arXiv:2007.09846v4 [math.MG] 20 Sep 2021 uemti geometry: metric Pure nrdcoylectures introductory no Petrunin Anton We discuss only domestic affairs of metric spaces; applications are given only as illustrations. These notes could be used as an introductory part to virtually any course in metric geometry. It is based on a part of the course in metric geometry given at Penn State, Spring 2020. The complete lectures can be found on the author’s website; it includes an introduction to Alexandrov geometry based on [1] and metric geometry on manifolds [28] based on a simplified proof of Gromov’s systolic inequality given by Alexander Nabutovsky [22]. A part of the text is a compilation from [1, 2, 24, 26, 27] and its drafts. I want to thank Sergio Zamora Barrera for help. Contents

1 Definitions 5 A Metricspaces ...... 5 B Variationsofdefinition...... 6 C Completeness ...... 7 D Compactspaces...... 8 E Properspaces...... 9 F Geodesics ...... 9 G Geodesicspacesandmetrictrees ...... 9 H Length...... 10 I Lengthspaces...... 11

2 Universal spaces 15 A Embeddinginanormedspace...... 15 B Extensionproperty...... 17 C Universality...... 19 D Uniquenessandhomogeneity ...... 20 E Remarks...... 22

3 Injective spaces 23 A Admissible and extremal functions ...... 23 B Injectivespaces ...... 25 C Spaceofextremalfunctions ...... 27 D Injectiveenvelope...... 29 E Remarks...... 30

4 Space of sets 31 A Hausdorffdistance ...... 31 B Hausdorffconvergence ...... 32 C Anapplication ...... 34 D Remarks...... 35

3 5 Space of spaces 37 A Gromov–Hausdorffmetric ...... 37 B Approximations...... 39 C Almostisometries...... 39 D Convergence...... 41 E Uniformly totally bonded families ...... 42 F Gromov’sselectiontheorem ...... 43 G Universalambientspace ...... 45 H Remarks...... 46

6 Ultralimits 49 A Facesofultrafilters ...... 49 B Ultralimitsofpoints ...... 50 C Ultralimitsofspaces ...... 52 D Ultrapower ...... 53 E Tangentandasymptoticspaces ...... 54 F Remarks...... 55

A Semisolutions 57

Bibliography 79 Lecture 1

Definitions

In this lecture we give some conventions used further and remind some definitions related to metric spaces. We assume some prior knowledge of metric spaces. For a more detailed introduction, we recommend the first couple of chapters in the book by Dmitri Burago, Yuri Burago, and Sergei Ivanov [7].

A Metric spaces

The distance between two points x and y in a metric space will be X denoted by x y or x y X . The latter notation is used if we need to emphasize| that− | the| distance− | is taken in the space . X Let us recall the definition of metric. 1.1. Definition. A metric on a set is a real-valued function (x, y) X 7→ x y X that satisfies the following conditions for any three points x,y,z7→ | − | : ∈ X (a) x y > 0, | − |X (b) x y =0 x = y, | − |X ⇐⇒ (c) x y = y x , | − |X | − |X (d) x y + y z > x z . | − |X | − |X | − |X Recall that a metric space is a set with a metric on it. The elements of the set are called points. Most of the time we keep the same notation for the metric space and its underlying set. The function distx : y x y 7→ | − | is called the distance function from x.

5 6 LECTURE 1. DEFINITIONS

Given R [0, ] and x , the sets ∈ ∞ ∈ X B(x, R)= y x y < R , { ∈ X | | − | } B[x, R]= y x y 6 R { ∈ X | | − | } are called, respectively, the open and the closed balls of radius R with center x. If we need to emphasize that these balls are taken in the metric space , we write X

B(x, R)X and B[x, R]X .

1.2. Exercise. Show that

p q + x y 6 p x + p y + q x + q y | − |X | − |X | − |X | − |X | − |X | − |X for any four points p, q, x, and y in a metric space . X B Variations of definition

Pseudometris. A metric for which the distance between two distinct points can be zero is called a pseudometric. In other words, to define pseudometric, we need to remove condition (b) from 1.1. The following observation shows that nearly any question about pseudometric spaces can be reduced to a question about genuine metric spaces. Assume is a pseudometric space. Consider an equivalence rela- tion on Xdefined by ∼ X x y x y =0. ∼ ⇐⇒ | − | Note that if x x′, then y x = y x′ for any y . Thus, defines a metric∼ on the quotient| − | set| −/ .| This way we∈ X obtain a metric|∗− ∗| space ′. The space ′ is called theX ∼corresponding metric space for the pseudometricX spaceX . Often we do not distinguish between ′ and . X X X ∞-metrics. One may also consider metrics with values in R ; we might call them -metrics, but most of the time we use the∪ {∞} term metric. ∞ Again nearly any question about -metric spaces can be reduced to a question about genuine metric spaces.∞ Set x y x y < ; ≈ ⇐⇒ | − | ∞ C. COMPLETENESS 7 it defines another equivalence relation on . The equivalence class of a point x will be called the metricX component of x; it will be ∈ X denoted by x. One could think of x as B(x, )X  the open ball centered at xX and radius in . X ∞ It follows that any ∞-metricX space is a disjoint union of genuine metric spaces  the metric∞ components of the original -metric space. ∞ 1.3. Exercise. Given two sets A and B on the plane, set

A B = µ(A B), | − | △ where µ denotes the Lebesgue measure and denotes symmetric dif- ference △ A B := (A B) (B A). △ ∪ \ ∩ (a) Show that is a pseudometric on the set of bounded closed subsets. |∗ − ∗| (b) Show that is an -metric on the set of all open subsets. |∗ − ∗| ∞ C Completeness

A metric space is called complete if every Cauchy sequence of points in convergesX in . X X 1.4. Exercise. Suppose that ρ is a positive continuous function on a complete metric space and ε> 0. Show that there is a point x such that X ∈ X ρ(x) < (1 + ε) ρ(y) · for any point y B(x, ρ(x)). ∈ Most of the time we will assume that a metric space is complete. The following construction produces a complete metric space ¯ for any given metric space . X X Completion. Given a metric space , consider the set of all Cauchy X C sequences in . Note that for any two Cauchy sequences (xn) and (yn) the right-handX side in ➊ is defined; moreover, it defines a pseudometric on C

➊ (xn) (yn) := lim xn yn . | − |C n→∞ | − |X The corresponding metric space ¯ is called completion of . Note that the original space X forms a dense subset inX its com- pletion ¯. More precisely, for eachX point x one can consider a X ∈ X constant sequence xn = x which is Cauchy. It defines a natural map 8 LECTURE 1. DEFINITIONS

¯. It is easy to check that this map is distance-preserving. In particular,X → X we can (and will) consider as a subset of ¯. X X 1.5. Exercise. Show that the completion of a metric space is com- plete.

D Compact spaces

Let us recall few equivalent definitions of compact metric spaces. 1.6. Definition. A metric space is compact if and only if one of the following equivalent conditionsK hold: (a) Every open cover of has a finite subcover. (b) For any open coverK of there is ε > 0 such that any ε-ball in lies in one elementK of the cover. (The value ε is called a LebesgueK number of the covering.) (c) Every sequence of points in has a subsequence that converges in . K (d) TheK space is complete and totally bounded; that is, for any ε> 0, the spaceK admits a finite cover by open ε-balls. K A subset N of a metric space is called ε-net if any point x lies on the distance less than ε fromK a point in N. Note that totally∈ K bounded spaces can be defined as spaces that admit a finite ε-net for any ε> 0. 1.7. Exercise. Show that a space is totally bounded if and only if it contains a compact ε-net for any Kε> 0. Let pack be the exact upper bound on the number of points ε X x ,...,xn such that xi xj > ε if i = j. 1 ∈ X | − | 6 If n = packε < , then the collection of points x1,...,xn is called a maximalXε-packing∞ . Note that n is the maximal number of open disjoint ε -balls in . 2 X 1.8. Exercise. Show that any maximal ε-packing is an ε-net. Conclude that a complete space is compact if and only if packε < < for any ε> 0. X X ∞ 1.9. Exercise. Let be a compact metric space and K f : K → K be a distance-nondecreasing map. Prove that f is an isometry; that is, f is a distance-preserving bijection. A metric space is called locally compact if any point in admits a compact neighborhood;X in other words, for any point x Xa closed ball B[x, r] is compact for some r> 0. ∈ X E. PROPER SPACES 9 E Proper spaces

A metric space is called proper if all closed bounded sets in are compact. ThisX condition is equivalent to each of the followingX statements: For some (and therefore any) point p and any R< , the ⋄ ∈ X ∞ closed ball B[p, R]X is compact. The function distp : R is proper for some (and therefore ⋄ any) point p ;X that → is, for any compact set K R, its inverse image ∈ X ⊂

dist−1(K)= x : p x K p { ∈ X | − |X ∈ } is compact. 1.10. Exercise. Give an example of space which is locally compact but not proper.

F Geodesics

Let be a metric space and I a real interval. A globally isometric mapXγ : I is called a geodesic1; in other words, γ : I is a geodesic if→ X → X γ(s) γ(t) = s t | − |X | − | for any pair s,t I. ∈ If γ :[a,b] is a geodesic and p = γ(a), q = γ(b), then we say that γ is a geodesic→ X from p to q. In this case, the image of γ is denoted by [pq], and, with abuse of notations, we also call it a geodesic. We may write [pq] to emphasize that the geodesic [pq] is in the space . X X In general, a geodesic from p to q need not exist and if it exists, it need not be unique. However, once we write [pq] we assume that we have chosen such geodesic. A geodesic path is a geodesic with constant-speed parameterization by the unit interval [0, 1].

G Geodesic spaces and metric trees

A metric space is called geodesic if any pair of its points can be joined by a geodesic.

1Different authors call it differently: shortest path, minimizing geodesic. Also, note that the meaning of the term geodesic is different from what is used in Rie- mannian geometry, altho they are closely related. 10 LECTURE 1. DEFINITIONS

1.11. Exercise. Let f be a centrally symmetric positive continuous function on S2. Given two points x, y S2, set ∈ x y = w f. k − k π π B(x, 2 )\B(y, 2 )

Show that (S2, ) is a geodesic space and the geodesics in (S2, ) run alongk∗ − great ∗k circles of S2. k∗−∗k A geodesic space is called a metric tree if any pair of points in are connected byT a unique geodesic, and the union of any two T geodesics [xy]T , and [yz]T contain the geodesic [xz]T . In other words any triangle in is a tripod; that is, for any three geodesics [xy], [yz], and [zx] have aT common point. 1.12. Exercise. Let p,x,y,z be points in a metric tree. Consider three numbers

a = p x + y z , b = p y + z x , c = p z + x y . | − | | − | | − | | − | | − | | − | Suppose that a 6 b 6 c. Show that b = c. Recall that the set

Sr(p) = x : p x = r X { ∈ X | − |X } is called a sphere with center p and radius r in a metric space . X 1.13. Exercise. Show that spheres in metric trees are ultrametric spaces. That is,

x z 6 max x y , y z | − | { | − | | − |} for any x,y,z Sr(p) . ∈ T H Length

A curve is defined as a continuous map from a real interval I to a metric space. If I = [0, 1], that is, if the interval is unit, then the curve is called a path. 1.14. Definition. Let be a metric space and α: I be a curve. We define the length of αX as → X

length α := sup X α(ti) α(ti−1) . 06 16 6 | − | t t ... tn i A curve α is called rectifiable if length α< . ∞ I. LENGTH SPACES 11

1.15. Theorem. Length is a lower semi-continuous with respect to the pointwise convergence of curves. More precisely, assume that a sequence of curves γn : I in a metric space converges pointwise to a curve γ : I ; that→ X is, for X ∞ → X any fixed t I, γn(t) γ (t) as n . Then ∈ → ∞ → ∞

➊ lim length γn > length γ∞. n→∞

Note that the inequality ➊ might be strict. For example, the diagonal γ∞ of the unit square can be approximated by stairs-like polygonal curves γn with sides parallel to the sides of the square (γ6 is on the picture). In this case

length γ∞ = √2 and length γn =2 for any n.

Proof. Fix a sequence t0 6 t1 6 ... 6 tk in I. Set

Σn := γn(t ) γn(t ) + + γn(tk ) γn(tk) . | 0 − 1 | ··· | −1 − | Σ := γ (t ) γ (t ) + + γ (tk ) γ (tk) . ∞ | ∞ 0 − ∞ 1 | ··· | ∞ −1 − ∞ | Note that for each i we have

γn(ti ) γn(ti) γ (ti ) γ (ti) | −1 − | → | ∞ −1 − ∞ | and therefore Σn Σ → ∞ as n . Note that → ∞ Σn 6 length γn for each n. Hence, lim length γn > Σ∞. n→∞ Since the partition was arbitrary, by the definition of length, the inequality ➊ is obtained.

I Length spaces

If for any ε > 0 and any pair of points x and y in a metric space , there is a path α connecting x to y such that X

length α< x y + ε, | − | 12 LECTURE 1. DEFINITIONS then is called a length space and the metric on is called a length metricX. X An -metric space is a length space if each of its metric compo- nents is∞ a length space. In other words, if is an -metric space, X ∞ then in the above definition we assume in addition that x y X < . Note that any geodesic space is a length space. The| − following| ∞ example shows that the converse does not hold. 1.16. Example. Suppose a space is obtained by gluing a countable I X 1 I collection of disjoint intervals n of length 1+ n , where for each n the left end is glued to p and the{ right} end to q. Observe that the space carries a natural complete length metric X with respect to which p q X =1 but there is no geodesic connecting p to q. | − |

1.17. Exercise. Give an example of a complete length space such that no pair of distinct points in can be joined by a geodesic.X X Directly from the definition, it follows that if α: [0, 1] is a path from x to y (that is, α(0) = x and α(1) = y), then → X length α > x y . | − | Set x y = inf length α k − k { } where the greatest lower bound is taken for all paths from x to y. It is straightforward to check that (x, y) x y is an -metric; moreover, ( , ) is a length space. The7→ k metric− k ∞is called the inducedX lengthk∗−∗k metric. k∗−∗k 1.18. Exercise. Let be a complete length space. Show that for any compact subset K in X there is a compact path-connected subset K′ that contains K. X

1.19. Exercise. Suppose ( , ) is a complete metric space. Show that ( , ) is complete.X |∗−∗| X k∗−∗k Let A be a subset of a metric space . Given two points x, y A, consider the value X ∈

x y A = inf length α , | − | α { } where the greatest lower bound is taken for all paths α from x to y in A. In other words A denotes the induced length metric on the subspace A.2 |∗ − ∗| 2 The notation |∗−∗|A conflicts with the previously defined notation for distance |x−y|X in a metric space X . However, most of the time we will work with ambient length spaces where the meaning will be unambiguous. I. LENGTH SPACES 13

Let x and y be points in a metric space . (i) A point z is called a midpoint betweenX x and y if ∈ X x z = y z = 1 x y . | − | | − | 2 ·| − | (ii) Assume ε > 0. A point z is called an ε-midpoint between x and y if ∈ X x z , y z 6 1 x y + ε. | − | | − | 2 ·| − | Note that a 0-midpoint is the same as a midpoint. 1.20. Lemma. Let be a complete metric space. (a) Assume that forX any pair of points x, y , and any ε > 0, there is an ε-midpoint z. Then is a length∈ X space. X (b) Assume that for any pair of points x, y , there is a mid- point z. Then is a geodesic space. ∈ X X Proof. We first prove (a). Let x, y be a pair of points. ε ∈ X Set εn = 4n , α(0) = x and α(1) = y. 1 Let α( 2 ) be an ε1-midpoint between α(0) and α(1). Further, let 1 3 1 α( 4 ) and α( 4 ) be ε2-midpoints between the pairs (α(0), α( 2 )) and 1 (α( 2 ), α(1)) respectively. Applying the above procedure recursively, k on the n-th step we define α( 2n ), for every odd integer k such that k k−1 0 < 2n < 1, as an εn-midpoint of the already defined α( 2n ) and k+1 α( 2n ). In this way we define α(t) for t W , where W denotes the set of dyadic rationals in [0, 1]. Since ∈is complete, the map α can be extended continuously to [0, 1]. Moreover,X

∞ n−1 length α 6 x y + X 2 εn 6 ➊ | − | · n=1 6 x y + ε . | − | 2 Since ε> 0 is arbitrary, we get (a). To prove (b), one should repeat the same argument taking mid- points instead of εn-midpoints. In this case, ➊ holds for εn = ε = =0. 1 Since in a compact space a sequence of n -midpoints zn contains a convergent subsequence, 1.20 immediately implies the following. 1.21. Proposition. Any proper length space is geodesic.

1.22. Hopf–Rinow theorem. Any complete, locally compact length space is proper. 14 LECTURE 1. DEFINITIONS

Before reading the proof, it is instructive to solve 1.10. Proof. Let be a locally compact length space. Given x , denote by ρ(x) theX least upper bound of all R > 0 such that the∈ closed X ball B[x, R] is compact. Since is locally compact, X ➋ ρ(x) > 0 for any x . ∈ X It is sufficient to show that ρ(x) = for some (and therefore any) point x . ∞ ∈ X ➌ If ρ(x) < , then B = B[x, ρ(x)] is compact. ∞ Indeed, is a length space; therefore for any ε > 0, the set B[x, ρ(x) εX] is a compact ε-net in B. Since B is closed and hence complete,− it must be compact. △

➍ R ρ(x) ρ(y) 6 x y X for any x, y ; in particular, ρ: is a| continuous− | function.| − | ∈ X X → Indeed, assume the contrary; that is, ρ(x) + x y < ρ(y) for some x, y . Then B[x, ρ(x)+ ε] is a closed subset| − of B[| y,ρ(y)] for some ε > ∈0. X Then compactness of B[y,ρ(y)] implies compactness of B[x, ρ(x)+ ε], a contradiction. △ Set ε = min ρ(y): y B ; the minimum is defined since B is compact and ρ is{ continuous.∈ From} ➋, we have ε> 0. ε Choose a finite 10 -net a1,a2,...,an in B = B[x, ρ(x)]. The union { } ε W of the closed balls B[ai,ε] is compact. Clearly, B[x, ρ(x)+ 10 ] W . ε ⊂ Therefore, B[x, ρ(x)+ 10 ] is compact, a contradiction.

1.23. Exercise. Construct a geodesic space that is locally compact, but whose completion ¯ is neither geodesic norX locally compact. X 1.24. Advanced exercise. Show that for any compact length-metric space there is a number ℓ such that for any finite collection of points thereX is a point z that lies on average distance ℓ from the collection; that is, for any x ,...,xn there is z such that 1 ∈ X ∈ X 1 X xi z X = ℓ. n · | − | i Lecture 2

Universal spaces

This lecture is based on the discussion of Urysohn space in the book of Mikhael Gromov [12].

A Embedding in a normed space

Recall that a function v v on a vector space is called norm if it satisfies the following condition7→ | | for any two vectorsV v, w and a scalar α: ∈ V v > 0; ⋄ | | α v = α v ; ⋄ | · | | |·| | v + w > v + w . ⋄ | | | | | | As an example, consider the space of real sequences equipped with ∞ ∞ sup norm denoted by the ℓ ; that is, ℓ -norm of a = a1,a2,... is defined by a ℓ∞ = sup an . | | n { | |} It is straightforward to check that for any normed space the func- tion (v, w) v w defines a metric on it. Therefore, any normed space is an example7→ | − of| metric space (in fact, it is a geodesic space). Of- ten we do not distinguish between normed space and the corresponding metric space.1 The following lemma implies that any compact metric space is isometric to a subset of a fixed normed space. Recall that diameter of a metric space (briefly diam ) is defined as least upper bound on the distances betweenX pairs of itsX points; that

1By Mazur–Ulam theorem, the metric remembers the linear structure of the space; a slick proof of this statement was given by Jussi V¨ais¨al¨a[32].

15 16 LECTURE 2. UNIVERSAL SPACES is, diam = sup x y : x, y . X { | − |X ∈X} 2.1. Lemma. Suppose is a bounded separable metric space; that X is, diam is finite and contains a countable, dense set wn . Given X X { } x , set an(x)= wn x . Then ∈ X | − |X ι: x (a (x),a (x),... ) 7→ 1 2 .∞defines a distance-preserving embedding ι: ֒ ℓ X → Proof. By the triangle inequality

➊ an(x) an(y) 6 x y . | − | | − |X Therefore, ι is short (in other words, ι is distance non-increasing). Again by triangle inequality we have

an(x) an(y) > x y 2 wn x . | − | | − |X − ·| − |X

Since the set wn is dense, we can choose wn arbitrarily close to x. { } Whence the value an(x) an(y) can be chosen arbitrarily close to x y . In other words| − | | − |X

sup wn x X wn y X > x y X . n { || − | − | − | |} | − | Hence ➋ sup an(x) an(y) > x y X ; n { | − |} | − | that is, ι is distance non-contracting. Finally, observe that ➊ and ➋ imply the lemma.

2.2. Exercise. Show that any compact metric space is isometric to a subspace of a compact geodesic space. K The following exercise generalizes the lemma to arbitrary separable spaces.

2.3. Exercise. Suppose wn is a countable, dense set in a metric space . Choose x ; given{ } x , set X 0 ∈ X ∈ X

an(x)= wn x wn x . | − |X − | − 0|X

Show that ι: x (a1(x),a2(x),... ) defines a distance-preserving em- .∞bedding ι: ֒ 7→ℓ X → B. EXTENSION PROPERTY 17

The following lemma implies that any metric space is isometric to a subset of a normed vector space; its proof is nearly identical to the proof of 2.3. 2.4. Lemma. Let be arbitrary metric space. Denote by ℓ∞( ) the space of all boundedX functions on equipped with sup-norm. X X Then for any point x , the map ι: ℓ∞( ) defined by 0 ∈ X X → X

ι: x (distx distx0 ) 7→ − is distance-preserving.

B Extension property

If a metric space is a subspace of a pseudometric space ′, then we say that ′ is anXextension of . If in addition diam ′ 6Xd, then we say that X ′ is a d-extension. X X X If the complement ′ contains a single point, say p, we say that ′ is a one-point extensionX \X of . In this case, to define metric on X ′, it is sufficient to specify theX distance function from p; that is, a functionX f : R defined by X →

f(x)= p x ′ . | − |X Any function f of that type will be called extension function or d- extension function respectively. The extension function f cannot be taken arbitrary  the triangle inequality implies that

f(x)+ f(y) > x y > f(x) f(y) | − |X | − | for any x, y . In particular, f is a non-negative 1-Lipschitz function on . Fora∈d-extension, X we need to assume in addition that diam 6 6 dXand f(x) 6 d for any x . These conditions are necessaryX and sufficient. ∈ X 2.5. Definition. A metric space meets the extension property if for any finite subspace andU any extension function f : R there is a point p suchF ⊂U that p x = f(x) for any x .F → ∈U | − | ∈F If we assume in addition that diam 6 d and instead of extension functions we consider only d-extensionU functions, then we arrive at a definition of d-extension property. If in addition is separable and complete, then it is called Urysohn space or d-UrysohnU space respectively. 18 LECTURE 2. UNIVERSAL SPACES

2.6. Proposition. There is a separable metric space with the (d-) extension property (for any d > 0).

Proof. Choose d > 0. Let us construct a separable metric space with the d-extension property. Let be a compact metric space such that diam 6 d. Denote by d theX space of all d-extension functions on equippedX with the metricX defined by the sup-norm. Note that the mapX d defined X → X by x distx is a distance-preserving embedding, so we can (and will) treat7→ as a subspace of d, or, equivalently, d is an extension of . X X X X Let us iterate this construction. Start with a one-point space 0 d X and consider a sequence of spaces ( n) defined by n+1 = n . Note that the sequence is nested; that is,X ... Xand theX union X0 ⊂ X1 ⊂

∞ = [ n; X n X comes with metric such that x y = x y if x, y n. | − |X∞ | − |Xn ∈ X Note that if is compact, then so is d. It follows that each space X X n is compact. In particular, ∞ is a countable union of compact Xspaces; therefore is separable.X X∞ Any finite subspace of lies in some n for n < . By F X∞ X ∞ construction, there is a point p n that meets the condition in 2.5 ∈ X +1 for any extension function f : R. That is, ∞ has the d-extension property. F → X The construction of a separable metric space with the extension property requires only minor changes. First, the sequence should be dn defined by n = , where dn is an increasing sequence such that X +1 Xn dn . Second, the point p should be taken in n k for sufficiently → ∞ X + large k, so that dn k > max f(x) . + { } 2.7. Proposition. If a metric space meets the (d-) extension prop- erty, then so does its completion. V

Proof. Let us assume meets the extension property. We will show that its completion Vmeets the extension property as well. The d- extension case can beU proved along the same lines. Note that is a dense subset in a complete space . Observe that has the approximateV extension property; that is, if U is a finite set,U ε > 0, and f : R is an extension function, thenF⊂U there exists p such that F → ∈U ➊ p x >< f(x) ε | − | ± C. UNIVERSALITY 19 for any x . Indeed, let us extend f to the whole by setting ∈F X f¯(z) = inf f(x)+ x z : x . { | − | ∈F} Observe that f¯ is an extension function. Since is dense in , we can choose a finite set ′ such that for any xV there isU x′ ′ with x x′ < ε . ItF remains∈ V to observe that the∈ point F p provided∈ Fby | − | 2 the extension property for the restriction f¯ ′ meets ➊. |F Therefore, there is a sequence of points pn such that for any x , ∈ U ∈F 1 pn x >< f(x) n . | − | ± 2 Moreover, we can assume that ➋ 1 pn pn < n | − +1| 2 for all large n. Indeed, consider the sets n = pn and the F F∪{ } functions fn : n R defined by fn(x)= f(x) if x = pn and F → 6 fn(pn) = max  pn x f(x) : x . | − |− ∈F 1 Observe that fn is an extension function for large n and fn(pn) < 2n . Therefore, applying the approximate extension property recursively we get ➋. By ➋, (pn) is a Cauchy sequence and its limit meets the condition in the definition of extension property (2.5). Note that 2.6 and 2.7 imply the following: 2.8. Theorem. Urysohn space and d-Urysohn space for any d > 0 exist.

C Universality

A metric space will be called universal if it includes as a subspace an isometric copy of any separable metric space. In 2.3, we proved that ℓ∞ is a universal space. The following proposition shows that an Urysohn space is universal as well. Unlike ℓ∞, Urysohn spaces are separable; so it might be considered as a better universal space. Theorem 2.17 will give another reason why Urysohn spaces are better. 2.9. Proposition. An Urysohn space is universal. That is, if is an Urysohn space, then any separable metric space admits a distance-U preserving embedding ֒ . S Moreover, for anyS finite→U subspace , any distance-preserving embedding ֒ can be extended toF a distance-preserving ⊂ S embedding F →U . ֒ S →U 20 LECTURE 2. UNIVERSAL SPACES

A d-Urysohn space is d-universal; that is, the above statements hold provided that diam 6 d. S Proof. We will prove the second statement; the first statement is its partial case for = ∅. F The required isometry will be denoted by x x′. 7→ Choose a dense sequence of points s1,s2,... . We may assume ′ ∈ S that = s1,...,sn , so si are defined for i 6 n. F { ′ } ∈U The sequence si for i>n can be defined recursively using the ′ ′ extension property in . Namely, suppose that s1,...,si−1 are already U ′ defined. Since meets the extension property, there is a point si such that U ∈U ′ ′ s s = si sj | i − j |U | − |S for any j

2.11. Exercise. Modify the proofs of 2.7 and 2.9 to prove the follow- ing theorem.

2.12. Theorem. Let K be a compact set in a separable space . Then any distance-preserving map from K to an Urysohn space canS be extended to a distance-preserving map on whole . S 2.13. Exercise. Show that (d-) Urysohn space is simply connected.

D Uniqueness and homogeneity

2.14. Theorem. Suppose and ′ ′ be finite isometric sub- spaces in a pair of (d-)UrysohnF⊂U spaces F and⊂U ′. Then any isometry ι: ′ can be extended to an isometryU U ′. F↔F U ↔U In particular, (d-)Urysohn space is unique up to isometry. While 2.9 implies that there are distance-preserving maps ′ and ′ , it does not solely imply the existence of an isometryU→U U ′.→ Its U construction uses the idea of 2.9, but it is applied back- and-forthU ↔U to ensure that the obtained distance-preserving map is onto. D. UNIQUENESS AND HOMOGENEITY 21

′ ′ ′ Proof. Choose dense sequences a1,a2, and b1,b2, . We ···∈U′ ′ ′ ···∈U can assume that = a1,...,an , = b1,...,bn and ι(ai) = bi for i 6 n. F { } F { } The required isometry ′ will be denoted by u u′. Set ′ ′ U ↔U ↔ ai = bi if i 6 n. ′ ′ Let us define recursively an+1,bn+1,an+2,bn+2,...  on the odd step we define the images of an+1,an+2,... and on the even steps we ′ ′ define inverse images of bn+1,bn+2,... . The same argument as in the ′ ′ ′ proof of 2.9 shows that we can construct two sequences a1,a2, and b ,b , such that ···∈U 1 2 ···∈U ′ ′ ai aj = a a ′ | − |U | i − j |U ′ ′ ai bj = a b ′ | − |U | i − j |U ′ ′ bi bj = b b ′ | − |U | i − j |U for all i and j. It remains to observe that the constructed distance-preserving bi- ′ ′ jection defined by ai ai and bi bi extends continuously to an isometry ′. ↔ ↔ U↔U Observe that 2.14 implies that the Urysohn space (as well as the d-Urysohn space) is finite-set homogeneous; that is, any distance-preserving map from a finite subset to the whole ⋄ space can be extended to an isometry. 2.15. Open question. Is there a noncomplete finite-set homoge- neous metric space that meets the extension property? This is a question of Pavel Urysohn; it appeared already in [31, §2(6)] and reappeared in [12, p. 83] with a missing keyword. In fact, I do not see an example of a 1-point homogeneous space that meets the extension property. Recall that Sr(p)X denotes the sphere of radius r centered at p in a metric space ; that is, X

Sr(p) = x : p x = r . X { ∈ X | − |X }

2.16. Exercise. Choose d [0, ]. Denote by d the d-Urysohn ∈ ∞ U space, so ∞ is the Urysohn space. U ∅ (a) Assume that L = Sr(p)Ud = . Show that L is isometric to ℓ; find ℓ in terms of r and d.6 U

(b) Let ℓ = p q . Show that the subset M d of midpoints | − |Ud ⊂ U between p and q is isometric to ℓ. U 22 LECTURE 2. UNIVERSAL SPACES

(c) Show that d is not countable-set homogeneous; that is, there is U a distance-preserving map from a countable subset of d to d U U that cannot be extended to an isometry of d. U In fact, the Urysohn space is compact-set homogeneous; more pre- cisely the following theorem holds. 2.17. Theorem. Let K be a compact set in a (d-)Urysohn space . Then any distance-preserving map K can be extended to an Uisometry of . → U U A proof can be obtained by modifying the proofs of 2.7 and 2.14 the same way as it is done in 2.11. 2.18. Exercise. Which of the following metric spaces are 1-point set homogeneous, finite set homogeneous, compact set homogeneous, countable homogeneous? (a) Euclidean plane, (b) Hilbert space ℓ2, (c) ℓ∞, (d) ℓ1

E Remarks

The statement in 2.3 was proved by Maurice Ren´eFr´echet in the paper where he first defined metric spaces [10]; its extension 2.4 was given by Kazimierz Kuratowski [20]. The question about the existence of a separable universal space was posted by Maurice Ren´eFr´echet and answered by Pavel Urysohn [31]. The idea of Urysohn’s construction was reused in graph theory; it produces the so-called Rado graph, also known as Erd˝os–R´enyi graph or random graph; a good survey on the subject is given by Peter Ca- meron [8]. Lecture 3

Injective spaces

Injective spaces (also known as hyperconvex spaces) are the metric analog of convex sets. This lecture is based on a paper by John Isbell [17].

A Admissible and extremal functions

Let be a metric space. A function r : R is called admissible if the followingX inequality X →

➊ r(x)+ r(y) > x y | − |X holds for any x, y . ∈ X 3.1. Observation. (a) Any admissible function is nonnegative. (b) If is a geodesic space, then a function r : R is admissible if andX only if X → B[x, r(x)] B[y, r(y)] = ∅ ∩ 6 for any x, y . ∈ X Proof. For (a), take x = y in ➊. Part (b) follows from the triangle inequality and the existence of a geodesic [xy]. A minimal admissible function will be called extremal. More pre- cisely, an admissible function r : R is extremal if for any admis- sible function s: R we haveX → X → s 6 r = s = r. ⇒ 23 24 LECTURE 3. INJECTIVE SPACES

3.2. Key exercise. Let r be an extremal function and s an admis- sible function on a metric space . Suppose that r > s c for some constant c. Show that c > 0 and rX6 s + c. −

3.3. Observations. Let be a metric space. X (a) For any point p the distance function r = distp is extremal. ∈ X (b) Any extremal function r on is 1-Lipschitz; that is, X r(p) r(q) 6 p q | − | | − | for any p, q . In other words, any extremal function is an extension function;∈ X see the definition on page 17. (c) Let r be an extremal function on . Then for any point p and any δ > 0, there is a point q X such that ∈ X ∈ X r(p)+ r(q) < p q + δ. | − |X Moreover, if is compact, then there is q such that X r(p)+ r(q)= p q . | − |X (d) For any admissible function s there is an extremal function r such that r 6 s.

Proof; (a). By the triangle inequality, ➊ holds; that is, r = distp is an admissible function. Further, if s 6 r is another admissible function, then s(p)=0 and ➊ implies that s(x) > p x . Whence s = r. | − | (b). By (a), distp is admissible. Since r is admissible, we have that

r > distp r(p). − Since r is extremal, 3.2 implies that

r 6 distp + r(p), or, equivalently, r(q) r(p) 6 p q − | − | for any p, q . The same way we can show that r(p) r(q) 6 p q . Whence the∈ statement X follows. − | − |

(c). Again, by (a), distp is an extremal function. Arguing by contra- diction, assume r(q) > distp(q) r(p)+ δ − B. INJECTIVE SPACES 25 for any q. By 3.2, we get that

r(q) 6 distp(q)+ r(p) δ − for any q. Taking q = p, we get r(p) 6 r(p) δ, a contradiction. − Suppose is compact. Denote by qn the point provided by the X 1 first part of (c) for δ = n . Let q be a partial limit of qn. Then r(p)+ r(q) 6 p q . | − |X Since r is admissible, the opposite inequality holds; whence the second part of (c) follows. (d). Follows by Zorn’s lemma.

B Injective spaces

3.4. Definition. A metric space is called injective if for any metric space and any of its subspaces Y any short map f : can be extendedX to a short map F : A; that is, f = F . A → Y X → Y |A 3.5. Exercise. Show that any injective space is (a) complete, (b) geodesic, and (c) contractible.

3.6. Exercise. Show that the following spaces are injective: (a) the real line; (b) complete metric tree; (c) coordinate plane with the metric induced by the ℓ∞-norm. The following exercise deals with metric spaces that in some sense dual to the injective spaces. 3.7. Exercise. Suppose that a metric space satisfies the following property: For any subspace in and any otherX metric space , any short map f : can beA extendedX to a short map F : Y. Show that A →is Y an ultrametric space; that is, the followingX → Y strong version of theX triangle inequality

x z 6 max x y , y z | − |X { | − |X | − |X } holds for any three points x,y,z . ∈ X 3.8. Theorem. For any metric space the following condition are equivalent: Y 26 LECTURE 3. INJECTIVE SPACES

(a) is injective Y (b) If r : R is an extremal function, then there is a point p suchY that → ∈ Y p x 6 r(x) | − | for any x . ∈ Y (c) is hyperconvex; that is, if B[xα, rα] α∈A is a family of closed ballsY in such that { } Y

rα + rβ > xα xβ | − |

for any α, β , then all the balls in the family B[xα, rα] α∈A have a common∈ A point. { }

Proof. We will prove implications (a) (b) (c) (a). ⇒ ⇒ ⇒ (a) (b). Let us apply the definition of injective space to a one-point extension⇒ of . It follows that for any extension function r : R there is a pointY p such that Y → ∈ Y p x 6 r(x) | − | for any x . By 3.3b, any extremal function is an extension function, whence the∈ Y implication follows. (b) (c). By 3.1b, part (c) is equivalent to the following statement: ⇒ If r : R is an admissible function, then there is a point p ⋄ suchY that → ∈ Y

➊ p x 6 r(x) | − | for any x . ∈ Y Indeed, set r(x) := inf rα : xα = x . (If xα = x for any α, then r(x) = .) The condition{ in (c) implies} that6 r is admissible. It ∞ remains to observe that p B[xα, rα] for every α if and only if ➊ holds. ∈ By 3.3d, for any admissible function r there is an extremal function r¯ 6 r; whence (b) (c). ⇒ (c) (a). Arguing by contradiction, suppose is not injective; that is,⇒ there is a metric space with a subset Ysuch that a short map f : cannot be extendedX to a short mapA F : . By Zorn’s lemma,A → we Y may assume that is a maximal subset;X that → is, Y the domain of f cannot be enlarged by aA single point.1

1In this case, A must be closed, but we will not use it. C. SPACE OF EXTREMAL FUNCTIONS 27

Fix a point p in the complement . To extend f to p, we need to choose f(p) in the intersection ofX \A the balls B[f(x), r(x)], where r(x) = p x . Therefore, this intersection for all x has to be empty. | − | ∈ A Since f is short, we have that

r(x)+ r(y) > x y > | − |X > f(x) f(y) . | − |Y Therefore, by (c) the balls B[f(x), r(x)] have a common point  a contradiction.

3.9. Exercise. Suppose a length space has two subspaces and such that = and is a one-pointW set. Assume X and Y are injective.X ∪ ShowY W that Xis ∩ injective Y X Y W 3.10. Exercise. Show that the d-Urysohn space is finitely hypercon- vex but not countably hyperconvex; that is, the condition in 3.8c holds for any finite family of balls, but may not hold for a countable family. Conclude that the d-Urysohn space is not injective. Try to do the same for the Urysohn space.

C Space of extremal functions

Let be a metric space. Consider the space Ext of extremal func- tionsX on equipped with sup-norm; that is, X X f g := sup f(x) g(x) : x . | − |Ext X { | − | ∈X} Recall that by 3.3a, any distance function is extremal. It follows that the map x distx produces a distance-preserving embedding ,Ext . So we7→ can (and will) treat as a subspace of Ext , or ֒ Xequivalently,→ X Ext as an extension of X. X X X Since any extremal function is 1-Lipschitz, for any f Ext and ∈ X p , we have that f(x) 6 f(p) + distp(x). By 3.2, we also get ∈ X f(x) > f(p) + distp(x). Therefore −

f p = sup f(x) distp(x) : x = ➊ | − |Ext X { | − | ∈X} = f(p).

In particular, the statement in 3.3c can be written as

f p + f q < p q + δ. | − |Ext X | − |Ext X | − |Ext X 28 LECTURE 3. INJECTIVE SPACES

3.11. Exercise. Let be a metric space. Show that Ext is compact if and only if so is .X X X 3.12. Exercise. Describe the set of all extremal functions on a metric space and the metric space Ext in each of the following case: (a) X is a metric space with exactlyX three points a,b,c such that X a b = b c = c a =1. | − |X | − |X | − |X (b) is a metric space with exactly four points p,q,x,y such that X p x = p y = q x = q x =1 | − |X | − |X | − |X | − |X and p q = x y =2. | − |X | − |X

3.13. Proposition. For any metric space , its extension Ext is injective. X X

3.14. Lemma. Let be a metric space. Suppose that r is an ex- X tremal function on Ext . Then r X Ext ; that is, the restriction of r to is an extremalX function. | ∈ X X Proof. Arguing by contradiction, suppose that there is an admissible function s: R such that s(x) 6 r(x) for any x and s(p) < < r(p) for someX → point p . Consider another function∈r¯ X: Ext R such that r¯(f) := r(f) if∈f X= p and r¯(p) := s(p). X → Let us show that r¯ is admissible;6 that is,

➋ f g 6 r¯(f)+¯r(g) | − |Ext X for any f,g Ext . Since r is∈ admissibleX and r¯ = r on (Ext ) p , it is sufficient to ➋ ➊ X \{ } prove if f = g = p. By , we have f p Ext X = f(p). Therefore, ➋ boils down6 to the following inequality| − |

➌ r(f)+ s(p) > f(p). for any f Ext . Fix small∈ δ >X0. Let q be the point provided by 3.3c. Then ∈ X r(f)+ s(p) > [r(f) r(q)] + [r(q)+ s(p)] > − since r is 1-Lipschitz, and r(q) > s(q), we can continue

> q f + [s(q)+ s(p)] > −| − |Ext X D. INJECTIVE ENVELOPE 29 by ➊ and since s is admissible

> f(q)+ p q > − | − | by 3.3c

>f(p) δ. − Since δ > 0 is arbitrary, ➌ and ➋ follow. Summarizing: the function r¯ is admissible, r¯ 6 r and r¯(p) < r(p); that is, r is not extremal  a contradiction.

Proof of 3.13. Choose an extremal function r : Ext R. Set s := X → := r X . By 3.14, s Ext ; that is, s is extremal. By 3.8b, it is sufficient| to show that∈ X

➍ r(f) > s f | − |Ext X for any f Ext . Since r∈is 1-LipschitzX (3.3b) we have that

s(x) f(x)= r(x) f x 6 r(f). − − | − |Ext X for any x . Since r is admissible we have that ∈ X s(x) f(x)= r(x) f x > r(f). − − | − |Ext X − for any x . That is, s(x) f(x) 6 r(f) for any x . Recall that ∈ X | − | ∈ X s f := sup s(x) f(x) : x ; | − |Ext X { | − | ∈X} hence ➍ follows.

3.15. Exercise. Let be a compact metric space. Show that for any two points f,g Ext X lie on a geodesic [pq] with p, q . ∈ X ∈ X D Injective envelope

An extension of a metric space will be called its injective envelope if is an injectiveE space and thereX is no proper injective subspace of Ethat contains . -E Two injectiveX envelopes e: ֒ and f : ֒ are called equiv alent if there is an isometry ι:X →E such thatX →Ff = ι e. E→F ◦ 3.16. Theorem. For any metric space , its extension Ext is an injective envelope. X X 30 LECTURE 3. INJECTIVE SPACES

Moreover, any other injective envelope of is equivalent to Ext . X X Proof. Suppose S Ext is an injective subspace containing . Since S is injective,⊂ there isX a short map w : Ext S that fixesX all points in . X → SupposeX that w : f f ′; observe that f(x) > f ′(x) for any x . Since f is extremal, f =7→f ′; that is, w is the identity map and therefore∈ X S = Ext . AssumeX we have another injective envelope e: ֒ . Then there are short maps v : Ext and w : Ext suchX →E that x = v e(x) and e(x)= w(x) forE → any x X . From above,X →E the composition v ◦w is the identity on Ext . In particular,∈ X w is distance-preserving. ◦ The compositionXw v : is a short map that fixes points in e( ). Since e: ֒ is◦ anE injective →E envelope, the composition w v andX therefore wXare→E onto. Whence w is an isometry. ◦

3.17. Exercise. Suppose is a subspace of a metric space . Show that the inclusion ֒ Xcan be extended to a distance-preservingU inclusion Ext ֒ XExt→. U X → U E Remarks

Injective spaces were introduced by Nachman Aronszajn and Prom Panitchpakdi [3]. The injective envelope was introduced by John Isbell [17]. It was rediscovered a couple of times since then; as a result the injective envelope has many other names including tight span and hyperconvex hull. Lecture 4

Space of sets

A Hausdorff distance

Let be a metric space. Given a subset A , consider the distance functionX to A ⊂ X distA : [0, ) X → ∞ defined as distA(x) := inf a x X . a∈A{ | − | } 4.1. Definition. Let A and B be two compact subsets of a metric space . Then the Hausdorff distance between A and B is defined as X A B Haus X := sup distA(x) distB(x) . | − | x∈X{ | − |}

The following observation gives a useful reformulation of the defi- nition: 4.2. Observation. Suppose A and B be two compact subsets of a metric space . Then A B Haus X < R if and only if and only if B lies in an XR-neighborhood| − of| A, and A lies in an R-neighborhood of B. Note that the set of all nonempty compact subsets of a metric space equipped with the Hausdorff metric forms a metric space. This new metricX space will be denoted as Haus . X 4.3. Exercise. Let be a metric space. Given a subset A define its diameter as X ⊂ X

diam A := sup a b . a,b∈A | − |

31 32 LECTURE 4. SPACE OF SETS

Show that diam: Haus R X → is a 2-Lipschitz function; that is,

diam A diam B 6 2 A B | − | ·| − |Haus X for any two compact nonempty sets A, B . ⊂ X 4.4. Exercise. Let A and B be two compact subsets in the Euclidean 2 plane R . Assume A B R2 <ε. | − |Haus (a) Show that Conv A Conv B Haus R2 <ε, where Conv A denoted the convex| hull of A−. | (b) Is it true that ∂A ∂B Haus R2 <ε, where ∂A denotes the bound- ary of A. | − | Does the converse hold? That is, assume A and B be two com- 2 pact subsets in R and ∂A ∂B R2 < ε; is it true that | − |Haus A B R2 <ε? | − |Haus Note that part (a) implies that A Conv A defines a short map Haus R2 Haus R2. 7→ → 4.5. Exercise. Let A and B be two compact subsets in metric space . Show that X

A B Haus X = sup max f(a) max f(b) , | − | f { a∈A { }− b∈B { } where the least upper bound is taken for all 1-Lipschitz functions f.

B Hausdorff convergence

4.6. Blaschke selection theorem. A metric space is compact if and only if so is Haus . X X The Hausdorff metric can be used to define convergence. Namely, suppose K ,K ,... , and K are compact sets in a metric space . If 1 2 ∞ X K Kn 0 as n , then we say that the sequence (Kn) | ∞ − |Haus X → → ∞ converges to K∞ in the sense of Hausdorff ; or we can say that K∞ is Hausdorff limit of the sequence (Kn). Note that the theorem implies that from any sequence of compact sets in one can select a subsequence that converges in the sense of Hausdorff;X for that reason, it is called a selection theorem. Proof; “only if” part. Consider the map ι that sends point x to the one-point subset x of . Note that ι: Haus is distance-∈ X preserving. { } X X → X B. HAUSDORFF CONVERGENCE 33

Suppose that A . Note that diam A = 0 if and only if A is a one-point set. Therefore,⊂ X from Exercise 4.3, it follows that ι( ) is a closed subset of the compact space Haus . Whence ι( ), andX therefore , are compact. X X X To prove the “if” part we will need the following two lemmas.

4.7. Monotone convergence. Let K1 K2 ... be a nested se- quence of nonempty compact sets in a metric⊃ space⊃ . Then K = X ∞ = Tn Kn is the Hausdorff limit of Kn; that is, K∞ Kn Haus X 0 as n . | − | → → ∞

Proof. By finite intersection property, K∞ is a nonempty compact set. If the assertion were false, then there is ε> 0 such that for each n one can choose xn Kn such that distK∞ (xn) > ε. Note that xn K1 ∈ 1 ∈ for each n. Since K1 is compact, there is a partial limit x∞ of xn.

Clearly, distK∞ (x∞) > ε. On the other hand, since Kn is closed and xm Kn for m > n, ∈ we get x Kn for each n. It follows that x K and therefore ∞ ∈ ∞ ∈ ∞ distK∞ (x∞)=0  a contradiction.

4.8. Lemma. If is a compact metric space, then Haus is com- plete. X X

Proof. Let (Qn) be a Cauchy sequence in Haus . Passing to a sub- X sequence of Qn we may assume that

➊ 1 Qn Qn 6 n | − +1|Haus X 10 for each n. 1 Denote by Kn the closed 10n -neighborhood of Qn; that is,

1 Kn = x : distQ (x) 6 n  ∈ X n 10

Since is compact so is each Kn. X 1 ➊ By 4.2, Qn Kn Haus X 6 10n . From , we get Kn Kn+1 for each n. Set| − | ⊃ ∞ K∞ = \ Kn. n=1

By the monotone convergence (4.7), Kn K∞ Haus X 0 as n . 1 | − | → → ∞ Since Qn Kn Haus X 6 10n , we get Qn K∞ Haus X 0 as n  hence| the− lemma.| | − | → → ∞

1Partial limit is a limit of a subsequence. 34 LECTURE 4. SPACE OF SETS

4.9. Exercise. Let be a complete metric space and K1,K2,... be a sequence of compactX sets that converges in the sense of Hausdorff. Show that the union K1 K2 ... is a compact closure. Use this statement to∪ show∪ that in Lemma 4.8 compactness of can be exchanged to completeness. X

Proof of “if” part in 4.6. According to Lemma 4.8, Haus is complete. It remains to show that Haus is totally bounded (1.6d);X that is, given ε> 0 there is a finite ε-net inXHaus . Choose a finite ε-net A in . DenoteX by B the set of all subsets of A. Note that B is a finite set inXHaus . For each compact set K , ′ X ⊂ X consider the subset K of all points a A such that distK (a) 6 ε. ′ ′ ∈ Observe that K B and K K Haus X 6 ε. In other words, B is a finite ε-net in Haus∈ . | − | X 4.10. Exercise. Let be a complete metric space. Show that is a length space if and onlyX if so is Haus . X X C An application

The following statement is called isoperimetric inequality in the plane. 4.11. Theorem. Among the plane figures bounded by closed curves of length at most ℓ the round disk has the maximal area. In this section, we will sketch a proof of the isoperimetric inequality that uses the Hausdorff convergence. It is based on the following exercise. 4.12. Exercise. Let be a subspace of Haus R2 formed by all compact convex subsets in R2.C Show that perimeter2 and area are continuous on . That is, if a sequence of convex compact plane sets Xn converges C to X∞ in the sense of Hausdorff, then

perim Xn perim X and area Xn area X → ∞ → ∞ as n . → ∞ Semiproof of 4.11. It is sufficient to consider only convex figures of the given perimeter; if a figure is not convex, pass to its convex hull and observe that it has a larger area and smaller perimeter. Note that the selection theorem (4.6) together with the exercise imply the existence of figure D with perimeter ℓ and maximal area.

2If the set degenerates to a line segment of length ℓ, then its perimeter is defined as 2·ℓ. D. REMARKS 35

It remains to show that D is a round disk. This is a problem in elementary geometry. Let us cut D along a chord [ab] into two lenses, L1 and L2. Denote ′ by L1 the reflection of L1 across the perpendicular bisector of [ab]. ′ ′ Note that D and D = L1 L2 have the same perimeter and area. That is, D′ has perimeter ℓ ∪and maximal possible area; in particular, D′ is convex. The following exercise will finish the proof.

4.13. Exercise. Suppose D is a convex figure such that for ′′′ L11 L1 any chord [ab] of D the above b a b 11 a L L construction produces a con- 22 22 ′ vex figure D . Show that D is D D′ a round disk. Another popular way to prove that D is a round disk is given by the so-called Steiner’s 4-joint method [5].

D Remarks

It seems that Hausdorff convergence was first introduced by Felix Hausdorff [14]. A couple of years later an equivalent definition was given by Wilhelm Blaschke [5]. The following refinement of the definition was introduced by Zdenˇek Frol´ık [11], later it was rediscovered by Robert Wijsman [33]. This re- finement is also called Hausdorff convergence; in fact, it takes an inter- mediate place between the original Hausdorff convergence and closed convergence, also introduced by Hausdorff in [14].

4.14. Definition. Let (An) be a sequence of closed sets in a metric space . We say that (An) converges to a closed set A in the sense X ∞ of Hausdorff if for any x , we have distAn (x) distA∞ (x) as n . ∈ X → → ∞ For example, suppose is the Euclidean plane and An is the circle with radius n and centerX at the point (n, 0). If we use the standard definition (4.1), then the sequence (An) diverges, but it converges to the y-axis in the sense of Definition 4.14. The following exercise is analogous to the Blaschke selection theo- rem (4.6) for the modified Hausdorff convergence. ∞ 4.15. Exercise. Let be a proper metric space and (An)n=1 be a sequence of closed setsX in . Assume that for some (and therefore X any) point x , the sequence an = distA (x) is bounded. Show ∈ X n 36 LECTURE 4. SPACE OF SETS

∞ that the sequence (An)n=1 has a convergent subsequence in the sense of Definition 4.14. Lecture 5

Space of spaces

A Gromov–Hausdorff metric

The goal of this section is to cook up a metric space out of met- ric spaces. More precisely, we want to define the so-called Gromov– Hausdorff metric on the set of isometry classes of compact metric spaces. (Being isometric is an equivalence relation, and an isometry class is an equivalence class with respect to this equivalence relation.) The obtained metric space will be denoted by GH. Given two metric spaces and , denote by [ ] and [ ] their isometry classes; X Y isoX Y that is, ′ [ ] if and only if ′ == . Pedantically, the Gromov– X ∈ X X X Hausdorff distance from [ ] to [ ] should be denoted as [ ] [ ] GH; but we will write it as X Y and say (not quite correctly)| X − Y “ | |X − Y|GH |X − GH is the Gromov–Hausdorff distance from to ”. In other words,− Y| from now on the term metric space mightX alsoY stand for its isometry class. The metric on GH is defined as the maximal metric such that the distance between subspaces in a metric space is not greater than the Hausdorff distance between them. Here is a formal definition: 5.1. Definition. Let and be compact metric spaces. The Gromov– Hausdorff distance X Y is defined by the following relation. |X − Y|GH Given r> 0, we have that GH < r if and only if there exist a metric space and subspaces|X − Y|′ and ′ in that are isometric Z X Y′ Z′ to and respectively and such that Haus Z < r. (Here ′X ′ Y |X − Y | ′ ′ Haus Z denotes the Hausdorff distance between sets and in|X −.) Y | X Y Z Note that passing to the subspace ′ ′ of does not affect the definition. Therefore we can always assumeX ∪ Y thatZ is compact. Z 37 38 LECTURE 5. SPACE OF SPACES

5.2. Theorem. The set of isometry classes of compact metric spaces equipped with Gromov–Hausdorff metric forms a metric space (which is denoted by GH). In other words, for arbitrary compact metric spaces , and the following conditions hold: X Y Z (a) > 0; |X − Y|GH (b) =0 if and only if is isometric to ; |X − Y|GH X Y (c) = ; |X − Y|GH |Y − X |GH (d) + > . |X − Y|GH |Y − Z|GH |X − Z|GH Note that (a), (c), and the “if”-part of (b) follow directly from Definition 5.1. Part (d) will be proved in Section 5B. The “only-if”- part of (b) will be proved in Section 5C. Recall that a denotes scaled by factor a > 0; that is, a is a metric space with·X the underlyingX set of and the metric defined·X by X x y := a x y . | − |a·X ·| − |X

5.3. Exercise. Let be a compact metric space, be the one-point metric space. X P Prove that (a) = 1 diam . |X − P|GH 2 · X (b) a b = 1 a b diam . | ·X − ·X |GH 2 ·| − |· X

5.4. Exercise. Let r be a rectangle 1 by r in the Euclidean plane A and r be a closed line interval of length r. Show that B 1 r r > |A −B |GH 10 for all large r.

5.5. Advanced exercise. Let and be compact metric spaces; denote by ˆ and ˆ their injectiveX envelopesY (see the definition on page 27). ShowX that Y ˆ ˆ 6 2 . |X − Y|GH ·|X − Y|GH B. APPROXIMATIONS 39 B Approximations

5.6. Definition. Let and be two metric spaces. A relation between points in andX isY called ε-approximation if the following≈ conditions hold: X Y For any x there is y such that x y. ⋄ ∈ X ∈ Y ≈ For any y there is x such that x y. ⋄ ∈ Y ∈ X ≈ If for some x, x′ and y,y′ we have x y and x′ y′, ⋄ then ∈ X ∈ Y ≈ ≈ ′ ′ x x X y y Y < 2 ε. | − | − | − | ·

5.7. Exercise. Let and be two compact metric spaces. Show that X Y <ε |X − Y|GH if and only if there is an ε-approximation between and . X Y In other words GH is the greatest lower bound of values ε> 0 such that there|X is − an Y|ε-approximation between and . X Y Proof of 5.2d. Suppose that is a relation between points in and , ⋄ ≈1 X Y is a relation between points in and . ⋄ ≈2 Y Z Consider the relation between points in and such that x z ≈3 X Z ≈3 if and only if there is y such that x 1 y and y 2 z. It is straightforward∈ to Y check that if ≈ is an ε -approximation≈ and ≈1 1 2 is an ε2-approximation, then 3 is an (ε1 + ε2)-approximation. ≈ Applying 5.7, we get that if ≈

<ε and <ε , |X − Y|GH 1 |Y − Z|GH 2 then <ε + ε . |X − Z|GH 1 2 Hence 5.2d follows.

C Almost isometries

5.8. Definition. Let and be metric spaces and ε> 0. A map1 f : is called an εX-isometryY if f( ) is an ε-net in and X → Y X Y ′ ′ x x X f(x) f(x ) Y < ε. | − | − | − | 1possibly noncontinuous 40 LECTURE 5. SPACE OF SPACES for any x, x′ . ∈ X 5.9. Exercise. Let and be compact metric spaces. X Y (a) If GH <ε, then there is a 2 ε-isometry f : . (b) If|X there − Y| is an ε-isometry f : ·, then X →<ε Y. X → Y |X − Y|GH Proof of the “only if”-part in 5.2b. Let and be compact metric X Y spaces. Suppose that GH < ε for any ε > 0; we need to show that there is an isometry|X − Y| . X → Y 1 By 5.9a, for each positive integer n, we can choose a n -isometry fn : . SinceX → Y is compact, we can choose a countable dense set S in . ApplyingX the diagonal procedure if necessary, we can assume thatX for every x S the sequence (fn(x)) converges in . Consider the pointwise limit∈ map f : S , Y ∞ → Y

f∞(x) := lim fn(x) n→∞ for every x S. Since ∈ ′ ′ 1 fn(x) fn(x ) Y >< x x X , | − | | − | ± n we have

′ ′ ′ f∞(x) f∞(x ) Y = lim fn(x) fn(x ) Y = x x X | − | n→∞ | − | | − | for all x, x′ S; that is, the map f : S is distance-preserving. ∈ ∞ → Y Therefore, f∞ can be extended to a distance-preserving map from the whole to . TheX latterY can be done by setting

f∞(x) = lim f∞(xn) n→∞ for some sequence of points (xn) in S that converges to x in . Indeed, X if xn x, then (xn) is Cauchy. Since f is distance-preserving, → ∞ yn = f∞(xn) is also a Cauchy sequence in ; therefore it converges. It remains to observe that this constructionY does not depend on the choice of the sequence (xn). This way we obtain a distance-preserving map f∞ : . It remains to show that f is surjective; that is, f ( )= X. → Y ∞ ∞ X Y The same argument produces a distance-preserving map g∞ : . If f is not surjective, then neither is the composition fY → → X ∞ ∞ ◦ g∞ : . So f∞ g∞ is a distance-preserving map from a compact space◦ toY → itself Y which is◦ not an isometry. The latter contradicts 1.9. D. CONVERGENCE 41 D Convergence

The Gromov–Hausdorff metric is used to define Gromov–Hausdorff convergence. Namely, a sequence of compact metric spaces n con- X verges to compact metric spaces ∞ in the sense of Gromov–Hausdorff if X

n 0 as n . |X − X∞|GH → → ∞ This convergence is more important than the metric  in all appli- cations, we use only the topology on GH and we do not care about the particular value of Gromov–Hausdorff distance between spaces. The following observation follows from 5.9:

5.10. Observation. A sequence of compact metric spaces ( n) con- verges to in the sense of Gromov–Hausdorff if and only ifX there is X∞ a sequence εn 0+ and an εn-isometry fn : n for each n. → X → X∞ In the following exercises converge means converge in the sense of Gromov–Hausdorff. 5.11. Exercise. (a) Show that a sequence of compact simply connected length spaces cannot converge to a circle. (b) Construct a sequence of compact simply connected length spaces that converges to a compact non-simply connected space.

5.12. Exercise. (a) Show that a sequence of length metrics on the 2-sphere cannot converge to the unit disk. (b) Construct a sequence of length metrics on the 3-sphere that con- verges to a unit 3-ball.

Given two metric spaces and , we will write 6 if there is a noncontracting map f : X ; thatY is, if X Y X → Y x x′ 6 f(x) f(x′) | − |X | − |Y for any x, x′ . ∈ X Further, given ε > 0, we will write 6 + ε if there is a map f : such that X Y X → Y x x′ 6 f(x) f(x′) + ε | − |X | − |Y for any x, x′ . ∈ X 42 LECTURE 5. SPACE OF SPACES E Uniformly totally bonded families

5.13. Definition. A family of (isometry classes) of compact metric spaces is called uniformly totallyQ bonded if it meets the following two conditions: (a) spaces in have uniformly bounded diameters; that is, there is D R suchQ that ∈ diam 6 D X for any space in . X Q (b) For any ε> 0 there is n N such that any space in admits an ε-net with at most n points.∈ X Q

5.14. Exercise. Let be a family of compact spaces with uniformly bounded diameters. ShowQ that is uniformly totally bonded if for any ε> 0 there is n N such that Q ∈ pack 6 n ε X for any space in . X Q Fix a real constant C. A Borel measure µ on a metric space is called C-doubling if X

µ[B(p, 2 r)] < C µ[B(p, r)] · · for any point p and any r> 0. A Borel measure is called doubling if it is C-doubling∈ Xfor some real constant C.

5.15. Exercise. Let (C,D) be the set of all the compact metric spaces with diameter atQ most D that admit a C-doubling measure. Show that (C,D) is totally bounded. Q Recall that we write 6 if there is a distance-nondecreasing map . X Y X → Y 5.16. Exercise. (a) Let be a compact metric space. Show that the set of all spaces suchY that 6 is uniformly totally bounded. X X Y (b) Show that for any uniformly totally bounded set GH there is a compact space such that 6 for any Qin ⊂ . Y X Y X Q F. GROMOV’S SELECTION THEOREM 43 F Gromov’s selection theorem

The following theorem is analogous to Blaschke selection theorems (4.6). 5.17. Gromov selection theorem. Let be a closed subset of GH. Then is compact if and only if it is totallyQ bounded. Q 5.18. Lemma. The space GH is complete. Let us define gluing of metric spaces that will be used in the proof of the lemma. Suppose and are metric spaces with isometric closed sets A and A′ ;U let ι:VA A′ be an isometry. Consider the space ⊂Uof all equivalence⊂ V classes→ in with the equivalence relation givenW by a ι(a) for any a A. U⊔V ∼ ∈ It is straightforward to check that the following defines a metric on : W u u′ := u u′ | − |W | − |U v v′ := v v′ | − |W | − |V u v := min u a + v ι(a) : a A | − |W { | − |U | − |V ∈ } where u,u′ and v, v′ . ∈U ∈V The space is called the gluing of and along ι; briefly, we W U V can write = ι . If one applies this construction to two copies of one spaceW withU ⊔ aV set A and the identity map ι: A A, then the obtainedU space is called the⊂Udouble of along A; this space→ can be denoted by 2 . U ⊔AU -Note that the inclusions ֒ and ֒ are distance pre serving. Therefore we can andU will→ W consideV →and W as the subspaces of ; this way the subsets A and A′ will beU identifiedV and denoted furtherW by A. Note that A = . U∩V⊂W Proof. Let ( n) be a Cauchy sequence in GH. Passing to a sub- X 1 sequence if necessary, we can assume that n n < n for |X − X +1|GH 2 each n. In particular, for each n there is a metric space n with V distance preserving inclusions n ֒ n and n ֒ n such that X →V X +1 →V 1 n n < n |X − X +1|Haus Vn 2 for each n. Moreover, we may assume that n = n n . V X ∪ X +1 Let us glue 1 to 2 along 2; to the obtained space glue 3 along , and so on.V TheV obtainedX metric space has an underlyingV set X3 W 44 LECTURE 5. SPACE OF SPACES

֒ formed by the disjoint union of all n such that each inclusion n → is distance preserving and X X ֒ →W 1 n n < n |X − X +1|Haus W 2 for each n. In particular,

➊ 1 m n < n−1 |X − X |Haus W 2 if m>n. Denote by ¯ the completion of . Observe that the union 1 W ➊ W 1 X ∪ 2 ... n is compact and implies that it forms a 2n−1 -net in ∪¯ X. Whence∪ ∪ X¯ is compact; see 1.6d and 1.7. W W Applying Blaschke selection theorem (4.6), we can pass to a subse- quence of ( n) that converges in Haus ¯ ; denote its limit by . Itre- X W X∞ mains to observe that is the Gromov–Hausdorff limit of ( n). X∞ X

Proof of 5.17; “only if” part. Suppose that there is no sequence εn 0 as described in 5.13. Observe that in this case there is a sequence→ of spaces n such that X ∈Q

pack n as n δ X → ∞ → ∞ for some fixed δ > 0. Since is compact, this sequence has a partial limit, say . Q X∞ ∈Q Observe that packδ ∞ = . Therefore, ∞ is not compact  a contradiction. X ∞ X

“If” part. Suppose sequence (εn) as in the definition of uniformly totally bonded families (5.13). Note that diam 6 ε for any . Given a positive integer n X 1 X ∈Q consider the set of all metric spaces n with the number of points at W most n and diameter 6 ε1. Note that n is a compact set in GH for each n. W Further, a subspace formed by a maximal εn-net of any X ∈Q belongs to n. Therefore, n is a compact εn-net in . That is, has a compactW ε-net forW any∩Qε> 0. Since is closed inQ a complete Qspace GH, it implies that is compact. Q Q 5.19. Exercise. Show that the space GH is (a) length, (b) geodesic.

5.20. Exercise. G. UNIVERSAL AMBIENT SPACE 45

(a) Show that

′ = inf ε> 0 : 6 + ε and 6 + ε |X − Y|GH { X Y Y X } defines a metric on the space of (isometry classes) of compact metric spaces.

(b) Moreover GH′ is equivalent to the Gromov–Hausdorff met- ric; that is,|∗ − ∗|

n 0 n ′ 0 |X − X∞|GH → ⇐⇒ |X − X∞|GH → as n . → ∞ G Universal ambient space

Recall that a metric space is called universal if it contains an isometric copy of any separable metric space (in particular, any compact metric space). Examples of universal spaces include Urysohn space and ℓ∞  the space of bounded infinite sequences with the metric defined by sup-norm; see 2.9 and 2.3. The following proposition says that the space in Definition 5.1 can be exchanged to a fixed universal space. W 5.21. Proposition. Let be a universal space. Then for any com- pact metric spaces and U we have X Y = inf ′ ′ |X − Y|GH {|X − Y |Haus U } where the greatest lower bound is taken over all pairs of sets ′ and ′ in which isometric to and respectively. X Y U X Y Proof of 5.21. By the definition (5.1), we have that

6 inf ′ ′ ; |X − Y|GH {|X − Y |Haus U } it remains to prove the opposite inequality. ′ ′ Suppse GH < ε; let , and be as in 5.1. We can assume that|X −= Y|′ ′; otherwiseX passY to theZ subspace ′ ′ of . In this case, Z isX compact;∪Y in particular, it is separable. X ∪Y Z Since isZ universal, there is a distance-preserving embedding of in ; letU us keep the same notation for ′, ′, and their images.Z It followsU that X Y ′ ′ < ε, |X − Y |Haus U  hence the result. 46 LECTURE 5. SPACE OF SPACES

5.22. Exercise. Let ∞ be the Urysohn space. Given two compact set A and B in defineU U∞

A B = inf A ι(B) ∞ , k − k {| − |Haus U } where the greatest lower bound is taken for all isometrics ι of ∞. Show that defines a pseudometric2 on nonempty compactU sub- sets of k∗−∗kand its corresponding metric space is isometric to GH. U∞ H Remarks

GH Suppose n , then there is a metric on the disjoint union X −−→X∞ X = G n X n∈N∪{∞} that satisfies the following property:

5.23. Property. The restriction of metric on each n and ∞ co- H X X incides with its original metric and n as subsets in X. X −→X∞ GH Indeed, since n , there is a metric on n = n such X −−→X∞ V X ⊔ X∞ that the restriction of metric on each n and coincides with its X X∞ original metric and n ∞ <εn for some sequence εn 0. |X − X |Haus Vn → Gluing all n along ∞, we obtain the required space X. V X GH In other words, the metric on X defines the convergence n GH X −−→ . This metric makes it possible to talk about limits of se- −−→X∞ quences xn n as n , as well as weak limits of a sequence of ∈ X → ∞ Borel measures µn on n and so on. For that reason, it isX useful to define convergence by specifying the metric on X that satisfies the property for the variation of Hausdorff convergence described in Section 4D. This approach is very flexible; in particular, it can be used to define Gromov–Hausdorff convergence of arbitrary metric spaces (net necessarily compact). In this case, a limit space for this generalized convergence is not uniquely defined. For example, if each space n in the sequence is isometric to the half-line, then its limit might beX isometric to the half- line or the whole line. The first convergence is evident and the second could be guessed from the diagram. Often the isometry class of the limit can be fixed by marking a point pn in each space n, it is called pointed Gromov–Hausdorff convergence X  we say that ( n,pn) converges to ( ,p ) if there is a metric on X X∞ ∞ 2The value kA − Bk is called Hausdorff distance up to isometry from A to B in U∞. H. REMARKS 47

X1

X2

...

X∞

H X such that n and pn p . For example, the sequence X −→X∞ → ∞ ( n,pn) = (R+, 0) converges to (R+, 0), while ( n,pn) = (R+,n) convergesX to (R, 0). X The pointed convergence works nicely only for proper metric spaces; the following theorem is an analog of Gromov’s selection theorem for this convergence. 5.24. Theorem. Let be a set of isometry classes of pointed proper metric spaces ( ,p). AssumeQ that for any R > 0, the R-balls in the spaces centeredX at the marked points form a uniformly totally bounded family of spaces. Then is precompact with respect to pointed Gromov– Hausdorff convergence.Q 48 LECTURE 5. SPACE OF SPACES Lecture 6

Ultralimits

Ultralimits provide a very general way to pass to a limit. This proce- dure works for any sequence of metric spaces, its result reminds limit in the sense of Gromov Gausdorff, but has some strange features; for example, the limit of a constant sequence does not coincide with this constant (see 6.10b). In geometry, ultralimits are used only as a canonical way to pass to a convergent subsequence. It is a useful thing in the proofs where one needs to repeat “pass to convergent subsequence” too many times. This lecture is based on the introduction to the paper by Bruce Kleiner and Bernhard Leeb [19].

A Faces of ultrafilters

Recall that N denotes the set of natural numbers, N = 1, 2,... { } 6.1. Definition. A finitely additive measure ω on N is called an ultrafilter if it satisfies the following condition: (a) ω(N)=1 and ω(S)=0 or 1 for any subset S N. An ultrafilter ω is called nonprincipal if in addition ⊂ (b) ω(F )=0 for any finite subset F N. ⊂ If ω(S)=0 for some subset S N, we say that S is ω-small. If ω(S)=1, we say that S contains ω⊂-almost all elements of N. Classical definition. More commonly, a nonprincipal ultrafilter is defined as a collection, say F, of sets in N such that 1. if P F and Q P , then Q F, 2. if P,Q∈ F, then⊃P Q F, ∈ 3. for any∈ subset P ∩N, either∈ P or its complement is an element of F. ⊂

49 50 LECTURE 6. ULTRALIMITS

4. if F N is finite, then F/ F. ⊂ ∈ Setting P F ω(P )=1 makes these two definitions equivalent. ∈ ⇔ A nonempty collection of sets F that does not include the empty set and satisfies only conditions 1 and 2 is called a filter; if in addition F satisfies condition 3 it is called an ultrafilter. From Zorn’s lemma, it follows that every filter contains an ultrafilter. Thus there is an ultrafilter F contained in the filter of all complements of finite sets; clearly, this ultrafilter F is nonprincipal. Stone–Cechˇ compactification. Given a set S N, consider subset ⊂ ΩS of all ultrafilters ω such that ω(S)=1. It is straightforward to check that the sets ΩS for all S N form a topology on the set of ultra- filters on N. The obtained space⊂ is called Stone–Cechˇ compactification of N; it is usually denoted as βN. Let ωn denotes the principal ultrafilter such that ωn( n )=1; that { } is, ωn(S)=1 if and only if n S. Note that n ωn defines a natural embedding N ֒ βN. Using∈ the described embedding,7→ we can (and will) consider N→as a subset of βN. The space βN is the maximal compact Hausdorff space that con- tains N as an everywhere dense subset. More precisely, for any compact Hausdorff space and a map f : N there is a unique continuous X → X map f¯: βN X such that the restriction f¯ N coincides with f. → | B Ultralimits of points

Further, we will need the existence of a nonprincipal ultrafilter ω, which we fix once and for all. Assume (xn) is a sequence of points in a metric space . Let us X define the ω-limit of (xn) as the point xω such that for any ε > 0, ω-almost all elements of (xn) lie in B(xω,ε); that is,

ω n N : xω xn <ε =1. { ∈ | − | } In this case, we will write

xω = lim xn or xn xω as n ω. n→ω → → For example, if ω is the principal ultrafilter such that ω n = 1 { } for some n N, then xω = xn. ∈ Alternatively, the sequence (xn) can be regarded as a map N . In this case, the map N can be extended to a continuous→ map X βN from the Stone–→Cechˇ X compactification βN of N. Then the → X ω-limit xω can be regarded as the image of ω. B. ULTRALIMITS OF POINTS 51

Note that ω-limits of a sequence and its subsequence may differ. n n For example, sequence yn = ( 1) is a subsequence of xn = ( 1) , but for any ultrafilter ω, we have− − −

lim xn = lim yn. n→ω 6 n→ω

6.2. Proposition. Let ω be a nonprincipal ultrafilter. Assume (xn) is a sequence of points in a metric space and xn xω as n ω. X → → Then xω is a partial limit of the sequence (xn); that is, there is a subsequence (xn)n∈S that converges to xω in the usual sense.

Proof. Given ε> 0, set Sε = n N : xn xω <ε . { ∈ | − | } Note that ω(Sε)=1 for any ε> 0. Since ω is nonprincipal, the set Sε is infinite. Therefore, we can choose an increasing sequence (nk) such that nk S 1 for each k N. Clearly, xn xω as k . ∈ k ∈ k → → ∞ The following proposition is analogous to the statement that any sequence in a compact metric space has a convergent subsequence; it can be proved the same way. 6.3. Proposition. Let be a compact metric space. Then any se- X quence of points (xn) in has a unique ω-limit xω. X In particular, a bounded sequence of real numbers has a unique ω-limit. The following lemma is an ultralimit analog of the Cauchy conver- gence test.

6.4. Lemma. Let (xn) be a sequence of points in a complete space . X Assume for each subsequence (yn) of (xn), the ω-limit

yω = lim yn n→ω ∈ X is defined and does not depend on the choice of subsequence, then the sequence (xn) converges in the usual sense.

Proof. If (xn) is not a Cauchy sequence, then for some ε> 0, there is a subsequence (yn) of (xn) such that xn yn > ε for all n. | − | It follows that xω yω > ε, a contradiction. | − | Ultralimits could shorten some proofs in the previous lecture. The following exercise provides an example. 6.5. Exercise. Use ultralimits to give a shorter proof of 5.2b (page 40). 52 LECTURE 6. ULTRALIMITS

6.6. Exercise. Denote by S the space of bounded sequences of real numbers. Show that there is a linear functional L: S R such that for any sequence s = (s ,s ,... ) S the image L(s) is→ a partial limit 1 2 ∈ of s1,s2,... .

C Ultralimits of spaces

Recall that ω denotes a nonprincipal ultrafilter on the set of natural numbers. Let n be a sequence of metric spaces. Consider all sequences of X points xn n. On the set of all such sequences, define a pseudometric by ∈ X ➊ (xn) (yn) = lim xn yn X . | − | n→ω | − | n Note that the ω-limit on the right-hand side is always defined and takes a value in [0, ]. (The ω-convergence to is defined analogously to the usual convergence∞ to ). ∞ ∞ Set ω to be the corresponding metric space; that is, the underlying X set of ω is formed by classes of equivalence of sequences of points X xn n defined by ∈ X

(xn) (yn) lim xn yn =0 ∼ ⇔ n→ω | − | and the distance is defined by ➊. The space ω is called the ω-limit of n. Typically ω will denote X X X the ω-limit of sequence n; we may also write X

n ω as n ω or ω = lim n. X → X → X n→ω X

Given a sequence xn n, we will denote by xω its equivalence ∈ X class which is a point in ω; it can be written as X

xn xω as n ω, or xω = lim xn. → → n→ω

6.7. Observation. The ω-limit of any sequence of metric spaces is complete.

Proof. Let n be a sequence of metric spaces and n ω as n ω. X X → X → Choose a Cauchy sequence x1, x2, ω. Passing to a subsequence, ···∈X1 we can assume that xk xm < for any k

1 inequality xn,k xn,m < k holds for ω-almost all n. It follows that we can choose| a− nested| sequence of sets

N = S S ... 1 ⊃ 2 ⊃ such that ω(Sm)=1 for each m, ⋄ ∅ Tm Sm = , and ⋄ 1 xn,k xn,l < for k

6.8. Observation. The ω-limit of any sequence of length spaces is geodesic.

Proof. If n is a sequence of length spaces, then for any sequence of X 1 pairs xn,yn Xn there is a sequence of -midpoints zn. ∈ n Let xn xω, yn yω and zn zω as n ω. Note that zω is a → → → → midpoint of xω and yω in ω. X By Observation 6.7, ω is complete. Applying Lemma 1.20 we get the statement. X

6.9. Exercise. Show that an ultralimit of metric trees is a metric tree.

D Ultrapower

If all the metric spaces in the sequence are identical n = , its ω-limit ω X X limn ω n is denoted by and called ω-power of . → X X X 6.10. Exercise. For any point x , consider the constant sequence ∈ Xω xn = x and set ι(x) = limn ω xn . → ∈ X (a) Show that ι: ω is distance-preserving embedding. (So we can and will considerX → X as a subset of ω.) X X (b) Show that ι is onto if and only if compact. X (c) Show that if is proper, then ι( ) forms a metric component of ω; that is,X a subset of ω thatX lie at a finite distance from a givenX point. X 54 LECTURE 6. ULTRALIMITS

Note that (b) implies that the inclusion ֒ ω is not onto if the space is not compact. However, the spacesX → Xand ω might be isometric;X here is an example: X X 6.11. Exercise. Let be a countable set with discrete metric; that is x y =1 if x = yX. Show that | − |X 6 (a) ω is not isometric to . X X (b) ω is isometric to ( ω)ω. X X 6.12. Observation. Let be a complete metric space. Then ω is geodesic space if and only ifX is a length space. X X Proof. The “if”-part follows from 6.8; it remains to prove the “only-if”- part Assume ω is geodesic space. Then any pair of points x, y X ω ∈ X has a midpoint zω . Fix a sequence of points zn such that ∈ X ∈ X zn zω as n ω. → → 1 1 Note that x zn x y and y zn x y as | − |X → 2 ·| − |X | − |X → 2 ·| − |X n ω. In particular, for any ε> 0, the point zn is an ε-midpoint of x and→ y for ω-almost all n. It remains to apply 1.20.

6.13. Exercise. Assume is a complete length space and p, q cannot be joined by a geodesicX in . Then there are at least two distinct∈ X geodesics between p and q in theX ultrapower ω. X 6.14. Exercise. Construct a proper metric space such that ω is not proper; that is, there is a point p ω and R

E Tangent and asymptotic spaces

Choose a space and a sequence of λn > 0. Consider the sequence X of scalings n = λn = ( , λn ). X ·X X ·|∗ − ∗|X Choose a point p and denote by pn the corresponding point ∈ X in n. Consider the ω-limit ω of n (one may denote it by λω ); X X X ·X set pω to be the ω-limit of pn. If λn 0 as n ω, then the metric component of pω in ω is → → λω ω X called λω-tangent space at p and denoted by T (or T if λn = n). p X p X If λn as n ω, then the metric component of pω is called → ∞ 1→ λω λω-asymptotic space and denoted by Asym or Asym . Note X X 1Often it is called asymptotic cone despite that it is not a cone in general; this name is used since in good cases it has a cone structure. F. REMARKS 55

that the space Asym and its point pω does not depend on the choice of p . X The∈ X following exercise states that the tangent and asymptotic spaces depend on the sequence λn and a nonprincipal ultrafilter ω. 6.15. Exercise. Construct a metric space with a point p such that λω X the tangent space Tp depends on the sequence λn and/or ultrafil- ter ω. X For nice spaces, different choices may give the same space. 6.16. Exercise. Let be the Lobachevsky plane; = Asym . (a) Show that is aL complete metric tree. T L (b) Show that T is one-point homogeneous; that is, given two points s,t thereT is an isometry of that maps s to t. (c) Show∈ Tthat has continuum degreeT at any point; that is, for any point t T the set of connected components of the complement t has∈ T cardinality continuum. (d)T\{ Prove} (a)–(c) if is Lobachevsky space and/or for the infinite 3-regular2 metricL tree with unit edge length.

F Remarks

A nonprincipal ultrafilter ω is called selective if for any partition of N into sets Cα α such that ω(Cα)=0 for each α, there is a set { } ∈A S N such that ω(S)=1 and S Cα is a one-point set for each α ⊂ . ∩ ∈The A existence of a selective ultrafilter follows from the continuum hypothesis [30]. For a selective ultrafilter ω, there is a stronger version of Propo- sition 6.2; namely we can assume that the subsequence (xn)n∈S can be chosen so that ω(S)=1. So, if needed, one may assume that the ultrafilter ω is chosen to be selective and use this stronger version of the proposition.

2that is, the degree of any vertex is 3. 56 LECTURE 6. ULTRALIMITS Appendix A

Semisolutions

1.2. Add four triangle inequalities (1.1d). 1.3;(a). Note that if µ(A)= µ(B)=0, then A B =0. Therefore, 1.1b does not hold for bounded closed subsets.| − It is| straightforward to check that for bounded measurable sets the remaining conditions in 1.1 hold true. (b). Note that distance from the empty set to the whole plane is infinite; so the value A B might be infinite. It is straightforward to check the remaining| conditions− | in 1.1. 1.4. Assume the statement is wrong. Then for any point x , there is a point x′ such that ∈ X ∈ X ρ(x) x x′ <ρ(x) and ρ(x′) 6 . | − | 1+ ε

′ Consider a sequence of points (xn) such that xn+1 = xn. Clearly,

ρ(x0) ρ(x0) xn xn 6 and ρ(xn) 6 . | +1 − | ε (1 + ε)n (1 + ε)n ·

Therefore, (xn) is Cauchy. Since is complete, the sequence (xn) X converges; denote its limit by x∞. Since ρ is a continuous function we get

ρ(x∞) = lim ρ(xn)= n→∞ =0.

The latter contradicts that ρ> 0.

57 58 APPENDIX A. SEMISOLUTIONS

1.5. Let ¯ be completion of . By the definition, for any y ¯ there X X ∈ X is a Cauchy sequence (xn) in that converges to y. X Choose a Cauchy sequence (ym) in ¯. From above, we can choose X points xn,m such that xn,m ym for any m. Choose zm = xnm,m ∈ X 1 → such that ym zm < . Observe that zm is Cauchy. Therefore, its | − | m limit z lie in ¯. Finally, show that xm z . ∞ X → ∞ 1.7. A compact ε-net N in contains a finite ε net F . Show and use that F is a 2 ε-net of . K · K 1.9. Given a pair of points x ,y , consider two sequences x , x ,... 0 0 ∈ K 0 1 and y0,y1,... such that xn+1 = f(xn) and yn+1 = f(yn) for each n. Since is compact, we can choose an increasing sequence of inte- K ∞ ∞ gers nk such that both sequences (xni )i=1 and (yni )i=1 converge. In particular, both are Cauchy; that is,

xn xn , yn yn 0 as min i, j . | i − j |K | i − j |K → { } → ∞ Since f is non-contracting, we get

x x 6 xn xn . | 0 − |ni−nj || | i − j |

It follows that there is a sequence mi such that → ∞

( ) xm x and ym y as i . ∗ i → 0 i → 0 → ∞ Set ℓn = xn yn . | − |K Since f is non-contracting, the sequence (ℓn) is nondecreasing.

By ( ), ℓmi ℓ0 as mi . It follows that (ℓn) is a constant sequence.∗ → → ∞ In particular,

x y = ℓ = ℓ = f(x ) f(y ) | 0 − 0|K 0 1 | 0 − 0 |K for any pair of points (x0,y0) in . That is, the map f is distance- preserving and, in particular, injective.K From ( ), we also get that f( ) is everywhere dense. Since is compact f∗: is surjective K hence the result. K K → K Remarks. This is a basic lemma in the introduction to Gromov– Hausdorff distance [see 7.3.30 in 7]. The presented proof is not quite standard, I learned it from Travis Morrison, a student in my MASS class at Penn State, Fall 2011. Note that this exercise implies that any surjective non-expanding map from a compact metric space to itself is an isometry . 59

1.10. Consider an infinite discrete space. 1.11. The conditions (a)–(c) in Definition 1.1 are evident. The triangle inequality (d) follows since ( )[B(x, π ) B(y, π )] [B(y, π ) B(z, π )] B(x, π ) B(z, π ). ∗ 2 \ 2 ∪ 2 \ 2 ⊇ 2 \ 2 π π Observe that B(x, 2 ) B(y, 2 ) does not π π\ overlap with B(y, 2 ) B(z, 2 ) and we get equal- ity in ( ) if and only\ if y lies on the great circle arc from∗ x to z. Therefore, the second state- ment follows. Remarks. This construction was given by Alek- sei Pogorelov [29]. It is closely related to the construction given by David Hilbert [15] which was the motivating example for his 4-th problem. 1.12. Without loss of generality, we may assume that the points p,x,y,z are distinct. Let K be the set in the tree covered by all six geodesics with the given endpoints. Show that K looks like an H or like an X; make a conclusion. 1.13. Apply 1.12. 1.17. Formally speaking, one-point space is an example, but there is not a trivial example as well. Consider the unit ball (B,ρ0) in the space c0 of all sequences con- verging to zero equipped with the sup-norm. Consider another metric ρ1 which is different from ρ0 by the con- formal factor ϕ(x)=2+ 1 x + 1 x + 1 x + ..., 2 · 1 4 · 2 8 · 3 where x = (x , x ... ) B. That is, if t x(t) for t [0,ℓ] is a curve 1 2 ∈ 7→ ∈ parametrized by ρ0-length, then its ρ1-length is defined by

ℓ length x := ϕ x(t) dt. ρ1 w ◦ · 0

Note that the metric ρ1 is bilipschitz to ρ0. Assume t x(t) and t x′(t) are two curves parametrized by 7→ 7→ ρ0-length that differ only in the m-th coordinate, denote by xm(t) ′ ′ and xm(t) respectively. Note that if xm(t) 6 xm(t) for any t and the ′ ′ function xm(t) is locally 1-Lipschitz at all t such that xm(t) < xm(t), then x′ x lengthρ1 6 lengthρ1 . 60 APPENDIX A. SEMISOLUTIONS

′ Moreover, this inequality is strict if xm(t) < xm(t) for some t. Fix a curve x(t), t [0,ℓ], parametrized by ρ -length. We can ∈ 0 choose large m so that xm(t) is sufficiently close to 0 for any t. In this case, it is easy to construct a function t x′ that meets the above 7→ m properties. It follows that for any curve x(t) in (B,ρ1), we can find ′ a shorter curve x (t) with the same endpoints. In particular, (B,ρ1) has no geodesics. Remarks. This solution was suggested by Fedor Nazarov [23].

1.18. Choose a sequence of positive numbers εn 0 and an εn-net → Nn of K for each n. Assume N0 is a one-point set, so ε0 > diam K. Connect each point x Nk to a point y Nk by a curve of length ∈ +1 ∈ at most εk. Consider the union K′ of all these curves with K; observe that K′ is compact and path-connected. Source: This problem was suggested by Eugene Bilokopytov [4].

1.19. Choose a Cauchy sequence (xn) in ( , ); it is sufficient X k∗−∗k to show that a subsequence of (xn) converges. Note that the sequence (xn) is Cauchy in ( , ); denote its X | ∗ − ∗ | limit by x∞. 1 Passing to a subsequence, we can assume that xn xn+1 < 2n . It follows that there is a 1-Lipschitz path γ in ( , k − ) suchk that 1 X k∗−∗k xn = γ( 2n ) for each n and x∞ = γ(0). It follows that

x∞ xn 6 length γ 1 6 k − k |[0, 2n ] 1 6 2n .

In particular, xn converges to x in ( , ). ∞ X k∗−∗k Source: [16, Corollary]; see also [25, Lemma 2.3]. 1.23. Consider the following subset of R2 equipped q with the induced length metric ... = (0, 1] 0, 1 1, 1 , 1 ,... [0, 1] X ×{ }
∪ { 2 3 }×
p Note that is locally compact and geodesic. Its completionX ¯ is isometric to the closure of equipped with the induced lengthX metric. Note that ¯ is obtained fromX by adding two points p = (0, 0) and q = (0, 1). X X Observe that the point p admits no compact neighborhood in ¯ and there is no geodesic connecting p to q in ¯. X X Source: [6, I.3.6(4)]. 61

1.24. If such a number does not exist, then the ranges of average distance functions have an empty intersection. Since is a compact length-metric space, the range of any continuous functionX on is a closed interval. By 1-dimensional Helly’s theorem, there is a pairX of such range intervals that do not intersect. That is, for two point-arrays (x1,...,xn) and (y1,...,ym) and their average distance functions

1 1 f(z)= X xi z X and h(z)= X yj z X , n · | − | m · | − | i j we have

( ) min f(z): z > max h(z): z . ∗ { ∈X} { ∈X} Note that

1 1 1 X f(yj )= X xi yj X = X h(xi); m · m·n · | − | n · j i,j i that is, the average value of f(yj ) coincides with the average value of h(xi), which contradicts ( ). ∗ Remarks. The value ℓ is uniquely defined; it is called the rendezvous value of . This is a result of Oliver Gross [13]. X 2.2. By Fr´echet lemma (2.1) we can identify with a compact subset of ℓ∞. K Denote by = Conv  it is defined as the minimal convex closed set in ℓ∞ thatL contains K. (In other words, is the minimal closed set containing such thatK if x, y , then t xL+ (1 t) y for any t [0, 1].) K ∈ L · − · ∈ L ∈ Observe that is a length space. It remains to show that is compact. L L By construction, is a closed subset of ℓ∞; in particular, it is a complete space. By 1.6Ld, it remains to show that is totally bounded. Recall that Minkowski sum A + B of two setsLA and B in a vector space is defined by

A + B = a + b : a A, b B . { ∈ ∈ } Observe that the Minkowski sum of two convex sets is convex. ∞ Denote by B¯ε the closed ε-ball in ℓ centered at the origin. Choose a finite ε-net N in for some ε> 0. Note that P = Conv N is a convex polyhedron; in particular,K Conv N is compact. Observe that N +B¯ε is closed ε-neighborhood of N. It follows that N + B¯ε K and therefore P + B¯ε . In particular, P is a 2 ε-net ⊃ ⊃ L · 62 APPENDIX A. SEMISOLUTIONS in ; since P is compact and ε> 0 is arbitrary, is totally bounded (seeL 1.7). L Remark. Alternatively, one may use that the injective envelope of a compact space is compact; see 3.5b, 3.11, and 3.13. 2.3. Modify the proof of 2.1. 2.10. Choose a separable space that has an infinite number of geo- desics between a pair of points withX the given distance between them; say a square in R2 with ℓ∞-metric will do. Apply to universality of Urysohn space (2.9). X 2.11. First let us prove the following claim: Suppose f : K R is an extension function defined on a compact ⋄ subset K of the→ Urysohn space . Then there is a point p such that p x = f(x) for anyUx K. ∈ U | − | ∈ Without loss of generality, we may assume that f(x) > 0 for any x K. Since K is compact, we may fix ε> 0 such that f(x) >ε. ∈ ε Consider the sequence εn = 100·2n . Choose a sequence of εn-nets Nn K. Applying universality of recursively, we may choose a ⊂ U point pn such that pn x = f(x) for any x Nn and pn pn = | − | ∈ | − −1| = 10 εn−1. Observe that the sequence (pn) is Cauchy and its limit p meets· p x = f(x) for any x K. | − | ∈ Now, choose a sequence of points (xn) in . Applying the claim, S we may extend the map from K to K x1 , further to K x1, x2 , and so on. As a result, we extend the∪{ distance-preserving} ∪{ map f to} the whole sequence (xn). It remains to extend it continuously to the whole space . S 2.13. It is sufficient to show that any compact subspace of the Urysohn space can be contracted to a point. K U Note that any compact space can be extended to a contractible compact space ′; for example, weK may embed into ℓ∞ and pass to its convex hull,K as it was done in 2.2. K By 2.17, there is an isometric embedding of ′ that agrees with the inclusion ֒ . Since is contractible in K′, it is contractible in . K → U K K U A better way. One can contract the whole Urysohn space using the following construction. Note that points in ∞ constructed in the proof of 2.6 can be multiplied number t [0X, 1]  simply multiply each function by t. That defines a map ∈

λt : X∞ → X∞ 63

that scales all distances by factor t. The map λt can be extended to the completion of , which is isometric to d (or ). X∞ U U Observe that the map λ1 is the identity and λ0 maps the whole space to a single point, say x  this is the only point of . Further, 0 X0 note that (t,p) λt(p) is a continuous map; in particular, d and are contractible.7→ U U As a bonus, observe that for any point p d the curve t λt(p) ∈U 7→ is a geodesic path from p to x0. Source: [12, (d) on page 82]. 2.16; (a) and (b). Observe that L and M satisfy the definition of d-Urysohn space and apply the uniqueness (2.14). Note that ℓ = diam L = min 2 r, d . { · } (c). Use (a), maybe twice. 2.18; (a). The euclidean plane is homogeneous in every sense. (b). The Hilbert space ℓ2 is finite-set homogeneous, but not compact set homogeneous, nor countable homogeneous. (c). The space ℓ∞ is 1-point homogeneous, but not 2-point homoge- neous. Try to show that there is no isometry of ℓ∞ such that (0, 0, 0,... ) (0, 0, 0,... ), 7→ (1, 1, 1,... ) (1, 0, 0,... ). 7→ (d). The space ℓ1 is 1-point homogeneous, but not 2-point homoge- neous. Try to show that there is no isometry of ℓ∞ such that (0, 0, 0,... ) (0, 0, 0,... ), 7→ (2, 0, 0 ... ) (1, 1, 0,... ). 7→ 3.2. Note that if c< 0, then r>s. The latter is impossible since r is extremal and s is admissible. Observe that the function r¯ = min r, s + c is admissible. Indeed, choose x, y . If r¯(x)= r(x) and r¯({y)= r(y}), then ∈ X r¯(x)+¯r(y)= r(x)+ r(y) > x y . | − | Further, if r¯(x)= s(x)+ c, then r¯(x)+¯r(y) > [s(x)+ c] + [s(y) c]= − = s(x)+ s(y) > > x y . | − | 64 APPENDIX A. SEMISOLUTIONS

Since r is extremal, we have r =r ¯; that is, r 6 s + c. 3.5. Choose an injective space . Y (a). Fix a Cauchy sequence (xn) in ; we need to show that it has a limit x . Consider metric on Y= N defined by ∞ ∈ Y X ∪ {∞}

m n := xm xn , | − |X | − |Y m X := lim xm xn Y . | − ∞| n→∞ | − |

Since the sequence is Cauchy, so is the sequence ℓn = xm xn Y for any m. Therefore, the last limit is defined. | − | By construction, the map n xn is distance-preserving on N . Since is injective, this map can7→ be extended to as a short⊂ map; X Y ∞ set x . Since xn x 6 n and n 0, we ∞ 7→ ∞ | − ∞|Y | − ∞|X | − ∞|X → get that xn x as n . → ∞ → ∞ (b). Applying the definition of injective space, we get a midpoint for any pair of points in . By (a), is a complete space. It remains to apply 1.20b. Y Y -c). Let k : ֒ ℓ∞( ) be the Kuratowski embedding (2.4). Ob) serve that ℓ∞( Y) →is contractible;Y in particular, there is a homotopy ∞ Y ,kt : ֒ ℓ ( ) such that k = k and k is a constant map. (In fact Y → Y 0 1 one can take kt = (1 t) k.) − · Since k is distance-preserving and is injective, there is a short map f : ℓ∞( ) such that the compositionY f k is the identity Y → Y ◦ .map on . The composition f kt : ֒ is a needed homotopy Y ◦ Y → Y 3.6. Suppose that a short map f : A is defined on a subset A of a metric space . We need to construct→ Y a short extension F of f. Without loss ofX generality, we may assume that A = ∅, otherwise map the whole to a single point. By Zorn’s lemma,6 it is sufficient to enlarge A byX a single point x / A. ∈ (a). Suppose = R. Set Y F (x) = inf f(a) a x : a A . { − | − | ∈ } Observe that F is short and F (a)= f(a) for any a A. ∈ (b). Suppose is a complete metric tree. Fix points p and q . Y ∈ X ∈ Y Given a point a A, let xa B[f(a), a p ] be the point closest to ∈ ∈ | − | f(x). Note that xa [qf(a)] and either xa = q or xa lies on distance a p from f(a). ∈ | − | Note that the geodesics [q xa] are nested; that is, for any a,b A ∈ we have either [q xa] [q xb] or [q xb] [q xa]. Moreover, in the first ⊂ ⊂ case we have xb f(a) 6 p a and in the second xa f(b) 6 p b . | − | | − | | − | | − | 65

It follows that the closure of the union of all geodesics [q xa] for a is a geodesic. Denote by x its endpoint; it exists since is complete.∈ A It remains to observe that x f(a) 6 p a forY any a ; that is, one can take f(p)= x. | − | | − | ∈ A (c). In this case, = (R2,ℓ∞). Note that (R2,ℓ∞) is a short map if and only ifY both of its coordinate projectionsX → are short. It remains to apply (a). More generally, any ℓ∞-product of injective spaces is injective; in particular, if and are injective then the product equipped with the metricY Z Y × Z

(y,z) (y′,z′) = max y y′ , z z′ | − |Y×Z { | − |Y | − |Z } is injective as well. 3.7. Choose three points x,y,z and set = x, z . Let f : be an isometry. Then F (y∈) X= f(x) or AF (y){ = f}(z). If f(Ay) → = =→f Y(x), then

y z > F (y) f(z) = | − |X | − |Y = x z . | − |X

Analogously if f(y) = f(z), then x y X > x z X . The strong triangle inequality holds in both cases.| − | | − | 3.9. Apply 3.8c.

3.10. Denote by d the d-Urysohn space, so ∞ is the Urysohn space. The extensionU property implies finite hyperconvexity.U It remains to show that d is not countably hyperconvex. U Suppose that d< . Then diam d = d and for any point x d ∞ U ∈U there is a point y d such that x y = d. It follows that there ∈ U | − |Ud is no point z such that z x 6 d for any x . Whence d Ud 2 d d is not countably∈U hyperconvex.| − | ∈U U Use 2.16b to reduce the case d = to the case d< . ∞ ∞ 3.11. Observe and use that the functions in Ext are 1-Lipschitz and uniformly bounded functions. X 3.12; (a). Let f be an extremal function. Observe that at least two of the numbers f(a)+ f(b), f(b)+ f(c), and f(c)+ f(a) are 1. It follows that for some x [0, 1 ], we have ∈ 2 f(a)=1 x, f(b)=1 x, f(c)=1 x, ± ± ± where we have one “minus” and two “pluses” in these three formulas. 66 APPENDIX A. SEMISOLUTIONS

Suppose that

g(a)=1 y, g(b)=1 y, g(c)=1 y ± ± ± is another extremal function. Then f g = x y if g has “minus” at the same place as f and f g =| x−+ y| otherwise.| − | It| follows− | that| Ext| is isometric to a tripod a X 1  three segments of length 2 glued at one end. (b). Assume f is an extremal function. Observe that f(x)+ f(y) = f(p)+ f(q)=2; in partic- c b ular, two values a = f(x) 1 and b = f(p) 1 completely describe the function− f. Since f−is x extremal, we also have that

(1 a)+(1 b) > 1 q p ± ± y for all 4 choices of signs; equivalently, a + b 6 1. | | | | It follows that Ext is isometric to the rhombus a + b 6 1 in the (a,b)-plane with theX metric induced by the ℓ∞-norm.| | | | 3.15. Recall that

f g = sup f(x) g(x) : x | − |Ext X { | − | ∈X} and f p = f(p) | − |Ext X for any f,g Ext and p . Since ∈is compactX we can∈ X find a point p such that X ∈ X f g = f(p) g(p) = f p g p . | − |Ext X | − | || − |Ext X − | − |Ext X | Without loss of generality, we may assume that

f p = g p + f g . | − |Ext X | − |Ext X | − |Ext X Applying 3.3c, we can find a point q such that ∈ X q p = f p + f q , | − |Ext X | − |Ext X | − |Ext X whence the result. Since Ext is injective (3.13), by 3.5b it has to be geodesic. It remains to noteX that the concatenation of geodesics [pq], [gf], and [fq] is a required geodesic [pq]. 67

3.17. Show that there is a pair of short maps Ext Ext Ext such that their composition is identity of . MakeX → a conclusion.U → X X ′ 4.3. Suppose that A B Haus X < r. Choose a pair of points a,a A on maximal distance| − from| each other. Observe that there are points∈ b,b′ B such that a b , a′ b′ < r. Whence ∈ | − |X | − |X a a′ b b′ 6 2 r | − |X − | − |X · and therefore

diam A diam B 6 2 A B . − ·| − |Haus X It remains to swap A and B and repeat the argument. 4.4; (a). Denote by Ar the closed r-neighborhood of a set A R2. Observe that ⊂ (Conv A)r = Conv(Ar), and try to use it. (b). The answer is “no” in both parts. For the first part let A be a unit disk and B a finite ε-net in A. Evidently, A B R2 <ε, but ∂A ∂B R2 1. | − |Haus | − |Haus ≈ For the second part take A to be a unit disk and B = ∂A to be its boundary circle. Note that ∂A = ∂B; in particular, ∂A | − ∂B 2 =0 while A B 2 =1. − |Haus R | − |Haus R Remark. There is the so-called lakes of Wada  an example of three (and more) open bounded topological disks in the plane that have identical boundaries. It can be used to construct more interesting examples for (b). 4.9. Show that for any ε > 0 there is a positive integer N such that Sn6N Kn is an ε-net in the union Sn Kn. Observe that Sn6N Kn is compact and apply 1.7. 4.10; “if” part. Choose two compact sets A, B ; suppose that A B < r. ⊂ X | − |Haus X Choose finite ε-nets a ,...am A and b ,...bn B. For { 1 } ⊂ { 1 } ⊂ each pair ai,bj construct a constant-speed path γi,j from ai to bj such that length γi,j < ai bj + ε. | − | Set C(t)= γi,j (t): ai bj < r + ε . { | − |X } Observe that C(t) is finite; in particular, it is compact. 68 APPENDIX A. SEMISOLUTIONS

Show and use that

A C(t)

area Ar = area A + r perim A + π r2. · · Taking derivative and applying the coarea formula, we get

perim Ar = perim A +2 π r. · · Observe that if A lies in a compact set B bounded by a closed curve, then perim A 6 perim B. Indeed the closest-point projection R2 A is short and it maps ∂B onto ∂A. → It remains to use the following observation: if An A∞, then for any r> 0 we have that →

r r A An and A A ∞ ⊃ ∞ ⊂ n for all large n. 4.13. Note that almost all points on ∂D have a defined tangent line. In particular, for almost all D b a pairs of points a,b ∂D the two angles α and β α between the chord [∈ab] and ∂D are defined. β The convexity of D′ implies that α = β; here we measure the angles α and β on one side from [ab]. Show that if the identity α = β holds for almost all chords, then D is a round disk.

4.15. Observe that all functions distAn are Lipschitz and uniformly bounded on compact sets. Therefore, passing to a subsequence, we may assume that the sequence distAn converges to some function f. −1 Set A = f 0 . It remains to show that f = distA∞ . ∞ { } 5.3; (a). Apply the definition for space obtained from by adding one point on distance 1 diam to eachZ point of . X 2 · X X 69

(b). Given a point x , denote by a x and b x the corresponding points in a and b ∈respectively. X Show· that· there is a metric on = a ·Xb such·X that Z ·X ⊔ ·X a x b x = |b−a| diam | · − · |Z 2 · X -for any x and the inclusions a ֒ , b ֒ are distance pre serving. ·X → Z ·X → Z

5.4. Arguing by contradiction, we can identify r and r with sub- spaces of a space such that A B Z 1 r r < |A −B |Haus Z 10 for large r; see the definition of Gromov–Hausdorff metric (5.1). Set n = r . Note that there are 2 n integer points in r: a = ⌈ ⌉ · A 1 = (0, 0), a2 = (1, 0),...,a2·n = (n, 1). Choose a point bi r that ∈4 B lies at the minimal distance from ai. Note that bi bj > 5 if i = j. 4 | − | 6 It follows that r > 5 (2 n 1). The latter contradicts n = r for large r. · · − ⌈ ⌉

1 Remark. Try to show that r r = for all large r. |A −B |GH 2 5.5. Suppose that <ε; we need to show that |X − Y|U ˆ ˆ < 2 ε. |X − Y|GH · Denote by ˆ the injective envelope of . Recall that , , and can be consideredU as subspaces of ˆ, ˆ, andU ˆ respectively.U X Y U X Y According to 3.17, the inclusions ֒ and ֒ can be .ˆ ֒ ˆextended to a distance-preserving inclusionsX → Uˆ ֒ Yˆ and→ U Therefore, we can and will consider ˆ and ˆ asX subspaces→ U ofYˆ.→ U X Y U Given f ˆ, let us find g ˆ such that ∈ U ∈ X ➊ f(u) g(u) < 2 ε | − | · for any u . Note that the restriction f X is admissible on . By 3.3d, there∈ is Ug ˆ such that | X ∈ X ➋ g(x) 6 f(x) for any x . ∈ X Recall that any extremal function is 1-Lipschitz; in particular, f and g are 1-Lipschitz on . Therefore, ➋ and <ε imply that U |X −Y|U g(u) f(u) 2 ε − · for any u . Whence ➊ follows. It follows∈U that ˆ lies in a 2 ε-neighborhood of ˆ in ˆ. The same way we show that Yˆ lies in a 2·ε-neighborhood of Xˆ in Uˆ. The latter means that ˆ ˆX < 2 ε·, and therefore ˆ Y ˆ U < 2 ε. |X − Y|Haus Uˆ · |X − Y|GH · Remark. This problem was discussed by Urs Lang, Ma¨el Pav´on, and Roger Z¨ust [21, 3.1]. They also show that the constant 2 is opti- 1 1 s 1 1

s

4 ℓℓ11

mal. To see this, look at the injective envelopes of two 4-point metric spaces shown on the diagram and observe that the Gromov–Hausdorff distance between the 4-point metric spaces is 1, while the distance between their injective envelopes approaches 2 as s . → ∞ 5.7; only-if part. Let us identify and with subspaces of a metric space such that X Y Z < ε. |X − Y|Haus Z Set x y if and only if x y Z <ε. It remains to check that is an ε-approximation.≈ | − | ≈ If part. Show that we can assume that

R = (x, y) : x y { ∈X×Y ≈ } is a compact subset of . Conclude that X × Y ′ ′ ′ x x X y y Y < 2 ε | − | − | − | · for some ε′ <ε. Show that there is a metric on = such that the inclusions Z X ⊔ Y ′ .and ֒ are distance preserving and x y Z = ε if x y ֒ ConcludeX → Z thatY → Z | − | ≈ 6 ε′ < ε. |X − Y|Haus Z 71

5.9; (a). Let be an ε-approximation provided by 5.7. For any x choose a point≈f(x) such that x f(x). Show that x f(x∈) Xis an 2 ε-isometry. ∈ Y ≈ 7→ ·

(b). Let x and y . Set x y if y f(x) Y <ε. Show that is a ε-approximation.∈ X ∈ Apply Y 5.7. ≈ | − | ≈

GH 5.11; (a). Suppose n and n are simply connected length metric space. It is sufficientX −−→X to show thatX any nontrivial covering map f : ˜ corresponds to a nontrivial covering map fn : ˜n n for largeXn →. X X → X The latter can be constructed by covering n by small balls that lie close to sets in evenly covered by f, prepareX a few copies of these sets and glue themX the same way as the inverse images of the evenly covered sets in glued to obtain ˜. X X (b). Let be a cone over Hawaiian ear- rings. ConsiderV the doubled cone  two copies of with glued theirW base points (see theV diagram). The space can be equipped with a length metricW (for example, the induced length metric from the shown embed- ding). Show that is simply connected, but is not; the latterV is a good exercise in topology. W If we delete from the earrings all small circles, then the obtained double cone becomes simply connected and it remains to be close to . That is is a Gromov–Hausdorff limit of simply connected spaces.W W Remark. Note that from part (b), the limit does not admit a nontrivial covering. So, if we define the fundamental group as the inverse image of groups of deck transformations for all the coverings of the given space, then one may say that Gromov–Hausdorff limit of simply connected length spaces is simply connected. 5.12, (a). Suppose that a metric on S2 is close to the disk D2. Note that S2 contains a circle γ that is close to the boundary curve of D2. 2 By the Jordan curve theorem, γ divides S into two disks, say D1 and D2. By 5.11b, the Gromov–Hausdorff limits of D1 and D2 have to con- tain the whole D2, otherwise the limit would admit a nontrivial cov- ering. Consider points p1 D1 and p2 D2 that are close to the center of D2. If n is large, the∈ distance p ∈ p has to be very small. On | 1 − 2|n 72 APPENDIX A. SEMISOLUTIONS

the other hand, any curve from p1 to p2 must cross γ, so it has length about 2  a contradiction. 5.14. Apply 1.8. 5.15. Choose a space in (C,D), denote a C-doubling measure by µ. Without loss of generality,X Q we may assume that µ( )=1. The doubling condition implies that X

D 1 µ[B(p, 2n )] > Cn for any point x . It follows that ∈ X n pack D 6 C . 2n X D By 1.8, for any ε > 2n−1 , the space admits an ε-net with at most Cn points. Whence (C,D) is uniformlyX totally bounded. Q 5.16; (a). Choose ε > 0. Since is compact, we can choose a finite Y ε-net y1,...,yn in . Suppose{ f : } Y be a distance-nondecreasing map. Choose one X → Y −1 point xi in each nonempty subset Bi = f [B(yi,ε)]. Note that the subset Bi has diameter at most 2 ε and ·

= [ Bi. X i

Therefore, the set of points xi is a 2 ε-net in . { } · X (b). Let be a uniformly totally bounded family of spaces. Suppose Q 1 that each space in has an n -net with at most Mn points; we may Q 2 assume that M0 =1. Consider the space of all infinite integer sequences m ,m ,... Y 0 1 such that 1 6 mn 6 Mn for any n. Given two sequences (ℓn), and (mn) of points in , set Y C (ℓn) (mn) = n , | − |Y 2 where n is the minimal index such that ℓn = mn and C is a positive constant. 6 Observe that is compact. Indeed it is complete and the sequences Y C with constant tails, starting from index n, form a finite 2n -net in . 1 Y Given a space in , choose a sequence of n nets Nn X Q 2 ⊂ X for each n. We can assume that Nn 6 Mn; let us label the points | | in Nn by 1,...,Mn . Consider the map f : defined by { } X → Y f : x (m1(x),m2(x),... ) where mn(x) is the label of a point in Nn → 1 that lies on the distance < 2n from x. 73

1 ′ 1 ′ If 2n−2 > x x X > 2n−1 , then mn(x) = mn(x ). It follows that ′ | − C| 6 f(x) f(x ) > n . In particular, if C > 10, then | − |Y 2 f(x) f(x′) > x x′ | − |Y | − |X for any x, x′ . That is, f is a distance-nondecreasing map . ∈ X X → Y 5.19; (a) Apply 4.10, 5.21, 5.18, and 1.20. (b). Choose two compact metric spaces and . Show that there are subsets ′, and ′ in the Urysohn spaceX thatY isometric to and respectivelyX andY such that U X Y

= ′ ′ . |X − Y|GH |X − Y |Haus U

Further, construct a sequence of compact sets n such that 1 ′ ′ Z ⊂ U n is an n -midpoint of , and in Haus and Z 2 X Y U 1 n n < n |Z − Z +1|Haus U 2 for any n. Observe that the sequence n converges in GH, and its limit by is a midpoint of and . Finally,Z apply 5.18 and 1.20. Z X Y Source: [18]. (b). Make fine burrows in the standard 3-ball without changing its topology, but at the same time come sufficiently close to any point in the ball. Consider the doubling of the obtained ball along its boundary; that is, two copies of the ball with identified corresponding points on their boundaries. The obtained space is homeomorphic to S3. Note that the burrows can be made so that the obtained space is sufficiently close to the original ball in the Gromov–Hausdorff metric. Source: [7, Exercises 7.5.13 and 7.5.17].

5.20; (a). To check that GH′ is a metric, it is sufficient to show that |∗ − ∗| iso ′ =0 = == ; |X − Y|GH ⇒ X Y the remaining conditions are trivial.

If GH′ = 0, then there is a sequence of maps fn : such that|X − Y| X → Y ′ ′ 1 fn(x) fn(x ) > x x . | − |Y | − |X − n Arguing the same way as in the proof of the “only if”-part in 5.2b (page 40), we get a distance-nondecreasing map f : . ∞ X → Y 74 APPENDIX A. SEMISOLUTIONS

The same way we can construct a distance-nondecreasing map g : . ∞ Y → X By 1.9, the compositions f g : and g f : ∞ ◦ ∞ Y → Y ∞ ◦ ∞ X → X are isometries. Therefore, f∞ and g∞ are isometries as well. (b). The implication

n 0 n ′ 0 |X − X∞|GH → ⇒ |X − X∞|GH → follows from 5.9a.

Now suppose n ∞ GH′ 0. Show that n is a uniformly totally bonded family.|X − X | → {X } If n 0, then we can pass to a subsequence such that |X − X∞|GH 6→ n > ε for some ε > 0. By Gromov selection theorem, |X − X∞|GH we can assume that n converges in the sense of Gromov–Hausdorff. X ′ From the first implication, the limit ∞ has to be isometric to ∞; on the other hand, ′ > εX a contradiction. X |X∞ − X∞|GH 5.22. Apply 2.17 and 5.21.

1 6.5. Suppose that fn : is a -isometry between compact spaces X → Y n for each n N. Consider the ω-limit fω of fn, ∈

fω(x) = lim fn(x); n→ω according to 6.3, fω is defined. Since

′ ′ 1 fn(x) fn(x ) ≶ x x | − | | − |± n we get that ′ ′ fω(x) fω(x ) = x x | − | | − | ′ for any x, x ; that is, fω is distance-preserving. Further, since fn 1 ∈ X 1 is a n -isometry, for any y there is xm such that fn(xn) y 6 n . Therefore, ∈ Y | − | fω(xω)= y, where xω is the ω-limit of xn; that is, fω is onto. It follows that fω : is an isometry. X → Y 6.6. Choose a nonprincipal ultrafilter ω and set L(s)= sω. It remains to observe that L is linear. 6.9. Let γ be a path from p to q in a metric tree . Assume that γ passes thru a point x on distance ℓ from [pq]. ThenT

➌ length γ > p q +2 ℓ. | − | · 75

Suppose that n is a sequence of metric trees that ω-converges to T ω. By 6.8, the space ω is geodesic. T The uniqueness of geodesicsT follows from ➌. Indeed, if for a geodesic [pωqω] there is another geodesic γω connecting its ends, then it has to pass thru a point xω / [pωqω]. Choose sequences pn, qn, xn n such ∈ ∈ T that pn pω, qn qω, and xn xω as n ω. Then → → → →

pω qω = length γ > lim ( pn xn + qn xn ) > | − | n→ω | − | | − | > lim ( pn qn +2 ℓn)= n→ω | − | · = pω qω +2 ℓω. | − | ·

Since xω / [pωqω], we have that ℓω > 0  a contradiction. ∈ It remains to show that any geodesic triangle ω is a tripod. Con- T sider the sequence of centers of tripods mn for a given sequences of points xn,yn,zn n. Observe that its ultralimit mω is the center of ∈ T the tripod with ends at xω,yω,zω ω. ∈ T 6.10. Further, we consider as a subset of ω. X X (a). Follows directly from the definitions.

(b). Suppose compact. Given a sequence (xn) in , denote its ω X ω X ω-limit in by x and its ω-limit in by xω. X ω X Observe that x = ι(xω ). Therefore, ι is onto. If is not compact, we can choose a sequence (xn) such that X xm xn >ε for fixed ε> 0 and all m = n. Observe that | − | 6 ε lim xn y > n→ω | − |X 2 ε for any y . It follows that xω lies on the distance at least from . ∈ X 2 X (c). A sequence of points (xn) in will be called ω-bounded if there is a real constant C such that X

p xn 6 C | − |X for ω-almost all n. The same argument as in (b) shows that any ω-bounded sequence has its ω-limit in . Further, if (xn) is not ω-bounded, then X

lim p xn X = ; n→ω | − | ∞ ω that is, xω does not lie in the metric component of p in . X 6.11. Show and use that the spaces ω and ( ω)ω have discrete metric and both have cardinality continuum.X X 76 APPENDIX A. SEMISOLUTIONS

... v0 v1 v2 v3 v4

6.13. Apply 6.4 and 6.12. 6.14. Consider the infinite metric tree with unit edges shown on the diagram. Observe that is proper.T T ω Consider the vertex vω = limn→ω vn in the ultrapower . Observe that ω has an infinite degree. Conclude that ω is not locallyT compact. T 6.15. Consider a product space [0, 1] [0, 1 ] [0, 1 ] ... . × 2 × 4 × 6.16; (a). Show that there is δ > 0 such that sides of any geodesic tri- angle intersect a disk of radius δ. Conclude that any geodesic triangle in Asym is a tripod. Make a conclusion. L (b). Observe that L is one-point homogeneous and use it.

(c). By (b), it is sufficient to show that pω has a continuum degree. Choose distinct geodesics γ ,γ : [0, ) L that start at a point 1 2 ∞ → p. Show that the limits of γ1 and γ2 run in the different connected components of (Asym ) pω . Since there is a continuum of distinct L \{ } geodesics starting at p, we get that the degree of pω is at least contin- uum. On the other hand, the set of sequences of points in L has car- dinality continuum. In particular, the set of points in Asym has cardinality at most continuum. It follows that the degree of anyL ver- tex is at most continuum. (d). The proof for the Lobachevsky space goes along the same lines. For the infinite 3-regular tree, part (a) follows from 6.9. The 3- regular tree is not one-point homogeneous, but it is vertex homoge- neous; the latter is sufficient to prove (b). No changes are needed in (c). Remark. Anna Dyubina and Iosif Polterovich [9] proved that the prop- erties (b) and (c) describe the tree up to isometry. In particular, the asymptotic space of the LobachevskyT plane does not depend on the choice of ultrafilter and the sequence λn . → ∞ Index

[∗∗], 9 geodesic path, 9 I, 9 gluing, 43 ε-midpoint, 13 λω-asymptotic space, 54 Hausdorff convergence, 35 λω-tangent space, 54 Hausdorff distance, 31 ω-almost all, 49 homogeneous, 21 ω-limit, 50 hyperconvex hull, 30 ω-limit space, 52 hyperconvex space, 26 ω-small, 49 1-Lipschitz function, 24 induced length metric, 12 injective envelope, 29 admissible function, 23 injective space, 25 almost isometry, 39 isometry, 8 isoperimetric inequality, 34 bounded space, 16 Lebesgue number, 8 closed ball, 6 length, 10 complete space, 7 length metric, 12 completion, 7 length space, 12 convergence in the sense of Haus- Lipschitz function, 32 dorff, 32 locally compact space, 8 curve, 10 maximal packing, 8 diameter, 15, 31 metric, 5 distance function, 5 ∞-metric, 6 double, 43 metric component, 7 doubling, 73 metric space, 5 doubling measure, 42 metric tree, 10 doubling space, 42 midpoint, 13 extension, 17 net, 8 extension function, 17 nonprincipal ultrafilter, 49 extension property, 17 norm, 15 extremal function, 23 one-point extension, 17 filter, 50 open ball, 6 geodesic, 9 partial limit, 33

77 78 INDEX path, 10 pointed convergence, 46 proper function, 9 proper space, 9 pseudometric, 6 rectifiable curve, 10 rendezvous value, 61 scaled space, 38, 54 selective ultrafilter, 55 separable space, 16 short map, 16 Stone–Cechˇ compactification, 50 sup norm, 15 tight span, 30 totally bounded space, 8 ultrafilter, 49, 50 nonprincipal ultrafilter, 49 selective ultrafilter, 55 ultralimit, 50 ultrametric space, 25 uniformly totally bonded family, 42 Urysohn space, 17 Bibliography

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