18.dvi @ Applied General TopologyUniversidad Polit�ecnica de ValenciaVolume 2, No. 1, 2001pp. 51 - 61 New results on topological dynamics ofantitriangular mapsF. Balibrea, J. S. C�anovas and A. Linero�Abstract. We present some results concerning the topologicaldynamics of antitriangular maps, F : X2 ! X2 with the formF(x; y) = (g(y); f(x)); where (X; d) is a compact metric spaceand f; g : X ! X are continuous maps. We make an specialanalysis in the case of X = [0; 1].2000 AMS Classi�cation: 54H20, 37B20, 37B10, 37E99.Keywords: Cournot duopoly, antitriangular maps, recurrence, topologicaldynamics. 1. IntroductionLet (X;d) be a compact metric space and let ' : X ! X be a continuousmap, ' 2 C(X;X): The pair (X;') is called a discrete dynamical system,whose orbits are given by the sequence f'n(x)g1n=0, x 2 X, where 'n = ' �'n�1, n � 1 and '0 = Identity. In general, the full knowledge of all orbitsof the system is a di�cult problem and it is only known in some particularcases. Nevertheless, good approximations can be given. These approachescan be probabilistic (invariant measures, metric entropy, ...) or topological(periodic structure, topological entropy, ...). In this paper we will follow thislast approach.A point x 2 X is periodic when, for some n > 0; is 'n(x) = x: If n = 1;the periodic point is called �xed point. The order or period of a periodic pointis precisely the smallest of the values m for which 'm(x) = x: We denote byPer(') the set of periods that the continuous map ' has.We use A(') to denote one of the following sets: the set of periodic points,P('); the set AP(') of almost periodic points, that is, the points x 2 X suchthat for any neighborhood V = V (x) of x; there is N = N(V ) 2 N such that'kN(x) 2 V; for every k � 0; the set UR(') of uniformly recurrent points, x 2 Xsuch that for any neighborhood V = V (x) of x; there is N = N(V ) such that�This paper has been partially supported by the D.G.I.C.Y.T. Grant PB98{0374{C03{01. 52 F. Balibrea, J. S. C�anovas and A. Linerofor all q > 0 it holds 'r(x) 2 V for some q � r < q + N; R(') = fx 2 X : x 2!'(x)g is the set of recurrent points, with !'(x) the omega-limit set of the pointx; that is, the points y 2 X such that there is a subsequence fnigi � N with'ni(x) ! y as ni ! 1; C(') = R(') is the centre of '; where A denotes theclosure of a set of A � X; !(') is the (global) !-limit set, !(') = Sx2X !'(x); (') is the set of non-wandering points, those points x 2 X such that forany neighborhood U = U(x) of x there exists N = N(U) 2 N in a way that'N(U)\U 6= ?; and �nally the set CR(') of chain-recurrent points, the pointsx 2 X for which given any " > 0; there is fxigni=0 � X such that x0 = x;d(xi+1;'(xi)) < " for i = 0;1; :::;n � 1 and d(x0;'(xn)) < ":It follows from the de�nitions that(1.1) P(') � AP(') � UR(') � R(') � C(')and(1.2) !(') � (') � CR('):In this paper, we are devoted to topological dynamics of antitriangular maps,that is, continuous maps F : X � X ! X � X of the form(1.3) F(x;y) = (g(y);f(x));with (x;y) 2 X � X.Antitriangular maps appear in some economical models, particularly with theso{called Cournot duopoly (see [12] or [8]). The Cournot duopoly consists in aneconomy in which two �rms are competitors in the same sector. This situationis modelled by a map F having the form of (1.3) and such that X = I = [0;1].From (1.3) it is clear that(1.4) F2n(x;y) = ((g � f)n(x);(f � g)n(y))and(1.5) F2n+1(x;y) = (g � (f � g)n(y);f � (g � f)n(x))for any (x;y) 2 X � X and for any n 2 N. So, it is natural to expect thedynamics of F to be strongly connected to the dynamics of g � f and f � g:In [10] a program is developed for triangular maps, that is, continuous mapsT : I2 ! I2 of the form T(x;y) = (f(x);g(x;y)). This is made investigatingthe relationship between the sets A(T) � I2 and A(f) � I, whereA(�) 2 fP(�);AP(�);UR(�);R(�);C(�);!(�); (�);CR(�)g:When X = I, a �rst step to follow a similar program for antitriangular mapsis to do the same with the sets A(F) � I2 and A(g � f);A(f � g) � I.For A(�) 2 fP(�);AP(�);C(�);CR(�)g we see that A(F) = A(g � f) �A(f � g)and when A(F) = R(F) the two situations, R(F) = R(g � f) � R(f � g) andR(F) $ R(g�f)�R(f �g) are possible (see [3]). We also see that in the case ofthe set of uniformly recurrent points the same result is true. For A(F) = (F)the situation is more complicated and the case (F) * (g � f) � (f � g)can happen. It remains open what is the situation when A(F) = !(F). We Topological dynamics of antitriangular maps 53conjecture that the cases !(F) = !(g�f)�!(f�g) and !(F) $ !(g�f)�!(f�g)could be found.We underline that similarly to the interval case in antitriangular maps on I2it is held C(F) = P(F); which is not true in general in the triangular case (see[10]).Similarly to the interval case, we construct examples proving that the follow-ing chain is possible(1.6) P(F) 6= AP(F) 6= UR(F) 6= R(F) 6= C(F) 6= !(F) 6= (F) 6= CR(F):The paper is organized as follows. In the next section, we study the rela-tionship between the sets A(F) and A(g � f) � A(f � g). The last section isconcerned with the introduction of the chain (1.6).2. Projection of the topological dynamicsIf (X;d) is a compact metric space, we denote the product space by (X�X;�),where �((x1;y1);(x2;y2)) = maxfd(x1;x2);d(y1;y2)gfor all (x1;y1);(x2;y2) 2 X � X. If f;g 2 C(X;X) we de�ne the product mapf � g : X � X ! X � X by (f � g)(x;y) = (f(x);g(y)) for all (x;y) 2 X � X.So, if F(x;y) = (g(y);f(x)) is an antitriangular map, we obtain that F2 =(g � f) � (f � g).In this section we consider an antitriangular map F : X � X ! X � X andwe study if the equality A(F) = A(g � f) � A(f � g) holds, where A(�) denotesone of the subsets P(�), AP(�), UR(�), R(�), C(�), !(�), (�) or CR(�). Beforestudying this problem, we need the following result.Proposition 2.1. Let (X;d) be a compact metric space. Let f;g 2 C(X;X).Then(a) P(f2) = P(f) and P(f � g) = P(f) � P(g).(b) AP(f2) = AP(f) and AP(f � g) = AP(f) � AP(g).(c) UR(f2) = UR(f) and UR(f � g) � UR(f) � UR(g).(d) R(f2) = R(f) and R(f � g) � R(f) � R(g).(e) C(f2) = C(f) and C(f � g) � C(f) � C(g).(f) !(f2) = !(f) and !(f � g) � !(f) � !(g).(g) (f2) � (f) and (f � g) � (f) � (g).(h) CR(f2) = CR(f) and CR(f � g) = CR(f) � CR(g).Additionally, if X = [0;1], then C(f � g) = C(f) � C(g).Proof. For the �rst part of (a)-(h) see [4]. The second part of properties (a){(h)follow from de�nitions. For instance, we prove here the equality CR(f � g) =CR(f) � CR(g).Let (x0;y0) 2 CR(f �g): Given an arbitrary " > 0; we must prove that thereis an "-chain for x and f, and an "-chain for y and g. For " > 0; there is an"-chain for (x0;y0) and f � g;(x0;y0);(x1;y1); :::;(xn;yn);(xn+1;yn+1) = (x0;y0); 54 F. Balibrea, J. S. C�anovas and A. Linerosuch that d((f � g)(xi;yi);(xi+1;yi+1)) < " for i = 0;1; :::;n: Clearlyx0;x1; :::;xn;xn+1 = x0is an "{chain for x0 and f andy0;y1; :::;yn;yn+1 = y0is an "{chain for y0 and g. So CR(f � g) � CR(f) � CR(g).Now, let x0 2 CR(f), y0 2 CR(g) and prove that (x0;y0) 2 CR(f � g).Fix " > 0 and let x0;x1; :::;xn;xn+1 = x0 be an "{chain for x0 and f; andy0;y1; :::;ym;ym+1 = y0 an "{chain for y0 and g. We can clearly assume thatn = m (repeating the chains if necessary). Then,(x0;y0);(x1;y1); :::;(xn;yn); :::;(xn+1;yn+1) = (x0;y0)is an "{chain for (x0;y0) and f � g. Hence (x0;y0) 2 CR(f � g).To �nish the proof, assume that X = [0;1] and prove that C(f � g) =C(f) � C(g). Since C(f) = P(f) = R(f) (cf. [6]), we obtain thatR(f) � R(g) = P(f) � P(g) = P(f) � P(g) = P(f � g) � C(f � g);and jointly with (e) we conclude the proof. �Theorem 2.2. Let (X;d) be a compact metric space. Consider f;g 2 C(X;X)and let F(x;y) = (g(y);f(x)). Then(a) P(F) = P(g � f) � P(f � g).(b) AP(F) = AP(g � f) � AP(f � g).(c) UR(F) � UR(g � f) � UR(f � g).(d) R(F) � R(g � f) � R(f � g).(e) C(F) � C(g � f) � C(f � g).(f) !(F) � !(g � f) � !(f � g):(g) (F2) � (g � f) � (f � g):(h) CR(F) = CR(g � f) � CR(f � g).If in addition X = I, then C(F) = C(g � f) � C(f � g).Proof. Just notice that F2 = (g � f) � (f � g) and apply Proposition 2.1. �Now we �x X = I. It was proved in [3] that the inclusion (d) of Theorem2.2 can be strict, that is, there is an antitriangular map F holding(2.7) R(F) $ R(g � f) � R(f � g):We are able to give an example showing that(2.8) UR(F) $ UR(g � f) � UR(f � g);that is, the inclusion (c) of Theorem 2.2 can be strict. To this end, considerthe trapezoidal tent map f(x) = maxf1 �j2x �1j;�g (� = 0:8249:::) from [11].The idea for constructing the example is the following: any in�nite !{limitset of f is contained in a solenoidal structure. This structure can be labelledby codes which characterizes the elements of in�nite !{limit sets (see below).We take two uniformly recurrent points x0;y0 belonging to the same in�nite Topological dynamics of antitriangular maps 55!{limit set and such that x0;y0 are labelled by the same code. We prove that(x0;y0) =2 UR(F) for the map F(x;y) = (y;f(x)).Now, we need some de�nitions. For any Z � Z let Z1 = f� = (�i)1i=1 : �i 2Z; i 2 Ng. For n 2 N let Zn = f(�1;�2; :::;�n) : �i 2 Z; 1 � i � ng. If � 2 Znand # 2 Zm, n 2 N, m 2 N [f1g, then � � # 2 Zn+m (where n +1 means 1)will denote the sequence � de�ned by �i = �i if 1 � i � n and �i = #i�n for anyi > n. In what follows we denote 0 = (0;0; : : : ;0; : : :) and 1 = (1;1; : : : ;1; : : :),while if � 2 Z1 then �jn 2 Zn is de�ned by �jn = (�1;�2; : : : ;�n).Proposition 2.3. Let f be the trapezoidal map de�ned above. Consider theantitriangular map F(x;y) = (y;f(x)): ThenUR(F) $ UR(f) � UR(f):Proof. By [11], f has periodic points of periods 2n, n 2 N [f0g. By [9, Propo-sition 1], there is a family fK�g�2Z1 of pairwise disjoint (possibly degenerate)compact subintervals of [0;1] satisfying the following properties.(P1) The interval K0 contains all absolute maxima of f.(P2) De�ne in Z1 the following total ordering: if �;� 2 Z1, � 6= � and k isthe �rst integer such that �k 6= �k then � < � if either Cardf1 � i k: Then f(K�) = K�. Also f(K0) � K1:(P4) For any n and � 2 Zn, let K� be the least interval including all intervalsK�, � 2 Z1, such that �jn = �. Then, for any � 2 Z1; K� =T1n=1 K�jn.(P5) If !(x;f) is an in�nite !-limit set of f; then !(x;f) � f0;1g1, wheref0;1g1 denotes the set of in�nite sequences (�i)1i=1 with �i 2 f0;1g forall i 2 N.(P2) gives us information on the positions of fK�g�2Z1 in [0;1] while (P3)gives us information on the dynamics of the family of intervals fK�g�2Z1. Onthe other hand, since f has an interval of absolute maxima, (P1) gives us thatK0 is non{degenerate. Let [x0;y0] = K0. By (P2) it is straightforward to provethat(2.9) f2n(1+2k)(K0) � K0jn�(1)for all n;k 2 N. Then (P3) gives us that(2.10) f22n�1(1+2k)(K0) < K0 < f22n(1+2k)(K0)for n;k 2 N. We claim that x0;y0 2 UR(f). In order to see this �x jK0j >" > 0 small enough and consider the open interval (x0 � ";x0 + "). By (P4)K0j2n�1�(1) � (x0�";x0 +") if 2n�1 � m and K0j2n�(1)\(x0�";x0+") = ? forall n 2 N. By (2.9) and (2.10), f22n�1(1+2k)(x0) 2 K0j2n�1�(1) � (x0 � ";x0 + ") 56 F. Balibrea, J. S. C�anovas and A. Lineroif 2n � 1 � m for all k 2 N. This gives us x0 2 UR(f) (similarly it can beproved that y0 2 UR(f)). Letx0(f) := fn 2 N : f2n(x0) < K0gand y0(f) := fn 2 N : f2n(y0) > K0g:Consider the antitriangular map F(x;y) = (y;f(x)). Then, it is clear thatF2(x;y) = (f(x);f(y)). By [5, Proposition 3.5], (x0;y0) 2 UR(F) if and onlyif x0(f) \ y0(f) is in�nite. However, by (2.10), x0(f) \ y0(f) = ?. Therefore(x0;y0) =2 UR(F) while (x0;y0) 2 UR(f)�UR(f). This concludes the proof. �We are not able to say anything about the inclusions (f){(g) of Theorem2.2, but we are able to give an example of an antitriangular map F such that (F) * (g � f) � (f � g) (compare with (a){(f)). To this end, we prove thefollowing lemma.Lemma 2.4. Let F(x;y) = (f(y);f(x)) be an antitriangular map with f 2C(I;I). Then x 2 (fn) if and only if (x;x) 2 (Fn) for any integer n � 1:Proof. First assume that x 2 (fn) and let V � I2 be an open neighborhoodof (x;x). Let U � I be an open set such that (x;x) 2 U � U � V . Sincex 2 (fn), there is a positive integer m such that (fn)m(U) \ U 6= ?. Henceaccording to (1.4) and (1.5) we haveFnm(U � U) \ (U � U) 6= ?;which gives us (x;x) 2 (Fn). On the other hand, let (x;x) 2 (Fn). LetU � I be an open set such that x 2 U. Since (x;x) 2 (Fn), there is anm 2 N such that (Fn)m(U � U) \ (U � U) 6= ?. Then fnm(U) \ U 6= ? andx 2 (fn). �Lemma 2.4 allows us to show that in general(2.11) (F) * (g � f) � (f � g):To see this, let ef be a continuous interval map such that (ef) n (ef2) 6= ?(see [7]) and de�ne eF(x;y) = (ef(y); ef(x)) (in this case g � f = f � g = ef2).Let x 2 (ef) n (ef2): By Lemma 2.4 we clearly obtain that x 2 (ef) implies(x;x) 2 (F), while (x;x) =2 (ef2)� (ef2) = (g �f)� (f �g). Additionally,we obtain that (x;x) 2 (F) n (F2).In [7] it is shown that in the case of continuous interval maps every successionof equalities and strict inclusions is possible in the chain(2.12) (f) % (f2) % (f22) % (f23) % ::: % (f2n) % (f2n+1) % :::According to Lemma 2.4, the same happens in the case of antitriangular mapsfor the chain of inclusions (F) % (F2) % (F22) % (F23) % ::: % (F2n) % (F2n+1) % :::It su�ces to take F(x;y) = (f(y);f(x)) where f holds (2.12). Topological dynamics of antitriangular maps 57Again concerning the non-wandering set, (2.11) gives us that it can happen (g � f) $ �1( (F)) and (f � g) $ �2( (F));where �i represents the canonical projection, i = 1;2. Notice that it is straight-forward to see that�1(A(F)) = A(g � f) and �2(A(F)) = A(f � g)for A(�) 2 fP(�);AP(�);UR(�);R(�);C(�);!(�);CR(�)g:Finally, we are able to prove that equalities are possible in Theorem 2.2 undersome particular assumptions. We will see this in the next section.3. Chain of inclusions3.1. General properties about the chain of inclusions. Let f : I ! I becontinuous. Then [6] provides that C(f) � !(f) and the inclusions from (1.1)and (1.2) can be rewritten as follows:(3.13) P(f) � AP(f) � UR(f) � R(f) � C(f) � !(f) � (f) � CR(f):Moreover the above inclusions can be strict.Proposition 3.1. There exists a continuous map f0 : I ! I such thatP(f0) 6= AP(f0) 6= UR(f0) 6= R(f0)6= C(f0) 6= !(f0) 6= (f0) 6= CR(f0):Proof. In [14, Theorem 4.6] we can �nd an interval map ef such thatP(ef) 6= AP(ef) 6= UR(ef) = R(ef) 6= C(ef) 6= !(ef) 6= (ef) 6= CR(ef):Then we de�ne a new continuous map byf0(x) = 8<: ef(3x); if x 2 [0; 13];a�ne in [13; 23];f(3x � 2), if x 2 [23;1];where f : I ! I is a continuous map with positive topological entropy (see[1] for de�nition). Then UR(f) 6= R(f) (see [14, Theorem 4.19]) and by anstandard argument (see e.g. [7]) f0 holds the statement. �Here we investigate if (3.13) and Proposition 3.1 are true in the setting ofantitriangular maps. From de�nitions it is clear that(3.14) C(F) � (F);but we are unable to say nothing about the inclusion(3.15) C(F) � !(F):For instance, this inclusion does not work for triangular maps, that is, two{dimensional maps with the form T(x;y) = (f(x);g(x;y)) (see [10]).Clearly, C(F) = R(F) � !(F). However, it is not known if !(F) is closed.This would give us !(F) = !(F), which would prove (3.15). 58 F. Balibrea, J. S. C�anovas and A. LineroIt is well known that C(f) = R(f) = P(f) in the case of interval maps(see [6]). However this is false for triangular maps ([10]). Now in the case ofantitriangular maps we obtain the following result.Theorem 3.2. Let F(x;y) = (g(y);f(x)) be an antitriangular map. ThenC(F) = P(F).Proof. Since P(F) � R(F), it is obvious that P(F) � R(F). In order to provethe converse inclusion, we use [6] and Theorem 2.2 to writeR(F) � R(g � f) � R(f � g) = P(g � f) � P(f � g) = P(F);which ends the proof. �In order to prove more results, we need some additional hypothesis on f. Acontinuous interval map f : I ! I is called a piecewise monotone map whenthere are 0 = a1 < a2 < ::: < an = 1 such that fj[ai;ai+1] is either decreasing orincreasing for 1 � i < n. Then we can prove the following result.Proposition 3.3. Let F(x;y) = (g(y);f(x)) be an antitriangular map suchthat f and g are piecewise monotone maps. Then!(F) � C(F) = !(F):Proof. If f;g are piecewise monotone maps then g�f and f�g are also piecewisemonotone maps. According to [4, Proposition 22, Chapter IV], P(g � f) =!(g � f) and P(f � g) = !(f � g): By Theorem 2.2 and [6]!(F) � !(g � f) � !(f � g) = P(g � f) � P(f � g) == R(g � f) � R(f � g) = R(F) = C(F):On the other hand, since P(F) � !(F); by Theorem 3.2 we obtainC(F) = P(F) � !(F) � !(g � f) � !(f � g)= P(g � f) � P(f � g) = P(F) = C(F): �For antitriangular maps it is possible to �nd an interesting periodic structure,similar to the �Sarkovski��'s ordering (see [13]). It is known that Per(g � f) =Per(f � g) and either Per(F) = } or Per(F) = } [ f2g ; with} = 2 (Per(g � f) n f1g) [ fk 2 Per(g � f) : k odd, k � 1g ;where 2A = f2a : a 2 Ag ; for A � N (see [2]). Then we say that F has typeless, equal or bigger than 21 if the related one-dimensional map g � f has thecorresponding type. Then the following result makes sense.Theorem 3.4. Let F(x;y) = (g(y);f(x)) be an antitriangular map such thatPer(F) � f2n : n 2 N [ f0gg. Assume that P(F) is a closed set. If A(�);B(�) 2fP(�);AP(�);UR(�);R(�);C(�);!(�); (�);CR(�)g it holds(a) A(F) = B(F).(b) A(F) = A(g � f) � A(f � g): Moreover (F2) = (F): Topological dynamics of antitriangular maps 59Proof. If P(F) is closed, according to Theorem 2.2 we �nd P(F) = P(g � f) �P(f � g) = P(g�f)�P(f �g) = P(F); so P(g�f) and P(f �g) are closed, henceP(g�f) = A(g�f); P(f �g) = A(f �g); where A(:) represents one of the othersseven sets ([14, Theorem 4.11]). ThereforeCR(F) = CR(g � f) � CR(f � g) = P(g � f) � P(f � g) = P(F):Now (1.1), (1.2) and Theorem 2.2 end the proof. �3.2. An example of strict inclusions. Now we study if Proposition 3.1 holdsin the case of antitriangular maps. To this end, �rst we prove the followinglemma.Lemma 3.5. Let F(x;y) = (y;f(x)) be an antitriangular map de�ned onI2. Let x 2 I. Then x 2 A(f) if and only if (x;x) 2 A(F) where A(�) 2fP(�);AP(�);UR(�);R(�);C(�);!(�); (�);CR(�)g:Proof. The cases P(�); AP(�), C(�), CR(�) hold by Theorem 2.2 and UR(�); R(�),!(�) follow easily from de�nitions. So, we prove the case (�). First assumethat x 2 (f) and let V � I2 be an open neighborhood of (x;x). Let U � I bean open set such that (x;x) 2 U � U � V . Since x 2 (f), there is a positiveinteger n such that fn(U) \ U 6= ?. ThenF2n(U � U) \ (U � U) = (fn(U) � fn(U)) \ (U � U) 6= ?;which provides (x;x) 2 (F). Second, assume that (x;x) 2 (F) and let U � Ibe an open neighborhood of x. Since (x;x) 2 (F), there is a positive integerm such that Fm(U �U)\(U �U) 6= ?. We have two possibilities: (1) m = 2nfor n 2 N and (2) m = 2n + 1 for n 2 N. If (1) happens, thenF2n(U � U) \ (U � U) = (fn(U) � fn(U)) \ (U � U) 6= ?and fn(U) \ U 6= ?. If (2) happens, thenF2n+1(U � U) \ (U � U) = (fn(U) � fn+1(U)) \ (U � U) 6= ?;and fn(U)\U 6= ? and fn+1(U)\U 6= ?. In both cases x 2 (f), which endsthe proof. �From Lemma 3.5 the following result follows.Theorem 3.6. 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Math.J. 16 (1964), 61-71 (in Russian); English version in \Thirty years after Sharkovski��'s Theorem:New Perspectives", World Scienti�c Series on Nonlinear Science, Series B 8 (1995), 1-11.[14] A. N. Sharkovsky, S. F. Kolyada and A. G. Sivak, and V.V. Fedorenko, Dynamics of One-Dimensional Maps, Mathematics and Its Applications, Volume 407 (Kluwer Academic Publish-ers, 1997). Received October 2000Revised version May 2001 F. BalibreaDepartamento de Matem�aticasUniversidad de Murcia30100 Campus de Espinardo, MurciSpainE-mail address: balibrea@um.es J. S. C�anovas Topological dynamics of antitriangular maps 61Departamento de Matem�atica Aplicada y Estad��sticaUniversidad Polit�ecnica de Cartagena30203 Cartagena, MurciaSpainE-mail address: Jose.Canovas@upct.es A. LineroDepartamento de Matem�aticasUniversidad de Murcia30100 Campus de Espinardo, MurciaSpainE-mail address: lineroba@um.es