Metric spaces

Metric spaces N. W. Sauer University of Calgary, Department of Mathematics and Statistics Calgary, Alberta, Canada T2N 1N4 [email protected] Abstract Lecture notes for a course on combinatorial aspects of metric spaces at the University of Calgary in the Fall of 2007.

1

Basic notions

A metric space on a set M is a function d : M 2 → R≥0 so that for all x, y, z ∈ M : 1. d(x, y) = 0 if and only if x = y. 2. d(x, y) = d(y, x). 3. d(x, y) + d(y, z) ≥ d(x, z). It is usual to speak of a metric space M and assume tacitly that the distance function is denoted by d. Actually, a metric space is a function d with domain some set M 2 and range R≥0 for which the three axioms above hold.

1.1

Examples of metric spaces

Any set M of numbers with d(x, y) = |x − y|. The space Rn with pthe Pn Euclidean 2distance d(x, y) = kx − yk = i=1 (xi − yi ) . n The space R with the supremum norm ds (x, y) = max{|xi − yi | : 1 ≤ i ≤ n}. 1

More generally, if (di ; i ∈ I) is a family of metric spaces all with the same base set M , then d : M → R≥0 given by d(x, y) = {sup{di (x, y) : i ∈ I} for all x, y ∈ M is a metric on M . Let (di : (Mi )2 → R≥0 ; i ∈ I) be a family of metric spaces. Let d :
2
Q M → R be given by d (x ; i ∈ I), (y ; i ∈ I) i ≥0 i i i∈I Q = sup{di (xi , yi ) : i ∈ I} for xi , yi ∈ Mi all i ∈ I. Then d is a metric on i∈I Mi . Let N = NN be the set of all sequences s = (si ; i ∈ N) whose entries 1 are non negative integers. For any two sequences s, t ∈ N let d(s, t) = δ(s,t) where δ(s, t) is the smallest number i with si 6= ti . The set N with the topology induced by this metric is the Baire space. The subspace C = 2N is he Cantor space. Let M be a set and d : M 2 → R≥0 . The function d is an ultrametric on M if for all x, y, z ∈ M : 1. d(x, y) = 0 if and only if x = y. 2. d(x, y) = d(y, x). 3. max{d(x, y), d(y, z)} ≥ d(x, z). Every ultrametric space is a metric space. The metric space N above is an ultrametric space. The length of a shortest path between vertices of a graph is a metric distance function on the graph. If M is a metric space with distance function d and N ⊆ M then the restriction of d to N 2 is a metric distance function on N . We say N is a subspace of M . An inner product on a vector space V over R is a function h , i : V ×V −→ R such that, for all k1 , k2 ∈ R and v1 , v2 , v, w ∈ V , the following properties hold: 1. hk1 v1 + k2 v2 , wi = k1 hv1 , wi + k2 hv2 , wi. 2. hv, wi = hw, vi. 3. hv, vi ≥ 0, and hv, vi = 0 if and only if v = 0. The standard example of an inner product is the dot product on Rn : ((x1 , . . . , xn ), (y1 , . . . , yn )) :=

n X i=1

2

xi yi .

An inner product space (or pre-Hilbert space) is a vector space R with an inner product h·, ·i. A norm on a real vector space V is a function k · k : X → R such that for all v, w ∈ V and a ∈ R. 1. kvk ≥ 0 and kvk = 0 if and only if v = 0. 2. kavk = |a|kvk. 3. kv + wk ≤ kvk + kwk. Let V be a vector space with norm k · k. Then d with d(v, w) = kv − wk is a metric on V . Every inner product p space is also a normed vector space, with the norm defined by kvk = hv, vi. This norm of an inner product space satisfies the parallelogram law, that is 2kvk2 + 2kwk2 = kv + wk2 + kv − wk2 . (Rewrite in terms of inner product and use axioms (i) and (ii) of inner product repeatedly.) The Cauchy-Schwarz inequality |hv, wi| ≤ kvk · kwk holds in any inner product space. Proof. Note kxv + wk2 = hxv + w, xv + wi = x2 kvk2 + 2xhv, wi + kwk2 . If v and w are linearly independent then kxv + wk2 > 0 and hence the quadratic on the right has no real roots which in turn implies that its discriminant is negative and hence hv, wi2 < kvk2 kwk2 . If w = xv then hv, wi2 = hv, xvi2 = x2 hv, vi2 = x2 kvk4 = kvk2 kwk2 . Let M be a metric space. The sequence (xn ) of elements of M is a Cauchy sequence if for every  > 0 there is N ∈ N so that d(xn , xm ) <  for all n, m ≥ N . A metric space is complete if every Cauchy sequence of elements of M converges. A Banach space (X, k · k) is a normed vector space such that X is complete under the metric induced by the norm k · k. A Hilbert space is an inner product space for which the metric kv−wk induced by the norm is complete. 3

Let `2 be the Hilbertspace of all square summable sequences. Then d : `2 → pP ∞ 2 R≥0 given by d(x, y) = i=1 (xi − yi ) is a metric with domain `2 . Remember that in the special case of normed spaces we have the following theorem: Theorem 1.1. Let X be a normed space. Then P X is complete if and only if every sequence (vn ) of elements in X with ∞ n=1 kvn k < ∞ the series Pfor ∞ n=1 vn converges. Proof. with elements in X with P∞ Let X be complete and (vn ) aPsequence n n=1 kvn k < ∞. Let  > 0 and wn = i=1 vi and n > m. Then:

n n

X

X

kwn − wm k = vi ≤ kvi k <  for all sufficiently large m.

i=m+1

i=m+1

P Suppose n=1 vn converges if ∞ n=1 kvn k < ∞. Let (wn ) be Cauchy. The sequence (wn ) contains a subsequence (ui ) with kui+1 − ui k < 21i . Let vi = ui+1 − ui . Then P∞

n X i=1

kvi k =

n X i=1

n X 1 . kui+1 − ui k < i 2 i=1

P∞

It follows that Pm i=1 kvi k < ∞ which in turn implies that Pm i=1 (ui+1 − ui ) = um+1 − u1 converges to some element v ∈ X. i=1 vi = Hence um converges to v + u1 . Hence, the Cauchy sequence (wn ) converges because it has a convergent subsequence. If Y is a Banach space and X is any normed vector space, then the set of continuous linear maps f : X → Y forms a Banach space, with norm given by the operator norm. In particular, since R and C are complete, the continuous linear functionals on a normed vector space B form a Banach space, known as the dual space of B. Examples: • Finite-dimensional normed vector spaces. • Lp spaces are by far the most common examples of Banach spaces. • `p spaces are Lp spaces for the counting measure on N. • Continuous functions on a compact set under the supremum norm. 4

1.2

Properties of Metric Spaces

Let M be a metric space. For a ∈ M and r > 0 we denote by Br (a) the ball in M with center a and radius r, that is: Br (a) = {x ∈ M : d(a, x) < r}. A ball with center a is a set Br (a) for some r > 0. The subset S ⊆ M is open if for every a ∈ S there exists 0 < r ∈ R with Br (a) ⊆ S. The closure S of a set S ⊆ M is the set of all a ∈ M with the poperty that Br (a) ∩ S 6= ∅ for all 0 < r ∈ R. The subset S of M is dense in M is S = M . The metric space M is separable if it contains a countable dense subset. The sequence (pi ; i ∈ N) of elements in M converges to the element p ∈ M if for every  > 0 there exists an n ∈ N so that {pi : i > n} ⊆ B (p). The sequence (pi ; i ∈ N) is a Cauchy T sequence of elements in M and hence has a limite, say p. Then p ∈ i∈N Si ∩ Br (q). The sequence (pi ; i ∈ N) of elements of M is a Cauchy sequence if for every  > 0 there is n ∈ N so that d(pi , pj ) <  for all i, j ≥ n. Every convergent sequence is Cauchy but a Cauchy sequence need not converge. A metric space M is complete if every Cauchy sequence of elements of M converges. The Cauchy completion of a metric space M is the intersection of all complete metric spaces which contain M as a subspace. Every metric space has an, up to isomorphism, unique Cauchy completion. Theorem 1.2 (Baire Category theorem). The intersection of a countable set of dense open subsets of a complete metric space M is dense in M . Proof. Let (Si ; i ∈ N) be open andTdense. Let q ∈ M and r > 0. We will show that there is an element p ∈ i∈N Si with p ∈ Br (q). There exists a sequence (Bsi (pi ); i ∈ ω) of balls in M with s0 = r and 0 < si ≤ 21i and p0 = q and Bs0 (p0 ) ⊇ Bs1 (p1 ) ⊇ Bs2 (p2 ) ⊇ Bs3 (p3 ) ⊇ · · · . Let M be a metric space. The set S ⊆ M is nowhere dense if no ball of M is a subset of the closure of S. A countable union of nowhere dense sets is a set of the first category and all the other subsets of M are of the second category. By taking complements in the statement of Theorem 1.2 we obtain the fact that no complete metric space is of the first category, the original result of Baire. 5

The spectrum Spec(M, p) of the element p in the metric space M is the S set {d(p, x) : x ∈ M } and the spectrum Spec(M ) of M is the set p∈M Spec(M, p). The spectrum of the ultrametric space N of the examples is the set { n1 : n ∈ N}. The diameter of the metric space M is the supremum of the spectrum of M . The function f of the metric space M to the metric space N is an isometry if dM (x, y) = dN (f (x), f (y)) for all x, y ∈ M . An isometry of a finite subspace of M to a subspace of N is a local isometry of M to N . A local isometry of M to M will usually just be called an isometry of M . An isometry of M onto M is an automorphism of M . Note that the set of automorphisms of M form a group under function composition. We denote the isomorphism group of M by Iso(M ). Definition 1.1. The metric space M is homogeneous if every local isometry of M can be extended to an automorphism of M . The metric space M is n-homogeneous if every local isomorphism of an n-element subset of M can be extended to an automorphism. Caution: In more analysis oriented areas of mathematics metric spaces which we call homogeneous are called ultrahomogeneous and 1-homogeneous spaces which we call transitive are called homogeneous. In model theory, homogeneous has still another, but somewhat related, meaning. For example, the metric space on N with d(n, m) = |n − m| is homogeneous. More generally the spaces Rn are homogeneous. The unit spere Sn := {x ∈ Rn : kxk = 1 in Rn+1 } is a homogeneous metric space. The ultrametric space N of the examples is homogeneous. The Hilbert space `2 and the unit sphere S∞ := {x ∈ `2 : kxk = 1}, the Hilbert sphere, are homogeneous metric spaces. In contrast to this it is known that: The Hilbert space `2 is the only separable infinite dimensional homogeneous Banach space. Every separable infinite dimensional 3-homogeneous Banach space is a Hilbert space. The problem whether the number 3 here can be replaced by 2 is known as the Banach-Mazur rotation problem. Mazur [10] verified it in the finite dimensional case and Pelczynski and Rolewicz [11] showed that the condition to be separable is necessary. The paper [12] is a survey on this problem. The image of a metric space M by an isometry is a copy of M . In particular an isometric image of the metric space M in the metric space M is a copy of M in M . The metric space M is indivisible if for every partition 6

(A0 , A1 , . . . , An−1 ) of M into finitely many parts there is i ∈ n and a copy M ∗ of M with M ∗ ⊆ Ai . Of the metric spaces listed in the examples only the Baire space is indivisible, [1]. Note that if for every partition (A0 , A1 ) of M into two parts there is i ∈ 2 and a copy M ∗ of M with M ∗ ⊆ Ai , then M is indivisible. For N ⊆ M and  > 0 let: (N ) = {p ∈ M : ∃q ∈ N (d(p, q) ≤ )}. The metric space M is -indivisible if for every partition (Ai ; i ∈ n ∈ ω) of M there is i ∈ n and a copy M ∗ of M with M ∗ ⊆ (Ai ) . The metric space M is approximately indivisible if it is -indivisible for every  > 0. Note that M is indivisible if it is 0-indivisible. Let f be a function of a metric space M to a metric space N . Then the oscillation of f on M is: osc(f |M ) = sup{dN (f (x), f (y)) : x, y ∈ M }. The function f is L-Lipschitz if dN (f (x), f (y)) ≤ LdM (x, y) for all x, y ∈ M . Note that every L-Lipschitz function is uniformly continuous. (For  > 0 let δ = L . Then for d(x, y) < δ: d(f (x), f (y)) < Ld(x, y) = L L = .) It follows that if there exists a 1-Lipschitz function f of M to a normed vector space N with osc(f |M ) >  then there exists a uniformly continuous function g of M to N with osc(g|M ) > 1. (g := 1 f . Then osc(g|M ) = sup{kg(x), g(y)k : x, y ∈ M } = sup{ 1 kf (x), f (y)k : x, y ∈ M }.) Observation 1.1. Let f be a uniformly continuous function of the approximately indivisible metric space M to the compact metric space N . Then there exists for every  > 0 a copy M ∗ of M so that osc(f |M ∗ ) < . Proof. For  > 0 let δ > 0 be such that d(f (x), f (y)) < /3 forSall x, y ∈ M with d(x, y) < δ. There is a finite subset Z ⊆ N with N ⊆ z∈Z B/3 (z). For z ∈ Z let Az = {x ∈ M : f (x) ∈ B/3 (z)}. There exists z ∈ Z and a copy M ∗ of M with M ∗ ⊆ (Az )δ . Hence for x, y ∈ M ∗ and x1 , y1 ∈ Az with d(x1 , x) < δ and d(y1 , y) < δ: d(f (x), f (y)) ≤ d(f (x), f (x1 )) + d(f (x1 ), f (y1 )) + d(f (y1 ), f (y)) < .

7

This Oberservation 1.1 has in the case where N is a closed interval of R a converse. There is a general notion for a topological space X acted upon by a a topological group G continuously to be oscillation stable, see [9] chapter 5 where the following result is stated. Theorem 1.3 (Pestov). For a complete homogeneous metric space M the following are equivalent: 1. Acted upon by the automorphism group Iso(M ) with the point open (pointwise convergent) topology, the metric space M is oscillation stable. 2. For every bounded 1-Lipschitz function f : M → R and every  > 0, there is a copy M ∗ of M such that osc(f |M ∗ ) < . 3. The space M is approximately indivisible. A metric space which is not oscillation stable is said to have distortion. Negating we obtain: Corollary 1.1. For a complete homogeneous metric space M the following are equivalent: 1. Acted upon by the automorphism group Iso(M ) with the point open (pointwise convergent) topology, the metric space M has distortion. 2. There is a bounded uniformly continuous function f : M → R whose oscillation on every copy M ∗ of M is at least 1. 3. There is a partion (A, B) of M and an  > 0 so that neither (A) nor (B) has a copy of M as a subset. We will not prove Theorem 1.3 but use it to state some historical results in terms of approximate indivisibility. The way those results have been stated have undergone a long development, mainly by Milman, who eventually stated them in terms of oscillation of uniformly bounded functions. The story began in 1959 with the following theorem of Dvoretzky [13]: Theorem 1.4 (Dvoretzky). For every k ∈ N and every  > 0 there exists a number N (k, ) so that if (X, k · k) is a Banach space of dimension larger than or equal to N (k, ), then there exists a subspace E ⊆ X of dimension 8

k and a positive quadratic form Q on E so that the corresponding Euclidean norm p | · | = Q( · ) on E satisfies: |v| ≤ kvk ≤ (1 + )|v|

for every v ∈ E.

Milman then extended and rerformulated Dvoretzky’s theorem and proofed [6]: Theorem 1.5 (Milman). Let (A0 , A1 , . . . , Ak−1 ) be a partition of S∞ . Then for every  > 0 and every n ∈ ω there is i ∈ k and a copy C of Sn in S∞ with C ⊆ (Ai ) . Theorem 1.5 led naturally to the conjecture that S∞ is approximately indivisible, but [8]: Theorem 1.6 (Odell-Schlumprecht). The Hilbert sphere S∞ is not approximately indivisible. We will construct various indivisible and approximately indiviseble metric spaces in the next section.

9

2

Urysohn spaces

2.1

The age of homogeneous metric spaces

Given a metric space M , a map f : M −→]0, +∞[ is Kat˘etov over M if ∀x, y ∈ M, |f (x) − f (y)| 6 d(x, y) 6 f (x) + f (y). Equivalently, one can extend the metric d to M ∪ {f } by defining, for every x, y in X, d(x, f ) = f (x) and d(x, y) = d(x, y). The corresponding metric space is then written M ∪ {f }. The set of all Kat˘etov maps over M is denoted by K(M ). For a metric subspace M of a metric space N and a Kat˘etov map f ∈ K(M ) the point y ∈ N realizes f over M if ∀x ∈ M d(y, x) = f (x). The set of all y ∈ N realizing f over M is denoted by T(f, N ) called the typeset of f in N . When N is clear from the context, the set T(f, N ) is simply written T(f ). The skeleton of a metric space M is the set of all finite subspaces of M . The age of M is the class of all metric spaces which are isometric to an element of the skeleton of M . A class A of finite metric spaces is an age if: 1. If B ∈ A and C is a metric space isometric to B then C ∈ A. 2. A contains a non empty metric space. 3. If B ∈ A and C is a subspace of B then C ∈ A. 4. If B ∈ A and C ∈ A with A ∩ C = ∅ then there exists a D ∈ A which has B and C as subspaces. (updirected.) Note that the age of a metric space is an age. The equivalence relation of being isomorphic partitions every age into equivalence classes. An age A is called countable if the cardinality of the cardinality of the set of isomorphism equivalence classes of A is countable. Items (ii) and (iii) imply that the empty metric space ∅ with distance function the empty function and the one point metric space are elements of every age. The empty function is Kat˘etov over the empty metric space and ∅ ∪ {∅} is the one point metric space. 10

S The spectrum Spec(A) of an age is the set A∈A Spec(A). Note that if the spectrum of A is countable then the age A is countable. The metric space M has the mapping extension property if for every finite subspace F ⊂ M and every Kat˘etov map f over F with F ∪ {f } an element of the age of M , the set T(f, M ) 6= ∅; that is, if there exists an isometry g of F ∪ {f } into M so that g restricted to F is the idendity map on F . Of course then g(f ) ∈ T(f, M ). Theorem 2.1. Let M and N be two countable metric spaces both having the mapping extension property and the same age. Let g be a local isomorphism of M to N . Then there exists an isometry of M onto N which extends g. Proof. Let the domain of g be the finite set F with |F | = k. Enumerate M into an ω-sequence m0 , m1 , m2 , m3 , dots so that F = {mi : i ∈ k}. Let n0 , n1 , n2 , n3 , . . . be an ω-enumeration of N so that ni = g(mi ) for all i ∈ k. Let f be the Kat˘etov map on g 00 (F ) with f (ni ) = d(mi , mk ) for all i ∈ k. It follows that every copy of F ∪mk in N is a copy of g 00 (F )∪f . Hence f can be realized in N by an element, say nt . We extend g to g ∗ by g ∗ = g ∪ {(mk , nt }. It follows from the construction that g ∗ is a local isomorphism. Similarly we find an element, say ms ∈ M so that g ∗ ∪ {(ms , nk )} is a local isomorphism. Continuing this process in a suitable way will produce an isometry of M onto N . The type or argument used in the proof of Theorem 2.1 is called a back and forth argument. The most common use of Theorem 2.1 is in the case in which g is the empty function. By just using the forth part of the argument we obtain: Theorem 2.2. Let M and N be countable metric spaces with the age of M a subset of the age of N . If N has the mapping extension property then there exists an isometry of M into N , that is there exists a copy of M in N . Theorem 2.3. Let M be a metric space. If M is homogeneous it has the mapping extension property. If M is countable and has the mapping extension property it is homogeneous. Proof. Let M be homogeneous and F a finite subspace of M and f a Kat˘etov map over F . Let F 0 ∪f 0 be a copy of F ∪f in M . There exists a local isometry

11

of F 0 to F which has an extension, say g, to an isometrie in iso(M ). Then g(f 0 ) realizes f over F . Let M be countable with the mapping extension property. Let F be a finite subspace of M and g a local isometry of F into M . It follows from Theorem 2.1 that g has an extension to an element of iso(M ). Definition 2.1. Let A be a class of metric spaces which is an age. The age A has amalgamation if for every B ∈ A and every subspace F of B and every Kat˘etov function f over F with F ∪ {f } ∈ A there exists a metric space D ∈ A, with D = B ∪ {f } and so that B and F ∪ {f } are subspaces of D. The triple (B, F, f ) is an amalgamation instance of A and the metric space D is the amalgam of the amalgamtion instanze (B, F, f ). Theorem 2.4. The age of every homogeneous metric space has amalgamation. Let A be a countable age of metric spaces which has amalgamation. Then there exists a homogeneous metric space MA whose age is A. The metric space MA unique up to isomorphism. Proof. Let M be a homogeneous metric space and B a finite subspace of M and F a subspace of B and f Kat˘etov over F with F ∪ {f } an element of the age of M . Because M has the mapping extension property according to Theorem 2.3 there exists an isometry g of F ∪ {f } into M which is the idendity on F . The subspace B ∪ {g(f )} is an amalgam of B and f . Let A be a countable age of metric spaces and let F be the set of all pairs (F, f ) with F ∈ A and f Kat˘etov over F and F ∪ {f } ∈ A. The pair (F, f ) ∈ F is isomorphic to the pair (F1 , f1 ) ∈ F if there exists an isometry g of F onto F1 so that f (x) = f1 (g(x)) for all x ∈ F . Let T be the set of all triples (g, F, f ) so that (F, f ) ∈ F and F ⊆ ω and g is an order preserving map of F into ω. The set T is countable. Let (gi , Fi , fi ) be an enumeration of T for which every element of T appears infinitely often. (Note that for every (F, f ) ∈ F there exists an i ∈ ω with (F, f ) being isomorphic to (Fi , fi ).) Let B ∈ A with B ⊆ ω and gi an isometry of Fi into B. Then f with f (gi (x) = fi (x) for all x ∈ Fi is Kat˘etov over gi00 (Fi ) = {gi (x) : x ∈ Fi } ⊆ B and gi00 (Fi ) ∪ {f } ∈ A. Because A has amalgamation there exists a metric space D so that gi00 ∪ {f } and B are subspaces of D. Indeed, there exists such a metric space D with D ⊆ ω. Such a metric space D will be called an extension of B via (gi , Fi , fi ). 12

We construct a sequence B0 ⊆ B1 ⊆ B2 ⊆ · · · of elements of A and a sequence n0 < n1 < n2 < · · · of numbers so that: 1. B0 = ∅ and n0 = 0. 2. ni+1 is the smallest number j larger than i so that gj is an isometry of Fj into Bi . 3. Bi+1 is an extension of Bi via (gi , Fi , fi ). S It follows that MA := i∈ω Bi has the mapping extension property.

2.2

Amalgamation

Definition 2.2. The metric spaces A and B are compatible if dA (x, y) = dB (x, y) for all x, y ∈ A∩B. For compatible metric spaces A and B we denote by A q B the set of all metric spaces D = A ∪ B with dD (x, y) = dA (x, y) if x, y ∈ A and dD (x, y) = dB (x, y) if x, y ∈ B. An amalgamation instance of the form (B ∪ {a}, B, f ) is an elementary amalgamation instance. Note that an age A of metric spaces has amalgamation if A ∩ A q (B ∪ {f }) 6= ∅ for every amalgamation instance (A, B, f ). Theorem 2.5. Let A be an age of metric spaces. The following are equivalent. 1. For every elementary amalgamation instance (B ∪{a}, B, f ) there exist an element of A in (B ∪ {a}) q (B ∪ {f }). 2. A has amalgamation. 3. The set A q B ∩ A 6= ∅ for all compatible metric spaces A, B ∈ A. Proof. Obviously, condition (iii) implies condition (ii) implies condition (i). We will prove that condition (i) implies (iii). We prove by induction on the cardinality of |A ∪ B| that if A satisfied condition (i) and A, B ∈ A are compatible then AqB∩A 6= ∅. If A ⊆ B then B ∈ AqB. If B ⊆ A then A ∈ AqB. Otherwise let a ∈ A\B and b ∈ B \A. Then A \ {a} ∈ A is compatible to B. Hence there is D ∈ (A \ {a} q B) ∩ A. Then A and D \ {b} are compatible. Hence there exists E ∈ A q (D \ b) ∩ A. 13

Let b play the role of the Kat˘etov function on E \ {a} with b(x) = dE (b, x) for all x ∈ E \ {a}. Then (E, E \ {a}, b) is an elementary amalgamation instance whose amalgam is an element of A q B ∩ A. Let A be an age of metric spaces. Because every subspace of a metric space in A is again an element of A and the elements of A are finite, there exists a set B of finite metric spaces so that: 1. There is no isometry of A to B for all A, B ∈ B. 2. A ∈ A if and only if for all B ∈ B there is no isometry of B to A. The set B is the boundary of A. For M a metric space, the boundary of M is the boundary of the age of M . The boundary of an age determines the age. It is often the most convenient way to define an age. That is after specifying the spectrum of the age on lists the boundary. For example, all finite metric spaces whose spectrum is an integer and which do not contain four points with all the six distances equal. The boundary of that age consists of all two element metric spaces with the distance between the two points not an integer and the set {{xi , yi , zi , ui } : i ∈ ω and d(x, y) = d(y, z) = d(z, u) = d(u, x) = d(x, z) = d(y, u)}. Of course another convenient way of defining an age is as the age of a metric space. Given a boundary B, that is a set of finite metric spaces which can not be embedded into each other, so that the age A determined by that boundary has amalgamation and is countable, there is then a unique up to isomorphism countable homogeneous metric space whose age is A. Hence the problem arises, given a boundary B, does the age determined by the boundary B have amalgamation. In many cases this is a very difficult problem in finite combinatorics. The following general observations can be made in case that the boundary consists of two element subsets only, that is if A consists of all finite metric spaces whose spectrum is a subset of the given spectrum for A. Definition 2.3. Let S ⊆ R. Then AS is the class of all finite metric spaces whose spectrum is a subset of S ∩ R≥0 ∪ {0} := S≥0 . Let S ⊆ R and (A, F, f ) an elementary amalgamation instance of AS with A = F ∪ {a}. In order to find an amalgam of this amalgamation instance we have to find a suitable distance r from a to f . Such a distance r will be suitable if and only if all triangles of the form a, x, f with x ∈ F satisfy 14

the triangle inequality. That is, if |dA (a, x) − f (x)| ≤ r ≤ dA (a, x) + f (x). Hence (A, F, f ) has an amalgam in A if and only if the basic amalgamation condition max{|dA (a, x) − f (x)|} ≤ r ≤ min{dA (a, x) + f (x)} x∈F

x∈F

(1)

is saisfied for some r ∈ S≥0 . Theorem 2.6. The set AS of metric spaces is an age for every set S. The age AS has amalgamation if every amalgamation instanz of AS of the form ({a, b, c}, {b, c}, f ) has an amalgam in AS . Proof. Items (i), (ii) and (iii) of the definition of age are easy to check. Let B, C ∈ AS with B ∩ C = ∅. Let r = max{diameter of B, diameter of C}. Then D ∈ B q C with dD (x, y) = dD (y, x) = r if x ∈ B and y ∈ C is an element of AS . The simplest non trivial case of an amalgamation instance is of the form ({a, b}, {b}, f ). Then D = {a, b}q({b}∪{f }) with dD (a, f ) = max{d(a, b), f (b)} is an amalgam in AS . Let a ∈ A ∈ AS and F = A\{a} and f kat˘etov over F with F ∪{f } ∈ AS . Then (A, F, f ) is an amalgamation instance of AS . We have to find a number r ∈ S so that dD (a, f ) = r for some D ∈ A q (F ∪ {f }). Such a number r has to satisfy the basic amalgamation condition (1). Let b ∈ F maximise the left hand side of condition (1) and c ∈ F minimise the right hand side of condition (1). For b = c we observed above that the amalgamation instance ({a, b}, {b}, f ) of AS has an amalgam in AS . If b 6= c the amalgamation instance ({a, b, c}, {b, c}, f ) has an amalgam in AS by assumption. Let (A, F, f ) be an amalgamation instanze of AS . We use induction on |A \ F | to find an amalgam of (A, F, f ) in AS . If |A \ F | = 1 we have the previous case. Let |A \ F | > 1 and a ∈ A \ F . There exists an amalgam D = (A \ {a}) ∪ F in AS . Extend f from F to A \ {a} by stipulating that f (x) = dD (x, f ) for all x ∈ A \ {a}. Then (A, A \ {a}, f ) is an amalgamation instance of AS . Let E be an amalgam of (A, A \ {a}, f ) in AS . Then E is an amalgam of (A, F, f ). Corollary 2.1. An age AS has amalgamation if and only if the set A q B contains an element of AS for all A, B ∈ AS with |A| = |B| = 3 and |A ∩ B| = 2. 15

Theorem 2.7. Let ({a, b, c}, {b, c}, f ) be an amalgamation instance of AR . Then |d(a, b) − f (b)| ≤ d(a, c) + f (c). Proof. We may assume without loss of generality that d(a, b) ≥ f (b). Assume for a contradiction that d(a, b) − f (b) > d(a, c) + f (c).

(2)

If d(a, b) ≥ d(a, c) then d(a, b) − d(a, c) ≤ d(b, c) ≤ f (b) + f (c) implies d(a, b) − f (b) ≤ d(a, c) + f (c) in contradiction to inequality (2). Hence d(a, b) < d(a, c) and d(a, c) − d(a, b) ≤ d(b, c) ≤ f (b) + f (c) implies d(a, b)+f (b) ≥ d(a, c)−f (c) which added into inequality (2) yields 2d(a, b) > 2d(a, c). Let S ⊆ R and s ∈ S with s > 0. Then s+ := min{x + y : 0 < x ∈ S and 0 < y ∈ S and s ≤ x + y} and s− := max{x − y : 0 < x ∈ S and 0 < y ∈ S and x ≥ y and s + y ≥ x}. It follows from Theorem 2.7 that s− ≤ s+ and hence from Theorem 2.6 that AS has amalgamation if and only if S ∩ [s+ − s− ] 6= ∅ for all 0 < s ∈ S. Corollary 2.2. The ages AR , AQ , An for n = {0, 1, 2, . . . , n − 1}, AN , AQ≤1 , A{1/(2n ):n∈N} have amalgamation. (Note that AR is not countable while all the others are countable.) Definition 2.4. A homogeneous metric space with age AS is denoted by US , the Urysohn space with spectrum S ∩ R≥0 ∪ {0}. It follows from Theorem 2.4 that a countable Urysohn space US exists if and only if the age AS is countable and has amalgamation.

2.3

The Urysohn space U

Let U, the ”Urysohn space” be the completion of the Urysohn space UQ . Lemma 2.1. Let M be a finite metric space and  > 0 and w ∈ M . Then there exists a Kat˘etov map f over M with 0 < f (w) <  and f (x) ∈ Q for all x ∈ M .

16

Proof. Let µ > 0 be smaller than  and smaller than min{d(w, x) : x ∈ M \ {w}} and smaller than: min{|d(w, x)−d(w, y)| : for all x, y ∈ M \ {w} for which d(w, x) 6= d(w, y)}. For every x ∈ M \ {w} let 0 < x < µ/2 be so that d(w, x) + x ∈ Q and x > y if d(w, x) > d(w, y) and x = y if d(w, x) = d(w, y). Then f with f (x) = d(w) + x and f (w) rational with µ/2 < f (w) < µ satisfies the conditions of being a Kat˘etov function over M with values in Q and with f (w) < . Corollary 2.3. For every finite metric space M = {xi : i ∈ n ∈ ω} and  > 0 there exists a metric space N = M ∪ {x0i : i ∈ ω} so that d(xi , x0i ) <  for all i ∈ n and d(x0i , x0j ) ∈ Q for all i, j ∈ n. Lemma 2.2. For every n ∈ ω and  > 0 there exists a δ > 0 so that if {a0 , a1 , a2 , . . . , an−1 } ∪ {b0 , b1 , b2 , . . . , bn−1 } is a metric space with distance function d so that if d(ai , bi ) < δ for all i ∈ n and f is a Kat˘etov function over {a0 , a1 , a2 , . . . , an−1 } then there exists a Kat˘etov function g over {b0 , b1 , b2 , . . . , bn−1 } so that g(bi ) ∈ Q and |f (ai ) − g(bi )| < . Proof. Let A = {ai : i ∈ n} and B = {bi : i ∈ n}. Assume that f (a0 ) ≥ f (a1 ) ≥ f (a2 ) ≥ · · · ≥ f (an−1 ) and let δ be smaller than /(2n + 1) and smaller than |f (x) − f (y)|/(2n + 1) for all x, y ∈ A with x 6= y. For i ∈ n let 0 < δi < δ be such that f (ai ) + 2iδ + δi ∈ Q and such that δi−1 < δi for all 1 ≤ i ≤ n − 1. For i ∈ n let g(bi ) = f (ai )+2iδ+δi . Noting that d(ai , aj )−2δ ≤ d(bi , bj ) ≤ d(ai , aj )+2δ for all i, j ∈ n it is easy to check that g is a Kat˘etov function over B. Also |g(bi )−f (ai )| = |f (ai )+2iδ+δi −f (ai )| = 2iδ+δi ≤ (2n+1)δ < . Lemma 2.3. Let M = A ∪ B with A = {ai : i ∈ n ∈ ω} and B = {bi : i ∈ n} be a metric space with distance function d so that d(ai , bi ) <  for  > 0. Let f be a Kat˘etov function over A and g a Kat˘etov function over B so that |f (ai ) − g(bi )| <  for all i ∈ n. Then every function h over B ∪ {f } with h(x) = g(x) for all x ∈ B and 2 ≤ g(f ) ≤ min{g(x) : x ∈ B} and g(f ) ≤ min{f (x) : x ∈ A} is a Kat˘etov function over B ∪ {f }. Proof. We have to varify that |g(x) − g(f )| ≤ d(f, x) ≤ g(x) + g(f ) for all x ∈ B. |g(bi ) − g(f )| = |g(bi ) − f (ai ) + f (ai ) − g(f )| ≤  + f (ai ) − g(f ) ≤ f (ai ) −  ≤ d(f, x) ≤ f (ai ) +  ≤ f (ai ) + 2 ≤ f (ai ) + g(f ). 17

Theorem 2.8. The Urysohn space U has the mapping extension property. Proof. Let F = {xi : i ∈ n ∈ ω} be a finite subset of U and f a Kat˘etov map over F . We will construct a Cauchy sequence c0 , c1 , c2 , c3 , . . . of elements in UQ so that for all x ∈ F the sequence (d(cn , x); n ∈ ω) is a Cauchy sequence converging to f (x). For a given /6 > 0 with 3 < min{f (x) : x ∈ F }, let {ai : i ∈ n} ⊆ UQ with d(ai , xi ) < δ given by Lemma 2.2. There exists a Kat˘etov function g with over {ai : i ∈ n} with rational values and |f (xi ) − g(ai )| < /6 for all i ∈ n. Because UQ has the mapping extension property, the function g has a realisation, say c in UQ . Let 0 /6 < /6 be given and {bi : i ∈ n} ⊆ UQ with d(bi , xi ) < δ 0 given by Lemma 2.2 for 0 /6. There exists a Kat˘etov function h over {bi : i ∈ n} with rational values and |f (xi ) − h(bi )| < /6 for all i ∈ n. Then d(ai , bi ) < /3. According to Lemma 2.3 there exists a Kat˘etov function k over {bi : i ∈ n} ∪ {g} with k(bi ) = g(bi ) for all i ∈ n and k(c) <  and with rational values. Again, because of the mapping exension property of UQ there exists a realisation c1 of k in UQ . Let P(j : j ∈ ω be a strictly decreasing sequence of positive numbers with j∈ω j < ∞. Let 0 determin c0 as above and 1 the point c1 with d(c0 , x1 ) < 0 and d(c0 , xi ) < 0 and d(c1 , xi ) < 1 for all i ∈ n and so on with d(cj , cj+1 ) < j and d(cj , xi ) < j for all j ∈ ω and i ∈ n. Theorem 2.9. The Urysohn space U is characterized up to isomorphism by the following properties: 1. U embeds every separable metric space isometrically. 2. U is homogeneous. 3. U is separable. Proof. Let V be a separable and homogeneous metric space which embeds every separable metric space isometrically. We will show that U and V are isomorphic. Let W be a countable dense subset of V. Enumerate W into an ωsequence (wi : i ∈ ω) in which every element of W appears infinitely often. 18

We will construct a sequence (ai :∈ ω) so that d(ai , aj ) ∈ Q and d(ai , wi ) < 1/i for all i, j ∈ ω. Let a0 ∈ V with d(a0 , w0 ) < 1. If a0 , a1 , . . . , an−1 are constructed, let f be a Kat˘etov function over {ai : i ∈ n} ∪ {wn } with values in Q and with f (wn ) < 1/n. Such a function f exists according to Lemma 2.1. The space V is homogeneous and hence satisfies the mapping extension property according to Theorem 2.3. Let an be the realisation of f in V. The set S = {ai : i ∈ ω} is dense in V and any two elements in S have a rational distance. Using a back and forth argument we construct an isometry g of UQ into V so that g −1 is an isometry of S into UQ . Hence V has an isometric copy of UQ as a subspace which is dense in V. It follows that V and UQ are isomorphic. It remains to prove that U the completion of UQ satisfies the three characterizing properties listed above. Let M be a separable metric space with W a countable dense subset. According to Theorem 2.2 there exists an embedding f of W into UQ . The extension of f to limits of sequences of elements in W is an embedding of M into U. Let F, H be finite subsets of U and f an isometry of F onto H. The Urysohn space U satisfies the mapping extension property according to Theorem 2.8 and hence we can use the back and forth argument to construct countable sets F1 ⊇ UQ ∪ F and H1 ⊇ UQ ∪ H and an isometry f1 of F1 onto H1 which is an extension of f . Extending f1 to the limits of F1 yields the automorphism of U extending f .

19

3

Some indivisibility and divisibility theorems

Much of the information in this section and the next is taken from [1]. Let S be a countable subset of [a, 2a] for a > 0. Then AS is countable and has the amalgamation property. According to Theorem 2.4 there exists a unique homogenous metric space US with age AS . The metric space US has the mapping extension property according to Theorem 2.3. In order to see that US is indivisible consider the metric
space M on the set of pairs (a, b) with a, b
∈ US and dM (a, b), (a1 , b1 ) = dUS (b, b1 ) if b 6= b1 and dM (a, b), (a1 , b1 ) = dUS (a, a1 ) if b = b1 . For b ∈ US let Mb := {(a, b) : a ∈ U. The metric space M contains for every partition A ∪ B = M of M a copy of US which is a subset of A or a copy of US which is a copy of B. For if each of the subspaces Mb conains an element ab ∈ A then {ab : b ∈ US } is a copy of US and otherwise there is a b ∈ US with Mb ⊆ B. The metric space M imbeds isometrically into US according to Theorem 2.2. It follows that US is indivisible. In particular, the metric spaces U{1} and U{1,2} are indivisible. That U{1,2,3} is indivisible follows from [14]. It has recently been proven that U1,2,3,...,n} is indivisible for all n ∈ N, [15]. More generally, if S is a finite set of positive real numbers for which there exists a number a with 0 6= a ∈ S and 2 min S ≤ a and min S + a ≤ max S it follows that AS is indivisible and then from [14] that US is indivisible. The sequence a0 , a1 , . . . , an−1 , an of elements in a metric space M is an -chain joining a0 and an if d(ai , ai+1 ) ≤  for all i ∈ n. The space M is Cantor connected if any two of its elements can be joined by an -chain for any  > 0. The Cantor connected component of an element a ∈ M is the largest Cantor connected subset of M containing a. The space M is totally Cantor disconnected if the Cantor connected component of every a reduces to a. See [16] for more details and references. For a ∈ M let λ (a) be the supremum of all reals l ≤ 1 for which there exists an -chain a0 , a1 , . . . , an−1 , an with d(a0 , an ) ≥ l containing a. (The condition l ≤ 1 saves us from having to consider the special case ∞.) Let λ(a) := sup{l ∈ R | ∀ > 0 (λ (a) ≥ l)}. A space (M ; d) is restricted if λ(a) = 0 for all a ∈ M . It follows that every restricted space is totally Cantor disconnected. There are totally Cantor disconnected spaces which are not restricted. Here is an example with a finite diameter : 20

Example 3.1. Let (M ; d) be the metric space so that: 1. M = {(0, 0)} ∪ {(m, n) ∈ N × N | m < n} 2. d((0, 0), (m, n)) =

m+1 n

3. d((m1 , n), (m2 , n)) =

|m1 −m2 | n

4. d((m1 , n1 ), (m2 , n2 )) =

m1 +1 n1

+

m2 +1 n2

when n1 6= n2 .

This example falls into the following category: Definition 3.1. A spider is a metric space (M ; d) so that 1. M = {(0, 0)} ∪ {(m, n) ∈ N × N | m < n} 2. d((0, 0), (n − 1, n)) ≥ r for some non-negative r and all n ∈ N∗ 3. d((m, n), (m + 1, n)) ≤ rn for all n ∈ N∗ where limn→∞ rn = 0. With this definition, we have easily Lemma 3.1. A metric space is restricted if it does not isometrically embed a spider. Definition 3.2. Let M be a metric space, a ∈ M and 0 ≤ r < s. Then Ra (r, s) := {x ∈ M | r ≤ d(a, x) < s}. Lemma 3.2. Let c ∈ M and 0 ≤ r0 < r1 < r2 < r3 and a ∈ Rc (r0 , r1 ) and b ∈ Rc (r2 , r3 ) then: 1. d(a, b) > r2 − r1 . 2. d(x, y) < 2r2 for all x, y ∈ Rc (r1 , r2 ). 3. If 0 <  < min{r1 − r0 , r3 − r2 } and x0 , x1 , x2 , . . . , xn−1 is an -sequence with xi 6∈ Rc (r0 , r1 ) ∪ Rc (r2 , r3 ) for all i ∈ n but with xi ∈ Rc (r1 , r2 ) for at least one i ∈ n, then xi ∈ Rc (r1 , r2 ) for all i ∈ n. 4. Let f be an isometry of M with f [M ] ∩ (Rc (r0 , r1 ) ∪ Rc (r2 , r3 )) = ∅ and let z ∈ M with λ(z) > 2r2 . Then f (z) 6∈ Rc (r1 , r2 ).

21

Proof. Items 1 and 2 follow from the triangle inequality. Item 3 follows from item 1 and item 4 follows from items 2 and 3. Definition 3.3. Let c ∈ M and 0 < l. Then [

Ec (l) :=

Rc

n≥2, n even

and Oc (l) :=

[ n odd

Rc

l(n − 1) ln
, n n+1

l(n − 1) ln
, . n n+1

Theorem 3.1. Let M be a countable metric space. If there exists an element a ∈ M with λ(a) > 0 then M is divisible. Proof. Since M is countable, it can be covered by a family of pairwise disjoint open balls with radius less that λ(a) . In fact, there exists a subset C of M 2 and for every c ∈ C a positive real lc so that: 1. lc 6= d(x, y) for every c ∈ C and x, y ∈ M . 2. 2lc < λ(a) for every c ∈ C. 3. For every element x ∈ M there is one and only one element c ∈ C with x ∈ Rc (0, lc ). (After enumerating M into an ω-sequence m0 , m1 , m2 , m3 , . . . such a set C and function l can be constructed step by step exhausting all of the elements of M .) Let [ [ E := Ec (lc ) and O := Oc (lc ). c∈C

c∈C

Then E ∪ O = M and E ∩ O = ∅. Assume for a contradiction that there is an isometry f which maps M into E. Then there is a c ∈ C so that f (a) ∈ Ec (lc ). But this is not possible according to Lemma 3.2 item 4. Similarly it is not possible that f maps M into O. Corollary 3.1. A countable metric space which is indivisible is restricted and hence totally Cantor disconnected. 22

The second part of the conclusion of the corollary above extends to uncountable metric spaces. Theorem 3.2. Let M be a metric space and r be a positive real, then there is a partition into two parts A0 and A1 which contains no Cantor connected subspace of diameter larger than r. Lemma 3.3. Le M be a metric space and r be a positive real number. Then there is a sequence (Eµ )µ 0 and every y ∈ M \ (BY ∪ Eµ ). Proof. Suppose S the sequence defined for all ν, ν < µ. If µ is a limit ordinal, set Eµ := {Eν : ν < µ}. If µ is a successor, say µ := ν + 1, pick x ∈ E 0 := M \ Eν , set R0x (0, r/2) := {y ∈ E 0 : d(x, y) < r/2} and set Eµ := Eν ∪ Rx0 (0, r/2). Decompose R0x (0, r/2) into countably many crowns rn/2
, as in the proof of Theorem 3.1 , the union of the even ones R0 x r(n−1)/2 n n+1 gives Aν,0 , the rest gives Aν,1 . Finally any Cantor connected subspace Y of Aµ,i must be included into
, rn/2 for some n, and therefore the required BY may be taken R0 x r(n−1)/2 n n+1
0 to be R x 0, sY with rn/2 < sY < r(n+1)/2 with Y := sY − rn/2 . n+1 n+2 n+1 S Proof of Theorem 3.2 Let Ai := {Aµ,i : µ < λ}. Then Ai contains no Cantor connected subspace X of diameter larger than r. Indeed, suppose the contrary. Let µ be minimum such that Eµ meets X. Clearly µ is a successor, say µ = ν + 1. Let x ∈ Xν := X ∩ Fν . Let Y be the Cantor connected component of x in Aν,i and let BY given by the above lemma. Claim X ⊆ BY . Indeed suppose not, let y ∈ X \ BY , let , 0 <  < Y and x0 := x, . . . , xk , . . . , xn = y be an  path contained in 23

X. Let ` be least index such that x` 6∈ BY . From x`−1 ∈ Aµ,i ∩ BY , we get d(x` , Aµ,i ∩ BY ) < Y . A contradiction. Since X ⊆ BY ⊆ Fµ , the diameter of X is at most r. The proof is complete. Definition 3.4. Let M be totally Cantor disconnected. Then d∗ (x, y) := inf{ > 0 | there exists an -sequence containing x and y}. Lemma 3.4. Let M be totally Cantor disconnected. Then M∗ := (M ; d∗ ) is an ultrametric space. Proof. Let x, y, z ∈ M with d∗ (x, y) ≥ d∗ (x, z) ≥ d∗ (y, z). Then for every  > d∗ (x, z) there are -sequences joining x to z and z to y, then one joining x to y, hence d∗ (x, y) ≤ . Thus d∗ (x, y) ≤ d∗ (x, z). (See [16], Theorem 1 and Lemma 8.) Theorem 3.3. Let M be a countable homogeneous indivisible metric space then M∗ is an homogeneous indivisible ultrametric space. Proof. Since M is indivisible it is totally Cantor disconnected, hence d∗ is well defined. Since M is homogeneous then d(x, y) = d(x0 , y 0 ) implies d∗ (x, y) = d∗ (x0 , y 0 ) for all x, y, x0 , y 0 ∈ M . From this property, every local isometry of M is a local isometry of M ∗ . Hence, since M is indivisible, M ∗ is indivisible. Since every automorphism of M is an automorphism of M ∗ , M ∗ is pointhomogeneous. According to Theorem 4.2, M ∗ is homogeneous. Theorem 3.4. Let M be a homogeneous metric space and V := Spec(M ). If M is totally Cantor disconnected and every three element metric space T with Spec(T) ⊆ V embeds into M then the set V \ {0} is either contained into an interval of the form [a → +∞) for some a ∈ R+ \ {0} or into an union of intervals of the form ∪{[a2(n+1) , a2n+1 ] : n < ω} ∪ [a0 → +∞) where . {an : n < ω} is a sequence such that a2n+1 ≤ a2n 2 Proof. Claim For every w ∈ V ∗ := Spec(M∗ ), ] w2 , w[∩V = ∅. Suppose the contrary. Pick r ∈] w2 , w[∩V = ∅. Since w ∈ V ∗ , we may find x, y such that d∗ (x, y) = w. Let n < ω and  := 2r, then there is an -sequence x0 , . . . , xn containing x, y. For i < n, let Ti := ({xi , xi+1 , zi }, di ) where di (xi , xi+1 ) := d(xi , xi+1 ), di (xi , zi ) = di (xi+1 ) := r. Each Ti is a metric space whith spectrum included into V , hence can be isometrically embedded 24

into M . Since M is homegenous, we may suppose that zi ∈ M and that the embedding is the inclusion. By adding the zi0 ’s to the x0i s we get a r-sequence containing x and y. Since r < w this gives a contradiction. Since every element of V ∗ is the infimum of elements of V it also follows that ] w2 , w[∩V ∗ = ∅. Let α := Inf (V \ {0}). If α 6= 0 set a := α; in this case V \ {0} ⊆ [a → +∞). If α = 0 then, since every element de V \ {0} majorizes some element of V ∗ \ {0} it follows that Inf (V ∗ \ {0}) = 0 too. Let {a2n : n < ω} be a strictly decreasing sequence of elements of V ∗ which converges to 0. Set a2n [∩V = ∅, hence a2n+2 ≤ a2n+1 . The a2n+1 := a2n . From the Claim ] a2n 2 rest follows. Theorem 3.5. Every unbounded metric space is divisible. Proof. Let M be an unbounded metric space. Construct a sequence of reals r0 , r1 , r2 , . . . and a sequence a0 , a1 , a2 , . . . of elements of M so that for every integer i ∈ N 1. d(a0 , ai+1 ) > 2ri . 2. d(a0 , ai+1 ) + ri < ri+1 . Let r0 := 0 and a0 ∈ M be arbitrary. Suppose that (ri : i ≤ n) and (ai : i ≤ n) have already been constructed. From the fact that M is unbounded, we can find an+1 ∈ M such that d(a0 , an+1 ) > 2rn . Next, choose rn+1 > d(a0 , an+1 )+rn . Note that the set {ri : i ∈ N} such constructed is unbounded. Let, given any c ∈ M , [ [ E := Rc (r2i , r2i+1 ) and O := Rc (r2i+1 , r2i+2 ). i∈N

i∈N

We prove that there is no isometric embedding of M into E or into O. Let f be an isometric embedding of M into M . Let i be minimal so that d(c, f (a0 )) < ri ; notice that i > 0 and f (a0 ) ∈ Rc (ri−1 , ri ). We have: d(c, f (ai+1 )) ≥ d(f (a0 ), f (ai+1 )) − d(c, f (a0 )) = = d(a0 , ai+1 ) − d(c, f (a0 )) > 2ri − ri = ri . Also: d(c, f (ai+1 )) ≤ d(c, f (a0 )) + d(f (a0 ), f (ai+1 )) ≤ ri + d(a0 , ai+1 ) < ri+1 . 25

It follows that f (ai+1 ) ∈ Rc (ri , ri+1 ). Therefore f [M ] intersects both E and O.

26

4

Ultrametric spaces

A metric space is an ultrametric space if it satisfies the strong triangle inequality d(x, z) ≤ max{d(x, y), d(y, z)}. See [16] for example. Note that a space is an ultrametric space if and only if d(x, y) ≥ d(y, z) ≥ d(x, z) implies d(x, y) = d(y, z). What we did in the previous section for general metric spaces work for ultrametric spaces. Let V be a set such that 0 ∈ V ⊆ R+ . Let MultV (resp. MultV,