c Pleiades Publishing, Ltd., 2010. ISSN 2070-0466, p-Adic Numbers, Ultrametric Analysis and Applications, 2010, Vol. 2, No. 2, pp. 100–113. 

RESEARCH ARTICLES

Ultrametricity and Metric Betweenness in Tangent Spaces to Metric Spaces∗ Oleksiy Dovgoshey** and Dmytro Dordovskyi*** Institute of Applied Mathematics and Mechanics of NASU, R.Luxemburg str. 74, Donetsk 83114, Ukraine Received September 10, 2009

Abstract—The paper deals with pretangent spaces to general metric spaces. An ultrametricity criterion for pretangent spaces is found and it is closely related to the metric betweenness in the pretangent spaces. DOI: 10.1134/S2070046610020020 Key words: metric spaces, pretangent spaces, ultrametric spaces, metric betweenness.

1. INTRODUCTION Analysis on metric spaces with no a priori smooth structure is in need of some generalized differentiations. Important examples of such generalizations and even an axiomatics of so-called "pseudogradients" can be found in [1, 3, 4, 10, 14, 16, 21] and, respectively, in [2]. A linear structure, and so a differentiation, for separable metric spaces can be obtained via their isometric embeddings into dual spaces of separable Banach spaces. For the application of this approach to develop a rather complete theory of rectifiable sets and currents on metric spaces see [3, 4]. Another natural way to obtain suitable differentiations on metric spaces is to induce some tangents at the points of these space. The Gromov– Hausdorff convergence and the ultra-convergence are, probably, the most widely applied today’s tools for the construction of such tangent spaces (see, for example, [8, 9] and, respectively, [7, 17]). Recently a new approach to the introduction of the tangent spaces at the points of general metric spaces was proposed in [13]. Our paper is devoted to the study of the last tangent spaces. We find necessary and sufficient conditions under which the tangent spaces are ultrametric, see Theorem 1 below. Our second main result is Theorem 2 that completely describes metric spaces for which tangents are, roughly speaking, the snowflaked versions of subsets of R. For convenience we recall the main notions from [13], see also [11]. Let (X, d) be a metric space. Fix a sequence r˜ of positive real numbers rn which tend to zero. In ˜ the set of all what follows this sequence r˜ will be called a normalizing sequence. Let us denote by X sequences of points from X. ˜ x Definition 3. Two sequences x ˜, y˜ ∈ X, ˜ = {xn }n∈N and y˜ = {yn }n∈N are mutually stable with respect to (w.r.t.) a normalizing sequence r˜ = {rn }n∈N if there is a finite limit d(xn , yn ) ˜ x, y˜). =: d˜r˜(˜ x, y˜) = d(˜ n→∞ rn lim

(1.1)

˜ is self-stable (w.r.t. r˜) if every two x We shall say that a family F˜ ⊆ X ˜, y˜ ∈ F˜ are mutually stable. A ˜ ˜ ˜ ˜ \ F˜ there is x family F ⊆ X is maximal self-stable if F is self-stable and for an arbitrary z˜ ∈ X ˜ ∈ F˜ such that x ˜ and z˜ are not mutually stable. A standard application of Zorn’s lemma leads to the following ∗

The text was submitted by the authors in English. E-mail: [email protected] *** E-mail: [email protected] **

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Proposition 1. Let (X, d) be a metric space and let p ∈ X. Then for every normalizing sequence ˜ p,˜r such that p˜ = {p, p, ...} ∈ X ˜ p,˜r . r˜ = {rn }n∈N there exists a maximal self-stable family X ˜ x, y˜) = d˜r˜(˜ ˜ p,˜r → R, where d(˜ ˜ p,˜r × X x, y˜) is defined by (1.1). Obviously, d˜ is Consider a function d˜ : X ˜ x, y˜) ≤ d(˜ ˜ x, z˜) + d(˜ ˜ z , y˜) for all symmetric and nonnegative. Moreover, the triangle inequality implies d(˜ ˜ ˜p,˜r , d) is a pseudometric space. ˜ p,˜r . Hence (X x ˜, y˜, z˜ from X ˜ p,˜r by x ˜ ∼ y˜ if and only if d˜r˜(˜ x, y˜) = 0. Then ∼ is an equivalence relation. Define a relation ∼ on X X ˜ p,˜r under the equivalence relation ∼. If a Let us denote by Ω = Ωp,˜r the set of equivalence classes in X p,˜ r

function ρ is defined on Ωp,˜r × Ωp,˜r by

˜ x, y˜) ρ(α, β) = d(˜

(1.2)

˜ is, by ˜p,˜r , d) for x ˜ ∈ α and y˜ ∈ β, then ρ is the well-defined metric on Ωp,˜r . The metric identification of (X definition, the metric space (Ωp,˜r , ρ). Definition 4. The space (ΩX p,˜ r , ρ) is pretangent to the space X at the point p w.r.t. a normalizing sequence r˜. ˜ p,˜r , see Proposition 1. Note that Ωp,˜r = ∅ because the constant sequence p˜ belongs to X Let {nk }k∈N be an infinite, strictly increasing sequence of natural numbers. Let us denote by r˜ the subsequence {rnk }k∈N of the normalizing sequence r˜ = {rn }n∈N and let x ˜ := {xnk }k∈N for every ˜ It is clear that if x ˜ and y˜ are mutually stable w.r.t. r˜, then x ˜ and y˜ are mutually stable x ˜ = {xn }n∈N ∈ X.  w.r.t. r˜ and x, y˜) = d˜r˜ (˜ x , y˜ ). (1.3) d˜r˜(˜ ˜ p,˜r is a maximal self-stable (w.r.t. r˜) family, then, by Zorn’s lemma, there exists a maximal selfIf X ˜ p,˜r such that {˜ ˜ p,˜r } ⊆ X ˜p,˜r . x : x ˜∈X stable (w.r.t. r˜ ) family X ˜ p,˜r to X ˜p,˜r with inr˜ (˜ ˜ p,˜r . It follows from (1.3) x) = x ˜ for all x ˜∈X Denote by inr˜ the mapping from X

X under which that, after the metric identifications, inr˜ pass to an isometric embedding em : ΩX p,˜ r → Ωp,˜ r the diagram  ˜ p,˜r ˜p,˜r −−in−r˜− → X X ⏐ ⏐ ⏐ ⏐  π π

(1.4)

em

ΩX −−−→ ΩX p,˜ r − p,˜ r is commutative. Here π, π  are canonical projection maps. Let X and Y be two metric spaces. Recall that a map f : X → Y is called an isometry if f is distance-preserving and onto.  X → ΩX is an isometry for every X ˜ p,˜r . Definition 5. A pretangent ΩX p,˜ r is tangent if em : Ωp,˜ r p,˜ r

To formulate the main results of the paper we define some functions on the Cartesian products X × X × X and X × X of the metric space (X, d). Let x, y and z be some points of (X, d). Let us define a quantity s(x, y, z) as a positive root of the equation (d(x, z))s + (d(z, y))s = (d(x, y))s

(1.5)

d(x, z) ∨ d(z, y) < d(x, y)

(1.6)

if and otherwise write s(x, y, z) = +∞. The function s : (x, y, z) → s(x, y, z) is correctly defined on X × X × X and s(x, y, z) ∈ [1, +∞] for all x, y, z ∈ X, see Lemma 1 in the next section of the paper. p-ADIC NUMBERS, ULTRAMETRIC ANALYSIS AND APPLICATIONS

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Let p be a marked point in X. Write  F (x, y) :=

d(x,y)(d(x,p)∧d(y,p)) (d(x,p)∨d(y,p))2

if

0

(x, y) = (p, p)

if

(x, y) = (p, p),

(1.7)

Φ(x, y, z) := F (x, y) ∨ F (x, z) ∨ F (y, z)

(1.8)

d(x, y) ∨ d(y, z) ∨ d(x, z) , d(x, y) ∧ d(y, z) ∧ d(z, x)

(1.9)

and Ψ(x, y, z) :=

where Ψ(x, y, z) := ∞ if d(x, y) ∧ d(y, z) ∧ d(z, x) = 0. Recall that a metric space (X, d) is ultrametric if the metric d satisfies the ultra-triangle inequality d(x, y)  d(x, z) ∨ d(y, z) for all x, y, z ∈ X. The following theorem is the first our main result. Theorem 1. Let (X, d) be a metric space with a marked point p. The following two statements are equivalent. (i) All pretangent spaces ΩX p,˜ r are ultrametric. (ii) We have the limit relation lim

s(x, y, z)

x,y,z→p Φ(x, y, z)

where

1 Φ(x,y,z)

Ψ(x, y, z) = ∞

(1.10)

=: ∞ if Φ(x, y, z) = 0.

Remark 1. It is easy to see that all pretangent spaces to ultrametric space X are ultrametric but the converse is not true. More precisely, there is a metric space X with a marked limit point p such that all ˜ we have pretangent spaces to X are ultrametric but for some sequences x ˜, y˜, z˜ ∈ X 1 d(xn , yn ) = d(xn , zn ) = d(zn , yn ) > 0 2 for all n ∈ N and lim d(xn , p) = lim d(yn , p) = lim d(zn , p) = 0.

n→∞

n→∞

n→∞

For the construction of such X see the proof of Proposition 3 below. Denote by M the class of all metric spaces (X, d) such that the equality d(x, z) = d(x, y) + d(y, z) holds whenever d(x, z)  d(x, y)  d(y, z). The next theorem, the second main result of the paper, can be considered as an analog of the Theorem 1 for the pretangent spaces which are not ultrametric. Theorem 2. Let (X, d) be a metric space, p a limit point of X and s1 a positive number. The membership relation s1 (ΩX p,˜ r, ρ ) ∈ M

(1.11)

holds for every pretangent space ΩX p,˜ r if and only if lim

x,y,z→p (x,y,z)∈X +3

Ψ(x, y, z)s2 (x, y, z) = ∞, Φ(x, y, z)(s1 − s(x, y, z))2

where X +3 = {(x, y, z) ∈ X × X × X : d(x, z)  d(x, y)  d(y, z) > 0} and  s(x, y, z) 1 if s(x, y, z) = ∞ = s1 − s(x, y, z) ∞ if s(x, y, z) = s1 . Membership relation (1.11) means, in particular, that ρs1 is a metric on ΩX p,˜ r. p-ADIC NUMBERS, ULTRAMETRIC ANALYSIS AND APPLICATIONS

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2. BETWEENNESS EXPONENT AND ULTRAMETRICITY OF PRETANGENT SPACES Let (X, d) be a metric spaces. Denote by t0 = t0 (X) = t0 (X, d) the supremum of positive numbers t for which the function dt : (x, y) → (d(x, y))t is a metric on X. It is clear that dt0 remains a metric if t0 = t0 (X) < ∞. The quantity t0 (X) will be called the betweenness exponent of the metric space (X, d). The proofs of the following two lemmas can be found in [12]. Lemma 1. Let x, y and z be points in a metric space X. If the inequality d(x, z) ∨ d(z, y) < d(x, y)

(2.1)

holds, then there exists a unique positive solution s0 of the equation (d(x, z))s + (d(z, y))s = (d(x, y))s ,

(2.2)

and this solution belongs to [1, ∞). Lemma 2. The equality t0 (X) = inf{s(x, y, z) : x, y, z ∈ X} holds for every nonvoid metric space X. Remark 2. For ultrametric spaces X we have the equality t0 (X) = ∞ since inequality (2.1) never holds in these spaces. In fact t0 (X) = ∞ is true if and only if X is ultrametric. For every metric space (Y, d) define the subset Y +3 of the Cartesian product Y × Y × Y by the rule (x, y, z) ∈ Y +3 ⇔ d(x, z)  d(x, y)  d(y, z) > 0. It is clear that Y

+3

(2.3)

= ∅ if and only if card Y  3.

Proposition 2. Let (X, d) be a metric space and let p be a limit point of X. If the equality lim

x,y,z→p

s(x, z, y) = s0 ∈ [1, ∞],

(2.4)

(x,y,z)∈X +3

+3 holds, then for every Ωp,˜r = ΩX p,˜ r and all (β, γ, δ) ∈ Ωp,˜ r we have the equality 1

ρ(β, δ) = ((ρ(β, γ))s0 + (ρ(δ, γ))s0 ) s0 with

(2.5)

1  1 ((ρ(β, γ))s0 + (ρ(δ, γ))s0 ) s0 := lim (ρ(β, γ))t + (ρ(δ, γ))t t = ρ(β, γ) ∨ ρ(δ, γ)

t→∞

if s0 = ∞. ˜ = {rn }n∈N and let {xn }n∈N ∈ β, {yn }n∈N ∈ γ, {zn }n∈N ∈ δ. Let Proof. Let (β, γ, δ) ∈ Ω+3 p,˜ r where r n n n (d[1] , d[2] , d[3] ) be a nonincreasing rearrangement of the vector dn := (d(xn , zn ), d(xn , yn ), d(yn , zn )) i.e., the vectors dn and (dn[1] , dn[2] , dn[3] ) have the same components but dn[1]  dn[2]  dn[3] . It is easy to see that there is a rearrangement (x∗n , yn∗ , zn∗ ) of (xn , yn , zn ) ∈ X 3 such that (dn[1] , dn[2] , dn[3] ) = (d(x∗n , zn∗ ), d(x∗n , yn∗ ), d(yn∗ , zn∗ )). ∗ ∗ ∗ +3 if n is taken large enough. The last equality and the relation (β, γ, δ) ∈ Ω+3 p,˜ r imply (xn , yn , zn ) ∈ X n Indeed, it is sufficient to show that d[3] > 0 which follows from

0 < ρ(γ, δ) = (ρ(γ, β) ∧ ρ(γ, δ) ∧ ρ(β, δ)) 1 (d(xn , zn ) n→∞ rn

= lim

dn [3] . n→∞ rn

∧ d(xn , yn ) ∧ d(yn , zn )) = lim

Consider first the case s0 < ∞. Write sn := s(x∗n , zn∗ , yn∗ ). Now using (2.4) and (2.5) we obtain ∗ ∗ 1 (d(xn , zn ) ∨ d(xn , yn ) ∨ d(yn , zn )) = lim d(xrnn,zn ) n→∞ rn n→∞  n  s  ∗ ∗ sn  ∗ ∗ sn  1 n d[2] sn d(xn ,yn ) d(yn ,zn ) lim + = lim + rn rn rn n→∞ n→∞

ρ(β, δ) = lim =

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rn

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that implies equality (2.5) for s0 < ∞. Suppose now that s0 = ∞. Let M be an arbitrary positive constant. Since the function f (t) = 1 t (a + bt ) t , a, b ∈ (0, ∞), is strictly decreasing in t ∈ (0, ∞), see Remark 4 below, and since equality (2.4) holds with s0 = ∞, we obtain the inequality 

1 d(x∗n , yn∗ ) M d(yn∗ , zn∗ ) M M d(x∗n , zn∗ )  + rn rn rn for sufficiently large n. Consequently, ∗) d(x∗n ,zn r n n→∞

ρ(β, δ) = lim  lim

n→∞



∗) d(x∗n ,yn rn



M +

∗ ,z ∗ ) d(yn n rn

M M1

 1  (ρ(β, γ))M + (ρ(γ, δ))M M .

(2.7)

Letting M → ∞ we have ρ(β, δ)  ρ(β, γ) ∨ ρ(γ, δ). The reverse inequality follows from the supposition (β, γ, δ) ∈ Ω+3 p,˜ r by (2.3). Thus (2.5) holds for all s0 ∈ [1, ∞]. Remark 3. Limit calculations in (2.6) - (2.7) are based on the following simple fact. If a sequence of vectors (xn1 , xn2 , xn3 ) tends to the vector (x1 , x2 , x3 ), then the sequence of their nonincreasing rearrangements (xn[1] , xn[2] , xn[3] ) tends to the rearrangement (x[1] , x[2] , x[3] ). Indeed, the classical HardyLittlewood-Polya inequality m m

ai bi  a[i] b[i] , i=1

ai , bi ∈ R, 1  i  m,

i=1

see, for example, [19, Chapter 6, A. 3], has as a consequence the estimation 3

(x[i] − xn[i] )2 

i=1

3

(xi − xni )2 . i=1

Corollary 1. If the metric space (X, d) is ultrametric, then all pretangent spaces ΩX p,˜ r are ultrametric for each p ∈ X. Remark 4. The sums

 n 1 t

t xi , xi ∈ (0, ∞), St (x) = i=1

decrease from +∞ to x1 ∨ . . . ∨ xn when t increases from 0 to +∞. The inequality St2 (x) < St1 (x), 0 < t1 < t2 < ∞, is sometimes referred to as the Jensen inequality. For the proof see, for example, [5]. As it was shown in Proposition 2 the condition lim

x,y,z→p

s(x, z, y) = ∞

(x,y,z)∈X +3

is sufficient for the ultrametricity of all pretangent spaces ΩX p,˜ r but it is not necessary as the following proposition shows. Proposition 3. For every s0 ∈ [1, ∞) there exists a metric space (X, d) with a marked point p such that lim

x,y,z→p

s(x, y, z) = s0

(x,y,z)∈X +3

but all pretangent spaces ΩX p,˜ r are ultrametric. p-ADIC NUMBERS, ULTRAMETRIC ANALYSIS AND APPLICATIONS

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Proof. Let ˜b = {bn }n∈N be a sequence of positive real numbers such that lim

bn n→∞ bn+1

105

= ∞ and bn >

bn+1 + 2bn+2 for each n ∈ N. Let us consider the metric space (X, d) with       {bn } ∪ {bn + bn+1 } ∪ {bn + 2bn+1 } X = {0} ∪ n∈N

n∈N

n∈N

and with d(x, y) = |x − y| and with a marked point 0. It simply follows from [13] that card ΩX 0,˜ r 2 X X for each pretangent space Ω0,˜r and that card Ωp,˜r = 1 if 0 = p ∈ X. Consequently all these pretangent spaces are ultrametric. Note now, that every triple (x, y, z) ∈ X 3 can be rearranged such that d(x, z) = d(x, y) + d(z, y). Consequently, for this (X, d), we have lim

x,y,z→p

s(x, y, z) = 1,

(x,y,z)∈X +3 1

1

so the proposition follows for s0 = 1. If s0 > 1, then d s0 is also a metric on X. The space (X, d s0 ), the snowflaked version of (X, d), is the desirable example. This proves the proposition. Let us prove now an ultrametricity criterion for pretangent spaces of general metric spaces. Proof of Theorem 1. (i) ⇒ (ii) Suppose statement (i) is true. If (1.10) does not hold, then there are ˜ such that α ∈ (0, ∞) and sequences x ˜ = {xn }n∈N , y˜ = {yn }n∈N , z˜ = {zn }n∈N from X lim xn = lim yn = lim zn = p

n→∞

n→∞

n→∞

and that s(xn , yn , zn ) Ψ(xn , yn , zn ) = α. n→∞ Φ(xn , yn , zn ) lim

1  ∞, 1  Ψ(x, y, z)  ∞ and 1  s(x, y, z)  ∞ hold for all Since the double inequalities 12  Φ(x,y,z) x, y, z ∈ X, we can suppose, proceeding to a subsequence if it is necessary, that there exist the following limits 1 =: φ0 , lim lim s(xn , yn , zn ) =: s0 (2.8) lim Ψ(xn , yn , zn ) =: ψ0 , n→∞ n→∞ Φ(xn , yn , zn ) n→∞

with ∞ > ψ0  1, ∞ > φ0  12 and with ∞ > s0  1. It follows from (1.8) that for every n ∈ N we have at least one of the following equalities F (xn , yn ) = Φ(xn , yn , zn ), F (yn , zn ) = Φ(xn , yn , zn ), F (zn , xn ) = Φ(xn , yn , zn ). Suppose the first equality (2.9)

F (xn , yn ) = Φ(xn , yn , zn )

holds on an infinite subset of N. Then, passing once again to a subsequence, we take that (2.9) is true for every n ∈ N. Hence the second equality in (2.8) can be rewritten as lim F (xn , yn ) = lim

n→∞

n→∞

1 d(xn , yn )(d(xn , p) ∧ d(yn , p)) = ∈ (0, 2]. (d(xn , p) ∨ d(yn , p))2 φ0

(2.10)

Analogously, we can suppose that the equality d(xn , p) = d(xn , p) ∨ d(yn , p)

(2.11)

holds on some infinite subset of N and passing to a subsequence, that this subset equals N. Relations (2.10)-(2.11) imply the inequality d(xn , p) > 0 for sufficiently large n. Write  1 if d(xn , p) = 0 (2.12) rn := d(xn , p) if d(xn , p) > 0 p-ADIC NUMBERS, ULTRAMETRIC ANALYSIS AND APPLICATIONS

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and r˜ := {rn }n∈N . We can now easily show that the quantities d(xn , p) , rn

d(yn , p) , rn

d(zn , p) , rn

are bounded above by a constant. Indeed, for d(xn ,yn ) rn

d(xn , yn ) , rn

d(xn ,p) rn

d(xn , zn ) , rn

d(yn ,p) rn

and

d(yn , zn ) rn

(2.13)

it follows from (2.11)–(2.12) and for

from the triangle inequality d(xn , p) d(yn , p) d(xn , yn )  + . rn rn rn

Since d(xrnn,yn ) is bounded above and the first limit in (2.8) is finite, the quantities d(yrnn,zn ) and d(xrnn,zn) are also bounded above. Finally, the inequality d(p, zn )  d(p, xn ) + d(xn , zn ) implies the desirable boundedness of d(zrnn,p) . Since all quantities in (2.13) are bounded, there is a sequence {nk }k∈N of natural numbers for which all limits d(xnk ,p) , k→∞ rnk

d(ynk ,p) , k→∞ rnk

lim

lim

d(xnk ,ynk ) , rnk k→∞

lim

d(znk ,p) , k→∞ rnk

lim

d(xnk ,znk ) , rnk k→∞

lim

(2.14)

d(ynk ,znk ) rnk k→∞

lim

are finite. Renaming x ˜ := {xnk }k∈N , y˜ := {ynk }k∈N , z˜ := {znk }k∈N and r˜ := {rnk }k∈N we obtain that x ˜, y˜, z˜ and p˜ are mutually stable w.r.t. r˜. We can now easily show that x, y˜) = 0 d˜r˜(˜

(2.15)

x, z˜) = 0 = d˜r˜(˜ y , z˜). d˜r˜(˜

(2.16)

and that For this purpose, note that (2.10) – (2.12) imply  lim F (xn , yn ) = lim d(xrnn,yn ) d(xrnn,p) ∧ n→∞

=

n→∞ n) lim d(xrnn,yn ) d(p,y rn n→∞

d(p,yn ) rn



= d˜r˜(˜ x, y˜)d˜r˜(˜ p, y˜) =

1 φ0

∈ (0, 2],

consequently relation (2.15) holds. Moreover (2.15) and the finiteness of the first limit in (2.8) imply (2.16). ˜ ∈ β, y˜ ∈ γ and z˜ ∈ δ for some β, γ, δ ∈ Ωp,˜r . The Let (ΩX p,˜ r , ρ) be a pretangent space such that x definition of the function s : (x, y, z) → s(x, y, z) and the finiteness of the last limit in (2.8) imply the equality ρ(β, γ) = ρ(β, γ) ∨ ρ(γ, δ) ∨ ρ(δ, β) and, in addition, it follows from (2.15) - (2.16) that ρ(β, γ) ∧ ρ(γ, δ) ∧ ρ(δ, β) > 0. We may assume, without loss of generality, that ρ(β, γ)  ρ(γ, δ)  ρ(δ, β) > 0, that is (γ, β, δ) ∈ Ω+3 p,˜ r . Using the last limit relation in (2.8) and reasoning as in the proof of Proposition 2 we obtain the equality 1

ρ(β, γ) = ((ρ(β, δ))s0 + (ρ(δ, γ))s0 ) s0 .

(2.17)

Since s0 ∈ [1, ∞), the last equality shows that (ΩX p,˜ r , ρ) is not an ultrametric space contrary to the assumption. To complete the proof of the implication (i) ⇒ (ii) it suffices to observe that (2.17) was derived from relations (2.15) and (2.16) and that these two relations remain valid if the pair (xn , yn ) in (2.9) is replaced by an arbitrary pair from the set {(xn , zn ), (yn , xn ), (yn , zn ), (zn , xn ), (zn , yn )}. (ii) ⇒ (i) Suppose now that (1.10) holds. We must prove that all pretangent spaces (ΩX p,˜ r , ρ) are ultrametric. To this end it suffices to show that for an arbitrary normalizing sequence r˜ the inequality x, y˜)  d˜r˜(˜ x, z˜) ∨ d˜r˜(˜ z , y˜) d˜r˜(˜ p-ADIC NUMBERS, ULTRAMETRIC ANALYSIS AND APPLICATIONS

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˜ whenever holds for all mutually stable (w.r.t. r˜) x ˜, y˜, z˜ ∈ X d˜r˜(˜ x, y˜)  d˜r˜(˜ x, z˜)  d˜r˜(˜ z , y˜) > 0 and whenever there are finite limits d(xn , p) d(yn , p) x, p˜) = lim , d˜r˜(˜ y , p˜) = lim , d˜r˜(˜ n→∞ n→∞ rn rn

(2.19)

d(zn , p) d˜r˜(˜ z , p˜) = lim , n→∞ rn

(2.20)

˜, {yn }n∈N = y˜, {zn }n∈N = z˜. Limit relation (1.10), the definition of Ψ and inequaliwhere {xn }n∈N = x ties (2.19) imply d˜r˜(˜ s(xn , yn , zn ) x, y˜) s(xn , yn , zn ) Ψ(xn , yn , zn ) = lim . n→∞ Φ(xn , yn , zn ) d˜r˜(˜ y , z˜) n→∞ Φ(xn , yn , zn )

∞ = lim Consequently we have

lim

s(xn , yn , zn )

n→∞ Φ(xn , yn , zn )

because, by (2.19), the quantity

x,˜ y) d˜r˜(˜ d˜r˜(˜ y ,˜ z)

=∞

(2.21)

are finite and positive. If, in addition, the equality lim s(xn , yn , zn ) = ∞

(2.22)

n→∞

holds, then reasoning as in the proof of the second part of Proposition 2 we obtain inequality (2.18). If (2.22) does not hold, then there are an infinite strictly increasing sequence {nk }k∈N of natural numbers and a constant s0 ∈ [1, ∞) such that lim s(xnk , ynk , znk ) = s0 .

k→∞

The last equality and (2.21) have as a consequence 1

lim

k→∞ Φ(xnk , ynk , znk )

= ∞.

(2.23)

It follows from this and (1.7) - (1.8) that ,p)

d(x

d(y

,p)

nk ∧ rnnk (d(xnk , p) ∧ d(ynk , p))d(xnk , ynk ) rnk k = d˜r˜(˜ x, y˜) lim  0 = lim  d(ynk ,p) 2 k→∞ k→∞ d(xnk ,p) (d(xnk , p) ∨ d(p, ynk ))2 ∨ rn rn k

k

and consequently lim 

k→∞

d(xnk ,p) rnk d(xnk ,p) rnk

∧ ∨

d(ynk ,p) rnk

 d(ynk ,p) 2 rnk

= 0.

Similarly we have lim 

k→∞

d(xnk ,p) rnk d(xnk ,p) rnk

∧ ∨

d(znk ,p) rnk

 d(znk ,p) 2 rnk

= lim  k→∞

d(ynk ,p) rnk d(ynk ,p) rnk

∧ ∨

d(znk ,p) rnk

 d(znk ,p) 2 rnk

= 0.

(2.24)

Note that x, p˜) ∨ d˜r˜(˜ y , p˜) ∨ d˜r˜(˜ z , p˜) > 0 d˜r˜(˜

(2.25)

x, y˜) = d˜r˜(˜ x, z˜) = d˜r˜(˜ z , y˜) = 0, contrary because in the opposite case the triangle inequality implies d˜r˜(˜ to (2.19). Suppose that x, p˜) ∨ d˜r˜(˜ y , p˜) ∨ d˜r˜(˜ z , p˜) = d˜r˜(˜ z , p˜). d˜r˜(˜ p-ADIC NUMBERS, ULTRAMETRIC ANALYSIS AND APPLICATIONS

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This equality and (2.24) imply 1 0 = lim k→∞ (d˜r˜(˜ z , p˜))2



d(xnk , p) d(znk , p) ∧ rnk rnk

=

d˜r˜(˜ x, p˜) , (d˜r˜(˜ z , p˜))2

x, p˜) = 0. Completely analogously we have d˜r˜(˜ y , p˜) = 0. It means that d˜r˜(˜ x, y˜) = 0 which implies i.e. d˜r˜(˜ x, p˜) (2.18). It still remains to note that similar arguments are applicable if the right side of (2.26) is d˜r˜(˜ ˜ ˜ y , p˜) instead of dr˜(˜ z , p˜). Hence in all cases (1.10) implies (2.18). or dr˜(˜ 3. METRIC BETWEENNESS IN PRETANGENT SPACES The purpose of this part of the paper is the proof of Theorem 2. Recall the following definition, see, for example, [20, p. 55]. Definition 6. Let (X, d) be a metric space and let x, y, z be distinct points of X. The point y lies between points x and z if d(x, z) = d(x, y) + d(y, z).

(3.1)

Remind that M is the class of all metric spaces (X, d) such that (3.1) holds for all x, y, z ∈ X whenever d(x, z)  d(x, y)  d(y, z). It is easy to see that X ∈ M if and only if the betweenness exponent t0 (Y ) = 1 for each Y ⊆ X with card Y  3. Proposition 2 shows that t0 (Θ) = s0 for every subspace Θ of pretangent space (Ωp,˜r , ρ) s0 provided that card Θ  3 and limit relation (2.4) holds with s0 < ∞. Thus the spaces (ΩX p,˜ r , ρ ) belong to M under these conditions. For the convenience we rewrite here the formulation of Theorem 1.7. Let (X, d) be a metric space, p a limit point of X and s1 a positive number. The membership relation s1 (ΩX p,˜ r, ρ ) ∈ M

(3.2)

holds for every pretangent space ΩX p,˜ r if and only if lim

x,y,z→p (x,y,z)∈X +3

Ψ(x, y, z)s2 (x, y, z) = ∞, Φ(x, y, z)(s1 − s(x, y, z))2

where s(x, y, z) = s1 − s(x, y, z)

(3.3)

 1 if s(x, y, z) = ∞ ∞ if s(x, y, z) = s1 .

Remark 5. If card ΩX p,˜ r  3, then using Lemma 2 we see that s1 equals the betweenness exponent of X (Ωp,˜r , ρ) whenever (3.2) holds. Moreover (3.2) holds for all s1 > 0 if and only if card ΩX p,˜ r  2. The following proof succeeds the scheme of the proof of Theorem 1. Proof of Theorem 2. Suppose that (3.2) holds for all pretangent spaces ΩX p,˜ r . We must prove that (3.3) holds. The direct calculations show that  1 if s1 = 1 (s(x, y, z))2 (3.4)  1 2 (s1 − s(x, y, z)) 1 ∧ (1−s1 )2 if s1 ∈ (0, ∞){1}. Hence if (3.3) does not hold, then as in the case of (2.8) there is a sequence of triples (xn , yn , zn ) ∈ X +3 , n ∈ N, such that lim xn = lim yn = lim zn = p

n→∞

n→∞

n→∞

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and that the following finite positive limits exist 1 = φ0 , n→∞ Φ(xn , yn , zn )

lim Ψ(xn , yn , zn ) = ψ0 ,

lim

n→∞

s2 (xn , yn , zn ) = s2 . n→∞ (s1 − s(xn , yn , zn ))2 lim

(3.5)

The condition (xn , yn , zn ) ∈ X +3 and (1.9) imply the equality d(xn , zn ) d(yn , zn )

Ψ(xn , yn , zn ) = and the membership relations

  d(xn , zn ) d(xn , zn ) d(xn , yn ) , ∈ 1, . d(xn , yn ) d(yn , zn ) d(yn , zn )

Consequently, replacing N by a suitable subset, we may suppose that there are finite constant ψ1 , ψ2 such that d(xn , zn ) d(xn , yn ) := ψ1 , lim := ψ2 (3.6) lim n→∞ d(xn , yn ) n→∞ d(yn , zn ) and ψ1 ψ2 = ψ0 and ψ1 ∧ ψ2  1. Using (3.5) and (3.6) and repeating the arguments from the first part of the proof of Theorem 1 we find a normalizing sequence r˜ = {rn }n∈N such that there are finite limits ˜ p, x d(˜ ˜) = lim

n→∞

d(xn ,p) rn ,

˜ p, y˜) = lim d(˜

n→∞

d(xn ,yn ) , rn n→∞

d(yn ,p) rn ,

˜ p, z˜) = lim d(˜

d(xn ,zn ) , rn n→∞

˜ x, z˜) = lim d(˜

˜ x, y˜) = lim d(˜

n→∞

d(zn ,p) rn ,

d(yn ,zn ) rn n→∞

˜ y , z˜) = lim d(˜

(3.7)

and, in addition, the inequalities ˜ x, z˜)  d(˜ ˜ x, y˜)  d(˜ ˜ y , z˜) > 0 d(˜

(3.8)

hold. Note that (3.7) is similar to (2.14) and (3.8) can be obtained as (2.15) — (2.16). Let us consider now the limit relation s2 (xn , yn , zn ) = s2 . n→∞ (s1 − s(xn , yn , zn ))2

(3.9)

lim

We first establish that s2 = 1. As usual, replacing N by a suitable infinite subset, we may suppose that there is the limit lim s(xn , yn , zn ) = s0 ∈ [1, ∞].

(3.10)

n→∞

If s0 = +∞, then, as in the proof of Proposition 2, we can obtain the equality ˜ x, z˜) = d(˜ ˜ x, y˜) ∨ d(˜ ˜ y , z˜). d(˜

(3.11)

Furthermore, it follows from (3.2) that  s  s  s ˜ x, y˜) 1 + d(˜ ˜ y , z˜) 1 . ˜ x, z˜) 1 = d(˜ d(˜

(3.12)

˜ y , z˜) = 0, contrary to (3.8). Hence the limit in (3.10) is Equalities (3.11) and (3.12) imply the equality d(˜ finite. If s2 = 1, then using (3.9) we have s20 = 1, (s0 − s1 )2 that is s0 = s1 /2. From this we obtain   s1   s1   s1 ˜ x, y˜) 2 + d(˜ ˜ y , z˜) 2 ˜ x, z˜) 2 = d(˜ d(˜ p-ADIC NUMBERS, ULTRAMETRIC ANALYSIS AND APPLICATIONS

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in the same manner as in the case s0 = ∞. It contradicts (3.12) because the function  1 ˜ x, y˜)s + (d(˜ ˜ y , z˜))s s , f (s) = (d(˜

(3.14)

is strictly decreasing in s. Thus s2 = 1 as it was indicated. Let us analyze now the possible value of s0 in (3.10). It was noted above that s0 = ∞ because s2 = 1. Hence s0 is a root of the equation x2 = s2 (x − s1 )2 where s2 ∈ (1, ∞). Since s2 = 1, two possible values of s0 are √ √ −s1 s2 s1 s2 and √ √ , 1 + s2 1 − s2 √ where we put s2 > 0. It follows from (3.15) that s0 = s1 because if s0 = s1 , then √ √ √ √ s1 (1 + s2 ) = s1 s2 or s1 (1 − s2 ) = −s1 s2 ,

(3.15)

i.e., s1 = 0 which contradicts the conditions of the theorem. As in (3.13) we obtain  s  s  s ˜ x, y˜) 0 + d(˜ ˜ y , z˜) 0 . ˜ x, z˜) 0 = d(˜ d(˜ The last equality contradicts (3.12) because function (3.14) is strictly decreasing on (0, ∞) and s0 = s1 . It follows that the positive constant s2 in (3.5) and (3.9) cannot be finite, contrary to the assumption. s1 Thus the limit relation (3.3) holds if all spaces (ΩX p,˜ r , ρ ) belong to M. Suppose now that limit relation (3.3) holds. We must prove that for every pretangent space ΩX p,˜ r the equality (ρ(β, δ))s1 = (ρ(β, γ))s1 + (ρ(γ, δ))s1

(3.16)

holds for all β, γ, δ ∈ ΩX p,˜ r whenever ρ(β, δ)  ρ(β, γ)  ρ(γ, δ).

(3.17)

Since (3.17) is trivial for ρ(γ, δ) = 0, we assume (3.18)

ρ(γ, δ) > 0. Note that (3.17) together with (3.18) are an equivalent of (β, γ, δ) ∈ Ω+3 p,˜ r. Let (β, γ, δ) ∈ Ω+3 p,˜ r is given and let x ˜∗ = {x∗n }n∈N ∈ β,

y˜∗ = {yn∗ }n∈N ∈ γ,

z˜∗ = {zn∗ }n∈N ∈ δ.

x∗n , yn∗ , zn∗

(3.19)

X +3

belongs to for n belonging to At least one of the six rearrangements of the points some infinite subsequence {nk }k∈N of natural numbers. We denote by xn the first element of this rearrangement, by yn the second and by zn the third one. Using, as usual, {nk }k∈N instead of the ˜ := sequence of all natural numbers we may suppose that (xk , yk , zk ) ∈ X +3 for each k ∈ N. Write x {xk }k∈N , y˜ := {yk }k∈N , z˜ := {zk }k∈N . Relations (3.19) imply the equalities x, z˜) = lim d˜r˜ (˜

k→∞

d(xk ,zk ) rnk

= ρ(β, δ),

d˜r˜ (˜ x, y˜) = lim

k→∞

d(yk ,zk ) k→∞ rnk

y , z˜) = lim d˜r˜ (˜

d(xk ,yk ) rnk

= ρ(β, γ), (3.20)

= ρ(γ, δ),

see Remark 3. Moreover, it follows directly from definitions of x ˜, y˜, z˜ that there are finite limits lim

k→∞

d(xk , p) , rnk

lim

k→∞

d(yk , p) rnk

and

lim

k→∞

d(zk , p) . rnk

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Consequently the family {˜ x, y˜, z˜, p˜ } is self-stable w.r.t. r˜ = {rnk }k∈N . Using (3.20) we can rewrite (3.16) in the equivalent form   s1   s1   s1 d˜r˜ (˜ x, z˜) = d˜r˜ (˜ x, y˜) + d˜r˜ (˜ y , z˜) .

(3.21)

Reasoning as in the proof of (2.21) we obtain s2 (xk , yk , zk ) = ∞. k→∞ Φ(xk , yk , zk )(s1 − s(xk , yk , zk ))2 lim

If, in addition, the equality lim s(xk , yk , zk ) = s1 is true, then, using the first part of the proof of k→∞

Proposition 2, we see that (3.21) holds. In the opposite case there is an infinite strictly increasing sequence of natural numbers km , m ∈ N, for which, similarly (2.23), we obtain lim

1

m→∞ Φ(xkm , ykm , zkm )

= ∞.

Now the equality (3.21) can be proved by the repetition of the arguments which stay after (2.23). In the following corollary and further we assume that R is the set of all real numbers with the usual metric d(x, y) = |x − y|. Corollary 2. Let (X, d) be a metric space with a limit point p, (ΩX p,˜ r , ρ) a pretangent space to X at the point p and s1 a positive number. If (3.3) holds for this s1 , then each one from the following conditions is sufficient that the space (ΩX p,˜ r , ρ) be tangent. s1 (i) There is no any isometric embeddings of (ΩX p,˜ r , ρ ) in R. s1 (ii) The space (ΩX p,˜ r , ρ ) is isometric to R. To prove this we will use the following particular case of the classical Menger’s result on the isometric embeddings into Euclidean spaces. Theorem 3 (K. Menger). Let Y ∈ M be a metric space with card Y  5. Then Y is isometric to some subset of R. Proof of Corollary 2. Suppose that condition (i) is fulfilled and (3.3) is true. Since (3.3) holds, we have, s1 by Theorem 2, that (ΩX p,˜ r , ρ ) ∈ M. Using Menger’s theorem 3 we see that (3.22)

card Ωp,˜r = 4. X such that If (ΩX p,˜ r , ρ) is not tangent, then there is a pretangent space Ωp,˜ r

Ωp,˜r em (Ωp,˜r ) = ∅

(3.23)

X , see Definition 5. Since (ΩX , ρs1 ) also belongs to M, for the isometric embedding em : ΩX p,˜ r → Ωp,˜ r p,˜ r relations (3.22) - (3.23) imply the inequality card Ωp,˜r  5. Hence, by Menger’s theorem, there exists an em

f

s1 X → ΩX → R is an isometric isometric embedding f of (ΩX r p,˜ r  , ρ ) into R. Now the superposition Ωp,˜ p,˜ r X s 1 embedding of (Ωp,˜r , ρ ) in R, contrary to (i).

Let condition (ii) and (3.3) be fulfilled. In order that (ΩX p,˜ r , ρ) be tangent, it suffices to show each isometric embedding f : R → Y is a bijection if Y ∈ M. Suppose an isometric embedding f : R → Y , Y ∈ M, is not bijective. Let a ∈ Y f (R) and let b ∈ f (R). Write s := ρ(a, b) where ρ is the metric on Y . Then there are two distinct points p1 , p2 ∈ R for which |f −1 (b) − p1 | = |f −1 (b) − p2 | = s. Consequently we have ρ(b, f (p1 )) = ρ(b, f (p2 )) = ρ(b, a) = s > 0 where all three points f (p1 ), f (p2 ) and b are distinct. Theorem 3 implies that (Y, ρ) can be isometrically embedded in R. Let g : Y → R be a such embedding. Then the points g(b), g(f (p1 )) and g(f (p2 )) are distinct points of the sphere {x ∈ R : |g(a) − x| = s}. It is impossible because every "sphere" in R contains only two points. p-ADIC NUMBERS, ULTRAMETRIC ANALYSIS AND APPLICATIONS

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Remark 6. The four-point metric spaces Y ∈ M which cannot be isometrically embedded in R are sometimes referred to as pseudo-linear quadruples. It is well known that (Y, d) is a pseudo-linear quadruples if and only if the points of Y can be labelled p0 , p1 , p2 , p3 such that d(p0 , p1 ) = d(p2 , p3 ) = s,

d(p1 , p2 ) = d(p0 , p3 ) = t,

d(p0 , p2 ) = d(p1 , p3 ) = s + t,

(3.24)

where s and t are some positive constants. See, for example, [6, p. 114]. Each pseudo-linear quadruples can be easily realized as a subset of the two-dimensional linear 2 where, as usual, the norm is (x , x ) = |x | ∨ |x |. To this end we put normed space l∞ 1 2 1 2 p0 = (0, 0), p1 = (s, s), p2 = (s + t, s − t), p3 = (t, −t)

(3.25)

Simple calculations show that equalities (3.24) hold in this case. Realization (3.25) leads us to the examples of metric spaces which have the pseudo-linear quadruples as tangent spaces. Example 1. Let r˜ = {rn }n∈N be a decreasing sequence of positive numbers such that lim rn /rn+1 = n→∞

2 which +∞ and let t, s ∈ (0, ∞). Write Y := {p0 , p1 , p2 , p3 } where pi , i = 0, . . . , 3 are the points of l∞ were defined by (3.25) and set  rn Y X= n∈N

with rn Y = {rn p : p ∈ Y }. Consider the metric space (X, d) with the metric d induced from the space 2 . It is easy to see that the sequences x ˜i := {rn pi }n∈N , i = 0, . . . , 3, are mutually stable w.r.t. the l∞ normalizing sequence r˜. Furthermore, it can be shown that there is a unique maximal self-stable family ˜ p ,˜r . Equalities ˜ p ,˜r and that card ΩX = 4, p0 = (0, 0), where ΩX is the metric identification of X X 0 0 p0 ,˜ p0 ,˜ r r (3.24) imply that ΩX p0 ,˜ r is isometric to the pseudo-linear quadruples and, consequently, is tangent by Corollary 2. Remark 7. Some details dropped under consideration of the above example can be easily regenerated from [13]. The next corollary of Theorem 2 clarifies "the geometrical sense" of the factor Φ(x, y, z) in limit relation (3.3). Corollary 3. Let (X, d) be a metric space and let p be a limit point of X. The following statements are equivalent: (i) (Ωp,˜r , ρs1 ) ∈ M for every pretangent (Ωp,˜r , ρ) and all s1 ∈ (0, ∞); (ii) card Ωp,˜r  2 for every pretangent (Ωp,˜r , ρ); (iii) lim F (x, y) = 0 where the function F was defined by (1.7); x,y→p

(iv)

lim

x,y,z→p

Φ(x, y, z) = 0 where the function Φ was defined by (1.8).

(x,y,z)∈X +3

Proof. (i) ⇒ (ii). It follows from the definition of M and from Remark 4. (ii) ⇒ (iii) was proved in [13]. (iii) ⇒ (iv) is immediate from the definitions of the functions Φ and F . (iv) ⇒ (i). To prove this note that (iv) implies (3.3) because the values of Ψ(x, y, z)s2 (x, y, z) (s1 − s(x, y, z))2 are bounded away from zero. Thus (i) follows by Theorem 2. ACKNOWLEDGMENT The first author was partially supported by the State Foundation for Basic Researches of Ukraine, Grant Φ 25.1/055. p-ADIC NUMBERS, ULTRAMETRIC ANALYSIS AND APPLICATIONS

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