TANGENT SPACES TO GENERAL METRIC SPACES

OLEKSIY DOVGOSHEY and OLLI MARTIO

We introduce a tangent space at an arbitrary point of a general metric space. Itis proved that all tangent spaces are complete. The conditions under which thesespaces have a finite cardinality are found.

AMS 2010 Subject Classification: 54E35.

Key words: metric space, finite tangent space, completeness of tangent spaces.

1. INTRODUCTION. MAIN DEFINITIONS

The recent achievements in the metric space theory are closely related tosome generalizations of differentiation. The concept of upper gradient [6] and[9], Cheeger’s notion of differentiability for Rademacher’s theorem in certainmetric measure spaces [4], the metric derivative in the studies of metric spacevalued functions of bounded variation [2], [3] and the Lipshitz type approachin [5] are interesting and important examples of such generalizations. Suchgeneralizations lead usually to some nontrivial results only if the metric spaceshave “sufficiently many” rectifiable curves.

We define “tangent” spaces at a point of a metric space as a certainquotient space of the sequences which converge to this point and after thatintroduce the “derivatives” of functions as corresponding quotient maps.

Let (X, d) be a metric space and let a be point of X. Fix a sequence rof positive real numbers rn which tend to zero. In what follows this sequencer will be called a normalizing sequence.

Definition 1.1. Two sequences x = xnn∈N and y = ynn∈N, xn, yn ∈X, are mutually stable (with respect to a normalizing sequence r = rnn∈N)if there is a finite limit

(1.1) limn→∞

d(xn, yn)rn

:= dr(x, y) = d(x, y).

We shall say that a family F of sequences of points from X is maximalmutually stable (with respect to a normalizing sequence r) if every two x, y ∈ F

are mutually stable and for an arbitrary z = znn∈N with zn ∈ X either z ∈ F

or there is x ∈ F such that x and z are not mutually stable.

REV. ROUMAINE MATH. PURES APPL., 56 (2011), 2, 137–155

138 Oleksiy Dovgoshey and Olli Martio 2

The standard application of Zorn’s Lemma leads to the following

Proposition 1.2. Let (X, d) be a metric space and let a ∈ X. Thenfor every normalizing sequence r = rnn∈N there exists a maximal mutuallystable family Xa = Xa,r such that a := a, a, . . . ∈ Xa.

Note that the condition a ∈ Xa implies the equality

limn→∞

d(xn, a) = 0

for every x = xnn∈N belonging to Xa.Consider the function d : Xa × Xa → R defined by (1.1). Obviously, d is

symmetric and nonnegative. Moreover, the triangle inequality for d implies

d(x, y) ≤ d(x, z) + d(z, y)

for all x, y, z from Xa. Hence (Xa, d) is a pseudometric space.

Definition 1.3. The pretangent space to the space X at the point a w.r.t.r is the metric identification of the pseudometric space (Xa,r, d).

Since the notion of pretangent space is an important step in the de-velopment of general machinery for the definition of tangent spaces and dif-ferentiability in arbitrary metric spaces, we recall this metric identificationconstruction.

Define a relation ∼ on Xa by x ∼ y if and only if d(x, y) = 0. Then∼ is an equivalence relation and let Ωa,r be the set of equivalence classes inXa under the equivalence relation ∼. It follows from general properties ofpseudometric spaces, see for example [7, Chapter 4, Theorem 15], that if ρ isdefined on Ωa by

(1.2) ρ(α, β) := d(x, y)

for some x ∈ α and y ∈ β, then ρ is a well-defined metric on Ωa. By definitionthe metric space (Ωa, ρ) is the metric identification of (Xa, d).

Remark that Ωa,r 6= ∅ for all r because the constant sequence a belongsto Xa,r in accordance with Proposition 1.2.

Let nkk∈N be an increasing sequence of natural numbers. Denote byr′ the subsequence rnk

k∈N of the normalizing sequence r = rnn∈N. It isclear that if x = xnn∈N and y = ynn∈N are mutually stable w.r.t. r, thenx′ := xnk

k∈N and y′ := ynkk∈N are mutually stable w.r.t. r′ and that

(1.3) dr(x, y) = dr′(x′, y′).

If Xa,r is a maximal mutually stable (w.r.t. r) family, then by Zorn’s Lemmathere exists a maximal mutually stable (w.r.t. r′) family Xa,r′ such that

x′ : x ∈ Xa,r ⊆ Xa,r′ .

3 Tangent spaces to general metric spaces 139

Denote by inr′ the mapping Xa,r → Xa,r′ with inr′(x) = x′ for all x ∈ Xa,r.If follows from (1.3) that after metric identifications the mapping inr′ inducesan isometric embedding in′: Ωa,r → Ωa,r′ , i.e., the diagram

(1.4)

Xa,rinr′−−−−−→ Xa,r′

p

yyp′

Ωa,rin′−−−−−→ Ωa,r′

is commutative. Here p, p′ are metric identification mappings p(x) = y ∈Xa,r : dr(x, y) = 0 and p′(x) = y′ ∈ Xa,r′ : dr′(x′, y′) = 0.

Let X and Y be two metric spaces. Recall that a map f : X → Y iscalled an isometry if f is distance-preserving and onto.

Definition 1.4. A pretangent Ωa,r is tangent if for every r′ the mappingin′: Ωa,r → Ωa,r′ is an isometry.

Remark 1.5. As has been stated above, the function in′ : Ωa,r → Ωa,r′

is an isometric embedding. Since every surjective, isometric embedding is anisometry, Ωa,r is tangent if and only if in′ is surjective for every r′.

The following question arieses naturally.

Problem 1.6. Let (X, d) be a metric space and let a ∈ X. Find theconditions under which all pretangent spaces Ωa,r are tangent.

Let (X1, d1) and (X2, d2) be metric spaces, and r1, r2 normalizing se-quences, and a1, a2 points of X1 and X2 respectively, and X1

a1,r1, X2

a2,r2maxi-

mal mutually stable families of sequences of points from X1 and X2 respec-tively. Let f : X1 → X2 be a function such that f(a1) = a2. For everyx1 := x1

nn∈N ∈ X1a1,r1

write

f(x1) := f(x1n)n∈N.

Definition 1.7. The function f is differentiable w.r.t. the pair (X1a1,r1

,

X2a2,r2

) if f(x1) ∈ X2a1,r2

and d1(x1, y1) = 0 implies d2(f(x1), f(y1)) = 0 for allx1, y1 ∈ X1

a1,r1.

Let pi : Xiairi

→ Ωai,ri , i = 1, 2, be metric identification mappings.

Definition 1.8. A function Df : Ωa1,r1 → Ωa2,r2 is a derivative of f w.r.t.pretangent spaces Ωai,ri , i = 1, 2, if f is differentiable w.r.t. (X1

a1,r1, X2

a2,r2)

140 Oleksiy Dovgoshey and Olli Martio 4

and the following diagram

X1a1,r1

f−−−−−−−→ X2a2,r2

p1

yyp2

Ωa1,r1

Df−−−−−−−−→ Ωa2,r2

is commutative.

Definitions 1.7 and 1.8 are a main goal of all previous constructions, so wepresent them here, although the properties of “metric space valued derivatives”Df are not discussed in this paper.

2. FINITE TANGENT SPACES

It is clear that Ωa is an one-point space if a is an isolated point of X.The converse proposition is also true.

Proposition 2.1. Let (X, d) be a metric space and let a ∈ X. Then a isan isolated point of X if and only if every pretangent space Ωa,r is a singleton.

Proof. If a is not an isolated point of X, then there is a sequence b =bnn∈N of points in X such that lim

n→∞d(a, bn) = 0 and d(a, bn) 6= 0 for all

n ∈ N. Consider the normalizing sequence r = rnn∈N with rn := d(a, bn).It follows immediately from (1.1) that dr(a, b) = 1 where a is the constantsequence a, a, . . .. The application of Zorn’s Lemma shows that there is amaximal mutually stable family Xa,r such that a, b ∈ Xa,r. Then the metricidentification of the pseudometric space (Xa,x, d) has at least two points.

The following proposition characterizes the points a ∈ X for which allpretangent spaces Ωa have cardinality at most two.

Define the function F : X ×X → R by the rule

(2.1) F (x, y) =

d(x, y)(d(x, a) ∧ d(y, a))

(d(x, a) ∨ d(y, a))2if (x, y) 6= (a, a),

0 if (x, y) = (a, a).

Theorem 2.2. Let (X, d) be a metric space and let a ∈ X. Then theinequality

(2.2) card(Ωa,r) ≤ 2

holds for every pretangent space Ωa,r if and only if

(2.3) limx→ay→a

F (x, y) = 0.

5 Tangent spaces to general metric spaces 141

Proof. Suppose that (2.3) does not hold. Then there are some sequencesx′nn∈N and y′nn∈N such that

limn→∞

x′n = limn→∞

y′n = a

and x′n 6= a 6= y′n for all n ∈ N and

(2.4) lim supn→∞

F (x′n, y′n) > 0.

Consider the normalizing sequence r = rnn∈N with rn := d(x′n, a)∨ d(y′n, a).Write

(xn, yn) :=

(x′n, y′n) if d(x′n, a) ≥ d(y′n, a),

(y′n, x′n) if d(y′n, a) > d(x′n, a).

Then we obtain

(2.5) F (x′n, y′n) = F (xn, yn) =d(xn, yn)

rn· d(yn, a)

rn.

Since

(2.6) d(yn, a) ≤ rn

and

(2.7) d(xn, yn) ≤ d(xn, a) + d(yn, a) ≤ 2rn,

we have

(2.8) F (xn, yn) ≤ 2

for all n ∈ N. Let xnk, ynk

k∈N be a subsequence of xn, ynN for which

lim supn→∞

F (x′n, y′n) = limk→∞

F (xnk, ynk

).

Inequalities (2.4) and (2.8) imply that

(2.9) 0 < limk→∞

F (xnk, ynk

) ≤ 2.

We may assume, without loss of generality, that xnk, ynk

k∈N and xn, ynn∈Ncoincide. Conditions (2.6) and (2.7) imply the existence of finite limits

limm→∞

d(xnm , ynm)rnm

≤ 2 and limm→∞

d(ynm , a)rnm

≤ 1

for some subsequence xnm , ynmm∈N of xn, ynn∈N. Now we can, once again,take nm = n. It follows from (2.5) and (2.9) that

0 < limn→∞

F (xn, yn) = limn→∞

d(xn, yn)rn

· limn→∞

d(yn, a)rn

≤ 2.

Consequently, we have

d(x, a) = 1, 0 < d(y, a) ≤ 1, 0 < d(x, y) ≤ 2

142 Oleksiy Dovgoshey and Olli Martio 6

for x = xnn∈N, y = ynn∈N. Hence, a, y, x correspond to distinct points ofthe pretangent space Ωa,r, contrary to (2.2). The implication (2.2) ⇒ (2.3) isestablished.

To prove that (2.3) implies (2.2), suppose

card(Ωa,r) ≥ 3

for some normalizing sequence r. Then there exist sequences x = xnn∈Nand y = ynn∈N such that all three quantities

d(x, a) = limn→∞

d(xn, a)rn

, d(y, a) = limn→∞

d(yn, a)rn

and

d(x, y) = limn→∞

d(xn, yn)rn

are finite and positive. Consider the sequence F (xn, yn)n∈N where F wasdefined by (2.1). Since

0 < limn→∞

d(xn, a) ∨ d(yn, a)rn

= d(x, a) ∨ d(y, a) < ∞

and

0 < limn→∞

d(xn, a) ∧ d(yn, a)rn

= d(x, a) ∧ d(y, a) < ∞,

we obtain

limn→∞

F (xn, yn) = limn→∞

d(xn,yn)rn

· d(xn,a)∧d(yn,a)rn(

d(xn,a)∨d(yn,a)rn

)2

=d(x, y)(d(x, a) ∧ d(y, a))(

d(x, a) ∨ d(y, a))2 ∈ (0,∞),

contrary to the limit relation (2.3). Hence, (2.3) implies (2.2).

Theorem 2.2 can be generalized as follows.For every natural n ≥ 2 let us denote by Xn the set of all n-tuples

x = (x1, . . . , xn) with terms xk ∈ X for k = 1, . . . , n. Define the functions

7 Tangent spaces to general metric spaces 143

Fn : Xn → R by the rule

(2.10) Fn(x1, . . . , xn) :=

:=

n∏k,l=1k<l

d(xk, xl)

(n∧

k=1

d(xk, a))

(n∨

k=1

d(xk, a))n(n−1)

2+1

if (x1, . . . , xn) 6= (a, . . . , a),

0 if (x1, . . . , xn) = (a, . . . , a),

wheren∧

k=1

d(xk, a) := min1≤k≤n

d(xk, a) andn∨

k=1

d(xk, a) := max1≤k≤n

d(xk, a).

Theorem 2.3. Let (X, d) be a metric space, a ∈ X and let n ≥ 2 be anatural number. Then the inequality

(2.11) card(Ωa,r) ≤ n

holds for every pretangent space Ωa,r if and only if

(2.12) limx1→a···

xn→a

Fn(x1, . . . , xn) = 0.

The proof of this theorem can be obtained as a direct generalization ofthe proof of Theorem 2.2 so we omit it.

Remark 2.4. The valuen∏

k,l=1k<l

d(xk, xl) in (2.10) can be regarded as a

metric-space analog of the Vandermonde Determinant.

There exist other functions Φn : Xn → R which can be used instead ofthe function Fn in Theorem 2.3. As an example consider

(2.13) Φn(x1, . . . , xn) :=

:=

n∧k,l=1k<l

d(xk, xl)

(n∧

k=1

d(xn, a))

(n∨

k=1

d(xk, a))2 if (x1, . . . , xn) 6= (a, . . . , a),

0 if (x1, . . . , xn) = (a, . . . , a).

144 Oleksiy Dovgoshey and Olli Martio 8

Then (2.12) holds if and only if

(2.14) limx1→a···

xn→a

Φn(x1, . . . , xn) = 0.

Indeed, it follows from the simple inequality

d(xk, xl) ≤ 2n∨

k=1

d(xk, a), k, l ∈ 1, . . . , n

that

Fn(x1, . . . , xn) ≤ 2n(n−1)

2−1Φn(x1, . . . , xn).

On the other hand, we have

Fn(x1, . . . , xn) ≥

n∧k,lk<l

d(xk, xl)

n∨k=1

d(xk, a)

n(n−1)

2

·

n∧

k=1

d(xk, a)

n∨k=1

d(xk, a)

≥

≥

n∧k,lk<l

d(xk, xl)

n∧k=1

d(xk, a)

n(n−1)

2

·

n∧

k=1

d(xk, a)

n∨k=1

d(xk, a)

n(n−1)

2

= (Φn(x1, . . . , xn))n(n−1)

2 .

Using Theorems 2.2 and 2.3 we can easily construct metric spaces withfinite tangent spaces. To this end, we first establish a lemma.

Let W be a family of some sequences of points from X. Suppose that a ∈W and that every two x, y ∈ W are mutually stable w.r.t. a fixed normalizingsequence r. Denote by Wm a maximal mutually stable family containing W ,by Ωa,r the pretangent space corresponding to (Wm, d) and by Ωw

a,r the metricidentification of the pseudometric space (W , d). Clearly, there is a unique“natural”, isometric embedding Em : Ωw

a,r → Ωa,r for which the diagram

(2.15)

Win−−−−−−→ Wm

p

yypm

Ωwa,r

Em−−−−−−→ Ωa,r

is commutative. Here p and pm are metric identification mappings and in(x) =x for all x ∈ W .

9 Tangent spaces to general metric spaces 145

Lemma 2.5. Let (X, d) be a metric space, a ∈ X and let n ≥ 2 be a na-tural number. Suppose that the limit relation (2.12) holds. Then the inequality

(2.16) card(Ωwa,r) ≥ n

implies that the embedding Em : Ωwa,r → Ωa,r is an isometry and that a pre-

tangent space Ωa,r is tangent.

Proof. By Theorem 2.3, the limit relation (2.12) implies that

(2.17) card(Ωa,r) ≤ n.

Since Em is an injective mapping, inequalities (2.16) and (2.17) imply

card(Ωa,r) = card(Ωwa,r) = n.

These equalities show that each isometric embedding Ωwa,r → Ωa,r is an isom-

etry because n is a natural number.Consider now the commutative diagram (1.4) with Xa,x = Wm. In com-

plete analogy with the above proof we can show that in′ the is an isometry forevery subsequence r′ of r. Hence, Ωa,r is tangent by Definition 1.4.

Example 2.6. Let H be a Hilbert space with the norm ‖ · ‖ and let

(2.18) dim H = m ≥ 1.

It follows from (2.18) that for every natural k ≤ m there are orthonormalvectors e1, . . . , ek in H. Let t = tjj∈N be a sequence of strictly decreasingpositive numbers tj for which

(2.19) limj→∞

tjtj+1

= ∞.

Write

(2.20) X := tjei : i = 1, . . . , k, j ∈ N ∪ 0,i.e., 0 ∈ X, and for all j ∈ N

X ∩ x ∈ H : ‖x‖ = t = ∅if t 6= tj and

X ∩ x ∈ H : ‖x‖ = tj = tje1, . . . , tjek.Consider the following sequences of elements of X

0 = 0, . . . , 0, . . ., x1 = tje1j∈N, . . . , xk = tjekj∈N.

It is easy to see that all these sequences are pairwise mutually stable w.r.t.the normalizing sequence r = t and that

d(xp, xq) := limj→∞

‖tjep − tjeq‖rj

=√

2

146 Oleksiy Dovgoshey and Olli Martio 10

for p 6= q, and that

d(0, xp) := limj→∞

‖tjep‖rj

= 1

for p ∈ 1, . . . , k.Let Wm be a maximal mutually stable family such that

Wm ⊇ W := 0, x1, . . . , xk.We claim the following:(a) For every y ∈ Wm there is x ∈ W such that d(y, x) = 0;(b) The space Ω0,r, corresponding to Wm, is tangent.Since card(Ωw

0,r) = k + 1, for the proof of (a) and (b) it is enough toshow that

(2.21) limx1→0···

xk+1→0

Φk+1(x1, . . . , xk+1) = 0,

where Φn is defined by (2.13) (see Remark 2.4 and Lemma 2.5).Let (x1, . . . , xk+1) be a k + 1-tuple with xi ∈ X, i ∈ 1, . . . , k + 1. We

may assume, without loss of generality, that xi 6= 0 for all i = 1, . . . , k + 1because Φk+1(x1, . . . , xk+1) = 0 in the opposite case. Using (2.20), we definethe natural numbers j = j(x1, . . . , xk+1) and s = s(x1, . . . , xk+1) such that

tj =k+1∨l=1

‖xl‖, tj+s =k+1∧l=1

‖xl‖.

Note that these numbers are well defined because the sequence t = tjj∈N isstrictly decreasing. It follows from the definition (2.20) that s ≥ 1 and that

limx1→0

...xk+1→0

j(x1, . . . , xk+1) = ∞. Now from (2.19) we obtain

0 ≤ lim supx1→0

...xk+1→0

Φ(x1, . . . , xk+1) ≤ 2 lim supx1→0

...xk+1→0

k+1∧l=1

‖xl‖

k+1∨l=1

‖xl‖

= 2 lim supx1→0

...xk+1→0

tj+1

tj· tj+2

tj+1· · · tj+s−1

tj+s≤ 2 lim

j→∞

tj+1

tj= 0.

Relation (2.21) is proved.

Let (X, ρ) and (Y, d) be metric spaces. Recall that a homeomorphismf : X → Y is a similarity if there is a positive number k such that

d(f(x), f(y)) := kρ(x, y)

11 Tangent spaces to general metric spaces 147

for all x, y ∈ X. The number k is the dilatation number of f . Two metricspaces X and Y are similar if there exists a similarity of X onto Y .

Proposition 2.7. Let H and X be the spaces defined in Example 2.6.Then, every tangent space at the point 0 ∈ X, is either a singleton or similarto 0, e1, . . . , ek ⊆ H.

Proof. Let Ω0,r be a tangent space to X at the point 0. Equality (2.21)implies that

card(Ω0,r) := m ≤ k + 1.

Let X0,r be a maximal mutually stable family for which p(X0,r) = Ω0,r (see dia-gram (1.4)). If Ω0,r is not one-point, then there are sequences xi = xi

nn∈N ∈X0,r such that

(2.22) limn→∞

‖xin‖

rn:= d(0, xi) ∈ (0,∞)

for i = 1, . . . ,m− 1 and

(2.23) limn→∞

‖xin − xj

n‖rn

:= d(xi, xj) ∈ (0,∞)

for i, j ∈ 1, . . . ,m− 1 if i 6= j. Relations (2.22) imply

(2.24) limn→∞

‖xjn‖

‖xin‖

=d(0, xj)d(0, xi)

∈ (0,∞).

It follows from (2.20), that for every x ∈ X \0 there is a unique l = l(x) ∈ Nfor which

‖x‖ = tl.

Substituting tl(x) in (2.24), we obtain

limn→∞

tl(xj

n)

tl(xin)

=d(0, xj)d(0, xj)

∈ (0,∞).

These relations and (2.19) imply the equality

‖xin‖ = ‖xj

n‖,for i, j ∈ 1, . . . ,m − 1, if n is taken large enough. Hence, using (2.23) and(2.20) for all sufficiently large n we have

(2.25) ‖xin − xj

n‖ = tl√

2δij , ‖xin‖ = tl

where δij is Kronecker’s delta, i, j ∈ 1, . . . ,m− 1 and l = l(xjn) = l(xi

n). If

card(Ω0,r) < k + 1,

then there exists xm = xmn n∈N such that

(2.26) ‖xin − xm

n ‖ = tl√

2, ‖xmn ‖ = tl

148 Oleksiy Dovgoshey and Olli Martio 12

for i = 1, . . . ,m−1 and for all n, l which satisfy (2.25). Relations (2.22), (2.23),(2.25) and (2.26) imply the existence of positive finite limits

limn→∞

‖xmn ‖

rnand lim

n→∞

‖xin − xm

n ‖rn

for i = 1, . . . ,m− 1 but it contradicts maximally of X0,r. Consequently,

card(Ω0,r) = k + 1.

It also follows from (2.25) that spaces Ω0,r and 0, e1, . . . , ek ⊆ H are similar.To construct an one-point tangent space Ω0,r consider the normalizing

sequence r = rnn∈N such that

(2.27) rn :=√

tntn+1,

where tjj∈N is the sequence from (2.20). For all x ∈ X \ 0 this definitionand (2.26) give either

(2.28)‖x‖rn

≥ tn√tntn+1

=√

tntn+1

,

if ‖x‖ ≥ tn, or

(2.29)‖x‖rn

≤ tn+1√tntn+1

=√

tn+1

tn

if ‖x‖ < tn. By (2.19),√

tntn+1

tends to infinity with n →∞. Hence, if there is

a finite limn→∞

‖xn‖rn

, then d(0, x) = 0, i.e., the pretangent space Ω0,r is one-point.

To complete the proof, it suffices to observe that (2.28) and (2.29) imply

‖x‖rnk

≥

√tnk

tnk+1

or, respectively,

‖x‖rnk

≤

√tnk

tnk+1

for every subsequence r′ = rnkk∈N of r. Hence, Ω0,r′ is also one-point.

Therefore, the one-point pretangent space Ω0,r is tangent.

The necessary and sufficient conditions under which there is an one-pointtangent space Ωa,r at an accumulation point a of (X, d) are found in [1].

Example 2.8. Let C be the complex plane with the usual distance d(z, w)=|z − w|. Fix the following three sequences:

zjj∈N with zj ∈ C and |zj | = 1;

13 Tangent spaces to general metric spaces 149

αjj∈N with αj ∈ R and limj→∞

αj = 0;

tjj∈N with tj ∈ R+, tj+1 < tj and limj→∞

tj+1

tj= 0.

Let X := tjzj : j ∈ N ∪ tjzjeiαj : j ∈ N ∪ 0. In the case underconsideration function (2.1) has the form

(2.30) F (z, w) =|z − w|(|z| ∧ |w|)

(|z| ∨ |w|)2

if (z, w) 6= (0, 0).We claim that the equality

(2.31) limz→0w→0

F (z, w) = 0

holds. Indeed, as in Example 2.7, for every (w, z) ∈ (X \ 0)2 we can find jand s such that j ≤ s and

|w| ∨ |z| = tj and |w| ∧ |z| = ts.

If j < s, then we have

F (z, w) ≥ |tj − ts|tst2j

=∣∣∣∣1− ts

tj

∣∣∣∣ tstj≥ ts

tj≥ tj+1

tj

but if j = s we obtain

F (z, w) =|tjzjeiαj − tjzj |tj

t2j= |eiαj − 1|.

Since

limj→∞

tj+1

tj= lim

j→∞|1− eiαj | = 0,

(2.31) holds. Theorem 2.2 implies that each pretangent space at the point0 ∈ X is either two-point or one-point. Note that two-point pretangent spacesat 0 ∈ X exist because 0 is not an isolated point of X. Moreover, all thesepretangent spaces are tangent by Lemma 2.6. To construct an one-point tan-gent space it is enough to take a normalizing sequence r with rj =

√tjtj+1

(see the end of the proof of Proposition 2.7).

3. COMPLETENESS OF THE TANGENT SPACES

In this section we shall prove that all tangent spaces to metric spaces arecomplete.

150 Oleksiy Dovgoshey and Olli Martio 14

Lemma 3.1. Let (X, d) be a metric space, let r = rnn∈N be a normali-zing sequence and let F 1

r , F 2r , F 3

r be mutually stable families of sequences x =xnn∈N with xn ∈ X for n ∈ N. Suppose that F 1

r is maximal, F 2r ⊆ F 1

r ∩ F 3r

and that the pseudometric space (F 2r , d) is a dense subspace of (F 3

r , d). Thenthe inclusion

(3.1) F 3r ⊆ F 1

r

holds.

Proof. Suppose that (3.1) does not hold. Since F 1r is maximal mutually

stable w.r.t. r, there are two sequences x = xnn∈N ∈ F 3r and y = ynn∈N ∈

F 1r such that either

(3.2) ∞ ≥ lim supn→∞

d(xn, yn)rn

> lim infn→∞

d(xn, yn)rn

≥ 0

or

(3.3) limn→∞

d(xn, yn)rn

= ∞.

For every ε > 0 we can find z = znn∈N ∈ F 2r such that

d(x, y) < ε

because F 2r is dense in F 3

r by the hypothesis of the lemma. The triangle in-equality implies ∣∣∣∣d(xn, yn)

rn− d(zn, yn)

rn

∣∣∣∣ ≤ d(xn, zn)rn

for all n ∈ N. Consequently, we obtain

(3.4)∣∣∣∣lim sup

n→∞

d(xn, yn)rn

− d(z, y)∣∣∣∣ ≤ d(x, y) < ε

and ∣∣∣∣lim infn→∞

d(xn, yn)rn

− d(z, y)∣∣∣∣ < ε.

Therefore, the inequality∣∣∣∣lim supn→∞

d(xn, yn)rn

− lim infn→∞

d(xn, yn)rn

∣∣∣∣ < 2ε

holds for all ε > 0, contrary to (3.2). To complete the proof, it suffices toobserve that (3.4) contradicts (3.3).

15 Tangent spaces to general metric spaces 151

Lemma 3.2. Let (X, d) be a metric space, let a ∈ X and let Ωa,r,r = rnn∈N, be a pretangent space with a corresponding maximal mutuallystable family Xa,r. Suppose that Λ is a bounded subset of Ωa,r. Then there is aconstant c > 0 such that for every α ∈ Λ we can find x = xnn∈N ∈ Xa,r with

(3.5) p(x) = α andd(xn, a)

rn≤ c

for n ∈ N, where p is the metric identification map.

Proof. WriteA := x ∈ Xa,r : p(x) ∈ Λ.

Since Λ is bounded in Ωa,r, we have

supd(x, a) : x ∈ A := k < ∞.

For every x = xnn∈N ∈ A introduce the sequence x∗ = x∗nn∈N by the rule

x∗n =

xn if d(xna) < (k + 1)rn,

a if d(xn, a) ≥ (k + 1)rn.

Note that for every x ∈ A there is n(x) ∈ N such that xn = x∗n if n ≥ n(x)because

d(x, a) = limn→∞

d(xn, a)rn

≤ k < k + 1.

Consequently, x∗ ∈ Xa,r for all x ∈ A. Moreover, relations (3.5) evidently holdwith x = x∗, xn = x∗n and c = k + 1, which is what had to be proved.

Theorem 3.3. Let (X, d) be a metric space, a ∈ X and let r ∈ rnn∈Nbe a normalizing sequence. Then a tangent space Ωa,r, if it exists, is complete.

Proof. Let αjj∈N be a fundamental sequence in the metric space Ωa,r.We must prove that αjj∈N is convergent. Let Xa,r be a maximal mutuallystable family with the pretangent space Ωa,r and let p : Xa,r → Ωa,r be themetric identification mapping. Every fundamental sequence of an arbitrarymetric space is bounded. Hence, by Lemma 3.5, there is c > 0 such that forevery j ∈ N there exists xj = xj

nn∈N ∈ Xa,r for which

(3.6) p(xj) = αj andd(xj

n, a)rn

≤ c

for all n ∈ N.In accordance with Definition 1.4 and Lemma 3.1 it suffices to prove

that there exists a sequence x = xnn∈N of points from X such that forsome subsequence of natural numbers nk and for all j ∈ N the sequences

152 Oleksiy Dovgoshey and Olli Martio 16

x′ = xnkk∈N and x′j = xj

nkk∈N are mutually stable w.r.t. r′ = rnkk∈N

and

(3.7) limj→∞

limk→∞

d(xnk, xj

nk)rnk

= 0.

Suppose now that (3.7) holds and, in addition, we have the followingcondition: for every k ∈ N there is j(k) ∈ N such that

(3.8) xnk= xj(k)

nk,

where xj(k)nk is the nkth element of xj(k). Then it follows from (3.6) that the

inequalityd(xj

nk , xnk)

rnk

≤ 2c

holds for all j, k ∈ N. In particular, we obtain

d(x1nk

, xnk)

rnk

≤ 2c

for all k ∈ N. Since every bounded infinite sequence of real numbers has a con-vergent subsequence, there is an increasing infinite subsequence n(1)

k k∈N of

the sequence nkk∈N such that limk→∞

d(xn(1)k

,x1

n(1)k

)

rn(1)k

is finite. Hence, the sequences

xn

(1)k

k∈N and x1

n(1)k

k∈N are mutually stable w.r.t. rn

(1)k

k∈N. Analogously,

by induction, we can prove that, for every integer i ≥ 2, there is a subsequencen(i)

k k∈N of the sequence n(i−1)k k∈N such that x

n(i)k

k∈N and xi

n(i)k

k∈N

are mutually stable w.r.t. rn

(i)k

k∈N. Consider now the diagonal sequence

n(k)k k∈N. For every i ∈ N the sequences xi

n(k)k

k∈N and xn

(k)k

k∈N are mutu-

ally stable w.r.t. rn

(k)k

k∈N. Moreover, (3.7) evidently holds with nk = j(k) =

n(k)k . Consequently, for the proof of the theorem it suffices to construct x such

that (3.8) and (3.7) are satisfied. In order to construct this x, we estimated(xj

n,xin)

rn.

Lemma 3.4. Let xjj∈N (xj = xjnn∈N) be a fundamental sequence in

a pseudometric space (Xa,r, d). Then there is a sequence x∗jj∈N ∈ Xa,r withthe following properties:

(i) The equality

(3.9) d(xj , x∗j ) = 0

holds for all j ∈ N.

17 Tangent spaces to general metric spaces 153

(ii) For every ε > 0 there exists j0(ε) ∈ N such that the inequality

(3.10) supn∈N

d(x∗jn , x∗in )rn

≤ ε

holds if i ∧ j ≥ j0(ε).

Proof. Let xjkk∈N be a subsequence of xjj∈N such that j1 < j2 <

· · · < jk · · · and

(3.11) d(xj , xjk) ≤

(12

)k+1

whenever j ≥ jk. For all j, n ∈ N write

(3.12) (1)xjn =

xjn if j ≤ j1,

xjn if j > j1 and

d(xj1n , xj

n)rn

≤ 12 ,

xj1n if j > j1 and

d(xj1n , xj

n)rn

> 12 ,

and put(1)xj := (1)xj

nn∈N.

Inequality (3.11) implies that, for every j ∈ N, there is n0(j) ∈ N such that(1)xj

n = xjn for all n ≥ n0(j). Hence, we have the equality

(3.13) d(xj ,(1)xj) = 0

for all j ∈ N. Note that (3.13) implies the inequality

d((1)xj , xjk) ≤

(12

)k+1

whenever j ≥ jk. Moreover, it follows from (3.12) that

supn∈N

d((1)xjn, (1)xi

n)rn

≤ supn∈N

d((1)xjn, xj1

n )rn

+ supn∈N

d((1)xin, xj1

n )rn

≤ 1

whenever i ∧ j > j1. Now we define (k)xjn by induction in k.

If k ≥ 2 and j, n ∈ N write

(k)xjn =

(k−1)xjn if j ≤ jk,

(k−1)xjn if j > jk and

d(xjkn , (k−1)xj

n)rn

≤(

12

)k

,

xjkn if j > jk and

d(xjkn , (k−1)xj

n)rn

>

(12

)k

,

154 Oleksiy Dovgoshey and Olli Martio 18

and put(k)xj := (k)xj

nn∈N.

In the same manner as in the case k = 1 we obtain the equality

(3.14) d(xj ,(k)xj) = 0

for all j, k ∈ N and the inequality

(3.15) supn∈N

d((k)xjn, (k)xi

n)rn

≤(

12

)k

whenewer i ∧ j > jk. Now set for all n ∈ N

x∗jn = (1)xjn if 1 ≤ j ≤ j1,

x∗jn = (2)xjn if j1 < j ≤ j2,

...x∗jn = (k)xj

n if jk−1 < j ≤ jk,

and so on. The sequence x∗j := x∗jn n∈N has the properties (i) and (ii) because(3.14) implies (3.9) and, moreover, (3.10) follows from (3.15).

Continuation of the proof of Theorem 3.3. Using Lemma 3.4, we mayassume that, for every ε > 0, there is j0 = j0(ε) ∈ N such that

(3.16) supn∈N

d(xin, xj

n)rn

≤ ε

if i ∧ j ≥ j0(ε). Set xn = xnn for all n ∈ N. It follows from (3.16) that

(3.17) lim supn→∞

d(xn, xjn)

rn≤ ε.

As was shown in the first part of the proof there is a subsequence r′ = rnkk∈N

of the sequence r such that the sequences x′j = xjnkk∈N and x′ = xnk

k∈Nare mutually stable w.r.t. r′ for all j ∈ N. Hence, by (3.17), we obtain

limn→∞

d(xnk, xj

nk)rnk

≤ ε

if j ≥ j0(ε). Letting j →∞ we obtain (3.7). To complete the proof, it sufficesto observe that (3.8) holds with j(k) = nk.

Recall that a map f : X → Y is called closed if the image of each setclosed in X is closed in Y .

Corollary 3.5. Let (X, d) be a metric space, let Y be a subspace ofX and let a ∈ Y . If a pretangent space Ωy

a,r is tangent, then iny : Ωya,r →

Ωa,r is closed.

19 Tangent spaces to general metric spaces 155

Proof. The map iny is an isometric embedding. Hence, iny is closedif and only if the set iny(Ω

ya,r) is a closed subset of Ωa,r. The space Ωy

a,r iscomplete by Theorem 3.6. Since a metric space is complete if and only if thisspace is closed in every its superspace (see, for example, [8, Theorem 10.2.1]),iny(Ω

ya,r) is closed in Ωa,r.

Acknowledgements. The first author is thankful to the Finnish Academy of Scienceand Letters for the support.

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Received 29 February 2009 Institute of Applied Mathematicsand Mechanics of NASU

R. Luxemburg str. 74, Donetsk 83114, Ukrainedovgoshey@iamm.ac.donetsk.ua

and

University of HelsinkiDepartment of Mathematics and Statistics

P.O. Box 68, FI-00014, Finlandolli.martio@helsinki.fi