# RELATIVE RANKS OF SEMIGROUPS OF MAPPINGS;

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### GENERATING CONTINUOUS MAPS WITH LIPSCHITZ FUNCTIONS

MICHA L MORAYNE

INSTYTUT MATEMATYKI I INFORMATYKI POLITECHNIKA WROC LAWSKA

UNIWERSYTET ´SLA¸ SKI 18-22/10/2010

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LECTURE 1 (3h) SIERPI ´NSKI’S THEOREM

A semigroup is an algebraic structure (S, ∗) where ∗ is an associative binary operation on elements of S. A monoid is a semigroup with an identity element.

If A ⊆ S, by hAi we denote the subsemigroup generated by A. In these series of lectures we will consider only monoids of mappings where the monoid operation will always be the composition of mappings. The starting point for the theory of monoids of mappings was the following Sierpi´nski’s theorem. It is still one of the most important facts of this theory.

Theorem 1.1 (Sierpi´nski) Let A be an infinite set. Let fi : A → A, i = 1, 2, . . .. There exist two mappings ϕ, ξ : A → A such that

{fi: i ∈ N} ⊆ h{ϕ, ξ}i .

Proof (Banach). Let

A =

[

i=0

Ai

be a partition of A into sets of the same cardinality. Let

A0=

[

i=1

A0,i

be a partition of A0 into sets of the same cardinality.

A0

A1

A2

..

. ..

.

A0,1 A0,2 A0,3 . . .

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First let us define ϕ. Namely, let it be any mapping that is a bijection from An onto An+1, for every n.

Now let us define ξ. First we define ξ on A \ A0= A1∪ A2∪ . . . . Namely, let ξ|An, n > 0, be any mapping that is a bijection from An onto A0,n.

Let us notice that so far none of the two mappings ϕ and ξ depends in any way on the mappings f1, f2, . . .. Thus the only place where it can happen is the remaining part of the definition of ξ, namely ξ|A0. Before we complete this definition let us notice that the mapping

γn= ξ ◦ ϕn◦ ξ ◦ ϕ is a bijection of A onto A0,n.

We want to encode the function fn on A0,n to ensure that ξγn = fn. We can do this defining

ξ|A0,n= fn◦ γ−1n .

(Note that we can do this because γn is a bijection.) Finally, fn = ξ2◦ ϕn◦ ξ ◦ ϕ

for each n = 1, 2, . . . . This completes the proof.  Remark In general, the conclusion of Sierpi´nski’s theorem cannot be strength- ened to get all mappings fn, n ∈ N, as iterations of a single mapping ψ : A → A.

Indeed, just let us consider two mappings f1 and f2where f1is a bijection and f2 is not. Assume that f1 = ψn and f2 = ψm. f1 = ψn implies that ψ is a bijection and f2= ψmimplies that ψ is not a bijection.

Let S be a semigroup and A ⊆ S. The relative rank of S with respect to A is the cardinal number

rank(S : A) = min{|B| : hA ∪ Bi = S}.

Corollary 1.2 Let A be any infinite set. Let V ⊆ AA. Then rank(AA : V) = 0, 1, 2 or is uncountable.

Proof. Assume that rank(AA : V) is countable. It means that there exists a countable family F = {f1, f2, . . .} of mappings from AA such that hV ∪ F i = AA. By Sierpi´nski’s theorem we can find two mappings ϕ, ξ ∈ AA such that

F ⊆ h{ϕ, ξ}i, whence hV ∪ {ϕ, ξ}i = AA. 

There are two groups of problems that arise naturally here. Once we know the rank is finite we may ask which one of these three possibilities 0, 1, 2 holds.

It is sometimes a difficult and challenging problem. The whole new realm of problems start with trying to make precise the word uncountable in the above conclusion. We will discuss several such problems for various monoids of map- pings, also going beyond the monoid AA.

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The following two propositions provide examples of submonoids of NN of cardinality continuum with very different relative ranks.

Let us define a submonoid ONto consist of all those mappings g ∈ NN that are weakly order preserving i.e. m ≤ n implies g(m) ≤ g(n).

Proposition 1.3 We have

rank(NN: ON) = 1.

Proof. Let φ : N → N be such a mapping that the set ϕ−1[{i}] is infinite for each i ∈ N. Let now f be any mapping from NN. Let us define g ∈ ON. We do it inductively. Let g(1) ∈ ϕ−1[{f (1)}]. Let g(2) ∈ ϕ−1[{f (2)}] and g(2) ≥ g(1).

It is possible to find such a g(2) because the set ϕ−1[{f (2)}] is infinite. Once g(1) < g(2) < . . . g(n − 1) have been defined we define g(n) ∈ ϕ−1[{f (n)}] and g(n) ≥ g(n − 1). It is possible because the set ϕ−1[{f (n)}] is infinite. Now we have

ϕ ◦ g(n) = f (n).

 Before we state and prove the next proposition let us recall the notion of an almost disjoint family of sets. We say that a family of sets A is almost disjoint if any two of its elements have finite intersection. We will need the following two lemmas.

Lemma 1.4 If |A| = ℵ0 then there is an almost disjoint family of subsets of A of cardinality continuum c = 20.

Proof. Of course, it is enough to provide a proof for any A satisfying the hypothesis. We choose

A =

[

n=1

{0, 1}{1,2,...,n},

(i.e. the set of all vertices of the infinite complete binary tree). Now our family is the family of all branches of the tree, i.e. the family whose ele- ments are sequences of functions (σn)n, σn : {1, 2, . . . , n} → {0, 1}, where

σn|{1, 2, . . . , m} = σmfor any m < n. 

Lemma 1.5 Let k ∈ N. Let A1, A2, . . . be subsets of N each of cardinality not exceeding k. Then there is no uncountable family F of functions f : N → N satisfying f (i) ∈ Ai for each i ∈ N, such that for any two functions f, g ∈ F the set {i : f (i) = g(i)} is finite.

Proof. The proof will be an induction with respect to k. For k = 1 the conclusion is obvious. Assume that the conclusion is true for some k ∈ N and that sets

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A1, A2, . . . are of cardinalities not exceeding k + 1. Assume for contradiction that such an uncountable family F exists. Let us take one function f ∈ F . Each function g from F has now diiferent values than f from some point on, say ng. Thus

F =

[

i=1

{g ∈ F : ng= i},

whence there must be i0 ∈ N such that the set G={g ∈ F : ng = i0} is uncountable. Of course, the family {g|{i0+ 1, i0+ 2, . . .} : g ∈ G} satisfies now the induction hypothesis for k and thus it cannot be uncountable. 

Let S2⊆ NN consist of all the functions f such that f [{2i − 1, 2i}] ⊆ {2i − 1, 2i}

for all i ∈ N.

Proposition 1.6 We have

rank(NN: S2) = c.

Proof. For contradiction assume that there exists F ∈ NNsuch that hS2∪ F i = NN and |F | = κ < c. Let f1, f2, . . . , fn be any functions from F . Let

F(f1,f2,...,fn)= {ϕn+1◦ fn◦ ϕn· · · ◦ f1◦ ϕ1: ϕ1, . . . , ϕn+1∈ S2}.

Now we are going to prove via induction with respect to n that the family F(f1,f2,...,fn)satisfies the hypothesis of Lemma 1.5. Let n = 0. Then the claim is satisfied with k = 2 because we deal only with ϕ1 ∈ S2. Assume now that the claim is satisfied for n ∈ N ∪ {0}. Thus

ϕn◦ fn−1· · · ◦ f1◦ ϕ1(i) ∈ A0i for some |Ai| ≤ k. Then

fn◦ ϕn· · · ◦ f1◦ ϕ1: ϕ1(i) ∈ fn[A0i] and |fn[A0i]| ≤ |A0i|. By the properties of mappings from S2

ϕn+1[fn[A0i]] ⊆ Ai, where

|Ai| ≤ 2|fn[A0i]| ≤ 2|A0i| ≤ 2k.

and the claim is proved. By Lemma 1.4 there exists an almost disjoint family A of infinite subsets of N. Let A ∈ A. Let fA: N → A be any strictly increasing function. The set {i : fA(i) = fB(i)} is finite for any sets A, B ∈ A and any

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i ∈ N, for otherwise the sets A and B would have infinite intersection which would contradict almost disjointness of A. Thus, by Lemma 2, only countably many functions fA, A ∈ A may belong to F(f1,f2,...,fn). As there are only κ families F(f1,f2,...,fn)and c functions fAsome functions fAdo not belong to any family F(f1,f2,...,fn) and thus to hS2∪ F i = NN. 

BIBLIOGRAPHY

1. S. Banach, Sur un th´eor`em de M. Sierpi´nski, Fundamenta Mathematicae 25 (1935), 5-6.

2. P.M. Higgins, J.M. Howie, J.D. Mitchell, N. Ruˇskuc, Countable versus un- countable ranks in infinite semigroups of transformations and relations, Pro- ceedings of the Edinburgh Mathematical Society 46 (2003), 531-544.

3. P.M. Higgins, J.D. Mitchell, M. Morayne, N. Ruˇskuc, Rank properties of endomorphisms of infinite partially ordered sets, Bulletin of the London Math- ematical society 38 (2006), 177-191.

4. J. D. Mitchell, M. Morayne, Y. P´eresse, and M. R. Quick, Generating trans- formation semigroups using endomorphisms of preorders, graphs, and toler- ances, Annals of Pure and Applied Logic 161 (2010), 1471-1485.

5. W. Sierpi´nski, Sur les suites infinies de functions d´efinies dans les ensembles quelconques, Fundamenta Mathematicae 24 (1935), 209-212.

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LECTURE 2 (6h)

UNIVERSAL SETS AND FUNCTIONS

The Baire space is the set N = NN endowed with the metric d defined as follows:

d((mi)i∈N, (ni)i∈N) = 1

min{i : mi6= ni}.

For considering the relative rank of the monoid of continuous mappings on the Baire space N = NNwith respect to the monoid of Lipschitz mappings on N we need several facts about the existence of so called universal functions (some of them standard and some not quite so standard).

Let us first introduce some notation.

Let X be any set. By P(A) we denote the power set of X i.e. the family of all subsets of X.

Let X, Y be any sets. For A ⊆ Y × X and y ∈ Y by Ay we denote the vertical section of A given by y i.e. the set {x ∈ X : (y, x) ∈ A}. Let A ⊆P(A).

We say that a set A ⊆ Y × X is universal for A if A ⊆ {Ay: y ∈ Y }.

Let now X be a topological space. By Σ01(X) we denote the family of all open subsets of X (the topology on X) and by Π01(X) the family of all closed subsets of X.

First we prove the following (classical and fundamental) lemma.

Lemma 2.1 Let X be a second countable topological space and Y be a topological space containing a topological copy C of a Cantor set {0, 1}N. Then there exists A ∈ Σ01(Y × X) universal for Σ01(X).

Proof. By the hypothesis there exists a homeomorphism ϕ : {0, 1}N→ C ⊆ Y . Let {U1, U2, . . .} be a base for the topology on X. We define now A ⊆ Y × X as follows:

(y, x) ∈ A ⇔ (y /∈ C) or x ∈[

{Ui: y = ϕ(t) and ti= 1} . We claim that A is open in Y ×X. Indeed, if y /∈ C then (y, x) ∈ (Y \C)×X ⊆ A and thus we have found an open neighbourhood of (y, x), namely (Y \ C) × X, that is in A. Now assume that (y, x) ∈ A and y ∈ C. Thus ϕ(t) = y, and x ∈ Uk

and tk = 1 for some k ∈ N. But then also sk = 1 for all s ∈ V , where V is a certain open neighbourhood of t. Then ϕ[V ] is an open neighbourhood of y in C and thus ϕ[V ] = C ∩ W for some set W open in Y . As (u, r) ∈ A for all u ∈ φ[V ] and r ∈ Uk and (u, r) ∈ A for all u ∈ Y \ C yield (y, x) ∈ W × Uk⊆ A we again found an open neighbourhood of (y, x) in A.  Let X, Y, Z be any sets. Let F ⊆ ZX. We say that a mapping Φ : Y × X → is universal for F if F ⊆ {Φ(y, ·) : y ∈ Y }.

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Let now X be a topological space. A function f : X → R is lower (upper) semicontinuous if for very α ∈ R the set f−1[(α, ∞)] (f−1[(−∞, α)]) is open in X. Let LSC(X) (U SC(X)) denote the class of all lower (upper) semicontinous functions from X into the interval [0, 1].

The following four lemmas will play an important technical role.

Lemma 2.2 Let X, Y be topological spaces, ϕ : X → Y be a continuous map- ping, and f : Y → R be a lower semicontinous function. Then ϕ ◦ f : X → R is also a lower semicontinuous function.

Proof. Let α ∈ R. We have

(ϕ ◦ f )−1[(α, ∞)] = f−1−1[(α, ∞)]].

Because ϕ is lower semicontinuous the set ϕ−1[(α, ∞)] is open in Y , and because f is continuous the set f−1−1[(α, ∞)]] is open in X.  Lemma 2.3 Let X be a topological space and F ⊆ LSC(X). Then sup F ∈ LSC(X).

Proof. Let α ∈ [0, 1]. We have

(sup F )−1[(α, 1]] = [

f ∈F

f−1[(α, 1]],

and this set is open in X because it is a union of open sets.  Lemma 2.4 Let X be a topological space. Let f ∈ LSC(X). Then f = sup F , where F is a family of functions of the form qnχUnwhere {q1, q2, . . .} is a certain enumeration of Q ∩ [0, 1] and Un are open sets.

Proof. It is easy to check that f = sup{qnχf−1[(qn,1]]}.  Lemma 2.5 Let X be a topological space and A ∈ X. Let f ∈ LSC(A), where A is equipped with the relative topology inherited from X. Then f can be extended to f∈ LSC(X).

Proof. Let U be an open set in A. Then U = A ∩ U, where U is an open set in X. By Lemma 2.4 the function f can be expressed as f = sup F , where F is a countable family of functions of the form qχU where q ∈ Q ∩ [0, 1] and U is open in A. Let

f= sup{qχU: qχB∈ F }.

Lemma 2.3 implies f∈ LSC(X). Simple checking shows that f|A = f .  Theorem 2.6 Let X be a second countable topological space and Y be any topological space containing a topological copy of the Cantor set {0, 1}N. Then there exists a function F ∈ LSC(Y × X) universal for LSC(X).

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Proof. The set ({0, 1}N)Nis homeomorphic to the Cantor set. To avoid techni- calities assume simply that ({0, 1}N)N⊆ Y . By Theorem 2.1 there exists a set V ∈ Σ01({0, 1}N× X) which is universal for Σ01(X). Let q1, q2, . . . be an enu- meration of the elements of Q ∩ [0, 1]. For t = (t(n))n ∈ ({0, 1}N)Nand x ∈ X, let Φ : ({0, 1}N)N× X :→ R be defined as follows:

Φ(t, x) = sup{qnχV

t(n)(x) : n ∈ N}.

The function Φ is lower semicontinuous. To show this it is enough, by Lemma 2.3, to show that (t, x) 7−→ χV

t(n) is lower semicontinuous. We have χV

t(n)(x) = χVn(t), x) = χV(Θ(t, x)),

where Θ(t, x) = (πn(t), x). Obviously, Θ : ({0, 1}N)N× X → {0, 1}N× X is a continuous mapping. Thus by Lemma 2.2 χU

t(n) is lower semicontinuous, and therefore so is Φ.

Now we will show that Φ is universal for LSC(X). Let f ∈ LSC(X). By Lemma 2.4 we can express f as

f = sup{qnχUn: n ∈ N},

where Un’s are some open sets in X. Let t = (t(n))n ∈ ({0, 1}N)Nsatisfy Vt(n)= Un.

Thus we have for this t:

f (·) = Φ(t, ·),

and this shows that Φ is indeed universal for LSC(X). Finally, by Lemma 2.5 Φ can be extended to F ∈ LSC(Y × X) which is, of course, also universal for

LSC(X). 

Let (X, d) be a metric space. Let C(X) denote the monoid of all continuous mappings from X to X, where the monoid operation is the composition of functions. Let L(X) denote the submonoid of C(X) consisting of all Lipschitz mappings from X to X, i.e. the mappings f satisfying d(f (x), f (y)) ≤ Cd(x, y), where C is a constant depending on f .

Proposition 2.7 There is no continuous mapping from N × N to N that is universal for C(N ).

Proof. Assume that there is such a mapping, call it F . Then the diagonal mapping f (x) = F (x, x) is continuous. For each mapping g ∈ C(N ) there exists x ∈ N such that g(·) = F (x, ·), and thus g(x) = F (x, x) = f (x). Consider now the function g ∈ C(N ) defined as g(x) = ((f (x))1+ 1, (f (x))2, . . .). it is obvious that g(x) 6= f (x) for any x ∈ N . It is a contradiction.  Let X, Y, Z be any sets. Let F ⊆ ZX. We say that a mapping F : A → Z, A ⊆ Y × X, is universal for F in a generalized sense if F ⊆ {F (y, ·) : y ∈ Y }.

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The word ”generalized” used here refers to the fact that the domain of F is not the whole Y × X. Note, however, that the domain of functions which we want to realize as sections of F is the whole X.

Theorem 2.8 There exists a Borel subset B of N ×N and a continuous mapping F : B → N universal for C(N ) in a generalized sense.

Proof. Let ψ : N → N × N and ξ : [0, 1] \ Q → N be homeomorphisms. Let L ∈ LSC(N × N ) and U ∈ U SC(N × N ) be universal for LSC(N ) and U SC(N ), respectively. Let (y, x) ∈ B if L((ψ(y))1, x) = U ((ψ(y))2, x) ∈ [0, 1] \ Q. In other words

B= G−1({0}) ∩ ¯L−1[[0, 1] \ Q],

where ¯L(y, x) = L((ψ(y))1, x) and G = ¯L − ¯U , for ¯U (y, x) = U ((ψ(y))2, x).

Thus it is obvious that B is a Borel subset of N × N . Finally, define F (y, x) = ξ ◦ ¯L(y, x)

for (y, x) ∈ B. It is obvious by construction that F is universal for C(N ) in a

generalized sense. 

Theorem 2.9 There exists a coanalytic set D ∈ Π11(N ) and a continuous mapping G : D × N → N universal for C(N ).

Proof. Let π1denote the projection onto the first axis. Let D = N \ π1(N2\B), where B is the set from Theorem 2.8. Note that D × N ⊆ B. By the properties of B and F from Theorem 2.9, we can define G as F |D × N .  Theorem 2.10 There exists a continuous mapping L : N × N → N universal for L(N ).

Proof. Let L(N , k) denote the class of Lipschitz mappings with constant k Let τk : Nk−1→ N be a bijection. For a sequence x = (x1, x2, . . .) ∈ N let x(i) = xi. We will define a mapping Φk : NN× N → N that is universal for L(N , k).

Let

k((c(i))i∈N, x))(n) = c(n)k(n+1)(x1, x2, . . . , xk(n+1)−1)).

It is easy to see that Φk is continuous.

Now we will show that Φk is universal for L(N , k). Let f ∈ L(N , k). Let us define c(i)f ∈ N as follows:

c(i)fk(i+1)(x1, x2, . . . , xk(i+1)−1)) = (f (x1, x2, . . . , xk(i+1)−1, 1, 1, . . . ))(i), for (x1, x2, . . . , xk(i+1)−1) ∈ Nk(i+1)−1.

We claim that f (x) = Φk((c(i)f )i∈N, x) for each x ∈ N . Fix x ∈ N and n ∈ N.

Then

d(x, (x1, . . . , xk(n+1)−1, 1, 1, . . .)) ≤ 1/k(n + 1),

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and so

d(f (x), f (x1, . . . , xk(n+1)−1, 1, 1, . . .)) ≤ 1/(n + 1).

Thus

(f (x))(n) = (f (x1, . . . , xk(n+1)−1, 1, 1, . . .))(n) = c(n)fk(n+1)(x1, . . . , xk(n+1)−1) = (Φk((c(i)f )i∈N, x))(n).

Next define Ψ : (N × NN) × N → N by

Ψ(k, ((c(i)f )i∈N, x)) = Φk((c(i)f )i∈N, x).

Note that the spaces N × NNand N are homeomorphic. Let θ : N → N × NN be such a homeomorphism. Finally define

F (y, x) = Ψ(θ(y), x).

F : N2→ N is the desired continuous mapping universal for L(N ).  Remark In fact, any section Φk(c, x) is Lipschitz with constant k + 1. Assume that 1/(k + 1)n < d(x, y) ≤ 1/kn. Thus xi= yi for all i ≤ kn. Hence for j < n

k((c(i))i∈N, x))(j) =

c(j)k(j+1)(x1, x2, . . . , xk(j+1)−1)) = c(j)k(j+1)(y1, x2, . . . , yk(j+1)−1)) = (Φk((c(i))i∈N, y))(j).

Thus

d(Φk((c(i))i∈N, x), Φk((c(i))i∈N, y) ≤ 1/n = (k + 1) 1

(k + 1)n ≤ (k + 1)d(x, y).

BIBLIOGRAPHY

1. J. Cicho´n, J.D. Mitchell, M. Morayne, Generating continuous mappings with Lipschitz mappings, Transactions of the American Mathematical Society 359 (2007), 2059-2074.

2. J. Cicho´n, M. Morayne, Universal functions and generalized classes of func- tions, Proceedings of the American Mathematical Society 102 (1988), 83-89.

3. K. Kuratowski, Topology Vol. I, Academic Press, New York, London, Pa´nstwowe Wydawnictwo Naukowe, Warsaw 1966.

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LECTURE 3 (3h)

CONTINUOUS FUNCTIONS vs LIPSCHITZ FUNCTIONS ON SOME CLASSICAL METRIC SPACES

We are now prepared to find rank(C(N ) : L(N )).

Theorem 3.1 rank(C(N ) : L(N )) = ℵ1.

Proof. We divide the proof into two parts. In the first part we prove that rank(C(N ) : L(N )) ≥ ℵ1. In the second part we prove that rank(C(N ) : L(N )) ≤ ℵ1.

Part 1. Assume for contradiction that the rank rank(C(N ) : L(N )) is countable.

Thus there exists a countable collection {f1, f2, . . .} such that C(N ) = hL(N ) ∪ {f1, f2, . . .}i . Hence every mapping g ∈ C(N ) is a composition

g = λk+1◦ fnk◦ λk◦ . . . ◦ λ2◦ fn1◦ λ1, for some λ1, . . . , λk+1∈ L(N ) and n = (n1, . . . , nk).

Let

An= {λk+1◦ fnk◦ λk◦ . . . ◦ λ2◦ fn1◦ λ1: λ1, . . . λk+1∈ L(N )}.

First we are going to construct a continuous mapping Hn: N ×N → N universal for An. Let τ : N → Nk+1be a homomorphism whose coordinate functions are denoted by τi,i = 1, 2, . . . , k + 1. Let F : N × N → N be a continuous mapping universal for L(N ), which exists by Theorem 2.10. Let us define Hnas follows:

Hn(y, x) = F (τk+1(y), ·) ◦ fnk◦ F (τk(y), ·) ◦ . . . ◦ F (τ2(y), ·) ◦ fn1◦ F (τ1(y), ·).

Let us enumerate all n’s as n(1), n(2), . . .. Let us now define a mapping H : N × N → N as:

H((y1, y2, . . .), x) = Hn(y1)((y2, y3, . . .), x).

It is easy to check that H is universal for C(N ) but this contradicts Proposition 2.7. And thus we have proved rank(C(N ) : L(N )) ≥ ℵ1.

Part 2. Let D ∈ Π11(N ) and G : D ×N → N be a continuous mapping universal for C(N ). They exist by Theorem 2.7. As every nonempty coanalytic set is a union of ℵ1 nonempty Borel sets, we can write

D = [

α<ℵ1

Bα,

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where Bα’s are Borel subsets of N . It is also known that every Borel subset of a Polish space, in particular of N , is a continuous image of N . Let then ξα: N → Bα be a continuous ’onto’ mapping.

Let us consider the following homeomorphism φ : N × N → N : φ((y1, y2, . . .), (x1, x2, . . .)) = (y1, x1, y2, x2, . . .).

The metric ρ((y, x), (y0, x0)) = d(y, y0) + d(x, x0) coincides with the product topology on N × N and it is easy to see that φ is a two ways Lipschitz homeo- morphism. It is also easy to notice that the mapping x 7−→ (y, x) from N into N2 is Lipschitz.

Now let f ∈ C(N ). Then there exist α < ℵ1 and z ∈ Bα such that f (·) = G(y, ·). Let z = ξα(y). Let us consider the following series of compositions which give f :

x 7−→ (x, y) 7−→ φ(y, x) 7−→ (y, x) 7−→ (ξα(y), x) 7−→ G(ξα(y), x) = f (x).

Notice that the mapping λ defined by

x 7−→ (x, y) 7−→ φ(y, x) is Lipschitz, and the mapping hα defined by

φ(y, x) 7−→ (y, x) 7−→ (ξα(y), x) 7−→ G(ξα(y), x)

is continuos and depends exclusively on α. Thus f = hα◦ λ. Hence we can ex- press every f ∈ C(N ) as a composition of a Lipschitz mapping and a continouos mapping that belongs to the family of cardinality not exceeding ℵ1. Thus we

have shown that rank(C(N ) : L(N )) ≤ ℵ1. 

Remark It turns out that if x is any point of N then rank(C(N \ {x}) : L(N \ {x})) = 1.

Because the spaces N and N \ {x} are homeomorphic this shows very strongly that it is the metric not topological structure of X that decides about rank(C(X) : L(C)). We will see further examples later.

A metric space X is concentric if it is unbounded and it is the union of countably many compact balls. Such spaces are common, for example every euclidean space Rn is concentric. The induced topology of a concentric metric space is noncompact, locally compact and second countable. On the other hand, for a topological space X which is noncompact, locally compact and second countable there exists a metric d such that (X, d) is concentric.

Let σ be a sequence in a metric space X and let Y be the set of elements that occur in σ, and its limit l, if it exists. Then σ is said to be extendible if Y is infinite and every continuous map from Y to Y that fixes l, if it exists, can be extended to an element of C(X).

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N<N denotes the set of all finite sequences of natural numbers. An element of N<N is denoted by i and |i| denotes the length of the sequence.

Thoerem 3.2 Let X be a concentric metric space which contains an extendible sequence with no convergent subsequence. Then

rank(C(X) : L(X)) > ℵ0.

Proof. Let L(X, n) denote the family of all Lipschitz mappings from X to X with constant n.

It suffices to prove that for every countable subset of C(X) there exists a continuous mapping that is not generated by the union of the Lipschitz mappings and this subset. We construct such a mapping γ recursively.

To this end, let µ1, µ2, . . . ∈ C(X) be arbitrary, let ρ = (x0, x1, x2, . . .) be an extendible sequence with no convergent subsequence, and let B(p, 1) ⊆ B(p, 2) ⊆ . . . be compact balls that comprise X for some p. Since all the balls we consider in this proof are centered on p, for brevity we write B(n) instead of B(p, n). The elements of hL(X) ∪ {µ1, µ2, . . .}i are finite compositions of the form

Φm+1◦ µim· · · ◦ Φ2◦ µi1◦ Φ1,

for some i1, i2, . . . , im∈ N and Φ1, Φ2, . . . , Φm+1∈ L(X). We represent the fam- ily hL(X) ∪ {µ1, µ2, . . .}i as a countable union of families of such compositions.

These families are determined, roughly speaking, according to the elements of {µ1, µ2, . . .} that appear, the Lipschitz constants involved, and the balls that contain images of x0. The mapping γ is initially defined on the sequence ρ. Thus there are countably many steps in the definition; at each of which, we ensure that γ does not belong to one of the given families. This process exhausts every possibility.

Precisely speaking, we consider the set

Σ = {[i, n] : i ∈ N<N, n ∈ N} (1) and for [i, n] ∈ Σ, with i = (i1, i2, . . . , ik) and n ∈ N, the family of compositions

F[i,n]= {Φk+1◦ µik· · · ◦ Φ2◦ µi1Φ1: Φj+1◦ µij◦ · · · ◦ µi1◦ Φ1(x0) ∈ B(n) , Φj+1∈ L(X, n) for each 0 ≤ j ≤ k}.

(2) Obviously,

[

[i,n]∈Σ

F[i,n] = hL(X) ∪ {µ1, µ2, . . .}i .

Since Σ is countable we may enumerate its elements as σ1, σ2, . . ..

Step 0 in the definition of γ is made by setting γ(x0) = x0.

At step r > 0 we define γ on xr in such a way that when the definition is complete γ will not be an element of Fσr. Assume that γ was defined on

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x1, x2, . . . , xr−1 in the previous steps such that any element of C(X) agreeing with γ on these points does not lie in Fσi for i < r. Also assume that σr= [i, n]

with i = (i1, . . . , ik) and n ∈ N.

This step is completed by proving that every element of Fσr maps xr into the same ball B(m) for some m. In fact, it suffices to prove that if for each 0 ≤ j ≤ k − 1 and Φj+1as in (2) we have

Φj+1◦ µij◦ · · · Φ2◦ µi1◦ Φ1(xr) ∈ B(mj+1), for some mj+1, then every

Φj+2◦ µij+1◦ · · · ◦ Φ2◦ µi1◦ Φ1(xr) ∈ B(mj+2), for some mj+2. Thus within step r we perform a finite induction.

For the base case, Φ1(x0) ∈ B(n) and d(Φ1(x0), Φ1(xr)) ≤ nd(x0, xr).

Therefore d(Φ1(xr), p) ≤ n + nd(x0, xr), and so Φ1(xr) ∈ B(m1) where m1 = n + nd(x0, xr). Note that the choice of m1 does not depend on Φ1, but on its Lipschitz constant.

The inductive hypothesis states that for every

Φj+1◦ µij◦ · · · ◦ Φ2◦ µi1◦ Φ1∈ F[(i1,...,ij),n]

we have

u = Φj+1◦ µij ◦ · · · ◦ Φ2◦ µi1◦ Φ1(xr) ∈ B(mj+1), for some mj+1. By the definition of F[i,n]

v = Φj+1◦ µij◦ · · · ◦ Φ2◦ µi1◦ Φ1(x0) ∈ B(n).

If M is the maximum of mj+1 and n then µij+1(u), µij+1(v) ∈ µij+1(B(M )).

Since the continuous image of a compact set is compact there exists M0 such that µij+1(B(M )) ⊆ B(M0). Thus d(µij+1(u), µij+1(v)) ≤ 2M0. But, again by the definition of F[i,n], Φj+2◦ µij+1(v) ∈ B(n). Therefore

d(p, Φj+2◦ µij+1(u)) ≤ n + 2nM0.

We deduce that Φj+2◦ µij+1◦ · · · ◦ Φ2◦ µi1◦ Φ1(xr) ∈ B(mj+2) where mj+2= n + 2nM0.

Define γ(xr) in such a way that γ(xr) /∈ B(mk+1).

When the recursion is complete, the extension of γ to an element of C(X) is not contained in Fσfor any σ ∈ Σ. This implies that γ 6∈ hL(X) ∪ {µ1, µ2, . . .}i.



Now we can derive the following corollary about euclidean spaces and their subsets.

Corollary 3.3 Let X = N, Z or R with the usual euclidean metric. Then rank(C(Xn) : L(Xn)) > ℵ0

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for any n ∈ N.

BIBLIOGRAPHY

1. J. Cicho´n, J.D. Mitchell, M. Morayne, Generating continuous mappings with Lipschitz mappings, Transactions of the American Mathematical Society 359 (2007), 2059-2074.

2. K. Kuratowski, Topology Vol. I, Academic Press, New York, London, Pa´nstwowe Wydawnictwo Naukowe, Warsaw 1966.

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LECTURE 4 (3h)

CONTINUOUS FUNCTIONS vs LIPSCHITZ FUNCTIONS ON COUNTABLE DISCRETE METRIC SPACES

In this lecture we return to monoids XX for infinite countable X’s. Namely, we consider countable discrete metric spaces (X, d). Because X is discrete we have C(X) = XX. We will be interested in the relative rank rank(XX: L(X)).

The monoid L(X) is the next example of a submonoid of XX. By Sierpi´nski’s theorem rank(XX : L(X)) = 0, 1, 2 or rank(XX: L(X)) > ℵ0. The full descrip- tion of these ranks for all countable discrete metric spaces is not known. In fact, it is not known even for every X ⊆ R with the euclidean metric. Nevertheless for the most classical examples of (X, d) the values of rank(XX : L(X)) have been found. In this last lecture we will give an account of this research.

A dominating family is any subfamily A of NN such that for any f ∈ NN there exists g ∈ A such that g(n) ≥ f (n) for all n’s greater than some N . We say that g eventually dominates f . The cardinal d is the minimal possible cardinality of a dominating family. It is known that both statements d < c and d= c are consistent with ZFC.

Note that if in the definition above we require that g(n) ≥ f (n) for each n (that g dominate f ) this gives an equivalent definition of d. It is also not important whether we use the sharp or weak inequality.

We start with the following theorem where we find rank(XX : L(X)) for a vast family of countable discrete metric spaces (X, d).

Theorem 4.1 Let (X, d) be a countable discrete metric space containing a Cauchy sequence. Then

rank(XX: L(X)) = 1.

Proof. Let y1, y2, . . . be a Cauchy sequence of different points of X. Let f ∈ XX be any function such that the set f−1[{x}] ∩ {y1, y2, . . .} is infinite for every x ∈ X. Let g ∈ XX. Let B(xi, ri) = {xi}, for ri> 0. Let N (1) be chosen in such a way that d(ym, yn) < r1 for m, n ≥ N (1). Let λ(x1) ∈ {yN (1), yN (1)+1, . . .}

and f (λ(x1)) = g(x1). By the assumption about f we can find such λ(x1).

Let N (2) be chosen in such a way that d(ym, yn) < r2 for m, n ≥ N (2). Let λ(x2) ∈ {yN (2), yN (2)+1, . . .} and f (λ(x2)) = g(x2). By the assumption about f we can find such λ(x2). We continue this process of choosing λ(xi)’s. Finally we obtain that d(ym, yn) < rifor m, n ≥ N (i) and and λ(xi) ∈ {yN (i), yN (i)+1, . . .}

for each i and f (λ(xi)) = g(xi). The last property means f ◦ λ = g.

Let us show that λ : X → X is Lipschitz with constant 1. Let i < j. We

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have

d(λ(xi), λ(xj)) < max{ri, rj} ≤ d(xi, xj).

Thus

hL(X) ∪ {f }i = XX, which gives

rank(XX: L(X)) = 1.

 We have proved (Corollary 3.3) that rank(NN : L(N)) > ℵ0. Here we will make this result more precise. Namely, we will prove the following theorem.

Theorem 4.2 rank(NN: L(N)) = d.

Proof. The proof consists of two parts. First we will prove that rank(NN : L(N)) ≥ d. Next, we will prove the opposite inequality.

Part 1. Let us assume that

hL(N) ∪ Fi = NN (3)

and |F | = rank(NN: L(N)).

Let f1, f2, . . . , fn ∈ F . Recall that LN(N, N ) denotes the family of functions from NNsatisfying the Lipschitz condition with constant N . Let

A(N )(f

1,...,fn)= {λn+1◦ fn◦ λn. . . ◦ f1◦ λ1: λ1, . . . , λn+1∈ L(N, N).

Of course,

hL(N) ∪ Fi = [

(f1,...,fn)∈Fn,N

A(N )(f

1,...,fn). (4)

We will use the following notation. Let ϕM(x) = M x. For f ∈ NN, the function ¯f is defined by

f (m) = max{f (1), f (2), . . . , f (m)}.¯

Obviously ¯f (m) is nondecreasing. We define F (f1, . . . , fn; M ) ∈ NNby F (f1, . . . , fn; M ) = ϕM◦ ¯fn◦ ϕM ◦ ¯fn−1◦ ¯f1◦ ϕM.

Let now g be any function from NN. It follows from (3) and (4) that g ∈ A(N )(f

1,...,fn)for some f1, f2, . . . , fn∈ F and N ∈ N which means that g = λn+1◦ fn◦ λn. . . ◦ f1◦ λ1,

λ1, . . . , λn+1∈ L(N, N).

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Let M = max{λ1(1), . . . λn+1(1), N }. It is easy to check that g ≤ F (f1, . . . , fn; M ).

Thus the family {F (f1, . . . , fn; M ) : f1, . . . , fn ∈ F , M ∈ M} is dominating for NN. Of course,

|{F (f1, . . . , fn; M ) : f1, . . . , fn∈ F , M ∈ M}| = |F|.

Hence |F | ≥ d and this ends the first part of the proof.

Part 2. Let G be any dominating family for NN and |G| = d. For g ∈ NN we define a new function g: g(1) = g(1), g(2) = 2g(1) + g(2), and, recursively, g(n + 1) = 2g(n) + g(n + 1). Of course, g≥ g, whence the family

G= {g: g ∈ G}

is also dominating for NNand |G| = d.

For f ∈ NNlet ˆf (n) = f (n + 1).

Let now f be any function from NN. Let g ∈ G dominate max{f, ˆf }. We have

g(n + 1) − g(n) ≥ g(n) ≥ g(n) and

g(n) ≥ f (n) > f (n) − f (n + 1) and

g(n) ≥ ˆf (n) = f (n + 1) > f (n + 1) − f (n).

Hence

g(n + 1) − g(n) > |f (n + 1) − f (n)|. (5) Let G = g[N]. We define λf : G → N by λf(g(i)) = f (i). By (5) λf

satisfies the Lipschitz condition with constant 1. We extend λf to a function Λf ∈ L(N, 2). Let r ∈ G, s ∈ G, r < s and (r, s) ∩ G = ∅. Let v ∈ (r, s). We define

Λf(v) = λf(s) − λf(r)

s − t (v − r) + λf(r)

 .

To check that Λf satisfies the Lipschitz conditition with constant 2 it is enough to check it for two consecutive points v, v + 1 ∈ [r, s]. Consider the case λf(s) − λf(r) ≥ 0; the other case is simillar. We have

Λf(v + 1) − Λf(v) = λf(s) − λf(r)

s − t (v + 1 − r)



− λf(s) − λf(r) s − t (v − r)



≤ λf(s) − λf(r)

s − t (v + 1 − r) −λf(s) − λf(r)

s − t (v − r) + 1 = λf(s) − λf(r)

s − t + 1 ≤ 2.

We also have f = Λf ◦ g. Hence NN= hL(N) ∪ Gi. 

BIBLIOGRAPHY

1. J. Cicho´n, J.D. Mitchell, M. Morayne, Y. P´eresse, Relative ranks of Lipschitz mappings on countable discrete metric spaces, submitted.

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APPENDIX

Let I = R \ Q. Let ρ be the the usual euclidean metric of the real line. We will prove the following theorem.

Theorem The spaces (NN, d) and (I, ρ) are homeomorphic.

Proof. It is easy to notice that the space (I, ρ) is homeomorhic to the space ((0, 1) ∩ I, ρ).

Let

Q ∩ (0, 1) = {sj: j = 1, 2, . . .}

and

I= (0, 1).

Let now

0 = q0< q1< q2< . . . → 1, where all qi are rational numbers.

Let us make sure that s1∈ {qi: i ∈ N}.

Let

Ii= (qi−1, qi),

for i ∈ N. We make sure that |Ii| < 1/2 for each i ∈ N.

Now we divide the intervals Ii = (qi−1, qi) into subintervals Ii,j = (qi,j, qi,j−1), j ∈ N, where

qi−1← . . . qi,2< qi,1< qi,0= qi.

At this second stage of the construction we make sure that |Ii,j| < 1/3 for each interval Ii,j and that

s2∈ {qi: i ∈ N} ∪ {qi,j : i, j ∈ N}.

We continue this construction and at the stage k, for k odd and ik−1∈ N, we have

qi1,...,ik−1 = qi1,...,ik−1,0< qi1,...,ik−1,1< ... → qi1,...,ik−1−1

and for k even and ik−1∈ N, we have

qi1,...,ik−1−1← · · · < qi1,...,ik−1,1< qi1,...,ik−1,0= qi1,...,ik−1.

We also make sure that sk is of the form qi1,...,ir, r ≤ k, i1, . . . ir ∈ N, and that for Ii1,...,ik being the open interval of the endpoints qi1,...,ik−1, qi1,...,ik,

|Ii1,...,ik| < 1

k + 1. (6)

We have

Ii1,...,ik,ik+1⊂ Ii1,...,ik (7)

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and

Ii1,...,ik∩ Ij1,...,jk = ∅ (8) unless i1= j1, . . . , ik= jk. We also have

qi1,...,ik ∈ ¯/ Ij1,...,jk+1,jk+2 (9) for any sequence j1, . . . , jk+1, jk+2. It follows now from (6) and (7) that for each fixed sequence (Ii1,...,ik)k the intersection of the closures of all the intervals from this sequence is a one-point set

\

k=1

i1,...,ik = {xi1,i2,...}.

We claim now that xi1,i2,... is an irrational number. Indeed, if it were rational, then it would be sk for some k ∈ N. Then, by the rules of the construction, it would be of the form qi1,...,ir, r ≤ k. Then, however, by (9) it could not be an element of ¯Ii1,...,ik+1,ik+2.

The argument above also shows that

\

k=1

i1,...,ik =

\

k=1

Ii1,...,ik.

Now we will define our homeomorphism f : N → I ∩ (0, 1). Let {xi1,i2,...} =

\

k=1

Ii1,...,ik, and

f ((ik)k) = xi1,i2,.... By (8) the mapping f is an injection.

Let now x ∈ I ∩ (0, 1). Then let x ∈ Ii1, x ∈ Ii1,i2, . . .. We have {x} =

\

k=1

Ii1,...,ik

and thus

f ((ik)k) = x.

Hence f is also a surjection.

Now we are going to prove that f is continuous. Actually, we will show more, namely, that f is a contraction. Let (ik)k, (jk)kbe two different elements of N . Let d((ik)k, (jk)k)) < 1/k0. Then the finite sequences (i1, . . . , ik0) and (j1, . . . , jk0) are identical and f ((ik)k), f ((jk)k) ∈ Ii1,...,ik0. By (6) we obtain

|f ((ik)k) − f ((jk)k)| < 1/k0.

In the last step we will show that the mapping f−1, inverse to f , is continuous as well.

Let us assume that yn ∈ (0, 1) ∩ I for n ∈ N and yn→ y ∈ (0, 1) ∩ I.

Let y ∈ Ii1,...,ik for some fixed k. Then for certain n0∈ N and all n ≥ n0we have yn∈ Ii1,...,ik. This implies that d(f−1(y), f−1(yn)) ≤ 1/k for n ≥ n0. This

concludes the proof. 

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