ≥ 1, there are an ∈ A and an ∈ A with 0 • d(an, an) ≤ 2n , 0 0 2+δ • d(an−1, an), d(an−1, an) ≤ 2n , 0 Pn δ • d(x, an), d(x, an) ≤ r + ∆n, where ∆n := i=1 2i , and 0 Pn • d(y, an), d(y, an) ≤ s + En, where En := i=1 2i .

We start by choosing δ a1 ∈ A ∩ B(x, r + 2 ), 0 0 δ a1 ∈ A ∩ B(x, r + 2 ), 0 with d(a1, a1) ≤ 2 and d(y, a1) ≤ s + 2 according to Step I. We then continue inductively as follows. First, since X is almost 3-hyperconvex, we can pick some δ 0 δ δ xn ∈ B(an−1, 2n + 2n+2 ) ∩ B(an−1, 2n + 2n+2 ) ∩ B(x, r + ∆n−1 − 2n + 2n+2 ). δ We denote rn := 2n + 2n+2 and sn := s + En−1. Observe that

d(xn, y) ≤ d(xn, an−1) + d(an−1, y) ≤ rn + sn. Now, by Step I, there are δ an ∈ A ∩ B(xn, rn + 2n+1 ), 0 0 δ an ∈ A ∩ B(xn, rn + 2n+1 ), 0 with d(an, an) ≤ 2n and d(y, an) ≤ sn + 2n = s + En. Moreover, we have δ 2+δ d(an−1, an) ≤ d(an−1, xn) + d(xn, an) ≤ 2rn + 2n+1 = 2n , d(x, an) ≤ d(x, xn) + d(xn, an) δ δ ≤ r + ∆n−1 − 2n + 2n+2 + rn + 2n+1 = r + ∆n, as desired.

0 Step III. Observe that the sequences (an)n∈N, (an)n∈N are Cauchy and since 0 the distance d(an, an) → 0, they have a common limit point a ∈ A ∩ A0 ∩ B(y, s + ) ∩ B(x, r + δ) 6= ∅. This concludes the proof.

58 V.2. ALMOST n-HYPERCONVEX METRIC SPACES

Lemma 2.6. Let X be a complete, almost 3-hyperconvex metric space and let A0 ∈ E˜3(X) and A1,A2 ∈ E˜2(X) be pairwise intersecting subsets. Then

A0 ∩ A1 ∩ A2 6= ∅.

Proof. Choose a point x0 ∈ A1 ∩ A2 and set r0 := d(x0,A0). By Lemma 2.5, there is r0 y0 ∈ A0 ∩ A1 ∩ B(x0, r0 + 12 ). 0 7 ˜ Let A0 := A0 ∩ B(y0, 6 r) ∈ E2(X). Using again Lemma 2.5, we have

0 13 r0 A0 ∩ A2 = A0 ∩ A2 ∩ B(y0, 12 r0 + 12 ) 6= ∅ and therefore, there is some

0 13 r0 7 7 z0 ∈ A0 ∩ A2 ∩ B(x0, 12 r0 + 12 ) = A0 ∩ A2 ∩ B(x0, 6 r0) ∩ B(y0, 6 r0).

Then, since A0 is almost externally 3-hyperconvex, there is some

r0 7 r0 7 r0 x¯0 ∈ B(x0, r0 + 12 ) ∩ B(y0, 12 r0 + 12 ) ∩ B(z0, 12 r0 + 12 ) ∩ A0 and using again Lemma 2.5, we ﬁnd

2 r0 5 r0 x1 ∈ A1 ∩ A2 ∩ B(¯x0, 3 r0 + 12 ) ∩ B(x0, 12 r0 + 12 ).

3 1 Note that d(x1,A0) ≤ d(x1, x¯0) ≤ 4 r0 =: r1 and d(x0, x1) ≤ 2 r0. Proceeding 3 n this way, we get some sequence (xn)n∈N ⊂ A1 ∩A2 with d(xn,A0) ≤ 4 r0 and 1 3 n d(xn, xn+1) ≤ 2 4 r0. Hence, (xn)n∈N is a Cauchy sequence and therefore converges to some x ∈ A0 ∩ A1 ∩ A2 6= ∅ since A0 is closed.

We are now able to give a proof of Theorem 2.1. In fact, we will show the following more general result.

Proposition 2.7. Let X be a complete metric space. Then, for every n ≥ 3, we have

(i) A ∈ E˜n(X) ⇒ A ∈ En(X),

(ii) A ∈ W˜ n(X) ⇒ A ∈ Wn(X), and

(iii) if X is an almost n-hyperconvex metric space, then X is n-hyperconvex.

Proof. First note that (ii) and (iii) directly follow from (i) by the same argument as in Lemma 2.2. To prove (i), we now consider two cases.

˜ 3 Case 1. Let A ∈ E3(X) and let {B(xi, ri)}i=1 be a family of closed balls with d(xi, xj) ≤ ri + rj. By Lemma 2.2, A is externally 2-hyperconvex and hence,

59 CHAPTER V. FINITE HYPERCONVEXITY

there is some y0 ∈ A ∩ B(x1, r1) ∩ B(x2, r2). Set s := d(y0, x3) − r3. By Lem- ma 2.3, we have Ai := A ∩ B(xi, ri) ∈ E˜2(X) and d(x3,Ai) ≤ r3 for i = 1, 2, since A∩B(xi, ri)∩B(x3, r3) 6= ∅. By Lemma 2.5, we therefore ﬁnd inductively

y1 ∈ A1 ∩ A2 ∩ B(x3, r3 + 2 ) ∩ B(y0, s + ), y2 ∈ A1 ∩ A2 ∩ B(x3, r3 + 4 ) ∩ B(y1, 2 + 2 ), . . yn ∈ A1 ∩ A2 ∩ B(x3, r3 + 2n ) ∩ B(yn−1, 2n−1 + 2n−1 ).

This is a Cauchy sequence which converges to some

3 \ y ∈ A1 ∩ A2 ∩ B(x3, r3) = A ∩ B(xi, ri) 6= ∅. i=1

n Case 2. Assume now that n ≥ 4 and let {B(xi, ri)}i=1 be a family of closed Tn balls with d(xi, xj) ≤ ri + rj, d(xi,A) ≤ ri. Then B := A ∩ j=3 B(xj, rj) ∈ E˜2(X) by Lemma 2.3. Moreover, for i = 1, 2, we have Ai := A ∩ B(xi, ri) ∈ ˜ Tn E3(X) by Lemma 2.3 and Ai ∩ B = A ∩ B(xi, ri) ∩ j=3 B(xj, rj) 6= ∅ by Lemma 2.2. Furthermore, observe that A1,A2,B ⊂ A and A is almost n- hyperconvex. Hence, by Lemma 2.6, we get

n \ A ∩ B(xi, ri) = A1 ∩ A2 ∩ B 6= ∅, i=1 as desired.

This proves Theorem 2.1 and as a consequence, we get the following corollary.

Corollary 2.8. Let X be an n-hyperconvex metric space for n ≥ 3. Then its metric completion is n-hyperconvex as well.

According to the previous propositions and the following lemma, in a complete, 4-hyperconvex metric space X, there is no diﬀerence between E˜n(X) and En(X) for n ≥ 2.

Lemma 2.9. Let X be a complete, 4-hyperconvex metric space. If A ∈ E˜2(X) then we have A ∈ E2(X).

2 Proof. Let {B(xi, ri)}i=1 with d(x1, x2) ≤ r1 + r2 and d(xi,A) ≤ ri. Then we have B(xi, ri) ∈ E˜3(X) by Lemma 2.3 and A ∩ B(xi, ri) 6= ∅ by Lemma 2.2. Therefore, we get B(x1, r1) ∩ B(x2, r2) ∩ A 6= ∅ by Lemma 2.6.

60 V.3. 4-HYPERCONVEX METRIC SPACES

V.3 4-Hyperconvex Metric Spaces

Let us now turn to 4-hyperconvex metric spaces. We will prove the following theorem step by step.

Theorem 3.1. Let X be a complete metric space and let A ⊂ X be an arbitrarily chosen non-empty subset. Then, the following hold:

(i) X is 4-hyperconvex if and only if X is n-hyperconvex for every n ∈ N.

(ii) A is externally 4-hyperconvex in X if and only if A is externally n-hyperconvex in X for every n ∈ N.

(iii) A is weakly externally 4-hyperconvex in X if and only if A is weakly externally n-hyperconvex in X for every n ∈ N.

Lemma 3.2. Let X be a complete, 4-hyperconvex metric space. If A0 ∈ E3(X), A1 ∈ E2(X) and A0 ∩ A1 6= ∅, then

A0 ∩ A1 ∈ E2(X).

Proof. Let B(x1, r1),B(x2, r2) be closed balls such that d(x1, x2) ≤ r1 + r2 and d(xi,A0 ∩ A1) ≤ ri. Deﬁne A2 := B(x1, r1) ∩ B(x2, r2) ∈ E2(X). Since, for k = 0, 1, the sets Ak are externally 2-hyperconvex, we have

A2 ∩ Ak = B(x1, r1) ∩ B(x2, r2) ∩ Ak 6= ∅.

Therefore, we get

A0 ∩ A1 ∩ B(x1, r1) ∩ B(x2, r2) = A0 ∩ A1 ∩ A2 6= ∅ by Lemma 2.6.

Proposition 3.3. Let X be a complete, 4-hyperconvex metric space. Then

(i) X is n-hyperconvex for every n ∈ N,

(ii) if A ∈ E2(X), we have A ∈ En(X) for every n ∈ N, and

(iii) if A1,...,Am ∈ E2(X) with Ai ∩ Aj 6= ∅ for all i, j ∈ {1, . . . , m}, then Tm i=1 Ai ∈ En(X) for every n ∈ N.

Proof. In order to show (i), we will prove by induction that the following claim is true. n Claim. For every family of closed balls {B(xi, ri)}i=1 with d(xi, xj) ≤ ri + rj, Tn we have i=1 B(xi, ri) ∈ E2(X). n+1 This clearly holds for n = 2. For n ≥ 2, consider {B(xi, ri)}i=1 with d(xi, xj) ≤ ri + rj. Observe that B(x1, r1),B(x2, r2) ∈ E3(X) and, by the

61 CHAPTER V. FINITE HYPERCONVEXITY

Tn+1 induction hypothesis, A := i=3 B(xi, ri) ∈ E2(X) and B(xi, ri) ∩ A 6= ∅ for i = 1, 2. Hence, by Lemma 2.6, we get

n+1 \ B(xi, ri) = B(x1, r1) ∩ B(x2, r2) ∩ A 6= ∅. i=1 0 Tn Again by the induction hypothesis, we have A := i=1 B(xi, ri) ∈ E2(X) and therefore, we conclude that

n+1 \ 0 B(xi, ri) = A ∩ B(xn+1, rn+1) ∈ E2(X) i=1 by Lemma 3.2. n+1 For (ii), we also do induction on n. Let {B(xi, ri)}i=1 be a collection of Tn balls with d(xi, xj) ≤ ri + rj and d(xi,A) ≤ ri. We have A0 := i=1 B(xi, ri), A1 := B(xn+1, rn+1) ∈ E3(X) by (i) and A0∩A 6= ∅ by the induction hypothesis. Hence, we get n+1 \ B(xi, ri) ∩ A = A0 ∩ A1 ∩ A 6= ∅ i=1 by Lemma 2.6. Finally, statement (iii) is a direct consequence of (ii), Lemma 2.6 and Lem- ma 3.2. Proof of Theorem 3.1. Statement (i) was shown in Proposition 3.3(i). Thus, n we will start proving (ii). Let A ∈ E4(X) and let {B(xi, ri)}i=1 be a collection of balls with d(xi, xj) ≤ ri + rj and d(xi,A) ≤ ri. Deﬁne Ai := B(xi, ri) ∩ A ∈ E3(A). We have

Ai ∩ Aj = B(xi, ri) ∩ B(xj, rj) ∩ A 6= ∅ and therefore, we get

n n \ \ B(xi, ri) ∩ A = Ai 6= ∅ i=1 i=1 by Proposition 3.3(iii). This is A ∈ En(X). n−1 Finally, let A ∈ W4(X) and let {B(xi, ri)}i=1 be a collection of balls in A with d(xi, xj) ≤ ri + rj and let x ∈ X, r ≥ 0 with d(x, A) ≤ r, d(x, xi) ≤ r + ri. 0 0 Then, we have A := A ∩ B(x, r) ∈ E3(A) and therefore A ∈ En−1(A) by 0 Proposition 3.3(ii). Moreover, B(xi, ri) ∩ A 6= ∅ and therefore

n−1 n−1 \ \ 0 B(xi, ri) ∩ B(x, r) ∩ A = B(xi, ri) ∩ A 6= ∅. i=1 i=1

This proves A ∈ Wn(X) and hence establishes (iii). Proposition 3.4. Let X be a complete, 4-hyperconvex metric space and let A ∈ W2(X). Then we have A ∈ Wn(X) for every n ∈ N.

62 V.3. 4-HYPERCONVEX METRIC SPACES

Proof. It is enough to show that A ∈ Wn(X) ⇒ A ∈ Wn+1(X) for n ≥ 2. For x ∈ X, let {x0, . . . , xn} ⊂ A ∪ {x} be a set of points with d(xi, xj) ≤ ri + rj and d(xi,A) ≤ ri. Without loss of generality, we may assume that x0 ∈ A. Deﬁne n \ B := B(xi, ri) ∈ E2(X). i=1

Since A ∈ Wn(X), there is some y ∈ A ∩ B 6= ∅. s Fix > 0 and let d := d(x0, y), s := d − r0 and n0 := ˜ . We start by choosing

a1 ∈ B(y, ) ∩ B(x0, d − ) ∩ A,

a2 ∈ B(a1, ) ∩ B(x0, d − 2) ∩ B.

Then, as long as n ≤ n0, we can inductively pick

an ∈ B(an−1, ) ∩ B(x0, d − n) ∩ A if n is odd, and an ∈ B(an−1, ) ∩ B(x0, d − n) ∩ B if n is even. Finally, assuming without loss of generality an0 ∈ B, there is some

a ∈ B(an0 , ) ∩ B(x0, r) ∩ A.

Especially, we have n \ a ∈ A ∩ B(xi, ri + ) 6= ∅ i=0 and thus A ∈ W˜ n+1(X). By Proposition 2.7, we conclude that A ∈ Wn+1(X), as desired.

Example 3.5. Consider c0 ⊂ l∞(N), the subspace of all null sequences. As it is mentioned in [Lin64], c0 is ﬁnitely hyperconvex but not hyperconvex. Moreover, c0 is externally n-hyperconvex in l∞(N) for every n ∈ N. Indeed, we will show that c0 ∈ E2(l∞(N)) and then use Proposition 3.3(ii). Let x := (xn)n∈N, y := (yn)n∈N ∈ l∞(N) with d∞(x, y) ≤ r + s, d∞(x, c0) ≤ r and d∞(y, c0) ≤ s. For every n ∈ N, choose some

zn ∈ B(xn, r) ∩ B(yn, s) with |zn| = inf{|ζ| : ζ ∈ B(xn, r) ∩ B(yn, s)}.

Deﬁne z := (zn)n∈N. Clearly, we have d(xn, zn) ≤ r and d(yn, zn) ≤ s. More- over, limn→∞ zn = 0, since lim supn→∞ |xn| ≤ r and lim supn→∞ |yn| ≤ s. Hence z ∈ B(x, r) ∩ B(y, s) ∩ c0 6= ∅.

Chapter VI

Convexity of Weakly Externally Hyperconvex Subsets

VI.1 Main Results

In this chapter, we relate weak notions of convexity of subsets of a metric space to the notions of (weak) external hyperconvexity. The ﬁrst challenge at this point consists in deﬁning convexity in general hyperconvex metric spaces. For doing this, we use again the notion of geodesic bicombings. Given a metric space X with a geodesic bicombing σ, we say that a subset A of X is σ-convex if, for every x, y ∈ A, we have σxy([0, 1]) ⊂ A. If X is a normed vector space with the linear geodesic bicombing, this notion coincides with the ordinary convexity. We will show that, under the appropriate assumptions the metric space X and on the geodesic bicombing σ, (weakly) externally hyperconvex subsets are σ-convex. We then combine these convexity results with local-to-global properties of (weakly) externally hyperconvex subsets with a geodesic bicombing. A subset A of a metric space X is uniformly locally (weakly) externally hyperconvex if there is some r > 0 such that, for every x ∈ A, the set A∩B(x, r) is (weakly) externally hyperconvex in B(x, r). Theorem 1.1. Let X be a hyperconvex metric space, let A ⊂ X be any subset and let σ denote a convex geodesic bicombing on X. (I) The following are equivalent: (i) A is externally hyperconvex in X. (ii) A is σ-convex and uniformly locally externally hyperconvex. (II) If straight curves in X are unique, the following are equivalent: (i) A is weakly externally hyperconvex and possesses a consistent, convex geodesic bicombing. (ii) A is σ-convex and uniformly locally weakly externally hyperconvex.

65 CHAPTER VI. CONVEXITY

Note that, if X is a proper hyperconvex metric space with finite combinatorial dimension, then a convex geodesic bicombing σ exists, the extra assumption in (II) holds and a weakly externally hyperconvex subset always possesses a consistent, convex geodesic bicombing. Corollary 1.2. Let X be a proper, hyperconvex metric space with finite combinatorial dimension and let σ be the unique consistent, convex geodesic bicombing on X. Then A ⊂ X is weakly externally hyperconvex in X if and only if A is σ-convex and locally weakly externally hyperconvex in X. Theorem 1.1 shows that there are substantial differences between hyperconvex subsets and externally hyperconvex or weakly externally hyperconvex subsets of a metric space since hyperconvex subsets are not convex in general. As an application, consider now the simple case of the n-dimensional normed n n space l∞ := (R , k · k∞), where k · k∞ denotes the maximum norm. If A is a n hyperconvex subset of l∞, then A does not need to be convex. However, as n soon as A is weakly externally 2-hyperconvex in l∞, it follows by Proposi- tion V.3.4 that A is weakly externally hyperconvex and thus A must be convex by Corollary 1.2. This leads to the following Helly type result. A family of subsets of a given set X is a Helly family of order k if every finite subfamily such that every k-fold intersection is non-empty has non-empty total intersection. For instance, for a hyperconvex metric space X, the set E(X) of externally hyperconvex subsets is a Helly family of order 2.

n Theorem 1.3. The collection W(l∞) of weakly externally hyperconvex subsets n of l∞ is a Helly family of order n + 1. Moreover, the order is optimal.

VI.2 σ-Convexity

The goal of this section is to prove σ-convexity for (weakly) externally hyperconvex subsets and therefore establish the implications (i) ⇒ (ii) in Theorem 1.1. Afterwards, we conclude this section with the proof of Theorem 1.3. A curve c: [0, 1] → X is straight if, for every x ∈ X, the function t 7→ d(x, c(t)) is convex. Note that, on the one hand, straight curves are geodesics and on the other hand, the geodesics of a convex geodesic bicombing are straight curves. Furthermore, we say that a metric space X has unique straight curves if for all x, y ∈ X, there is at most one straight curve c: [0, 1] → X with c(0) = x and c(1) = y. For instance, normed vector spaces and metric spaces with ﬁnite combinatorial dimension have unique straight curves, compare [DL15]. Proposition 2.1. Suppose that X is a metric space with a geodesic bicombing σ such that the geodesics σxy are straight curves. Moreover, let E ∈ E2(X). Then E is σ-convex. Proof. Assume, by contradiction, that E is not σ-convex. Then there are x, y ∈ E such that σxy([0, 1]) * E.

66 VI.2. σ-CONVEXITY

Hence, there are 0 ≤ t1 < t2 ≤ 1 such that σxy([t1, t2]) ∩ E = {x,¯ y¯} with x¯ = σxy(t1) andy ¯ = σxy(t2). Let

R := max d(σxy(t),E) > 0. t∈[t1,t2] 0 0 Deﬁne s := min{t ∈ [t1, t2]: d(σxy(t ),E) = R} and z := σxy(s). In particular, R − we have d(E, z) = R. For any ∈ (0, 2d(x,y) ), we set z := σxy(s − ) and + − − + + z := σxy(s + ). We then have R := d(z ,E) < R and R := d(z ,E) ≤ R. Moreover, by the choice of , we get

− − − R R = d(z ,E) ≥ d(z, E) − d(z, z ) = R − d(x, y) > 2 + R and similarly R > 2 . In particular, we have − + R R − + d(z , z ) = 2d(x, y) < 2 + 2 < R + R and therefore, by external 2-hyperconvexity of E, we can pick e ∈ E ∩ B(z−,R−) ∩ B(z+,R+).

Now, since the curve t 7→ σxy(t) is straight in X, it follows that 1 − + 1 − + d(E, σxy(s)) ≤ d(e, σxy(s)) ≤ 2 d(e, z ) + d(e, z ) ≤ 2 R + R < R, which is a contradiction to the deﬁnition of s. Hence, E is σ-convex. Proposition 2.2. Suppose that X is a metric space with unique straight curves and let σX be a convex geodesic bicombing on X. Moreover, we assume that W W ∈ W3(X) and W possesses a consistent geodesic bicombing σ . Then it W X X follows that σ = σ |W ×W ×[0,1]. In particular, W is σ -convex. Proof. We claim that the geodesics of σW are straight curves in X. Consider z ∈ X and let s := d(z, W ). For w, w0 ∈ W , letw, ¯ w¯0 ∈ E := B(z, s) ∩ W such that s + d(w, ¯ w) = d(z, w¯) + d(w, ¯ w) = d(z, w), (2.1) s + d(w ¯0, w0) = d(z, w¯0) + d(w ¯0, w0) = d(z, w0). We have W W W W d(z, σww0 (t)) ≤ d(z, σw¯w¯0 (t)) + d(σw¯w¯0 (t), σww0 (t)). (2.2)

To bound the ﬁrst term on the right-hand side of (2.2), note that E ∈ E2(W ) and W W thus E is in particular σ -convex by Proposition 2.1. Hence, d(z, σw¯w¯0 (t)) = s. On the other hand, to bound the second term on the right-hand side of (2.2), note that, by convexity of σW , we get

W W 0 0 d(σw¯w¯0 (t), σww0 (t)) ≤ (1 − t)d(w, ¯ w) + td(w ¯ , w ). Putting those two estimates together and using (2.1), we obtain that

W 0 d(z, σww0 (t)) ≤ (1 − t)d(z, w) + td(z, w ). It follows that the consistent geodesic bicombing σW consists of straight curves in X. Since straight curves in X are unique, σW must coincide with the re- X X striction σ |W ×W ×[0,1]. Therefore, W is σ -convex.

67 CHAPTER VI. CONVEXITY

Remark 2.3. In [Lan13], U. Lang proves that the injective hull of certain dis- cretely geodesic metric spaces (including hyperbolic groups) has the structure of a locally ﬁnite polyhedral complex of ﬁnite combinatorial dimension. The cells are weakly externally hyperconvex subsets of this complex by Remark III.3.4 and from Proposition 2.2 it follows now that the unique consistent, convex geodesic bicombing on this complex is linear inside the cells.

Lemma 2.4. Let X be a metric space with a conical geodesic bicombing σ and let A ⊂ X be σ-convex. Then the following holds:

(i) For all x, y ∈ X and t ∈ [0, 1], we have

d(σxy(t),A) ≤ (1 − t)d(x, A) + td(y, A).

(ii) If σ is consistent, then, for all x, y ∈ X, the function t 7→ d(σxy(t),A) is convex.

Proof. We prove (ii). For s1, s2 ∈ [0, 1] and > 0, there are p, q ∈ A with

d(σxy(s1), p) ≤ d(σxy(s1),A) + and d(σxy(s2), q) ≤ d(σxy(s2),A) + . We get

d(σxy((1 − t)s1 + ts2),A) ≤ d(σσxy(s1)σxy(s2)(t), σpq(t))

≤ (1 − t)d(σxy(s1),A) + td(σxy(s2),A) + for all > 0. Finally, let ↓ 0.

Corollary 2.5. Suppose that X is a proper metric space of ﬁnite combinatorial dimension which admits a consistent geodesic bicombing σ. Then, for all x, y ∈ X and W ∈ W3(X), the function t 7→ d(σxy(t),W ) is convex. Proof of Theorem 1.3. By Proposition 2.2, weakly externally hyperconvex sub- n sets of l∞ are convex in the usual sense. By the classical Helly theorem, it follows n that W(l∞) is a Helly family of order n + 1. It thus remains to be proved that n n+1 is optimal. In other words, we need to ﬁnd A := {A1,...,An+1} ⊂ W(l∞) such that Tn+1 (a) one has i=1 Ai = ∅ and (b) for any j ∈ {1, . . . , n + 1}, one has B := T A 6= ∅. j A∈A\{Aj }

68 VI.3. LOCALLY WEAKLY EXTERNALLY HYPERCONVEX SUBSETS

We can deﬁne the following weakly externally hyperconvex half-spaces n A1 := {x ∈ R : x1 ≥ 0 }, n A2 := {x ∈ R : x2 ≥ x1 }, n A3 := {x ∈ R : x3 ≥ x2 }, . . n An := {x ∈ R : xn ≥ xn−1}, n An+1 := {x ∈ R : −1 ≥ xn }. Tn Tn+1 Note, if x ∈ i=1 Ai, then x1, . . . , xn ≥ 0 and thus i=1 Ai = ∅ as desired. Moreover, note that

(−1, −1, −1,..., −1, −1, −1) ∈ B1,

( 0, −1, −1,..., −1, −1, −1) ∈ B2,

( 0, 0, −1,..., −1, −1, −1) ∈ B3, . .

( 0, 0, 0,..., 0, 0, −1) ∈ Bn,

( 0, 0, 0,..., 0, 0, 0) ∈ Bn+1. This concludes the proof.

VI.3 Locally Weakly Externally Hyperconvex Subsets

In Section IV.2, we showed that a uniformly locally hyperconvex metric space with a geodesic bicombing is hyperconvex. We will now extend this result to the classes of weakly externally hyperconvex and externally hyperconvex subsets. Let X be a metric space. A subset A ⊂ X is called locally (weakly) externally hyperconvex in X if, for all x ∈ A, there is some rx > 0 such that A ∩ B(x, rx) is (weakly) externally hyperconvex in B(x, rx). If we can choose rx = r > 0 for all x ∈ A, we call A uniformly locally (weakly) externally hyperconvex in X. Lemma 3.1. Let X be a hyperconvex metric space. Then A ⊂ X is locally (weakly) externally hyperconvex in X if and only if, for every x ∈ A, there is some rx > 0 such that A ∩ B(x, rx) is (weakly) externally hyperconvex in X.

Proof. If A ∩ B(x, rx) is (weakly) externally hyperconvex in B(x, rx), then

rx rx A ∩ B(x, 2 ) = (A ∩ B(x, rx)) ∩ B(x, 2 ) is (weakly) externally hyperconvex in

rx rx B(A ∩ B(x, 2 ), 2 ) ⊂ B(x, rx)

rx and therefore, by Lemma II.3.16 and Lemma II.3.17, B(x, 2 ) is (weakly) externally hyperconvex in X. The converse is obvious.

69 CHAPTER VI. CONVEXITY

Lemma 3.2. Let X be a metric space and A ⊂ X. Then A ∈ E(X) if and only if A ∩ B(z, R) ∈ E(B(z, R)) for all z ∈ X, R ≥ d(z, A).

Proof. If A ∈ E(X), then A ∩ B(z, R) ∈ E(B(z, R)) by Lemma II.1.2. For the other direction, we first observe that A is hyperconvex by Lem- ma IV.2.2. Let {xi}i∈I ⊂ X with d(xi,A) ≤ ri and d(xi, xj) ≤ ri + rj. We define Bi := A ∩ B(xi, ri). Since, for fixed i, j ∈ I and some s > 0, there are z ∈ X and R > 0 such that B(xi, ri + s),B(xj, rj + s) ⊂ B(z, R), we have Bi ∩ Bj 6= ∅ and Bi ∈ E(A) by Lemma II.3.17. Hence, we can conclude \ \ A ∩ B(xi, ri) = Bi 6= ∅ i∈I i∈I by Proposition II.2.1.

Lemma 3.3. Let X be a metric space and A ⊂ X. Then A ∈ W(X) if and only if A ∩ B(z, R) ∈ W(B(z, R)) for all z ∈ X,R ≥ d(z, A).

Proof. If A ∈ W(X), then we obviously have A ∩ B(z, R) ∈ W(B(z, R)) by Proposition II.3.9. For the other direction, we first observe that A is hyperconvex by Lem- ma IV.2.2. Let x ∈ X, {xi}i∈I ⊂ A with d(x, A) ≤ r, d(x, xi) ≤ r + ri and d(xi, xj) ≤ ri + rj. Define B¯ := A ∩ B(x, r) and Bi := A ∩ B(xi, ri). Since, for fixed i, j ∈ I and some s > 0, there are z ∈ X and R > 0 such that B(x, r + s),B(xi, ri + s),B(xj, rj + s) ⊂ B(z, R), we have B¯ ∩ Bi ∩ Bj 6= ∅ and B,B¯ i ∈ E(A) by Lemma II.3.17. Hence, we can conclude \ \ A ∩ B(x, r) ∩ B(xi, ri) = B¯ ∩ Bi 6= ∅ i∈I i∈I by Proposition II.2.1.

Lemma 3.4. Let X be a hyperconvex metric space and assume that we have A ∩ B(z, R) ∈ W(B(z, R)) for all z ∈ B(A, R). Then A must be proximinal in 5 5 5 B A, 4 R . Especially, for all z¯ ∈ B A, 4 R , we have A ∩ B z,¯ 4 R 6= ∅. 5 R Proof. Letz ¯ ∈ B(A, 4 R) and d(¯z, A) = R + t with t ∈ 0, 4 . Then, for R every ∈ 0, 4 , there is some z ∈ B(A, R) with d(z, z) ≤ t + . Especially, we have d(z, z R ) ≤ R, i.e. z ∈ B(z R ,R), and, by our assumption, there is 4 4 some proximinal non-expansive retraction ρ: B(z R ,R) → A ∩ B(z R ,R). Since 4 4 0 d(ρ(z), z¯) ≤ R + t + and d(ρ(z), ρ(z0 )) ≤ d(z, z0 ) ≤ 2t + + , there is some \ y ∈ B(¯z, R) ∩ B (ρ(z), t + ) . >0 Finally, we have d(y, A) = t ≤ R and hence, there is somey ¯ ∈ A with d(¯y, y) = t and thus d(¯y, z¯) ≤ d(¯y, y) + d(y, z¯) ≤ t + R = d(¯z, A), 5 i.e. A is proximinal in B A, 4 R .

70 VI.3. LOCALLY WEAKLY EXTERNALLY HYPERCONVEX SUBSETS

We are now able to prove our local-to-global result for (weakly) externally hyperconvex subsets of a metric space with a geodesic bicombing.

Proposition 3.5. Let X be a hyperconvex metric space with a geodesic bicombing σ. If A ⊂ X is σ-convex and uniformly locally (weakly) externally hyperconvex, then A is (weakly) externally hyperconvex in X.

Proof. The two cases are similar but we prove them separately to be precise.

Case 1. Let A be σ-convex and uniformly locally externally hyperconvex.

Let P(R) be the property given by the following expression: P(R): ∀z ∈ B(A, R): A ∩ B(z, R) ∈ E(B(z, R)). We want to show that this is true for every R > 0. Then, by Lemma 3.2, the set A is externally hyperconvex in X. Since A is uniformly locally externally hyperconvex, P(R) clearly holds for 5 some small R > 0. Hence, we need to show that P(R) ⇒ P( 4 R). 5 For z ∈ B A, 4 R , let 0 5 0 0 0 0 0 X := B z, 4 R , A := A ∩ X and B (x, r) := B(x, r) ∩ X for x ∈ X . By Lemma II.3.17, it is enough to show that

0 0 0 R A ∈ E B A , 4 . (3.1) First, note that A0 6= ∅ by Lemma 3.4. We establish now the following property for all R0 > 0: 0 0 0 0 R 0 0 P (R ): ∀{xi}i∈I ⊂ B (A , 4 ) with d(xi, xj) ≤ ri + rj, d(xi,A ) ≤ ri and ri ≤ R , 0 T 0 we have A ∩ i∈I B (xi, ri) 6= ∅. 0 5 5 Observe that P ( 2 R) implies (3.1) since balls with center in B(z, 4 R) and 5 5 radius bigger than 2 R contain B(z, 4 R).

0 R Step I. P ( 2 ) holds.

5 We have d(xi, z) ≤ 4 R, d(xi, xj) ≤ R and hence, there is some

\ R z¯ ∈ B(z, R) ∩ B(xi, 2 ). i∈I

Then we have d(¯z, A) ≤ d(¯z, xi)+d(xi,A) ≤ R and A∩B(¯z, R) ∈ E(B(¯z, R)) by P(R). Clearly, z, xi ∈ B(¯z, R) and therefore, if we have d(xi,A ∩ B(¯z, R)) ≤ ri 5 and d(z, A ∩ B(¯z, R)) ≤ 4 R, there is some

5 \ y ∈ A ∩ B(¯z, R) ∩ B z, 4 R ∩ B(xi, ri) 6= ∅ i∈I

0 T 0 and thus y ∈ A ∩ i∈I B (xi, ri) 6= ∅.

71 CHAPTER VI. CONVEXITY

R Indeed, since, for every ∈ 0, 4 , there is some

0 5 a ∈ A = A ∩ B z, 4 R

R with d(a, xi) ≤ min ri, 4 + . It follows

R R d(a, z¯) ≤ d(a, xi) + d(xi, z¯) ≤ 4 + + 2 ≤ R. and therefore a ∈ A ∩ B(¯z, R). Hence, we get

d(xi,A ∩ B(¯z, R)) ≤ d(xi, a) ≤ ri + for all > 0, i.e.

d(xi,A ∩ B(¯z, R)) ≤ ri, and 5 d(z, A ∩ B(¯z, R)) ≤ d(z, a) ≤ 4 R, as desired.

Step II. We establish P0(R0) ⇒ P0(2R0).

1 1 Let yij := σxixj 2 and zi := σxiz 2 . We get

5 R d(zi,A) ≤ d(zi, xi) + d(xi,A) ≤ 8 R + 4 ≤ R, 1 5 d(yij, zi) ≤ 2 d(xj, z) ≤ 8 R ≤ R, 1 5 d(z, zi) = 2 d(z, xi) ≤ 8 R ≤ R, i.e. yij, z ∈ B(zi,R), and A ∩ B(zi,R) ∈ E(B(zi,R)) by P(R). Note that there 0 5 0 R 3 is some a ∈ A = A ∩ B(z, 4 R) with d(a, xi) ≤ d(xi,A ) + 8 ≤ 8 R and thus

5 3 d(zi, a) ≤ d(zi, xi) + d(xi, a) ≤ 8 R + 8 R = R.

5 Hence, it follows a ∈ Ai := A ∩ B(zi,R) ∩ B(z, 4 R) 6= ∅ and therefore we get Ai ∈ E(B(zi,R)). Moreover, we have

1 rj rk d(yij, yik) ≤ 2 d(xj, xk) ≤ 2 + 2 , 0 1 0 0 ri rj R d(yij,A ) ≤ 2 d(xi,A ) + d(xj,A ) ≤ min{ 2 + 2 , 4 }

R 0 0 by Lemma 2.4. For ∈ 0, 8 , let a ∈ A with d(yij, a) ≤ d(yij,A ) + . Then R 5 we have d(a, zi) ≤ d(a, yij) + d(yij, zi) ≤ 4 + + 8 R ≤ R, i.e. a ∈ Ai, and therefore 0 ri rj R d(yij,Ai) = d(yij,A ) ≤ min{ 2 + 2 , 4 }. ri R 5 T rj Hence, there arex ¯i ∈ B(Ai, min{ 2 , 4 }) ∩ B(z, 4 R) ∩ j∈I B(yij, 2 ) with

ri rj d(¯xi, x¯j) ≤ d(¯xi, yij) + d(yij, x¯j) ≤ 2 + 2 , 0 ri R d(¯xi,A ) ≤ d(¯xi,Ai) ≤ min{ 2 , 4 }.

0 0 0 T ri 0 T Therefore, by P (R ), we get y ∈ A ∩ i∈I B(¯xi, 2 ) ⊂ A ∩ i∈I B(xi, ri) 6= ∅ as desired.

72 VI.3. LOCALLY WEAKLY EXTERNALLY HYPERCONVEX SUBSETS

Case 2. Let A be σ-convex and uniformly locally weakly externally hyperconvex.

Let P(R) be the property given by the following expression: P(R): ∀z ∈ B(A, R): A ∩ B(z, R) ∈ W(B(z, R)). We want to show that this is true for every R > 0. Then, by Lemma 3.3, the set A is weakly externally hyperconvex in X. Since A is uniformly locally externally hyperconvex, P(R) clearly holds for 5 some small R > 0. Hence, we need to show that P(R) ⇒ P( 4 R). 5 For z ∈ B A, 4 R , let 0 5 0 0 0 0 0 X := B z, 4 R , A := A ∩ X and B (x, r) := B(x, r) ∩ X for x ∈ X . By Lemma II.3.16, it is enough to show that

0 0 0 R A ∈ W B A , 4 . (3.2) First, note that A0 6= ∅ b