RECURRENT ITERATED FUNCTION SYSTEMS

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ALEXANDRU MIHAIL

The theory of fractal sets is an old one, but it also is a modern domain of re- search. One of the main source of the development of the theory of fractal sets is the discovery of a new types of fractal sets. One of the well-known examples are the attractors of iterated function systems. There are several generalization of iterated functions systems: directed-graph iterated function systems, random iterated function systems and so on. The aim of the paper is to present a generalization of iterated function systems (IFS) and of their attractors, the recurrent iterated function systems.

AMS 2000 Subject Classification: 28A80.

Key words: recurrent iterated function system, fractal set, Lipschitz function, contraction, autosimilar set.

1. INTRODUCTION

In this section we review concepts, notation and basic results concerning iterated function systems (IFS). Details can be found in [1], [2] or [3].

Let (X, d_X) and (Y, d_Y) be two metric spaces andC(X, Y) the set of continuous functions fromXin toY. OnC(X, Y) we will consider the generalized metric d:C(X, Y)×C(X, Y)→R+=R+∪ {+∞}defined by

d(f, g) = sup

x∈X

dY(f(x), g(x)).

For a sequence (f_n)_nof elements fromC(X, Y) andf ∈C(X, Y),f_n→^s f denotes pointwise convergence,fnu.c

→f uniform convergence on compact sets, and fn

→u f uniform convergence (that is, convergence in the generalized metric d).

For a function f : X → X we denote by Lip(f) ∈ R+ the Lipschitz constant associated with f, that is,

Lip(f) = sup

x,y∈X;x6=y

d(f(x), f(y)) d(x, y) .

A function f is said to be a Lipschitz function if Lip(f) < +∞ and a contraction if Lip(f)<1.

REV. ROUMAINE MATH. PURES APPL.,53(2008),1, 43–53

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Let

Con(X) ={f :X→X |Lip(f)<1}.

Let (X, d) be a metric space andK(X) the set of nonvoid compact subsets of X. ThenK(X) with the Hausdorff-Pompeiu distanceh:K(X)×K(X)→ [0,∞) defined by

h(E, F) = max(d(E, F), d(F, E)) = min{r |E⊂B(F, r) and F ⊂B(E, r)}, where d(E, F) = sup

x∈E

d(x, F) = sup

x∈E

( inf

y∈F d(x, y)),is a metric space.

(K(X), h) is a complete metric space if (X, d) is a complete metric space.

(K(X), h) is compact if (X, d) is compact (see [4]).

The Hausdorff-Pompeiu distance can be extended to a semidistance on P^∗(X) with values inR+ with the same definition, where P^∗(X) is the set of nonvoid subsets of X.

Concerning the Hausdorff-Pompeiu distance we have the following important properties.

Proposition 1.1. Let (X, d) be a metric space,f :X→X a Lipschitz function and K(X) the set of compact subsets of X. If (H_i)i∈I and (K_i)i∈I

are two families of subsets of X,then 1)h(S

i∈I

Hi, S

i∈I

Ki) =h(S

i∈I

Hi,S

i∈I

Ki)≤sup

i∈I

h(Hi, Ki);

2)h(f(K), f(H))≤Lip(f)h(K, H) for any K, H ∈K(X).

An iterated function system, (IFS for short) consists of a metric space (X, d) and a finite family of contractions fk : X → X, k ∈ {1,2, . . . , n}, n∈N^∗.We denote such an IFS byS = (X,(f_k)_k=1,n). Its contraction constant is cS = max

k=1,n

Lip(f_k).

There are three important mathematical results concerning contractions used in the IFS theory.

Theorem 1.1 (Banach Contraction Principle). Let (X, dX) be a complete metric space and f ∈Con(X). Then there exists a unique α ∈X such that f(α) =α. Moreover, lim

n→∞dX(f^[n](x), α) = 0 for any x∈X.

Theorem 1.2(Continuity Theorem). Let (X, dX) be a complete metric space, f, g∈Con(X), α the fixed point of f and β the fixed point of g. Then

d_X(α, β)≤d(f, g) 1

1−min{Lip(f),Lip(g)}.

Theorem 1.3 (Collage Theorem). Let (X, d_X) be a complete metric space, f ∈ Con(X) and x ∈ X. Let α ∈ X be the unique fixed point of f.

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Then

d_X(x, α)≤ d_X(f(x), x) 1−Lip(f).

With an IFS S = (X,(f_k)_k=1,n) with contraction constant cS = max

k=1,n

Lip(f_k) one can associate the function FS : K(X) → K(X) defined by FS(B) =

n

S

k=1

f_k(B), B ∈ K(X), which is a cS-contraction. Therefore, by the Banach contraction principle, we have

Theorem 1.4. Let (X, d_X) be a complete metric space and S = (X,(f_k)_k=1,n)an IFS with cS = max

k=1,n

Lip(f_k)<1. Then there exists a unique set A(S),such that FS(A(S)) =A(S), and for every H₀∈K(X)the sequence (Hn)n, where Hn+1 =FS(Hn), n∈N, is convergent to A(S) in K(X). As for the speed of convergence we have

h(Hn, A(S))≤ cⁿ_S 1−cS

h(H0, H1), for all n∈N. In particular,

h(H₀, A(S))≤ 1 1−cS

h(H₀, H₁).

We also have the following result concerning the continuity of the attractor of an IFS.

Theorem 1.5. Let (X, d_X) be a complete metric space and x∈X. Let S = (X,(fk)_k=1,n) and S⁰ = (X,(gk)_k=1,n) be two iterated function systems with contraction constants cS and c_S⁰. Then

h(A(S), A(S⁰))≤ max

k=1,n

d(f_k, g_k) 1

1−min(cS, c_S⁰).

We present now the idea of our generalization. Instead of considering contractions f from X in to X in the definition of an IFS, we shall consider contractions from X² = X ×X in to X. We shall call recurrent iterated function systems (RIFS for short) this kind of iterated function systems. On X² we consider the distance d2 : X² ×X² → [0,+∞) given by d2((x, x1),(y, y1)) = max{d(x, y), d(x₁, y1)}.

Definition 1.1. Let (X, d) be a complete metric space. For a function f :X×X→Xthe number

Lip(f) = sup

x,y,x1,y1∈X;x6=yorx16=y1

d(f(x, x₁), f(y, y₁)) max{d(x, y), d(x₁, y₁)}

is called the Lipschitz constant of f. We have Lip(f)∈[0,+∞].

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A functionf :X×X →Xis said to be a Lipschitz function if Lip(f)<∞ and a contraction if Lip(f)<1. Let

Con2(X) ={f :X×X→X |Lip(f)<1}.

Definition 1.2. Let (X, d) be a complete metric space. A recurrent iterated function systems (RIFS for short) on X consists of a finite family of contractions (f_k)_k=1,n,n∈N, with f_k :X×X→ X, and is denoted byS = (X,(f_k)_k=1,n). The contraction constant of such an RIFS iscS = max

k=1,n

Lip(f_k).

2. MAIN RESULT

In this section we prove the existence of the attractor of an RIFS and also present its main properties.

Definition 2.1. Let (X, d) be a metric space and f : X ×X → X a continuous function. The function F_f :K(X)×K(X)→K(X) defined by

F_f(K, H) =f(K, H) =f(K×H) ={f(x, y)|x∈K and y∈H}

is called the set function associated with f.

LetS= (X,(fk)_k=1,n) be an RIFS. The functionFS :K(X)×K(X)→ K(X) defined byFS(K, H) =

n

S

k=1

f_k(K, H) is called the set function associated with S.

The next result is obvious.

Lemma 2.1. If (X, d_X) is a complete metric space fn, f : X×X → Xare continuous functions, n ∈ N, such that fn

→u f and H, K ∈ K(X), then lim

n→∞fn(H, K) =f(H, K) in K(X).

The next result is a consequence of Lemma 2.1 and Proposition 1.1.

Lemma 2.2. Let (X, dX) be a complete metric space. Let Sm = (X,(f_k^m)_k=1,n) and S = (X,(f_k)_k=1,n), m ∈ N^∗, be recurrent iterated function systems. If f_k^m→^u fk on a dense set in X as m→+∞,k∈ {1,2, . . . , n}, then lim

n→∞FSn(K, H) =FS(K, H) for every H, K ∈K(X).

Lemma 2.3.Let f :X×X →Xbe a Lipschitz function. Then Lip(F_f)≤ Lip(f).

Proof. Let K, H, K₁, H₁ ∈ K(X) and r = max(h(K, K₁), h(H, H₁)).

Then K ⊂ B(K1, r) and H ⊂ B(H1, r). Let x ∈ K and y ∈ H. There are

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x1∈K1 and y1 ∈H1 such that d(x, x1)< rand d(y, y1)< r. Then d(f(x1, y1), f(x, y))≤Lip(f) max{d(x, x₁), d(y, y1)}<Lip(f)r.

This means thatf(x, y)∈B(f(x₁, y₁),Lip(f)r)⊂B(f(K₁, H₁),Lip(f)r) which implies thatf(K, H)⊂B(f(K1, H1),Lip(f)r).

In a similar way, we obtain the inclusionf(K₁, H₁)⊂B(f(K, H),Lip(f)r).

This means that h(f(K, H), f(K1, H1))≤Lip(f)r.

The next result is a key one.

Lemma 2.4.Let S= (X,(f_k)_k=1,n)be an RIFS with cS= max

k=1,n

Lip(f_k)<

<1.Then the set function FS :K(X)×K(X)→K(X) associated with S, is a cS contraction.

Proof. We have

h(FS(K, H), FS(K1, H1)) =h n

[

k=1

f_k(K, H),

n

[

k=1

f_k(K1, H1)

≤

≤ max

k=1,n

h(f_k(K, H), f_k(K₁, H₁))≤cSmax(h(K, K₁), h(H, H₁)).

For a contraction of typef :X×X →X we have results similar to those of type f :X →X. We shall sketch the proof and present some applications.

Theorem 2.1(Banach Contraction Principle). Let (X, d) be a complete metric space and f ∈Con2(X)with contractivity factor c∈[0,1). Then there exists a unique α ∈X such that f(α, α) = α. Moreover, for any x₀, x₁ ∈X the sequence (xn)n≥0 defined by xn+1=f(xn, xn−1),n∈N^∗,is convergent to α. As for the speed of convergence of (x_n)n≥0 we have

d(x_n, α)≤ 2c^[n/2]

1−c max{d(x₀, x₁), d(x₁, x₂)}.

In particular,

d(x0, α)≤ 2

1−cmax{d(x₀, x1), d(x1, x2)}.

Proof. Letzn= max{d(x_n+1, xn), d(xn, xn−1)}. We have d(x_n+1, x_n) =d(f(x_n, xn−1), f(xn−1, xn−2))≤

≤cmax{d(xn−1, xn−2), d(x_n, xn−1)}=czn−1

and

d(xn+2, xn+1) =d(f(xn+1, xn), f(xn, xn−1))≤

≤cmax{d(x_n+1, xn), d(xn, xn−1)} ≤cmax{cz_n−1, zn−1}=czn−1. Then zn+1≤czn−1.It follows that

d(xn+1, xn)≤c^[n/2]z1

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and

d(xn+p, xn)≤d(xn+p, xn+p−1) +d(xn+p−1, xn+p−2) +· · ·+d(xn+1, xn)≤

≤c[(n+p−1)/2]

z1+c[(n+p−2)/2]

z1+· · ·+c^[n/2]z1, hence

d(x_n+p, x_n)≤ 2c^[n/2]z1

1−c for every p∈N^∗.

The rest of the proof goes in the same manner as that for the Banach contraction principle.

Using Lemma 2.4 in the case of an RIFS, Theorem 2.1 becomes

Theorem 2.2. Let (X, dX) be a complete metric space and S = (X,(f_k)_k=1,n)an RIFS with contraction constant cS = max

k=1,n

Lip(f_k) < 1.

Then there exists a unique A(S)∈K(X) such that FS(A(S), A(S)) =A(S).

Moreover, for any H₀, H₁ ∈ K(X), the sequence (H_n)n≥1 defined by Hn+1 = FS(Hn, Hn−1), n ∈ N^∗, is convergent to A(S). As for the speed of convergence we have

h(Hn, A(S))≤ 2c^[n/2]_S 1−cS

max{h(H₀, H1), h(H2, H1)}.

In particular (Collage Theorem), h(H₀, A(S))≤ 2

1−cS

max{h(H₀, H₁), h(H₂, H₁)}.

We remark that A(S) =

n

[

k=1

f_k(A(S), A(S)) =

n

[

k=1

[

x∈A(S)

f_k({x}, A(S)) =

=

n

[

k=1

[

x∈A(S)

f_k(A(S),{x}) =

n

[

k=1

[

x,y∈A(S)

f_k({x},{y}).

Theorem 2.3(Continuity Theorem). Let (X, d_X) be a complete metric space and f, g∈Con2(X)with fixed points α and β and contraction constants c_f and c_g. Then

d_X(α, β)≤minn 1

1−c_fd_X(f(β, β), g(β, β)), 1

1−c_gd_X(f(α, α), g(α, α))o

≤

≤ 1

1−min(cf, cg)d(f, g).

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Proof. We have

d_X(α, β) =d_X(f(α, α), g(β, β))≤d_X(f(α, α), f(β, β)) +d_X(f(β, β), g(β, β))≤

≤c_fd_X(α, β) +d_X(f(β, β), g(β, β))≤c_fd_X(α, β) +d(f, g), whence

d_X(α, β)≤ 1

1−c_fd_X(f(β, β), g(β, β))≤ 1

1−c_f d(f, g).

Similarly, we have

d_X(α, β)≤ 1

1−c_gd_X(f(α, α), g(α, α))≤ 1

1−c_gd(f, g) and the conclusion follows.

Theorem 2.4(Continuity Theorem). Let (X, dX) be a complete metric space, and S = (X,(f_k)_k=1,n) and S⁰ = (X,(g_k)_k=1,n)be two RIFS with c=

max

k=1,n

Lip(f_k)<1 and c⁰ = max

k=1,n

Lip(g_k)<1. Then h(A(S), A(S⁰))≤ 1

1−min(c, c⁰) max

k=1,n

d(f_k, g_k).

Another version of the continuity theorem is

Theorem 2.5. Let (X, dX) be a complete metric space and a function f :X×X→X. Let us consider f_n∈Con₂(X), where n∈N, with fixed point αn. Assume c = sup

n≥1

cfn < 1, where cfn = Lip(fn), and fn

→s f on a dense set in X. Then f ∈ Con₂(X), Lip(f) ≤ c and if α is the fixed point of f, then lim

n→∞ α_n=α.

Proof. It follows from the hypotheses that f_n^u.c→ f on X×X and c_f = Lip(f)≤c <1. Hencef is a contraction. By Theorem 2.3 we have

dX(α, αn)≤ 1

1−c dX(fn(α, α), f(α, α)).

Theorem 2.6. Let (X, dX) be a complete metric space, n ∈ N^∗, and S_m = (X,(f_k^m)_k=1,n), m ∈ N^∗, and S = (X,(f_k)_k=1,n) recurrent iterated function systems with contraction constants c_n and c₀. If sup

n≥1

c_n=c <1and f_k^m →^s f_k on a dense set in X×X, k∈ {1,2, . . . , n}, then lim

n→∞A(S_n) =A(S).

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3. EXAMPLES AND REMARKS

Let (X, dX) be a complete metric space, S = (X,(fk)_k=1,n) an RIFS, A(S)∈K(X) the attractor ofS and (H_n)n∈N^∗ a sequence defined byH_n+1= FS(Hn, Hn−1), n∈N^∗, which is convergent to A(S). We make some remarks on the monotony of the sequence (H_n)n∈N^∗.

IfH₀ ⊂H₁ ⊂H₂, thenHn−1 ⊂H_n, for all n∈N^∗, andA(S) = S

n≥1

H_n. In particular, if H⊂FS(H, H) then H⊂A(S).

Let A = {x₀ | there exists k such that f_k(x₀, x₀) = x₀} be the set of fixed points of the functions (f_k)_k=1,n. ThenA⊂f(A, A), and so A⊂A(S).

In other words, every fixed point of a function from S is in A(S).

IfH0⊃H1 ⊃H2, thenHn−1 ⊃Hn, for all n∈N^∗, and A(S) = T

n≥1

Hn. Proposition 3.1. Let (X, dX) be a complete metric space, S = (X,(f_k)_k=1,n) a recurrent iterated function system with contraction constant c,and H∈K(X). If H⊂B(FS(H, H), r) then H⊂B A(S),_1−c^r

.

Proof. Let (H_n)n∈N^∗ be the sequence defined by H_n+1 = FS(H_n, H_n), n ∈ N^∗, and H0 = H. As in Lemmas 2.3 and 2.4, we have by induction that H_n ⊂ B(H_n+1, rcⁿ), so H ⊂ B

H_n, r

n−1

P

k=0

c^k

. Since H_n → A(S) and

n→∞limr

n−1

P

k=0

c^k = _1−c^r , we get H⊂B(A(S),_1−c^r ).

Proposition 3.2. Let (X, d_X) be a complete metric space, S = (X,(f_k)_k=1,n) an RIFS , x₁, x₂, . . . , x_l, y₁, y₂, . . . , y_m ∈A(S) and t:{1,2, . . . , l+m} → {1,2, . . . , n} a fixed function. Let S₁ = (X,(g_k)_k=1,l+m)be the IFS defined by

g_k(x) =

( f_t(k)(x_k, x) if k≤n f_t(k)(x, yk−n) if k≥n+ 1.

Then A(S₁)⊂A(S).

Proof. We have g_k(A(S)) = f_t(k)(x_k, A(S)) ⊂ A(S) if k ≤ n and g_k(A(S)) =f_t(k)(A(S), y_k−n)⊂A(S) if k≥n+ 1. Hence

FS₁(A(S)) =

l+m

[

k=1

g_k(A(S))⊂A(S).

It follows that

A(S₁)⊂A(S).

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Let (X, dX) be a metric space and a function. f :X×X →X. Let us consider the function ∆(f) : X → X defined by ∆(f)(x) = f(x, x), x ∈ X.

For an RIFS S = (X,(fk)_k=1,n), let us define ∆(S) = (X,(∆(fk))_k=1,n). IfS has contraction constantc, then ∆(S) has a contraction constant less than c.

Proposition 3.3. Let (X, dX) be a complete metric space and S = (X,(f_k)_k=1,n)an RIFS . Then A(∆(S))⊂A(S).

Proof. It is easy to see that F_∆(S)(H)⊂FS(H, H) for everyH ∈K(X).

Indeed

F_∆(S)(H) =

n

[

k=1

∆(f_k)(H) =

n

[

k=1

{f(x, x)|x∈H}=

=

n

[

k=1

{f(x, y)|x∈H andy ∈H}=

n

[

k=1

f_k(H, H) =FS(H, H).

Also, forH, K ∈K(X) such thatH ⊂Kwe haveF_∆(S)(H)⊂FS(K, K) since F_∆(S)(H)⊂F_∆(S)(K)⊂FS(K, K).

LetH ∈K(X) be fixed. Consider the sequences (Hn)n∈N^∗and (Kn)n∈N^∗

defined by H₀ =H, K₀ =H, H_n+1 =FS(H_n, H_n),and F_∆(S)(K_n) =K_n+1, where n∈N.

It is clear that K₀ ⊂H₀. IfK_n⊂H_n then

K_n+1 =F_∆(S)(K_n)⊂FS(H_n, H_n) =H_n+1, for all n∈N. Since lim

n→∞Kn=A(∆(S)) and lim

n→∞Hn=A(S), we have A(∆(S))⊂A(S).

Example 1. Every IFS can be seen as an RIFS. Indeed, let S = (X,(f_k)_k=1,n) be an IFS. Let S = (X,(f_k)_k=1,n) be the RIFS defined by f_k(x, y) =f_k(x). Then A(S) =A(S).

Example 2. LetX =Rand let f, g:R×R→Rbe defined byf(x, y) =

x

3 +^y₃ and g(x, y) = ^x₃ +²₃.

Let the RIFS S = (R,(f, g)), and let FS : K(R)×K(R)→K(R) be defined by FS(K, H) = f(K, H)∪g(K, H).The attractor of S is [0,1]. Indeed f([0,1],[0,1]) = [0,2/3]⊂[0,1] andg([0,1],[0,1]) = [2/3,1]⊂[0,1],so that

FS([0,1],[0,1]) = [0,1].

It follows that A(S) = [0,1].

Example 3. LetX =Rand let f, g:R×R→Rbe defined byf(x, y) =

x

5 +^y₅ and g(x, y) = ^x₅ +^y₅ +³₅.

Let the RIFSS = (R,(f, g)), and letF :K(R)×K(R)→K(R) be defined by FS(K, H) = f(K, H)∪g(K, H). The attractor of S is [0,2/5]∪[3/5,1].

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Indeed f([0,1],[0,1]) = [0,2/5] ⊂ [0,1] and g([0,1],[0,1]) = [3/5,1] ⊂ [0,1], so that

FS([0,1],[0,1]) = [0,2/5]∪[3/5,1]⊂[0,1].

It follows thatA(S)⊂[0,1] andA(S) = T

n≥1

H_n, where (H_n)n≥1is the sequence defined by Hn+1=f(Hn, Hn), n∈N, and H0= [0,1].

We remark thatf(x, y) =f(y, x),g(x, y) =g(y, x),F(K, H) =F(H, K), g(x, y) = f(x, y) + ³₅ and g(K, H) = f(K, H) + ³₅. Let also t : R→Rbe defined by t(x) = −x+ ¹₂. Then (t◦t)(x) = x, f(t(x), t(y)) = t(g(x, y)), g(t(x), t(y)) =t(f(x, y)) andt(A(S)) =A(S). In other words the attractor of the RIFS S = (R,(f, g)) is symmetric with respect to 1/2. Then

H1 =FS([0,1],[0,1]) = [0,2/5]∪[3/5,1], f([0,2/5]∪[3/5,1],[0,2/5]∪[3/5,1]) =

=f([0,2/5],[0,2/5])∪f([0,2/5],[3/5,1])∪f([3/5,1],[3/5,1]) = [0,4/25]∪[3/25,7/25]∪[6/25,2/5] = [0,2/5],

g([0,2/5]∪[3/5,1]) = [3/5,1],

H2 =F(H1, H1) =f(H1, H1)∪g(H1, H1) = [0,2/5]∪[3/5,1] =H1. It follows that

A(S) =H₁= [0,2/5]∪[3/5,1].

We remark that A(S) cannot be the attractor of a classical IFS with two injective functions. Such an attractor, if disconnected, must be totally disconnected. However, A(S) is the attractor of a classical IFS with four functions, namely f0 :R→R defined by

f₀(x) = min{|x|,2/5}

2 ,

f1 :R→R defined by

f₁(x) = min{2/5− |x−2/5|,2/5}

2 +4

5, f₂ :R→R defined by

f2(x) = min{3/5 +|x−3/5|,1}

2 +4

5 and f₃ :R→R defined by

f3(x) = min{1− |x−1|,2/5}

2 +1

2.

Example 4. Let X = R and f₀, f₁ : R²→R²be defined by f₀(x, y) = (^x₃,0) and f₁(x, y) = (^x₃ +²₃,0).

(11)

Let the IFS S = (R,(f₀, f₁)) and A(S) = C the Cantor set. Let C0 = f₀(C) and C1 = f₁(C). Let us note that C0 ⊂ C, C1 ⊂ C, f₀(C0) ⊂ C₀ and f₁(C₁) ⊂ C₁. Consider the RIFS S = (R,(f₀, f₁)), where f₀ : R²×R²→R²andf1 :R²×R²→R² are given byf0((x1, x2),(y1, y2)) = (^x₃¹,^y₃¹) and f₁((x₁, x₂),(y₁, y₂)) = (^x₃¹ +²₃,0).Then A(S) =C₀×C₀∪C₁× {0}.

Indeed, since

f0(C0×C0∪C1× {0}) =f0(C0×C0)∪f0(C1× {0}) =

=f₀(C0)×f₀(C0)∪f₀(C1)× {0} ⊂C0×C0∪C1× {0}

and

f₁(C₀×C₀∪C₁× {0}) =f₁(C₀×C₀)∪f₁(C₁× {0}) =

=f1(C0)× {0} ∪f1(C1)× {0}=C1× {0}, we have A(S)⊂C0×C0∪C1× {0}.

Since f₀((x₁, x₂),(0,0)) =f₀(x₁, x₂) and f₁((x₁, x₂),(0,0)) =f₁(x₁, x₂), we get C⊂A(S). Also,f₀(C×C) =C₀×C₀ and f₁(C×C) =C₁× {0}. It follows that

C₀×C₀∪C₁× {0} ⊂A(S), which completes the proof.

REFERENCES

[1] M.F. Barnsley,Fractals Everywhere. Academic Press, New York, 1998.

[2] K.J. Falconer,The Geometry of Fractal Sets. Cambridge Univ. Press, Cambridge, 1985.

[3] K.J. Falconer,Fractal Geometry-Foundations and Applications.Wiley, 1990.

[4] N Secelean,M˘asur˘a ¸si fractali. Ed. Univ. Sibiu, 2002.

Received 1st March 2006 University of Bucharest

Faculty of Mathematics and Computer Science Str. Academiei 14

010014 Bucharest, Romania mihail alex@yahoo.com