@ Appl. Gen. Topol. 20, no. 2 (2019), 325-347 doi:10.4995/agt.2019.11065 c© AGT, UPV, 2019 The function ωf on simple n-ods Ivon Vidal-Escobar and Salvador Garcia-Ferreira Centro de Ciencias Matemáticas, Universidad Nacional Autónoma de México, Campus Morelia, Apartado Postal 61-3, Santa Maŕıa, 58089, Morelia, Michoacán, México. (jpaula@matmor.unam.mx; sgarcia@matmor.unam.mx) Communicated by J. Galindo Abstract Given a discrete dynamical system (X, f), we consider the function ωf -limit set from X to 2 X as ωf (x) = {y ∈ X : there exists a sequence of positive integers n1 < n2 < . . . such that lim k→∞ f nk (x) = y}, for each x ∈ X. In the article [1], A. M. Bruckner and J. Ceder es- tablished several conditions which are equivalent to the continuity of the function ωf where f : [0, 1] → [0, 1] is continuous surjection. It is natural to ask whether or not some results of [1] can be extended to fi- nite graphs. In this direction, we study the function ωf when the phase space is a n-od simple T . We prove that if ωf is a continuous map, then Fix(f2) and Fix(f3) are connected sets. We will provide examples to show that the inverse implication fails when the phase space is a simple triod. However, we will prove that: Theorem A 2. If f : T → T is a continuous function where T is a simple triod, then ωf is a continuous set valued function iff the family {f0, f1, f2, . . . } is equicontinuous. As a consequence of our results concerning the ωf function on the simple triod, we obtain the following characterization of the unit interval. Theorem A 1. Let G be a finite graph. Then G is an arc iff for each continuous function f : G → G the following conditions are equivalent: (1) The function ωf is continuous. (2) The set of all fixed points of f2 is nonempty and connected. 2010 MSC: 54H20; 54E40; 37B45. Keywords: simple triod; equicontinuity; ω-limit set; fixed points; discrete dynamical system. Received 28 November 2018 – Accepted 15 April 2019 http://dx.doi.org/10.4995/agt.2019.11065 I. Vidal-Escobar and S. Garcia-Ferreira 1. Introduction In this article, a continuum is a nonempty compact connected metric space. We shall consider discrete dynamical systems (X,f) where the space X is a continuum and f : X → X is a continuous map. Given a dynamical system (X,f), we define: f0 as the identity map of X, fn = f◦fn−1 for every positive n ∈ N, and the function ωf -limit set from X to 2X as ωf (x) = {y ∈ X : there is a sequence of positive integers n1 < n2 < ... such that lim k→∞ fnk (x) = y}, for each x ∈ X. We remark that ωf (x) = ⋂ m≥0 {fn(x) : n ≥ m} for every x ∈ X. Given ([0, 1],f) a discrete dynamical system, the authors of [1] proved that the following conditions are equivalence: (1) ωf is a continuous function. (2) The set of fixed points of f2, Fix(f2), is connected and nonempty. (3) f is equicontinuous. We wonder whether or not this result can be extended to discrete dynamical systems (G,f), where G is finite graph. We answer this question in negative form, actually, we prove that the unique finite graph that satisfies the equiv- alence (1) ⇔ (2) is the arc. To obtain this result, we start proving some properties that satisfies ωf , when the phase space is a dendroid, a fan and finally a simple triod T, the latter continuum is the union of 3 arcs emanating from a point v such that the intersection of any two of the arcs is v. For a simple triod, we prove that if ωf is a continuous function, then Fix(f 2) and Fix(f3) are connected and nonempty. Moreover, we give examples to show that the connectivity of Fix(f2) and Fix(f3) does not imply the continuity of the map ωf . The proofs of these assertions will be given in the third Section. In the fourth Section, we will prove the equivalence (1) ⇔ (3) when the phase space is a simple triod, T. This assertion requires some properties of Fix(f) when f is a surjective map and ωf is a continuous function. More precisely, we prove that f−1(Fix(f)) = Fix(f) and Fix(f) coincides with one of the following sets: T , the vertex of T and some edge of T . We also show that each point of T is a periodic point of any continuous surjection f : T → T , with period at most 3. In general is interesting to find conditions equivalent to the equicontinuity of a map (the papers [1], [2], [7], and [9] contain results in this direction). 2. Preliminaries Given a discrete dynamical system (X,f). For a point x ∈ X, the orbit of x under f is the set Of (x) = {fn(x) : n ∈ N}; x is said to be a fixed point of f if f(x) = x, x is said to be n-periodic point, if fn(x) = x and fi(x) 6= x for every 1 ≤ i < n with n ≥ 1; x is said to be periodic point if there exists an n ∈ N such that x is an n-periodic point. The sets of fixed points, n-periodic c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 326 The function ωf on simple n-ods points and periodic points of f are denoted by Fix(f), Pern(f) and Per(f), respectively. A function f : X → X is said to be equicontinuous (relative to the metric d) if for each ε > 0, there exists δ > 0 such that d(fn(x),fn(y)) < ε for each x,y ∈ X with d(x,y) < δ and all n ∈ N. A function f : X → X is said to be topological transitivity if for every pair of nonempty open sets U and V in X, there exists n ∈ N such that fn(U) ∩V 6= ∅. For a continuum X we denote the collection of all nonempty compact subsets of X by 2X, we consider 2X equipped with the Hausdorff metric. An arc is a continuum homeomorphic to [0, 1]. Given an arc A and a homeomorphism h : [0, 1] → A, the points h(0) and h(1) are called the end points of A. A simple closed curve is a continuum homeomorphic to a circle. Given a point x ∈ X, x is said to be an end point of X, if for each arc A in X such that x ∈ A, then x is an end point of the arc A. Let n ∈ N ∪{ω}, x is said to be a point of order n in the classical sense, or here briefly a point of order n, ordX(x) = n, if x is a unique common end point of every two of exactly n arcs contained in X. If ordX(x) ≥ 3, then x is said to be a ramification point of X. The sets of end points and ramification points of X are denoted by E(X) and R(X), respectively. A continuum X is said to be arc-wise connected provided that for every two points a,b ∈ X there exists an arc in X with end points a and b. X is said to be unicoherent provided that for every two proper subcontinua A and B of X such that X = A∪B, we have that A∩B is connected. A continuum X is irreducible between a and b if no proper subcontinuum of X contains a and b. Given A, B and C subontinua of X, we said that C is irreducible from A to B if A ∩ C 6= ∅ 6= B ∩ C and no proper subcontinuum of C intersects both A and B. Given a property P, a continuum X is said to be hereditarily P , provided that every non degenerate subcontinuum of X has the property P. A dendroid is a hereditarily arc-wise connected and hereditarily unicoherent continuum. It is well know that a dendroid is unique arc-wise connected. If X is a dendroid we denote by [a,b] the arc in X with end points a and b. A fan is a dendroid with exactly one ramification point. A finite graph is a continuum which can be written as the union of finitely many arcs any two of which are either disjoint or intersect only in one or both of their end points. A tree is finite graph without simple closed curves. Given n ∈ N, n > 2 a simple n−od with vertex v is the union of n arcs emanating from the point v and such that v is the intersection of any two of the arcs, v is called de vertex of the simple n-od. Throughout this paper, T will be denote a simple n-od with vertex v and set of end points E(T) = {e1,e2, . . . ,en}, we consider T with convex metric d. In order to define some specific functions on a simple triod we will consider the special case when T = ([−1, 1] ×{0}) ∪ ({0}× [0, 1]), T is a subspace of the Euclidian plane R2, and its vertex will be v = (0, 0) and its end points will be e1 = (−1, 0), e2 = (1, 0), e3 = (0, 1). The following result is well known and will very useful throughout this work, we present a proof for the convenience of the reader. c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 327 I. Vidal-Escobar and S. Garcia-Ferreira Proposition 2.1. Let X be a dendroid. If F be a closed non-connected subset of X, then there exist points a,b ∈ X such that [a,b] ∩F = {a,b}. Proof. Let A and B components of F, and let Y be the irreducible continuum from A to B. By [8, 11.37(a)] there exist a0 ∈ A and b0 ∈ B such that Y be the irreducible continuum between a0 and b0. Since X is a dendroid, we have that Y = [a0,b0]. If [a0,b0] ∩ F = {a0,b0}, then the proposition follows. Suppose that [a0,b0] ∩ F 6= {a0,b0}. Let p ∈ [a0,b0] − F and let U ⊂ [a0,b0] − F be the component of [a0,b0] − F that contains p. We observe that U = (a,b). By [8, 5.6], we have that F ∩Bd(U) 6= ∅ and then F ∩{a,b} = {a,b}, hence [a,b] ∩F = {a,b}. � The implicit properties of continua that we will use in this article may be found in the book [8]. 3. Connectivity of the set of fixed points of an iterate To start this section, we shall study those discrete dynamical systems (X,f) where X is a fan and the function ωf is continuous. The next, four lemmas will provide the basic information that we need for our purposes. Lemma 3.1. Let (X,f) be a discrete dynamical system where X is a dendroid. If n ∈ N and a,x ∈ X such that [a,x] ∩Fix(fn) = {a} and fn(z) ∈ [a,z) for each z ∈ (a,x], then we have that a ∈ ωf (z) for each z ∈ [a,x]. Proof. Let z ∈ [a,x]. If z = a the result is immediately. Assume that z 6= a, it follows from the hypothesis that fnk(z) ∈ [a,fn(k−1)(z)) for all k ∈ N. Then we have that {fnk(z)}∞k=1 is a monotone sequence and hence this sequence converges. Set c := limk→∞f nk(z) and notice that limk→∞f n(fnk(z)) = c. Then we have that fn(c) = c. Since fnk(z) ∈ [a,x] for each k ∈ N, we must have that c ∈ [a,x] and since [a,x]∩Fix(fn) = {a}, c = a. Thus, a ∈ ωf (z). � Lemma 3.2. Let (X,f) be a discrete dynamical system where X is a dendroid such that ωf is a continuous function. If n ∈ N and a,b,c ∈ X such that b ∈ (a,c), [a,c] ∩ Fix(fn) = {a,c}, fn(b) = a, then ωf (y) = ωf (c) for each y ∈ [a,c]. Proof. Fix an arbitrary point y0 ∈ [a,c]. Since fn(b) = a and fn(c) = c, we have that y0 ∈ [a,c] ⊂ fn([b,c]). Hence, there exists y1 ∈ [b,c] such that fn(y1) = y0. Notice that y1 ∈ (y0,c]; otherwise, y0 ∈ [y1,c] and since fn(b) = a, [b,y1] ∩ Fix(fn) 6= ∅ which contradicts the hypothesis. As y1 ∈ [y0,c] ⊂ fn([y1,c]), then there exists y2 ∈ [y1,c] such that f(y2) = y1, and since y2 ∈ [y1,c] ⊂ fn([y2,c]), we can find y3 ∈ [y2,c] so that f(y3) = y2. By continuing with this process, we may construct a sequence {yk}∞k=1 such that yk+1 ∈ [yk,c] and fn(yk) = yk−1 for each k ∈ N. Since {yk}∞k=1 is a monotone sequence, it converges and ωf (yk) = ωf (y0), for each k ∈ N. Put limk→∞yk = d and notice that limk→∞f n(yk) = d ∈ [b,c]. Then fn(d) = d and since [b,c] ∩ Fix(fn) = {c}, we have that d = c. By the continuity of ωf , we obtain c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 328 The function ωf on simple n-ods that limk→∞ωf (yk) = ωf (c), and as ωf (yk) = ωf (y0) for each k ∈ N, then ωf (y0) = ωf (c). � Lemma 3.3. Let (X,f) be a discrete dynamical system where X is a dendroid such that ωf is a continuous function. If n ∈ N and a,b,c ∈ X such that b ∈ (a,c) and [a,b] ∩Fix(fn) = {a}, fn(b) = c, then ωf (y) = ωf (a) for each y ∈ [a,c]. Proof. By a procedure similar to the one used in the proof of Lemma 3.2, we may construct a sequence {yk}∞k=1 such that limk→∞yk = d exists, yk+1 ∈ [a,yk] and f n(yk) = yk−1 for every k ∈ N. Notice that limk→∞fn(yk) = d ∈ [a,b]. Then fn(d) = d and since [a,c] ∩ Fix(fn) = {a} we have that d = a. Since ωf is a continuous function we have that limk→∞ωf (yk) = ωf (a), and as ωf (yk) = ωf (y0) for each k ∈ N, then ωf (y0) = ωf (a). � Lemma 3.4. Let (X,f) be a discrete dynamical system where X is a dendroid such that ωf is a continuous function. If b,c,d ∈ X such that b 6= d, c ∈ (b,d), f(c) = d, f(d) = b, then ωf (c) ⊂ [c,d]. Proof. As f(c) = d and f(d) = b, then [b,d] ⊂ f([c,d]). Hence, there exist r1,s1 ∈ [c,b] and r1 6= s1 such that f(r1) = c, f(s1) = d and f((r1,s1)) = (c,d). It is easy to proof that r1 6= c and s1 6= d. Following with this procedure, we may construct two sequences {rn}∞n=1 and {sn}∞n=1 such that [rn+1,sn+1] ⊂ [rn,sn], rn 6= rn+1 6= sn, sn+1 6= sn, fn(rn) = c, fn(sn) = b, fn((rn,sn)) = (c,b) for each n ∈ N. Then, we have that ωf (rn) = ωf (c) and ωf (sn) = ωf (b) for each n ∈ N. Without loss of generality, suppose that limn→∞rn = r. Observe that r ∈ ⋂∞ n=1[rn,sn]. Since ωf is a continuous func- tion limn→∞ωf (rn) = ωf (r). On the other hand, since ωf (rn) = ωf (c) for each n ∈ N, ωf (r) = ωf (c). Since fn(r) ∈ fn([rn,sn]) = [c,d] for each n ∈ N, then we obtain that ωf (r) ⊂ [c,d]. Therefore, ωf (c) ⊂ [c,d]. � Theorem 3.5. Let (X,f) be a discrete dynamical system where X is a fan such that ωf is a continuous function. If n ∈ N and a,b ∈ Fix(fn) such that a 6= b and Fix(fn) ∩ [a,b] = {a,b} then ωf (a) = ωf (b). Proof. Let v be the ramification point of X. We shall consider the following cases: Case 1. fn(x) ∈ [a,b] for every x ∈ (a,b). Since Fix(fn) ∩ [a,b] = {a,b}, we have that either fn(x) ∈ [a,x) for every x ∈ (a,b) or fn(x) ∈ (x,b] for every x ∈ (a,b). Without loss of generality suppose that fn(x) ∈ [a,x) for every x ∈ (a,b). It follows from Lemma 3.1 that a ∈ ωf (x), for each x ∈ (a,b). As ωf is a continuous function, we obtain that a ∈ ωf (b). Since fn(b) = b and fn(a) = a, and a ∈ ωf (b) = {f(b), . . . ,fn(b)}, we must have that ωf (b) = ωf (a). Case 2. There exists x ∈ (a,b) such that fn(x) /∈ [a,b]. Choose e1,e2 ∈ E(X) such that [a,b] ⊂ [e1,e2] and a ∈ [e1,v]. Case 2.1. fn(x) ∈ [e1,e2]. Since fn(x) ∈ [e1,e2] − [a,b], without loss of generality, we may assume that fn(x) ∈ [e1,a). Then, it follows from the continuity of fn that there exists c ∈ [a,b] such that fn(c) = a. Thus, by c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 329 I. Vidal-Escobar and S. Garcia-Ferreira Lemma 3.2, we obtain that ωf (y) = ωf (b) for each y ∈ [a,b]. As a consequence, ωf (a) = ωf (b). Case 2.2. fn(x) ∈ X − [e1,e2]. Without loss of generality, we suppose that x ∈ [a,fn(x)]. It follows from Lemma 3.3 that ωf (y) = ωf (a) for each y ∈ [a,fn(x)]. If [a,b] ⊂ [a,v], then ωf (b) = ωf (a) since [a,v] ⊂ [a,fn(x)]. Next, assume that [a,b] 6⊂ [a,v]. Then, b 6= v and so either fn((v,b]) 6⊂ [v,e2] or fn((v,b]) ⊂ [v,e2]. Case 2.2.1. fn((v,b]) 6⊂ [v,e2]. Then there exists y ∈ [v,b] so that fn(y) = v. According to Lemma 3.3, we have that ωf (y) = ωf (b) for each y ∈ [v,b]. Since ωf (v) = ωf (a) and ωf (v) = ωf (b), we obtain that ωf (a) = ωf (b). Case 2.2.2. fn((v,b]) ⊂ [v,e2]. • If fn(y) ∈ (y,b], for each y ∈ [v,b], by Lemma 3.1, then we have that b ∈ ωf (y) for each y ∈ [v,b]. In particular, b ∈ ωf (v) = ωf (a). Therefore, ωf (a) = ωf (b). • There exists c ∈ [v,b] ⊂ [a,b] such that fn(c) = b. By Lemma 3.2, we know that ωf (y) = ωf (a) for each y ∈ [a,b]. Hence, ωf (b) = ωf (a). In both subcases, we conclude that ωf (a) = ωf (b). � In [1] the authors studied the relation between the continuity of ωf and the connectivity of the sets Fix(f) and Fix(f2), where f is a continuous map from [0, 1] to [0, 1] . The Corollary 3.6 is a generalization of [1, Lemma 1.1] when f is a continuous function from a fan to itself. Further the Theorem 3.8 generalizes [1, Theorem 1.2 (1) → (6)], in the case where f is a continuous function from a simple triod to itself. Corollary 3.6. Let (X,f) be a discrete dynamical system where X is a fan such that ωf is a continuous function. Then Fix(f) is connected. Proof. Suppose to the contrary, Fix(f) is not connected. Then, by Proposition 2.1, there are points a,b ∈ Fix(f), a 6= b such that [a,b] ∩Fix(f) = {a,b}. It follows from Theorem 3.5 that ωf (a) = ωf (b), but this is impossible because of ωf (a) = {a} and ωf (b) = {b}. � Remark 3.7. Let (X,f) be a discrete dynamical system where X is a fan. If a,b ∈ Fix(fn) are distinct, for some 1 < n ∈ N, and ωf (b) = ωf (a), then the following statements a 6= f(a) and b 6= f(b). From now on we will consider discrete dynamical systems on a simple triod. Our next task is to analyze the consequences when ωf is a continuous function. Theorem 3.8. Let (T,f) be a discrete dynamical system where T is a simple n-od and such that ωf is a continuous function, then Fix(f 2) is connected. Proof. Let v ∈ T the vertex of T. Suppose to the contrary that Fix(f2) is non- connected. By Proposition 2.1, there exist two points a,b ∈ Fix(f2) a 6= b such that [a,b]∩Fix(f2) = {a,b}. Theorem 3.5 asserts we have that ωf (a) = ωf (b). So, f(a) = b, and f(b) = a. c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 330 The function ωf on simple n-ods Case 1. v /∈ (a,b). Since [a,b] ⊂ f([a,b]) and v /∈ (a,b), then f would have a fixed point on (a,b), and hence (a,b) also would have a fixed point of f2, this is a contradiction. Case 2. v ∈ (a,b). Without loss of generality, we may assume that a ∈ [e1,v] and b ∈ [v,e2]. Notice that [a,b] ⊂ [e1,e2]. Case 2.1. f(v) ∈ [e1,e2]. Suppose that f(v) ∈ (v,e2]. Since [v,f(v)] ⊂ [a,f(v)] ⊂ f([v,b]), we have that [v,b) has a fixed point of f. Therefore [v,b) has a fixed point of f2, this contradicts our supposition. Similarly, we analyze the case when f(v) ∈ (v,e1]. Case 2.2. f(v) /∈ [e1,e2]. Since v ∈ [a,f(v)] = [f(b),f(v)] ⊂ f([v,b]) there exists v1 ∈ [v,b] such that f(v1) = v. Notice that v 6= v1, in other case v ∈ Fix(f2) which is impossible because of Fix(f2) ∩ [a,b] = {a,b}. As v1 ∈ [v,b] ⊂ [b,f(v)] and [v1,b] ∩ Fix(f2) = {b} it follows from Lemma 3.3 that ωf (y) = ωf (b), for each y ∈ [b,f(v)], in particular ωf (v) = ωf (a) = {a,b}. Let ε > 0 such that Bε(b) ⊂ (v1,e2] and Bε(a) ⊂ (v,e1], by continuity there exists 0 < δ < ε such that if d(x,y) < δ then d(f(x),f(y)) < ε. Since ωf (v) = ωf (b) = {a,b} there exists m ∈ N such that fm(v) ∈ Bδ(a). So, fm+1(v) ∈ Bε(b). Since [v,v1] ⊂ [fm(v),fm+1(v)] = [fm+1(v1),fm+1(v)] ⊂ fm+1([v,v1]) we have that [v,v1] ∩Fix(fm+1) 6= ∅. Given x ∈ [v,v1] ∩Fix(fm+1), we obtain that ωf (x) = Of (x). on the other hand as x ∈ [v,v1] ⊂ [b,f(v)], we have that ωf (x) = ωf (b) = {a,b}, then Of (x) = {a,b} which is impossible because of a 6= x 6= b and x ∈Of (x). � The following example show that the converse of Corollary 3.6 and Theorem 3.8 are not true. Therefore, we have that [1, Theorem 1.2 (6) ⇔ (1)] is not satisfied, when f is a continuous function from a simple triod to itself. Example 3.9. We will give a function f from T to itself such that the sets Fix(f) and Fix(f2) are connected but the function ωf is not continuous and Fix(f3) is not connected. In order to have this done, we will assume that T = ([−1, 1] ×{0}) ∪ ({0}× [0, 1]). Define f : T → T given by f((x,y)) =   (2x + 3 2 , 0) (x,y) ∈ [−1,−3 4 ] ×{0}, (3 + 4x, 0) (x,y) ∈ [−3 4 ,−1 2 ] ×{0}, (−2x, 0) (x,y) ∈ [−1 2 , 0] ×{0}, (0, 2x) (x,y) ∈ [0, 1 2 ] ×{0}, (0, 3 − 4x) (x,y) ∈ [ 1 2 ,−3 4 ] ×{0}, (2x− 3 2 , 0) (x,y) ∈ [−3 4 , 1] ×{0}, (−2y, 0) (x,y) ∈{0}× [0, 1 2 ], (3 − 4y, 0) (x,y) ∈{0}× [ 1 2 , 3 4 ], (0, 2y − 3 2 ) (x,y) ∈{0}× [ 3 4 , 1]. c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 331 I. Vidal-Escobar and S. Garcia-Ferreira In the Figure 1 we have the graph of f. Figure 1. In the Figure 2 we have the graph of f2. Figure 2. We will see that Fix(f) and Fix(f2) are connected sets however ωf is not a continuous function. Indeed, it follows from the definition that Fix(f) = {v} = Fix(f2). Further, if A is a subset of T , with T 6= A 6= {v}, then we have that f(A) 6= A. By [5, Theorem 2.2.2 (1) ⇔ (9)], we know that f is a transitive continuous function. Hence, according to [5, Theorem 2.2.2 (1) ⇔ (16)], we have that there exists x0 ∈ T such that Of (x0) = T . So, for every ε > 0, we can find y0 ∈ Of (x0) so that d(y0,v) < ε, but ωf (v) = {v} and ωf (y0) = ωf (x0) = T , which shows that ωf is not a continuous function at v. Finally, its easy to see that Fix(f3) is not a connected set (compare with Theorem 3.11 below). Next let see how Example 3.9 gives a new characterization of the arc in terms of the continuity of ωf and the connectivity of Fix(f 2). Theorem A 1. Let G be a finite graph. Then G is an arc if and only if for each continuous function f : G → G, the following conditions are equivalent: (1) ωf is a continuous function, (2) Fix(f2) is connected and nonempty. c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 332 The function ωf on simple n-ods Proof. Necessity. Suppose that G is an arc and that f : G → G is a continuous function, since an arc has the fixed point property, we have that Fix(f2) is nonempty. In accordance with [1, Theorem 1.2 (1) ↔ (6)], we have that the conditions (1) and (2) are equivalents. Sufficiency. Assume that conditions (1) and (2) are equivalents for each continuous function f : G → G. Suppose that G is not an arc, we need analyze the following two cases: Case 1. G has a simple closed curve. Let S ⊂ G be a simple closed curve. Since G is 1-dimensional, by [3, Theorem VI 4], there is retraction r : G → S and we consider an irrational rotation g : S → S. Then, the function f = g◦r is a continuous and satisfies that ωf (x) = S for each x ∈ G. That is, ωf is a continuous function. On the other hand, since g is irrational rotation we have that Fix(f2) = ∅, this is a contradiction. Case 2. G has not a simple closed curve. In this case, G contains a simple triod T with vertex v. According to [6, Theorem 2.1], there is a retraction r : G → T. If f : T → T is as in Example 3.9, then h = f ◦ r is a continuous function such that Fix(h2) = {v} and ωh is not a continuous function, but this is impossible. In any case we obtain a contradiction. Therefore, G is an arc. � The following theorem shows that the statement of Theorem 3.8 is valid for f3. Lemma 3.10. Let (X,f) be a discrete dynamical system. If a,b ∈ Fix(f3), a 6= b and ωf (a) = ωf (b). Then the following conditions hold: (1) a 6= fi(a), with i ∈{1, 2}, (2) b 6= fi(b), with i ∈{1, 2}. Proof. As ωf (a) = ωf (b), we know that either a = f(b) or a = f 2(b). (1). Suppose that a = f(a). If a = f(b), then a = f2(a) = f3(b) = b, but this is a contradiction. Now, if a = f2(b), then a = f(a) = f3(b) = b and so we obtain a contradiction. Assume that f2(a) = a. If a = f(b), then a = f2(a) = f3(b) = b, which is impossible. If a = f2(b), then f(a) = f3(b) = b and a = f2(a) = f(b), then b = f2(b), hence a = b and we have a contradiction. Clause (2) is established in a similar way. � The function f : [0, 1] → [0, 1], define by f(x) = 1 − x for each x ∈ [0, 1], witnesses that Lemma 3.10 does not work for n = 4. Theorem 3.11. Let (T,f) be a discrete dynamical system where T is a simple n-od and such that ωf is a continuous function, then Fix(f 3) is connected. Proof. Let v ∈ T the vertex of T. Suppose that Fix(f3) is not connected. By Proposition 2.1, there are points a,b ∈ Fix(f3) such that [a,b] ∩ Fix(f3) = {a,b}. As a consequence of Theorem 3.5, we obtain that ωf (a) = ωf (b). By Lemma 3.10, we can suppose that ωf (b) = {b,f(b),a}. Case 1. There exists an arc A ⊂ T such that ωf (b) ⊂ A. c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 333 I. Vidal-Escobar and S. Garcia-Ferreira Case 1.1. a ∈ (b,f(b)). Being that f(a) = b and f(b) = f(b), by Lemma 3.4, we obtain that ωf (a) ⊂ [b,a], which contradicts f(b) 6∈ [b,a]. Case 1.2. f(b) ∈ (a,b). Since f(f(b)) = a and f(a) = b, we obtain of Lemma 3.4 that ωf (f(b)) ⊂ [a,f(b)], this is impossible because of b 6∈ [a,f(b)]. Case 1.3. b ∈ (a,f(b)). As f(b) = f(b) and f(f(b)) = a, by Lemma 3.4 we have that ωf (b) ⊂ [b,f(b)], this is a contradiction to a 6∈ [b,f(b)]. Case 2. ωf (b) 6⊂ A for any arc A ⊂ T. Without loss of generality, suppose that a ∈ [v,e1], b ∈ [v,e2] and f(b) ∈ [v,e3]. Case 2.1. f(v) ∈ [e1,e2]. Assume that f(v) ∈ (v,e2]. As [f(v),v] ⊂ [f(v),f(b)] ⊂ f([v,b]), we must have that [v,b) has a fixed point of f; hence, (a,b) has a fixed point of f3 which is a contradiction. Similarly, we obtain a contradiction if f(v) ∈ (v,e1]. Case 2.2. f(v) /∈ [e1,e2]. Case 2.2.1. f(v) ∈ (v,e3]. Since [v,f(v)] ⊂ [a,f(v)] = [f(f(b)),f(v)] ⊂ f([v,f(b)]), we have that [v,f(b)] has a fixed point of f. Let z ∈ [v,f(b)] be a fixed point of f. As [v,f(b)] ⊂ [f(v),b] = [f(v),f(a)] ⊂ f([a,v]), there exits y ∈ [a,v] for which f(y) = z. Then, ωf (y) = {z}. On the other hand, as v ∈ [a,f(v)] = [f(f(b)),f(v)] ⊂ f([f(b),v]) there is v1 ∈ (v,f(b)) such that f(v1) = v. As v1 ∈ [b,f(b)] = [f(a),f(b)] ⊂ f([a,b]) there is v2 ∈ (a,b) such that f(v2) = v1 and so f 3(v2) = f(v). As [a,b]∩Fix(f3) = {a,b}, if v2 ∈ [a,v], then [a,v2] ∩Fix(f3) = {a} and v2 ∈ [a,f(v)] it follows from Lemma 3.3 that ωf (x) = ωf (a) for each x ∈ [a,f(v)] in particular ωf (y) = ωf (a). Hence, ωf (a) = {z} but this is impossible because of ωf (a) = {b,f(b),a}. Similarly in the case that v2 ∈ [v,b] we obtain a contradiction. Case 2.2.2. f(v) /∈ (v,e3]. Since v ∈ [f(b),f(v)] ⊂ f([v,b]) there exists v1 ∈ [v,b] such that f(v1) = v. Since [v,b] ∩ Fix(f) = ∅ we have that v 6= v1. As v1 ∈ [v,b] ⊂ [f(v),b] and [v1,b] ∩ Fix(f2) = {b} it follows from Lemma 3.3 that ωf (y) = ωf (b), for each y ∈ [b,f(v)]. Thus, ωf (v) = ωf (b) = {b,f(b),a}. Let ε > 0 such that Bε(b) ⊂ (v1,e2] and Bε(a) ⊂ (v,e1], since f is a continuous function there exists 0 < δ < ε such that if d(x,y) < δ then d(f(x),f(y)) < ε. Since ωf (v) = ωf (b) = {b,f(b),a} there exists m ∈ N such that fm(v) ∈ Bδ(a). Hence, as f(a) = b we have that fm+1(v) ∈ Bε(b). On the other hand [v,v1] ⊂ [fm(v),fm+1(v)] = [fm+1(v1),fm+1(v)] ⊂ fm+1([v,v1]) we have that [v,v1]∩Fix(fm+1) 6= ∅. Let x ∈ [v,v1]∩Fix(fm+1). We have that ωf (x) = Of (x). As x ∈ [v,v1] ⊂ [b,f(v)], we obtain that ωf (x) = ωf (b) = {b,f(b),a} which is not possible because of a 6= x 6= b, f(b) 6= x and x ∈Of (x) � The following example shows that the converse of Theorem 3.11 is not true. Example 3.12. We will give a continuous function f from T to itself such that the sets Fix(f) and Fix(f3) are connected but the function ωf is not continuous and Fix(f2) is not connected. Define f : T → T given by c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 334 The function ωf on simple n-ods f((x,y)) =   (−3 2 (x + 1 3 ), 0) (x,y) ∈ [−1,−1 3 ] ×{0}, (0, 1 3 + x) (x,y) ∈ [−1 3 , 0] ×{0}, (0, 1 3 −x) (x,y) ∈ [0, 1 3 ] ×{0}, ( 3 2 ( 1 3 −x), 0) (x,y) ∈ [ 1 3 , 1] ×{0}, (0, 1 3 ) (x,y) ∈{0}× [0, 1]. The graph of f, f2 and f3 appear in the Figure 3. Figure 3. It follows from the definition that Fix(f) = {(0, 1 3 )} = Fix(f3) which are both connected. Since Fix(f2) = {(0, 1 3 ),e1,e2}, then Fix(f2) cannot be con- nected. Hence, by Theorem 3.8 we conclude that the function ωf is not con- tinuous. By a procedure similar to the one used in the construction of Example 3.9, we can define a function f from a simple 4-od to itself, for which the sets Fix(f), Fix(f2) and Fix(f3) are connected, but Fix(f4) is not connected and ωf is not continuous. Question 3.13. Let T be a simple n-od and f : T → T be a continuous map. If there is m ∈ N such that Fix(fn) is connected, for all n > m, must ωf be a continuous map? In relation with Theorem 3.8 and Theorem 3.11 we have the following ques- tions. c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 335 I. Vidal-Escobar and S. Garcia-Ferreira Question 3.14. Let T be a simple n-od and let f : T → T be a continuous map. If ωf is a continuous map, must Fix(f m) be connected for all m ≥ 4? Question 3.15. Let X be a fan and let f : X → X be a continuous map. If ωf is a continuous map, must Fix(f m) be connected for all m ≥ 2? In connection with Question 3.13, the following example shows that, there is a fan X and a continuous function f : X → X that satisfy: Fix(fn) is a connected set, for all n ∈ N and ωf is not continuous. Example 3.16. Let L be the segment of the line that join (0, 0) with (1, 0) and for each n ∈ N let Ln be the segment of the line that join (0, 0) with (1, 1n). Define X = ⋃∞ n=1 Ln ∪ L. For each n ∈ N define fn : [0, 1] → Ln given by fn(t) = (t, t n ), fn is an homeomorphism. Define f : X → X given by f((x,y)) = { fn( n n+1 (f−1n (x,y))) (x,y) ∈ Ln, (x,y) (x,y) ∈ L. Figure 4. Then, we have that ωf ((x,y)) = { (0, 0) (x,y) ∈ Ln, (x,y) (x,y) ∈ L. Hence, ωf is not continuous. On the other hand Fix(f n) = L, so Fix(fn) is connected, for each n ∈ N. 4. Equicontinuity of functions on the simple triod Now, we address our attention to discrete dynamical systems on a simple triod. Through this section T will be denote a simple triod with vertex v and set of end points E(T) = {e1,e2,e3}, we consider T with convex metric d . Mainly we prove that ωf is a continuous function if and only if f is equicontinuous. To have this done we need the following preliminary results. Lemma 4.1. If (T,f) is a discrete dynamical system such that f is a surjective map and ωf is a continuous function. If x0 ∈ T −Fix(f) satisfies that f(x0) ∈ Fix(f), then the following conditions hold: c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 336 The function ωf on simple n-ods (1) There exists a sequence {xn}∞n=1 with the following properties: (a) xn ∈ T −Fix(f) for each n ∈ N, (b) xi 6= xj for each i,j ∈ N such that i 6= j, (c) f(xn) = xn−1 for each n ∈ N, and (d) limn→∞xn = f(x0). (2) f(x0) ∈ Fr(Fix(f)), (3) [f(x0),x0] ∩Fix(f) = {f(x0)}. Even more, when Fix(f) = {f(x0)} we have stronger properties: (4) If f(x0) ∈ (v,ei) for some i{1, 2, 3}, then there exists a strictly growing sequence {kn}∞n=1, when kn ∈ N, that satisfies the following conditions: (i) If n is odd xkn ∈ (ei,f(x0)) and xkj+1 ∈ (ei,xkj ) for each j ∈ {kn, . . . ,kn+1 − 2}, (ii) if n is even xkn ∈ (f(x0),x0) and xkj+1 ∈ (xkj,x0) for all j ∈ {kn, . . . ,kn+1 − 2}. (5) If f(x0) = v and the sequence {xn}∞n=1 satisfies that {xn : n ∈ N}∩ (v,ei] is infinite, for each i ∈{1, 2, 3}, then there are strictly increasing sequences {kn}∞n=1 and {ln}∞n=1 that satisfy kn ≤ ln < kn+1, for every n ∈ N, and the following conditions: (i) If n ∈ N∪{0} and i ∈{1, 2, 3}, then xk3n+i ∈ (ei,v), (ii) If n ∈ N, i ∈ {1, 2, 3}, ln > kn and xkn ∈ (ei,v), then xkj+1 ∈ (xkj,ei], for each j ∈{kn, . . . , ln − 1}, (iii) If n ∈ N and i ∈ {1, 2, 3}, then xl3n+i ∈ (v,ei) and xl3n+i+1 /∈ (v,ei). Proof. (1). Since f is surjective there exists x1 ∈ T such that f(x1) = x0, notice that x1 6= x0 and x1 ∈ T − Fix(f). Assume that we constructed x1,x2, . . . ,xn−1 that satisfy (a) − (c). Being as f is surjective there exists xn ∈ T such that f(xn) = xn−1, it is clear that xn satisfy (a) − (c). Thus, we construct our sequence {xn}∞n=1. It follows from (c) that fn+1(xn) = f(x0) for each n ∈ N. Hence, ωf (xn) = {f(x0)} for each n ∈ N. Now, we procedure to prove (d). Without loss of generality, suppose that limn→∞xn = z, then f(z) = limn→∞f(xn) = limn→∞xn−1 = z. On the other hand, since ωf is a continuous function, we have that limn→∞ωf (xn) = {f(x0)} for each n ∈ N, then ωf (z) = {f(x0)}. Then, z = f(x0) and (d) is proved. (2). By (a) and (d), we have that xn ∈ T − Fix(f) for each n ∈ N, and limn→∞xn = f(x0), further, by hypothesis f(x0) ∈ Fix(f). Therefore, f(x0) ∈ Fr(Fix(f)) and this shows (2). To prove (3) suppose that [f(x0),x0] ∩ Fix(f) 6= {f(x0)}. Since Fix(f) is connected, there is z ∈ Fix(f) −{f(x0)}, for which [f(x0),x0] ∩Fix(f) = [f(x0),z]. Let y ∈ (f(x0),z) ⊂ Fix(f). Since [f(x0),z] ⊂ f([z,x0]), there exists x ∈ (z,x0) such that f(x) = y and x /∈ Fix(f). According to (d) and (2), we have that there exists a sequence {yn}∞n=1 such that limn→∞yn = u, and y ∈ Fr(fix(f)), but this is a contradiction because of u ∈ (f(x0),z). So, (3) is proved. c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 337 I. Vidal-Escobar and S. Garcia-Ferreira (4) Without loss of generality, we assume that f(x0) ∈ (e1,v). Notice that [e1,f(x0)) ∩Fix(f) = ∅ = (f(x0),v] ∩Fix(f). Hence, we must have that • f(y) ∈ (y,v] ∪ [v,e2] ∪ [v,e3] for each y ∈ [e1,f(x0)) and • f(y) ∈ [e1,y), for each y ∈ (f(x0),v]. Let ε > 0 such that Bε(f(x0)) ⊂ (e1,v). Since limn→∞xn = f(x0), there is m ∈ N that satisfies that xn ∈ Bε(f(x0)) for each n > m. Fix k1 > m so that xk1 ∈ (e1,f(x0)). Since f(xk1+1) = xk1 we have that xk1+1 ∈ [e1,xk1 ) ∪ (f(x0),x0). If xk1+1 ∈ (f(x0),x0), then put k2 = k1 + 1. If not, then xk1+1 ∈ [e1,xk1 ), and since limn→∞xn = f(x0), there is k2 > k1 such that xk2 ∈ (f(x0),v) and xkj+1 ∈ (e1,xkj ) for each j ∈ {k1, . . . ,k2 − 2}. As f(xk2+1) = xk2 , we must have that xk2+1 ∈ [e1,f(x0)) ∪ (xk2,x0). If xk2+1 ∈ [e1,f(x0)), then we define k3 = k2 + 1. Assume that xk2+1 ∈ [e1,f(x0)). Since limn→∞xn = f(x0), there exists k3 > k1 such that xk3 ∈ (e1,f(x0)) and xkj+1 ∈ (xkj,x0) for all j ∈{k2, . . . ,k3 −2}. By following with this process we obtain our desired sequence. (5) Without loss of generality, we suppose that x0 ∈ [e1,v). Since Fix(f) = {v}, we have that f(y) ∈ (y,v]∪[v,e2]∪[v,e3] for each y ∈ (x0,v). Choose ε > 0 so that Bε(v) is connected and Bε(v) ⊂ T −{x0}. Since limn→∞xn = f(x0), there exist m ∈ N such that xn ∈ Bε(f(x0)) for each n > m. Fix k1 > m such that xk1 ∈ (x0,v). Since f(xk1+1) = xk1 , we obtain that xk1+1 ∈ (x0,xk1 ) ∪ (v,e2] ∪ (v,e3]. If xk1+1 ∈ (v,e2], then put l1 = k1 and k2 = k1 + 1. Then assume that xk1+1 ∈ (x0,xk1 ). Since limn→∞xn = f(x0), there is l1 > k1 so that xkj+1 ∈ (xkj,x0) ⊂ (xkj,e1] and xl1+1 /∈ (v,e1) for j ∈ {k1, . . . , l1 − 1}. If xl1+1 ∈ (v,e2], we set k2 = l1 + 1. Suppose that xl1+1 ∈ (v,e3]. Since {xn : n ∈ N}∩ (v,e2] is infinite and limn→∞xn = v, there is k2 > l1 + 1 such that xk2 ∈ (v,e2) and xj /∈ (v,e2) for each j ∈ {l1 + 1, . . . ,k2−1}. Thus, f(xk2+1) = xk2 and then xk2+1 ∈ (xk2,e2)∪(v,e1]∪(v,e3]. We have arrived to the conditions from the beginning. By this way we define our required sequences {kn}∞n=1 and {ln}∞n=1. � Theorem 4.2. Let (T,f) be a discrete dynamical system such that f is a surjective map and ωf is a continuous function. If Fix(f) 6= T , then f(T − Fix(f)) ⊂ T −Fix(f). Proof. Suppose that there is x0 ∈ T − Fix(f) such that f(x0) ∈ Fix(f). Without of generality, suppose that x0 ∈ [e1,v]. By (1) of Lemma 4.1, we obtain that there exists a sequence {xn}∞n=1 with the following properties: (a) xn ∈ T −Fix(f) for each n ∈ N, (b) xi 6= xj for each i,j ∈ N such that i 6= j, (c) f(xn) = xn−1 for each n ∈ N, (d) limn→∞xn = f(x0). We consider the following cases: Case 1. Either Fix(f) is not degenerate or Fix(f) ∈ E(T). Since ωf is a continuous function, by Corollary 3.6, we have that Fix(f) is connected and, then we have that either Fix(f) ⊂ [e1,x0) or Fix(f) ⊂ (x0,v]∪ [v,e2]∪ [v,e3]. c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 338 The function ωf on simple n-ods Case 1.1. Fix(f) ⊂ [e1,x0). Suppose that a,b ∈ [e1,x0), such that b ∈ [a,x0) and Fix(f) = [a,b]. By (2) and (3) of Lemma 4.1 we have that f(x0) ∈ Fr(Fix(f)) and [f(x0),x0] ∩ Fix(f) = {f(x0)}, then b = f(x0). Let ε > 0 such that Bε(f(x0)) ⊂ (a,x0). Since limn→∞xn = f(x0), then there exists k ∈ N such that xn ∈ Bε(f(x0)) ∩ (f(x0),x0) for each n > k, and there exists l > k such that xl+1 ∈ (f(x0),xl). Since [b,xl] ⊂ f([xl+1,x0]), [xl+1,x0] has a fixed point of f, but this is a contradiction. Case 1.2. Fix(f) ⊂ (x0,v] ∪ [v,e2] ∪ [v,e3]. Choose ε > 0 such that Bε(f(x0)) ∩ [e1,f(x0)] ⊂ (x0,f(x0)]. Since limn→∞xn = f(x0), we can find k ∈ N such that xn ∈ Bε(f(x0))∩ [x0,f(x0)] for each n > k, and there is l > k such that xl+1 ∈ (xl,f(x0)). Since [xl,f(x0)] ⊂ f([x0,xl+1]), we obtain that [x0,xl+1] has a fixed point of f, but this is impossible. Case 2. Fix(f) = {f(x0)}, f(x0) /∈ E(T) and f(x0) 6= v. Suppose that f(x0) ∈ (ei,v) for some i ∈ {1, 2, 3}, we obtain from (4) of Lemma 4.1 that there exists a strictly increasing sequence {kn}∞n=1, when kn ∈ N, that satisfies the following conditions: (i) If n is odd xkn ∈ (ei,f(x0)) and xkj+1 ∈ (ei,xkj ) for each j ∈{kn, . . . ,kn+1 − 2}, (ii) if n is even xkn ∈ (f(x0),x0) and xkj+1 ∈ (xkj,x0) for all j ∈{kn, . . . ,kn+1 − 2}. Pick an odd integer m so that xkm ∈ [ei,f(x0)) and xkm+2 ∈ (ei,xkm ). For definition, we have that f(xkm+1 ) = xkm+1−1 and xkm+1−1 ∈ (ei,xkm ) ⊂ [e1,f(x0)). Then [xkm+1−1,f(x0)] ⊂ f([f(x0),xkm+1 ]). On the other hand, since xkm+2 ∈ (ei,xkm ) and xkm+1−1 ∈ (ei,xkm ) ⊂ [ei,f(x0)), then xkm+2 ∈ [xkm+1−1,f(x0)] ⊂ f([f(x0),xkm+1 ]). Hence, there is z ∈ [f(x0),xkm+1 ] such that f(z) = xkm+2 . Observe that f 2(z) = f(xkm+2 ) = xkm+2−1. Since [f(x0),xkm+2−1] ⊂ f2([z,x0]), we must have that f2([z,x0]) has a fixed point of f2. That is, there is y ∈ [z,x0], such that f2(y) = y. By Theorem 3.8, we have that Fix(f2) is connected and, so [f(x0),y] ⊂ Fix(f2). Hence, f2(z) = z which is impossible because of f2(z) = xkm+2−1. Case 3. Fix(f) = {f(x0)} and f(x0) = v. If there exist i,j ∈{1, 2, 3} and k ∈ N such that xn ∈ [ei,ej] for each n > k, then the proof follows as in the Case 2. Thus, we may suppose that {xn : n ∈ N,n > K}∩ (v,ei] is infinite for each i ∈ {1, 2, 3}. It follows from (5) of Lemma 4.1 that there are strictly increasing sequences {kn}∞n=1 and {ln}∞n=1, that satisfy kn ≤ ln < kn+1, for every n ∈ N, and the following conditions: (i) for each n ∈ N∪{0} and i ∈{1, 2, 3}, xk3n+i ∈ (ei,v), (ii) for each n ∈ N and i ∈ {1, 2, 3}, if ln > kn and xkn ∈ (ei,v), then xkj+1 ∈ (xkj,ei], for each j ∈{kn, . . . , ln − 1}, (iii) for each n ∈ N and i ∈{1, 2, 3}, xl3n+i ∈ (v,ei) and xl3n+i+1 /∈ (v,ei). If there exists r ∈ N, such that lr + 1 6= kr+1 the proof follows as in the Case 2. Suppose that ln + 1 = kn+1 for each n ∈ N. Let m ∈ N such that xkm ∈ [x0,v], xkm+3 ∈ (xkm,v). Then we have that xkm+2 ∈ [e3,v) and xkm+1 ∈ [e2,v). By definition we obtain that f(xkm+3 ) = xkm+3−1, c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 339 I. Vidal-Escobar and S. Garcia-Ferreira then [xkm+3−1,v]) ⊂ f([xkm+3,v], by (ii) xkm+3 ∈ (xkm+3−1,v) ⊂ f([xkm+3,v]. Then there is z ∈ [xkm+1,v] such that f(z) = xkm+2 . Moreover f(xkm+2 ) = xkm+2−1 then [xkm+2−1,v] ⊂ f([xkm+2,v]) ⊂ f2([z,v]). By (ii) we have that xkm+1 ∈ (xkm+2−1,v), then there exists w ∈ [z,v] such that f2(w) = xkm+1 . So [xkm+1−1,v] ⊂ f([xkm+1,v]) ⊂ f3([w,v]), since f(xkm+1 ) = xkm+1−1. By (ii), we obtain that xk1 ∈ (xkm+1−1,v) and xkm+3 ∈ (xk1,v). Then, there exists y ∈ [w,v] such that f3(y) = xkm+3 . Hence, we obtain that [v,xkm+3 ] ⊂ f3([x0,y]). Thus, [x0,y] has a fixed point of f 3, says y′ ∈ [x0,y]. By Theorem 3.11, we know that Fix(f3) is connected and, then [y′,v] ⊂ Fix(f3), which implies that f3(y) = y, but this contradicts the equality f3(y) = xkm+3 . � By the Corollary 3.6, we know that Fix(f) is connected, when ωf is a continuous map and f is a surjective map of a simple triod to itself. So, it can be a point, an arc or a simple triod. The following result limits these possibilities and clarifies that set it is. Theorem 4.3. Let (T,f) be a discrete dynamical system such that f is a surjective map and ωf is a continuous function. Then, one of the following conditions holds: (1) Fix(f) = T , (2) Fix(f) = {v}, or (3) There exists i ∈{1, 2, 3}, such that Fix(T) = [v,ei]. Proof. Suppose that neither of the conditions, (1)-(3), is true. Since ωf is a continuous function, we obtain from Corollary 3.6 that Fix(f) is connected. We consider the following cases: Case 1. v ∈ Fix(f). By assumption, for each i ∈{1, 2, 3} there is ci ∈ [v,ei] so that Fix(f) = [c1,v]∪[v,c2]∪[v,c3]. As (2) and (3) fail, we have that ci 6= v and cj 6= ej for some i,j ∈ {1, 2, 3}. Suppose, without loss of generality, that e1 6= c1. Since f is a surjective map, we can find e ∈ T such that f(e) = e1. We may assume that e /∈ (e1,c1); otherwise, (e1,c1) would have a fixed point of f. Suppose, without loss of generality, that e ∈ (c2,e2]. Case 1.1. c1 6= v. Since [e1,c1] ⊂ [e1,v] ⊂ f([v,e]), there exists c ∈ [v,e] such that f(c) = c1. As c ∈ (v,e] ⊂ (v,e2], then c /∈ Fix(f). It follows from Theorem 4.2 that f(c) /∈ Fix(f). This contradicts the fact f(c) = c1 ∈ Fix(f). Case 1.2. c2 6= v. As v ∈ [e1,c2] ⊂ f([c2,e]), there is c ∈ [c2,e] such that f(c) = v. Since c ∈ [c2,e] and c2 6= v, we have that c 6= v. Hence, c /∈ Fix(f), but this contradicts Theorem 4.2. Case 1.3. c1 = v = c2. Since neither (2) nor (3) hold, we obtain that v 6= c3 6= e3. As f is a surjective function there exits d ∈ T such that f(d) = e3. Then we proceed as in Case 1.1. Thus, this case is impossible. Case 2. v /∈ Fix(f). Suppose, without of generality, that Fix(f) ⊂ [e1,v). Pick a,b ∈ [e1,v) such that b ∈ [a,v) and Fix(f) = [a,b]. Notice that f(v) ∈ (b,v); in other case, f(v) ∈ [e2,e3] and then [e2,e3] has a fixed point of f, but this impossible because of Fix(f) ⊂ [e1,v). Since f(v) ∈ (b,v), f(y) ∈ (b,y) for each y ∈ (b,v], we obtain from Lemma 3.1 that b ∈ ωf (v). c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 340 The function ωf on simple n-ods Case 2.1. a = e1. As f is a surjective map, there exists di ∈ T such that f(di) = ei for each i ∈ {2, 3}. Note that di /∈ (v,ei); otherwise, (v,ei) has a fixed point of f. Thus, d2 ∈ (v,e3) and d3 ∈ (v,e2). Since [e2,v] ⊂ f([v,d2]), we can choose c2 ∈ [v,d2] for which f(c3) = d3. So f2(c3) = e3 and, then (v,e3] has a fixed point of f 2, says y ∈ (v,e3]. We know from Theorem 3.5 that Fix(f2) is connected, and so [b,y] ⊂ Fix(f2) which implies that v ∈ Fix(f2). Therefore, ωf (v) = {v,f(v)}. As b ∈ ωf (v), we have that f(v) = b, but this contradicts Theorem 4.2 since v /∈ Fix(f). Case 2.2. a 6= e1. Since f is a surjective function, there is e ∈ T so that f(e) = e1. Notice that e /∈ [e1,a); if not, [e1,a) has a fixed point of f. Remember that f(y) ∈ (b,y) for each y ∈ (b,v]. Hence, we have that e ∈ [e2,e3]. As [e1,f(v)] ⊂ f([v,e]), there exists c ∈ [v,e] such that f(c) = a. Since c ∈ [v,e] ⊂ [e2,e3], c /∈ [a,b] ⊂ [e1,v). But this is impossible by the Theorem 4.2. In each case we obtain a contradiction. Thus, we obtain that one of the conditions either (1), (2) or (3) holds. � Proposition 4.4. Let (T,f) be a discrete dynamical system such that f is a surjective map and ωf is a continuous function. Then for each i ∈ {1, 2, 3} there exists j ∈{1, 2, 3} such that f([v,ei]) = [v,ej]. Proof. Fix i ∈{1, 2, 3} and assume that f([v,ei]) 6= [v,ej] for each j ∈{1, 2, 3}. It follows from Theorem 4.3 that v ∈ Fix(f). We consider the following cases. Case 1. There exist distinct j,k ∈{1, 2, 3} so that f([v,ei])∩(v,ej] 6= ∅ 6= f([v,ei]) ∩ (v,ek]. It follows from assumption that there exists c,d ∈ (v,ei] such that f(c) ∈ (v,ej] and f(d) ∈ (v,ek]. Then, v ∈ f([c,e]) ⊂ f((v,ei]) and, hence there is v′ ∈ [c,e] such that f(v′) = v. By Theorem 4.2, we have that v = f(v′) /∈ Fix(f), which is a contradiction. Case 2. There exists j ∈ {1, 2, 3} such that f([v,ei]) ⊂ [v,ej). As f is a surjective function there exists e ∈ T such that f(e) = ej. Since f([v,ei]) ⊂ [v,ej), e /∈ [v,ei]. Suppose that e ∈ [v,ek] for k ∈{1, 2, 3}\{i}. • If f([v,ek]) 6= [v,ej]. Then, there exits c ∈ (v,ek] such that f(c) /∈ (v,ej]. Assume that f(c) ∈ [v,el], where l 6= j. Then, f([v,ek]) ∩ (v,ej] 6= ∅ 6= f([v,ek]) ∩ (v,el]. • Suppose that f([v,ek]) = [v,ej]. Then, f([v,ei])∪f([v,ek]) = [v,ej]. First, if i 6= j 6= k, then [v,ek] ∪ [v,ei] ⊂ f([v,ej]) and so f([v,ej]) ∩ (v,ek] 6= ∅ 6= f([v,ej]) ∩ (v,ei]. Now, if i = j and l ∈{1, 2, 3}\{i,k}, then [v,ek] ∪ [v,el] ⊂ f([v,el]) and hence f([v,el]) ∩ (v,ek] 6= ∅ 6= f([v,el]) ∩ (v,el]. Finally if j = k and l ∈ {1, 2, 3} \ {i,k}, then [v,ei] ∪ [v,el] ⊂ f([v,el]) which implies that f([v,el]) ∩ (v,ei] 6= ∅ 6= f([v,el]) ∩ (v,el]. In each one of the previous case, we arrived to the conditions of Case 1. So we can obtain a contradiction. � Theorem 4.5. Let (T,f) be a discrete dynamical system such that f is a surjective map and ωf is a continuous function, then f|E(T) is a permutation. c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 341 I. Vidal-Escobar and S. Garcia-Ferreira Proof. Suppose that Fix(f) 6= T. By Proposition 4.4, we have that for each i ∈{1, 2, 3} there exists ji ∈{1, 2, 3} such that f([v,ei]) = [v,eji ]. By Theorem 4.3 we need consider the following cases. Case 1. There exists i ∈ {1, 2, 3} satisfying Fix(f) = [v,ei]. Assume, without of generality, that i = 1. Then, we have that f([v,e2]) = [v,e3] and f([v,e3]) = [v,e2]. We will prove, that f(e2) = e3 and f(e3) = e2. Suppose that f(e2) 6= e3, since f([v,e2]) = [v,e3], we have that there exists b2 ∈ (v,e2) such that f(b2) = e3. Case 1.1. f(e3) = e2. Since f(b2) = e3 we have that [v,e2] ⊂ f2([v,b2]). Then, (b2,e2] has a fixed point of f 2, says c ∈ (b2,e2]. By Theorem 3.8, we have that Fix(f2) is connected. Thus, [v,c] ⊂ Fix(f2) for which f2(b2) = b2, this is a contradiction, because of f2(b2) = e2. Case 1.2. f(e3) 6= e2. As f([v,e2]) = [v,e3] we have that, there exists b3 ∈ (v,e3) such that f(b3) = e2. Since [v,e3] ⊂ f([v,b2]), there is c2 ∈ [v,b2] such that f(c2) = b3. Then [v,e2] ⊂ f2([v,b3]) = and so f((c2,e2]) has a fixed point of f2, says c ∈ (c2,e2]. We know from the Theorem 3.8 that Fix(f2) is connected. Hence, [v,c] ⊂ Fix(f2) and then f2(c2) = c2. This contradicts the fact f2(c2) = e2. Case 2. Fix(f) = {v}. By Proposition 4.4, we can assume, without of generality, that f([v,e1]) = [v,e2], f([v,e2]) = [v,e3], f([v,e3]) = [v,e1]. We will show, that f(e1) = e2, f(e2) = e3 and f(e3) = e1. Suppose that f(e1) 6= e2. Since f([v,e1]) = [v,e2], there exists b1 ∈ (v,e1) that satisfies f(b1) = e2. Case 2.1. f(e2) = e3 and f(e3) = e1. Since f(b1) = e2, we obtain that [v,e1] = f([v,e3]) = f 2([v,e2]) = f 3([v,b1]). Hence, (b1,e1] has a fixed point of f3, says c ∈ (b1,e1]. It follows from Theorem 3.11 that Fix(f3) is connected. Thus, [v,c] ⊂ Fix(f3) for which f3(b1) = b1, this is a contradiction, because of f3(b1) = e1. Case 2.2. f(e2) 6= e3 and f(e3) = e1. By assumption f([v,e2]) = [v,e3], for which, there is b2 ∈ [v,e2] such that f(b2) = e3. By other hand as [v,e2] = f([v,b1]), we can find c1 ∈ [v,e1] such that f(c1) = b2. Hence, [v,e1] = f([v,e3]) = f 2([v,b2]) ⊂ f3([v,c1]) and then (c1,e1] has a fixed point of f3, says c ∈ (c1,e1]. We know from the Theorem 3.11 that Fix(f3) is connected. Hence, [v,c] ⊂ Fix(f3) and so f3(c1) = c1. This contradicts the fact f3(c1) = e1. Case 2.3. f(e2) = e3 and f(e3) 6= e1. Since f([v,e3]) = [v,e1]. So, we can choose b3 ∈ [v,e3] such that f(b3) = e1. By other hand as f(b1) = e2, then [v,e3] = f([v,e2]) = f 2([v,b1]), and so, we can pick c1 ∈ [v,b1] such that f2(c1) = b3. Hence, [v,e1] = f([v,b3]) ⊂ f3([v,c1]) (c1,e1] has a fixed point of f3, says c ∈ (c1,e1]. By Theorem 3.11, Fix(f3) is connected. Thus, [v,c] ⊂ Fix(f3) and then f3(c1) = c1, but this is impossible since f3(c1) = e1. Case 2.4. f(e2) 6= e3 and f(e3) 6= e1. By supposition f([v,e2]) = [v,e3] and f([v,e3]) = [v,e1], then we can find b2 ∈ (v,e2) and b3 ∈ (v,e3) such that f(b2) = e3 and f(b3) = e1. Since [v,e2] ⊂ f([v,b1]), there is c1 ∈ (v,b1) such that f(c1) = b2. So, [v,e3] ⊂ f2([v,c1]), then choose d1 ∈ [v,c1] such that f2(d1) = b3. Hence, [v,e1] ⊂ f3([v,d1]) for which (d1,e1] has a fixed point of f3, says c ∈ (d1,e1]. We know from the Theorem 3.11 that Fix(f3) is connected. c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 342 The function ωf on simple n-ods Hence, [v,c] ⊂ Fix(f3) and then f3(d1) = d1. This is a contradiction because of f3(d1) = e1. To proof f(e2) = e3 and f(e3) = e1, we proceed in the similar way. In each case we have that f|E(T) is a permutation. � Corollary 4.6. Let (T,f) be a discrete dynamical system such that f is a surjective map and ωf is a continuous function. Then T = Per(f) and each point of T is n−periodic, where n ∈{1, 2, 3}. Proof. It follows from Theorem 4.3 that one of the following conditions holds. (1) Fix(f) = T, (2) Fix(f) = {v}, (3) There exists i ∈{1, 2, 3}, such that Fix(T) = [v,ei]. If Fix(f) = T , then each point of T is a fixed point, so 1-periodic and T = Per(f). Case 1. There exists i ∈ {1, 2, 3} such that Fix(f) = [v,ei]. Suppose that i = 1. We obtain from Proposition 4.4 that f([v,e2]) = [v,e3] and f([v,e3]) = [v,e2]. It follows from Theorem 4.5 that f(e2) = e3 and f(e3) = e2. Thus, f2(e2) = e2, f 2(e3) = e3 and f 2(e1) = e1. By Theorem 3.8 we have that Fix(f2) is connected, hence Fix(f2) = T. Therefore, each point of T is at most 2-periodic and T = Per(f). Case 2. Fix(f) = {v}. By Proposition 4.4, we have that for each i ∈ {1, 2, 3} there exists ji ∈ {1, 2, 3} such that f([v,ei]) = [v,eji ]. Without of generality, suppose that f([v,e1]) = [v,e2], f([v,e2]) = [v,e3] and f([v,e3]) = [v,e1]. We obtain from the Theorem 4.5 that f(e1) = e2, f(e2) = e3 and f(e3) = e1. So, f 3(e1) = e1, f 3(e2) = e2 and f 3(e3) = e3. We know from the Theorem 3.11 that Fix(f3) is connected. Hence, Fix(f3) = T and so each point of T is 3-periodic and T = Per(f). � The following result is the version of the Theorem 4.5, when the phase space is the arc. Following the same way, of the proof, of Theorem 4.5 it is easy to prove it. Theorem 4.7. Let f : [0, 1] → [0, 1] be a surjective continuous function such that ωf is a continuous function. Then f(0) = 0 and f(1) = 1 or f(0) = 1 and f(1) = 0. Corollary 4.8. Let f : [0, 1] → [0, 1] be a surjective continuous function such that ωf is a continuous function. Then f = f 0 or f2 = f0 Corollary 4.9. Let f : [0, 1] → [0, 1] be a surjective continuous function such that ωf is a continuous function. Then each point of [0, 1] is n−periodic with n ∈{1, 2}. The following result shows that it is sufficient to request the equicontinuity of a function f, so that, the function ωf will be a continuous map, when the phase space is a compact metric space. c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 343 I. Vidal-Escobar and S. Garcia-Ferreira Proposition 4.10. Let (X,f) be a discrete dynamical system, where X is a metric compact space with metric d. If f is equicontinuous, then ωf is a continuous function. Proof. Let ε > 0. Since f is equicontinuous there is 0 < δ ≤ ε 2 such that d(fn(x),fn(y)) < ε 2 , for each x,y ∈ X with d(x,y) ≤ δ, and every n ∈ N. Fix x,y ∈ X such that d(x,y) < δ. Choose z ∈ ωf (x), by definition, there is a sequence of positive integers n1 < n2 < ..., such that limk→∞f nk (x) = z. As X is compact, we can assume that limk→∞f nk (y) = y′. Hence, y′ ∈ ωf (y). Since d(fnk (x),fnk (y)) < ε 2 , for each k ∈ N, we obtain that d(z,y′) ≤ ε 2 < ε. Thus, ωf (x) ⊂ Nε(ωf (y)). Similarly we obtain that ωf (y) ⊂ Nε(ωf (x)). It follows from [4, 2.9] that H(ωf (x),ωf (y)) < ε. Therefore, ωf is a continuous function. � Now we will prove the main result of this work, in which we show that the equicontinuity of a function f is equivalent to the continuity of ωf , when the phase space is the simple triod, this result was proved in [1], when the phases space is the arc. Theorem A 2. Let (T,f) be a discrete dynamical system. Then, ωf is a con- tinuous function if and only if f is equicontinuous. Proof. If f is equicontinuous by Proposition 4.10 we have that ωf is a contin- uous function. Suppose that ωf is a continuous function, we consider the following cases. Case 1. f is a surjective function. We know from Theorem 4.6 that T = Per(f). Thus, each point of T has periodic orbit and so ωf is a periodic orbit for every point of X. Moreover ωf is a continuous function, then follows of [9, Theorem 3.8] that f is equicontinuous. Case 2. f is not a surjective function. Define R = ⋂∞ n=1 f n(T). Notice that f(R) = R 6= ∅, then f|R : R → R is a surjective continuous function. • If R is a point, then f|R is equicontinuous. • If R is an arc, it follows from [1, Theorem 1.2] that f|R is equicontinuous. • If R is a simple triod, we know from Case 1 that f|R is equicontinuous. So, f|R is equicontinuous. Case 2.1. R is degenerate. Pick a ∈ T such that R = {a}. Let ε > 0, since {a} = ⋂∞ n=1 f n(T) = limn→∞f n(T), then there exists m ∈ N such that H(fn(T),{a}) < ε 2 for each n > m. For which d(fn(x),a) < ε 2 for each x ∈ T and n ∈ N. By other hand as fn is a continuous function for each n ∈ {1, . . . ,m}, there exists δ > 0 such that d(fn(x),fn(y)) < ε, for each x,y ∈ T with d(x,y) ≤ δ, and all n ∈ {1, . . . ,m}. It follows from the above that d(fn(x),fn(y)) < ε, for each x,y ∈ T with d(x,y) ≤ δ, and every n ∈ N. Therefore f is equicontiuous. Case 2.2. R is not degenerate. By assumption, either R is an arc or R a a simple triod. We suppose that R is a simple triod, the proof when R is an arc it follows similarly. Since R is a simple triod, by Corollary 4.6, we have that R = Per(f|R) and each point of R is n− periodic for some n ∈ {1, 2, 3}. c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 344 The function ωf on simple n-ods Thus, f6(x) = x for each x ∈ R. For i ∈ {1, 2, 3}, choose ci ∈ (v,ei] such that R = [c1,v] ∪ [v,c2] ∪ [v,c3]. Set si = diam([v,ci]) for each i ∈ {1, 2, 3}, and s = min{si : i ∈ {1, 2, 3}}. Let 0 < ε < s such that Nε(R) ⊆ T . Now, we will find positive numbers δ3 < δ2 < δ1 < ε 2 as follows: • Since f, f2, . . . ,f6 are continuous maps, there exists 0 < δ1 < ε2 such that, if x,y ∈ T and d(x,y) < δ1, then d(fn(x),fn(y)) < ε2 for each n ∈{1, 2, . . . , 6}. • f|R is equicontinuous, we can find 0 < δ2 < δ1 such that d(fn(x),fn(y)) < ε 2 for each x,y ∈ T with d(x,y) ≤ δ2, and every n ∈ N. • Since R = limn→∞fn(T) = R, there is m ∈ N, m = 6k for some k ∈ N, such that H(fn(T),R) < δ2 for each n > m. • Since f, f2, . . . ,fm+1 are continuous functions, then there exists 0 < δ3 < δ2 such that, if x,y ∈ T , d(x,y) < δ3 and n ∈ {1, 2, . . . ,m + 1}; then d(fn(x),fn(y)) < δ2. Fix x,y ∈ T such that d(x,y) < δ3. We have that d(fm+1(x),fm+1(y)) < δ2 and fm+1(x),fm+1(y) ∈Nδ2 (R). Case 2.2.1. fm+1(x),fm+1(y) ∈ R. Since d(fm+1(x),fm+1(y)) < δ2, we have that d(fn(fm+1(x)),fn(fm+1(y))) < ε 2 for each n ∈ N. Moreover, since d(x,y) < δ3, we know that d(f n(x),fn(y)) < δ2 < ε 2 for each n ∈ {1, 2, . . . ,m + 1}. Therefore, d(fn(x),fn(y)) < ε for each n ∈ N. Case 2.2.2. fm+1(x) ∈ R and fm+1(y) /∈ R. Suppose, without of general- ity, that c1 ∈ [fm+1(y),fm+1(x)]. Then we have that d(fm+1(y),c1) < δ2 and d(fm+1(x),c1) < δ2. For which d(f n(fm+1(x)),fn(c1)) < ε 2 , for each n ∈ N. Since d(fm+1(y),c1) < δ2 < δ1, we have that d(f m+1+i(y),fi(c1)) < ε 2 for each i ∈ {1, 2, . . . , 6}, but we know that H(fn(T),R) < δ2 for each n > m, then d(fm+1+i(y),fi(c1)) < δ2 for each i ∈ {1, 2, . . . , 6}. Now, if there is i ∈ {1, 2, . . . , 6} such that fm+1+i(y) ∈ R, then d(fn(fm+1+i(y)),fn(fi(c1))) < ε2 , for each n ∈ N, so d(fn(fm+1(x)),fn(fm+1(y))) < ε. Further, since d(x,y) < δ3, we have that d(f n(x),fn(y)) < ε 2 for each n ∈{1, 2, . . . ,m + 1}. Therefore, d(fn(x),fn(y)) < ε for each n ∈ N. Assume that fm+1+i(y) /∈ R for each i ∈ {1, 2, . . . , 6}. We know that d(fm+7(y),f6(c1)) = d(f m+7(y),c1) < δ2, then we can proceed in a way similar to what was done previously, and so d(fn(fm+1(y),fn(c1))) < ε 2 for each n ∈ N. Further, since d(fn(fm+1(x)),fn(c1))) < ε 2 and d(fn(fm+1(y)),fn(c1))) < ε 2 , we obtain d(fn(fm+1(x)),fn(fm+1(y))) < ε. Moreover, as d(x,y) < δ3, we have that d(fn(x),fn(y)) < δ2 < ε 2 for each n ∈ {1, 2, . . . ,m + 1}. Therefore, d(fn(x),fn(y)) < ε for each n ∈ N. Case 2.2.3. fm+1(x),fm+1(y) /∈ R. Since d(fm+1(x),fm+1(y)) < δ2 < s, and H(fm+1(T),R) < δ2 we can suppose, without loss of generality that d(fm+1(x),c1) < δ2, and d(f m+1(x),c1) < δ2. Following similarly to Case 2.2.2, we obtain the following inequality d(fn(fm+1(x),fn(c1))) < ε 2 and d(fn(fm+1(y),fn(c1))) < ε 2 for each n ∈ N, so d(fn(fm+1(x)),fn(fm+1(y))) < ε. As d(x,y) < δ3, d(f n(x),fn(y)) < δ2 < ε 2 for each n ∈ {1, 2, . . . ,m + 1}. Therefore, d(fn(x),fn(y)) < ε for each n ∈ N. � c© AGT, UPV, 2019 Appl. Gen. Topol. 20, no. 2 345 I. Vidal-Escobar and S. Garcia-Ferreira Corollary 4.11. Let (T,f) be a discrete dynamical system such that ωf is a continuous function. Then ωfn is a continuous function and Fix(f n) is connected, for each n ∈ N. Proof. By Theorem A 2 we have that f is equicontinuos, notice that if f es equicontinuos then fn is equicontinuos then by Theorem A 2 we have that ωfn is a continuous function, and so it follows from Corollary 3.6 that Fix(fn) is connected. � Question 4.12. Can Theorem A 2 be extended when the phase space is a n-od with n ≥ 4? To finish this paper, we give an example of a function f that is equicontin- uous, however the function ωf is not continuous. When the phase space is a harmonic fan. Example 4.13. Consider X, L, Ln and fn : [0, 1] → Ln as in Example 3.16. Define gn : [0, 1 2n ] → [0, n+1 2n ] given by gn(t) = (n + 1)(t); hn : [ 1 2n , 1 2n−1 ] → [0, n+1 2n ] given by hn(t) = (n + 1)( 1 2n−1 − t) and f((x,y)) =   fn+1(gn(f −1 n (x,y))) (x,y) ∈ Ln and f−1n (x,y) ∈ [0, 1 2n ], fn+1(hn(f −1 n (x,y))) (x,y) ∈ Ln and f−1n (x,y) ∈ [ 1 2n , 1 2n−1 ], (0, 0) (x,y) ∈ Ln and f−1n (x,y) ∈ [ 1 2n−1 , 1], (0, 0) (x,y) ∈ L. We have that for each x ∈ X, ωf (x) = {(0, 0)}, hence ωf is a continuous function. We will show that f is not equicontinuous. We consider X with the maximum metric, dM . Let ε > 0, given m ∈ N such that 1 2m < ε < 1 2m−1 . We consider the point (x,y) = ( 1 2mm! , 1 2mm! ) ∈ L1, then we have that fm−1((x,y)) = fm( 1 2m ) ∈ Lm, so fm((x,y)) = fm+1(m+12m ), then fm+1((x,y)) = (0, 0) because of m+1 2m > 1 2m−1 . Now, we have that dM ((x,y), (0, 0)) = 1 2mm! < 1 2m < ε, and we have that dM (f m(x,y),fm(0, 0)) = m+1 2m > 1 2m−1 > ε. Therefore, f is not equicontinuous. Acknowledgements. 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