induced mappings on symmetric products - auburn...

Volume 37, 2011

Pages 367–401

http://topology.auburn.edu/tp/

Induced mappings on symmetric

products

by

Galo Higuera and Alejandro Illanes

Electronically published on December 8, 2010

Topology Proceedings

Web: http://topology.auburn.edu/tp/Mail: Topology Proceedings

Department of Mathematics & StatisticsAuburn University, Alabama 36849, USA

E-mail: [email protected]: 0146-4124

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TOPOLOGYPROCEEDINGSVolume 37 (2011)Pages 367-401

http://topology.auburn.edu/tp/

E-Published on December 8, 2010

INDUCED MAPPINGS ON SYMMETRIC

PRODUCTS

GALO HIGUERA AND ALEJANDRO ILLANES

Abstract. Let X be a metric continuum. For a positiveinteger n, let Fn(X) be the hyperspace of nonempty sub-sets of X with at most n points. For a given mapping be-tween continua f : X → Y , we study the induced mappingfn : Fn(X) → Fn(Y ) given by fn(A) = f(A) (the image of Aunder f). Given a topological or dynamical property M thatmappings can have, we study under which conditions the factthat f has property M implies that fn has property M, andvice versa.

1. Introduction

A continuum is a nonempty compact connected metric space.A continuum is said to be nondegenerate if it has more than onepoint. Given a continuum X, consider the following hyperspaces ofX:

2X = {A ⊂ X : A is nonempty and closed},C(X) = {A ∈ 2X : A is connected}, and for each n ≥ 1,

Fn(X) = {A ∈ 2X : A has at most n points},Cn(X) =

{A ∈ 2X : A has at most n components

}.

All these hyperspaces are considered with the Hausdorff metricH. In this paper the word mapping stands for a continuous andsurjective function.

2010 Mathematics Subject Classification. Primary 54B20, 54H25; Secondary54F15.

Key words and phrases. Atomic, confluent, continuum, expansive homeo-morphism, homeomorphism, hyperspace, induced mapping, light, linking, map-ping, mixing, MO, monotone, open, OM, refinable, semi-confluent, solenoid,symmetric product, transitive, universal, weakly confluent, weakly mixing.

c⃝2010 Topology Proceedings.

367

368 GALO HIGUERA AND ALEJANDRO ILLANES

Every mapping between continua f : X → Y induces a mappingbetween each of the respective hyperspaces in the following way:2f : 2X → 2Y is defined as 2f (A) = {f(a) : a ∈ A}. The inducedmapping to the other hyperspaces is simply the restriction of 2f toeach of such hyperspaces.

The induced mappings to other hyperspaces such as 2X , C(X)and Cn(X) have been previously studied by several authors (see forexample [3]-[7], [9]-[11], [16]-[21] and [23]). In this paper we makea systematic study of the induced mapping fn : Fn(X) → Fn(Y ),for a mapping between continua f : X → Y . Given a topologicalor dynamical property M that mappings can have, we study underwhich conditions the fact that f has property M implies that fnhas property M, and vice versa.

Some previous basic result have been shown in the thesis [13]and [29].

The authors wish to thank Gerardo Acosta, Mauricio E. Chacon,Rodrigo J. Hernandez, Hector Mendez-Lango, Juan Mireles,Christopher Mouron and Norberto Ordonez for fruitful discussionson the topic of this paper.

Besides the introduction and a section of preliminaries, we dividethe paper in two big sections. In the first one, we study traditional(or topological) properties defined for mappings between continuasuch as openness, monoteneity, confluence, etc. In the second onewe study dynamical properties such as transitivity, mixing, chaos,etc.

2. Preliminaries

The symbol N denotes the set of positive integers. All spaces areassumed to be continua unless otherwise stated. Given a continuumX, a point a ∈ X and ε > 0, the ε-ball around a is denoted byB(ε, a). The ε-ball around A in Fn(X) is denoted by BH(ε,A).Also, for A ⊂ X, we denote the diameter of A by diam(A) andNε(A) =

∪{B(ε, a) : a ∈ A}. All the hyperspaces are considered

with the Hausdorff metric (see [22, 2.1, p. 11]). It is known that thistopology concides with the Vietoris Topology (see [22, 3.1, p. 16])defined as follows: Given a finite collection of subsets U1, . . . , Um

of X we define

INDUCED MAPPINGS ON SYMMETRIC PRODUCTS 369

⟨U1, . . . , Um⟩ =

{A ∈ 2X : A ⊂m∪i=1

Ui and A ∩ Ui = ∅ for each i ∈ {1, . . . ,m}} .

The family {⟨U1, . . . , Um⟩ : m ∈ N and Ui is open in X for eachi ∈ {1, . . . ,m}} is a basis for the Vietoris topology. We define⟨U1, . . . , Um⟩n = ⟨U1, . . . , Um⟩ ∩ Fn(X).

Proceeding as in [8, Lemma 2.1], the following lemma can beproved.

Lemma 2.1. Let A be a connected, closed subset of Fn(X) suchthat A ∩ Fm(X) = ∅, for some m ≤ n. Let A =

∪{B : B ∈ A}.

Then A ∈ Cm(X) and every component of A intersects every ele-ment of A ∩ Fm(X).

We will use the following particular case of [25, Lemma 1].

Lemma 2.2. Let C1, . . . , Cm be pairwise disjoint subcontinua ofX. Suppose that m ≤ n. Then ⟨C1, . . . , Cm⟩n is a subcontinuum ofFn(X).

Definition 2.3. Given A,B ∈ C(X), with A B, we say thata continuous function α : [0, 1] → C(X) is an order arc from Ato B in C(X) if α(0) = A, α(1) = B and α(s) α(t) for every0 ≤ s < t ≤ 1.

The existence of order arcs is a very well known fact of the theoryof hyperspaces and it is stated in the following theorem (see [22,14.6, p. 112]).

Theorem 2.4. Given A,B ∈ C(X) such that A B, there existsan order arc from A to B in C(X).

3. Traditional Properties

3.1. Homeomorphisms. The following theorem is immediate.

Theorem 3.1. The following statements for a mapping f : X → Yare equivalent:

a) f is a homeomorphism,b) fn is a homeomorphism, for some n ∈ N,c) fn is a homeomorphism, for every n ∈ N.


3.2. Monotone and Open Mappings.

Definition 3.2. We say that a map f : X → Y is monotone iff−1(y) is a connected subset of X for every y ∈ Y .

Theorem 3.3. The following statements for a mapping f : X → Yare equivalent:

a) f is monotone,b) fn is monotone for some n ∈ N,c) fn is monotone for every n ∈ N.

Proof. a) ⇒ c). Suppose that f : X → Y is monotone and letn ∈ N. Let B = {y1, . . . , ym} ∈ Fn(Y ) where m ≤ n. For everyi ∈ {1, . . . ,m}, let Ci = f−1(yi). Since f is monotone, Ci is asubcontinuum of X. In addition, the sets C1, . . . , Cm are pairwisedisjoint. It follows from Lemma 2.2 that the set A = ⟨C1, . . . , Cm⟩nis a subcontinuum of Fn(X). It is easy to show that A = f−1n (B).This concludes the proof that f−1n (B) is connected and shows thatfn is monotone.

b) ⇒ a). Suppose that fn is monotone for some n ∈ N. Lety ∈ Y . Then f−1n ({y}) is a subcontinuum of Fn(X). Let A =∪{B : B ∈ f−1n ({y})}. Since f is surjective, there exists x ∈ X

such that f(x) = y. Then {x} ∈ f−1n ({y})∩F1(X). By Lemma 2.1,A is connected. Clearly, A = f−1(y). Thus f−1(y) is connected.Hence f is monotone. �

Definition 3.4. A mapping f : X → Y is open if f(U) is open inY for every open subset U of X.

Theorem 3.5. A mapping f : X → Y is open if and only if themapping f2 : F2(X) → F2(Y ) is open.

Proof. (Necessity). Suppose that f is open. Consider an open sub-set U of F2(X). Pick an element f2(A) ∈ f2(U), where A ∈ U . Weput A = {p, q}, where possibly p = q. Since U is open, there existsε > 0 such that BH(ε,A) ⊂ U . Since f is open, there exists δ > 0such that B(δ, f(p)) ⊂ f(B(ε, p)) and B(δ, f(q)) ⊂ f(B(ε, q)). Iff(p) = f(q), we can also ask that B(δ, f(p)) ∩B(δ, f(q)) = ∅.

We claim that BH(δ, f2(A)) ⊂ f2(U). Let B = {w, z} ∈ BH

(δ, f2(A)). Then H({w, z}, {f(p), f(q)}) < δ. In the case thatf(p) = f(q), we may assume that w ∈ B(δ, f(p)) and z ∈ B(δ, f(q)).


In the case that f(p) = f(q), then w, z ∈ B(δ, f(p)) = B(δ, f(q)).In both cases, we may assume that w ∈ B(δ, f(p)) ⊂ f(B(ε, p))and z ∈ B(δ, f(q)) ⊂ f(B(ε, q)). Then there exist u ∈ B(ε, p)and x ∈ B(ε, q) such that f(u) = w and f(x) = z. Notice thatH({u, x}, {p, q}) < ε. Hence {u, x} ∈ U and f2({u, x}) = {w, z}.Thus B ∈ f2(U). This shows that BH(δ, f2(A)) ⊂ f2(U). Hencef2(U) is open.

(Sufficiency). Suppose that f2 is open and let U be an opensubset of X. Given p ∈ U , we have {p} ∈ ⟨U⟩2. Since f2 is open,f2(⟨U⟩2) is an open subset of F2(Y ) that has the element f2({p}) ={f(p)}, so there exists ε > 0 such that BH(ε, {f(p)}) ⊂ f2(⟨U⟩2).

We claim that B(ε, f(p)) ⊂ f(U). Let y ∈ B(ε, f(p)). Then{y} ⊂ f2(⟨U⟩2). So there exists B ∈ ⟨U⟩2 such that {y} = f2(B).Pick a point b ∈ B. Then b ∈ U and f(b) = y. This shows thaty ∈ f(U). We have shown that B(ε, f(p)) ⊂ f(U). Hence f(U) isopen in Y . Therefore f is open. �

Theorem 3.6. If f : X → Y is a mapping such that Y is nonde-generate and fn : Fn(X) → Fn(Y ) is open for some n ≥ 3, then fis a homeomorphism.

Proof. It suffices to show that f is one-to-one. Suppose, to thecontrary, that there exist two points x1 = x2 in X such thatf(x1) = f(x2). Since Y is nondegenerate and f is surjective, thereexists x3 ∈ X such that f(x3) = f(x1). Let ε > 0 be such thatB(ε, f(x1)) ∩ B(ε, f(x3)) = ∅. By the continuity of f , there existsδ > 0 such that the sets B(δ, x1), B(δ, x2) and B(δ, x3) are pairwisedisjoint, f(B(δ, x3)) ⊂ B(ε, f(x3)) and f(B(δ, x1)) ∪ f(B(δ, x2)) ⊂B(ε, f(x1)).

Since fn is open, the set fn(BH(δ, {x1, x2, x3})) is an open subsetof Fn(Y ) that has the element {f(x1), f(x3)}. Hence there existsη > 0 such that BH(η, {f(x1), f(x3)}) ⊂ fn(BH(δ, {x1, x2, x3})).We may assume that η < ε. Pick n−1 different points y1, . . . , yn−1 ∈B(η, f(x3)) \ {f(x3)}. Let B = {f(x1), y1, . . . , yn−1}. Notice thatB ∈ BH(η, {f(x1), f(x3)}) and since this set is contained infn(BH(δ, {x1, x2, x3})), there exists A ∈ BH(δ, {x1, x2, x3}) suchthat f(A) = B. Then there exist a1, a2 ∈ A such that a1 ∈ B(δ, x1)and a2 ∈ B(δ, x2). Then a1 = a2. In addition, there existsu1, . . . , un−1 ∈ A such that f(u1) = y1, . . . , f(un−1) = yn−1. Since


the points y1, . . . , yn−1 are pairwise different, the points u1, . . . , un−1are pairwise different. Given i ∈ {1, . . . , n − 1}, f(ui) = yi ∈B(η, f(x3)) ⊂ B(ε, f(x3)). Hence f(ui) /∈ B(ε, f(x1)) and f(ui) /∈f(B(δ, x1))∪f(B(δ, x2)). This implies that ui /∈ B(δ, x1)∪B(δ, x2).Hence ui = a1, a2. Then all of the points a1, a2, u1, . . . , un−1 aredifferent and all of them are elements of A. This is absurd sinceA ∈ Fn(X). This contradiction shows that f is one-to-one. There-fore f is a homeomorphism. �

Definition 3.7. A mapping f : X → Y is OM (respectively, MO)if there exist a continuum Z and mappings g : X → Z and h : Z →Y such that f = h ◦ g, g is monotone and h is open (respectively, gis open and h is monotone).

Definition 3.8. Given a sequence {Am}∞m=1 of subsets of X definelim supm→∞Am as the set of points x ∈ X such that there exists asequence of positive numbers m1 < m2 < · · · and there exist pointsxmk

∈ Amksuch that limxmk

= x.

The following characterization of OM mappings was showed in[24, 2.2, p. 102, and Corollary 3.1, p. 104] by A. Lelek and D. R.Read.

Lemma 3.9. A mapping f : X → Y is OM if and only if, for everysequence {ym}∞m=1 in Y that converges to a point y ∈ Y , we havethat lim supm→∞ f−1(ym) meets every component of f−1(y).

Theorem 3.10. If fn : Fn(X) → Fn(Y ) is OM for some n ∈ N,then f is OM.

Proof. Let {ym}∞m=1 be a sequence in Y that converges to a pointy ∈ Y . Let C be a component of f−1(y) and let C be the componentof f−1n ({y}) that contains F1(C). Then since C is connected andF1(C) ⊂ C, it follows from Lemma 2.1 that M =

∪{E : E ∈ C} is

connected. We will show that M = C.

Given x ∈ C, {x} ∈ F1(C) ⊂ C, so x ∈ M . Hence C ⊂ M . Ifx ∈ M there exists A ∈ C such that x ∈ A. Since A ∈ C ⊂f−1n ({y}),fn(A) = {y}, so f(x) = y. We have that M ⊂ f−1({y}) and M isconnected. Since C ⊂ M and C is a component of f−1({y}), wehave C = M .


Since fn is OM and the sequence {{ym}}∞m=1 converges to {y},there exists an element A ∈ (lim supm→∞ f−1n ({ym})) ∩ C. Thusthere exist positive integers m1 < m2 < · · · and elements Amk

∈f−1n ({ymk

}) such that limAmk= A. Fix x ∈ A. Then there exist

points xmk∈ Amk

such that limxmk= x ([22, 4.5, p. 25]). For

each k ∈ N, fn(Amk) = {ymk

}, then f(xmk) = ymk

, so xmk∈

f−1(ymk). Since limxmk

= x, x ∈ lim supm→∞ f−1(ym). We alsoknow that x ∈ A ∈ C and that M = C, therefore x ∈ C. Thenx ∈ (lim supm→∞ f−1(ym)) ∩ C.

We have shown that, for each component C of f−1(y) and eachsequence {ym}∞m=1 in Y that converges to y ∈ Y , we have thatC ∩ (lim supm→∞ f−1(ym)) = ∅. By Lemma 3.9 we conclude thatf is OM. �

The next natural question is if the converse of Theorem 3.10holds. For the case n = 2 we have already done all the necessarywork to assure it does.

Theorem 3.11. If f is OM (MO), then f2 is OM (MO).

Proof. If f is OM, then there exists a continuum Z and mappingsg : X → Z and h : Z → Y such that f = h ◦ g, g is monotone andh is open. Consider then the continuum F2(Z) and the mappingsg2 : F2(X) → F2(Z) and h2 : F2(Z) → F2(Y ). By Theorems 3.3and 3.5 we know that g2 is monotone and h2 is open. And it is clearthat f2 = h2◦g2, thus f2 is OM. The proof for MO is analogous. �

Corollary 3.12. A mapping f : X → Y is OM if and only iff2 : F2(X) → F2(Y ) is OM.

As in the case of open mappings, Corollary 3.12 cannot be ex-tended for the case n ≥ 3 (compare with Proposition 9 of [5]).

Example 3.13. There exist a continuumX and a mapping f : X →X that is OM and MO and, for each n ≥ 3, the induced mappingfn : Fn(X) → Fn(X) is neither OM nor MO.

Let X = [0, 1] and f : [0, 1] → [0, 1] be defined by:

f(x) =

{2x, if x ∈

[0, 12

],

2− 2x, if x ∈[12 , 1

].


The map f is known as the tent map. Since f is an open mappingand f = f ◦ id = id ◦f , where id : [0, 1] → [0, 1] is the identitymapping, we conclude that f is OM and MO.

Suppose that n ≥ 3 and that fn is OM. Then there exist a con-tinuum Z and mappings g : Fn([0, 1]) → Z and h : Z → Fn([0, 1])such that g is monotone, h is open and fn = h ◦ g. Notice thatf−1(y) has at most two points for each y ∈ [0, 1]. It follows that,for every finite subset M of [0, 1], the set f−1(M) is finite, and alsothe set P(f−1(M)) = {W : W ⊂ M} is finite. Given A ∈ Fn([0, 1]),it is clear that f−1n (A) ⊂ P(f−1(A)). Hence f−1n (A) is finite.

Given z ∈ Z we have fn(g−1(z)) = (h ◦ g)(g−1(z)) = h(z), so

g−1(z) ⊂ f−1n (h(z)). Since g is monotone and f−1n (h(z)) is finite,g−1(z) is a one-point set. Thus g is one-to-one, then g is a homeo-morphism. In particular, g is open. Since h is open, by hypothesis,we have that fn = h ◦ g is open. Since n ≥ 3, it follows fromTheorem 3.6 that f is a homeomorphism, a contradiction. Thisshows that f is not an OM mapping.

The following theorem shows that fn cannot be an MO map-ping. �Theorem 3.14. Let n ≥ 3 and f : X → Y be a mapping such thatY is nondegenerate and fn is MO, then f is monotone.

Proof. Suppose to the contrary that f is not monotone. Then thereexists y1 ∈ Y and nonempty disjoint compact subsets K and L ofX such that f−1(y1) = K ∪ L. Fix a point y2 ∈ Y − {y1} anda sequence {vm}∞m=1 of pairwise different elements in Y − {y1, y2}such that lim vm = y2. For each m ∈ N, choose a point um ∈ Xsuch that f(um) = vm. We may assume that limum = x2 for somepoint x2 ∈ X. Thus f(x2) = y2.

Let Z be a continuum and let g : Fn(X) → Z, h : Z → Fn(Y ) bemappings such that g is open, h is monotone and fn = h ◦ g. LetD = h−1({y1, y2}). Then D is a subcontinuum of Z. It is easy tosee that g−1(D) =

⟨f−1(y1), f

−1(y2)⟩n. Let E = f−1(y2).

Note that⟨f−1(y1), f

−1(y2)⟩n= ⟨K ∪ L,E⟩n = ⟨K,E⟩n∪⟨L,E⟩n

∪ ⟨K,L,E⟩n and ⟨K,L,E⟩n = ∅ (n ≥ 3). Thus ⟨K,L,E⟩nis a nonempty open and closed subset of g−1(D) (realtive to thetopology of g−1(D)). Choose points x1 ∈ K and x3 ∈ L. Let Cbe the component of g−1(D) containing the element {x1, x2, x3}.


Since {x1, x2, x3} ∈ ⟨K,L,E⟩n, C ⊂ ⟨K,L,E⟩n. Let V,W be dis-joint open subsets of Y such that {y2} ∪ {v1, v2, . . . } ⊂ V andy1 ∈ W . Then E ⊂ f−1(V ) and K ∪ L ⊂ f−1(W ).

Since g is open, g is confluent ([30, Theorem (7.5), p. 148],definition of confluence is in Definition 3.17). Thus g(C) = D. Since{x1, x2} ∈

⟨f−1(y1), f

−1(y2)⟩n= g−1(D), g({x1, x2}) ∈ D. Hence,

there exists C ∈ C ⊂ ⟨K,L,E⟩n such that g(C) = g({x1, x2}). Theset C is of the form C = {p1, . . . , pr, pr+1 . . . , ps, ps+1 . . . , pt}, wheret ≤ n, the points p1, . . . , pt are pairwise different, {p1, . . . , pr} ⊂ K,{pr+1 . . . , ps} ⊂ L and {ps+1 . . . , pt} ⊂ E. Let ε > 0 be such thatB(ε, p1), . . . , B(ε, pt) are pairwise disjoint, B(ε, p1)∪· · ·∪B(ε, ps) ⊂f−1(W ) and B(ε, ps+1) ∪ · · · ∪B(ε, pt) ⊂ f−1(V ).

Let U = ⟨B(ε, p1), . . . , B(ε, pt)⟩n. Then U is open in Fn(X),C ∈ U , g(U) is open in Z and g({x1, x2}) = g(C) ∈ g(U). Sincelim

m→∞{x1, um, . . . , um+n−2} = {x1, x2}, we have

limm→∞

g({x1, um, . . . , um+n−2}) = g({x1, x2}) ∈ g(U) .

So, there exists m ∈ N such that g({x1, um, . . . , um+n−2}) ∈ g(U).Thus, there existsA ∈ U such that g(A)= g({x1, um, . . . , um+n−2}).Therefore f(A) = fn(A) = h(g(A)) = h(g({x1, um, . . . , um+n−2})) =fn({x1, um, . . . , um+n−2})={y1, vm, . . . , vm+n−2}. Let q1, . . . , qn−1∈A be such that f(q1) = vm, . . . , f(qn−1) = vm+n−2. Notice that{q1, . . . , qn−1} ⊂ f−1(V ).

Notice that p1 ∈ K and pr+1 ∈ L, so there exist a1 ∈ A ∩B(ε, p1) ⊂ f−1(W ) and a2 ∈ A ∩ B(ε, pr+1) ⊂ f−1(W ). Thusthe points a1, a2, q1, . . . , qn−1 are pairwise different and they areelements of A. This is a contradiction since A has at most n points.

Therefore f is monotone. �Corollary 3.15. Suppose that Y is a nondegenerate continuum.Then the following statements for a mapping f : X → Y are equiv-alent:

a) f is monotone,b) fn is monotone for some n ∈ N,c) fn is monotone for every n ∈ N,d) fn is MO for some n ≥ 3.

Question 3.16. Does Theorem 3.14 hold for n = 2?


3.3. Confluent Mappings.

Definition 3.17. A mapping f : X → Y is said to be:

1) Confluent if for every subcontinuum B of Y and every com-ponent A of f−1(B) we have that f(A) = B.

2) Weakly confluent if for every subcontinuum B of Y , thereexists a component A of f−1(B) such that f(A) = B.

3) Semi-confluent if for every subcontinuum B of Y and everypair of components C andD of f−1(B) we have that f(C) ⊂f(D) or f(D) ⊂ f(C).

Clearly every confluent mapping is a weakly confluent and a semi-confluent mapping.

Example 3.18. (compare with [18, Example 5.1]). There existcontinua X and Y and a confluent (and thus, weakly confluent andsemi-confluent) mapping f : X → Y such that f2 : F2(X) → F2(Y )is neither a confluent, weakly conluent nor semi-confluent mapping.

Proof. Let C be the complex plane and S1 ⊂ C the unit circle cen-tered at the origin. Let X = S1∪ I ∪J , where I and J are two raysconverging each one to one half of S1 as it is shown in Figure 1.Notice that the continuum X is the union of two topological copiesof the sin( 1x)-continuum joined by the end points of the limit seg-

ments. Let f be the restriction of the complex function z → z2 toX. It is easy to show that f is confluent. Let Y = f(X). Then f(I)and f(J) are two rays converging to S1 as it is shown in Figure 1.We will construct a subcontinuum K of F2(Y ) which will be usefulto deny the three definitions of confluence stated above.

Let ε = 116 . Let α : [0,∞) → f(I) and β : [0,∞) → f(J) be

parametrizations of f(I) and f(J), respectively, by arc length. Let

A = {{α(t), α(t+ ε)} : t ∈ [0,∞)}and

B = {{β(t), β(t+ ε)} : t ∈ [0,∞)} .

Notice that A (respectively, B) consists of the pairs of points inf(I) (respectively, f(J)) such that the subarc of I joining them haslenght equal to ε. Since the function t → {α(t), α(t+ε)} from [0,∞)to F2(X) is continuous, we have that A is connected. Similarly, Bis connected.


Let

D={{e(−π+t)i, e(−π−t+ε)i}∈F2(S1) : t ∈ [0, ε]}∪{{e(π−t)i, e(π+t−ε)i}∈F2(S1) : t ∈ [0, ε]} and C={{eti, e(t+ε)i}∈F2(S1) : t ∈ [−π, π − ε]}.

It is easy to see that clF2(Y )(A)∪clF2(Y )(B) = A ∪ B ∪ C ∪ D andclF2(Y )(A) ∩ clF2(Y )(B) = C ∪ D. Thus, the set K = clF2(Y )(A) ∪clF2(Y )(B) is a subcontinuum of F2(Y ). We use K to deny the threedefinitions related to confluence.

Let S+ = {z ∈ S1 : Re(z) ≥ 0} and S− = {z ∈ S1 : Re(z) ≤ 0}.Let α0 : S1 → S+ be the map defined as α0(z) = |Re(z)| + i Im(z)and let β0 : S1 → S− be the map defined as β0(z) = − |Re(z)| +i Im(z). For each θ ∈ R, let γ(θ) = {α0(e

iθ), α0(ei(θ+ ε

2))}, δ(θ) =

{β0(eiθ), β0(ei(θ+ε2))}, λ(θ) = {α0(e

iθ),−α0(ei(θ+ ε

2))} and η(θ) =

{β0(eiθ),−β0(ei(θ+ ε

2))}. Then the maps γ, δ, λ and η are de-

fined from R to F2(S1) and they have the following properties:

{ei(π2− ε

4),−ei(

π2− ε

4)} ∈ λ(R)∩η(R), for each θ ∈ R, γ(θ) ∈ ⟨S+,S+−

{i,−i}⟩2, δ(θ) ∈ ⟨S−,S− − {i,−i}⟩2, γ(R) ∩ δ(R) = ∅, (γ(R) ∪δ(R))∩ (λ(R)∪ η(R)) = ∅ and the sets γ(R), δ(R) and λ(R)∪ η(R)are compact and connected.

Since f |I : I → f(I) is a homeomorphism and f−1(f(I)) = I,we have that f−12 (A) is a connected subset of F2(I) that is home-

omorphic to A. Similarly, f−12 (B) is a connected subset of F2(J)

that is homeomorphic to B. It is easy to show that f−12 (K) =

(f−12 (A)∪γ(R))∪(f−12 (B)∪δ(R))∪(λ(R)∪η(R)) and the components

of f−12 (K) are the sets f−12 (A)∪γ(R), f−12 (B)∪δ(R) and λ(R)∪η(R).Since f2(f

−12 (A)∪γ(R))∩f2(f−12 (B)∪δ(R)) ⊂ F2(S1) and f2(λ(R)∪

η(R)) ⊂ F2(S1), we have that f2(f−12 (A) ∪ γ(R)) ( f2(f

−12 (B) ∪

δ(R)), f2(f−12 (B) ∪ δ(R)) ( f2(f−12 (A) ∪ γ(R)) and no component

L of f−12 (K) has the property that f2(L) = K. Therefore, f2 isneither confluent, weakly confluent nor semi-confluent. �Theorem 3.19. If fn : Fn(X) → Fn(Y ) is confluent, for somen ∈ N, then f : X → Y is confluent.

Proof. Let B be a subcontinuum of Y and D a component off−1(B). Notice that F1(B) is a subcontinuum of Fn(Y ). Notethat F1(D) is a connected subset of f−1n (F1(B)). Let C be thecomponent of f−1n (F1(B)) that contains F1(D). By Lemma 2.1,M =

∪{E : E ∈ C} is connected. We will show that M = D.


X

f

Y

Figure 1

Clearly, D ⊂ M . Given x ∈ M there exists A ∈ C such thatx ∈ A. Since A ∈ C ⊂ f−1n (F1(B)), fn(A) = {b} for some b ∈ B,and in particular f(x) = b. We have shown that M ⊂ f−1(B).Since M is connected, D ⊂ M and D is a component of f−1(B),we obtain that D = M .

Now, we show that f(M) = B. Clearly, f(M) ⊂ B. Givenb ∈ B, {b} ∈ F1(B). Since fn is confluent and C is a componentof f−1n (F1(B)), fn(C) = F1(B). Then there exists A ∈ C suchthat fn(A) = {b}. Fix x ∈ A, then f(x) = b and x ∈ M . Thusf(x) ∈ f(M) and b ∈ f(M). This shows that B ⊂ f(M), thenf(M) = B. Hence f(D) = B. Therefore f is confluent. �

Theorem 3.20. If fn : Fn(X) → Fn(Y ) is weakly confluent, forsome n ∈ N, then f is weakly confluent.

Proof. Let B be a subcontinuum of Y . Since fn is weakly conflu-ent there exists a component D of f−1n (F1(B)) such that fn(D) =F1(B).


Let G =∪{E : E ∈ D}. Clearly, G ⊂ f−1(B). Choose a compo-

nent C of G. Let D be the component of f−1(B) that contains C.Given y ∈ B, since fn(D) = F1(B), there exists E ∈ D such thatfn(E) = {y}. Since D is closed and connected, by Lemma 2.1, Eintersects every component of G. So we can pick z ∈ C ∩E. Sincez ∈ E, f(z) = y. In addition, since z ∈ C ⊂ D, f(z) = y ∈ f(D).We have shown that B ⊂ f(D) and then f(D) = B. This ends theproof that f is weakly confluent. �Theorem 3.21. If fn : Fn(X) → Fn(Y ) is semi-confluent, forsome n ∈ N, then f is semi-confluent.

Proof. Let B be a subcontinuum of Y and let C and D be com-ponents of f−1(B). Notice that F1(B), F1(C) and F1(D) are con-nected, F1(C) ⊂ f−1n (F1(B)) and F1(D) ⊂ f−1n (F1(B)).

Take the components C and D of f−1n (F1(B)) that containF1(C) and F1(D), respectively. Let M =

∪{E : E ∈ C} and

N =∪{E : E ∈ D}. Proceeding as in the proof of Theorem 3.19,

it follows that M = C and that N = D.

Since fn is semi-confluent we may assume that fn(C) ⊂ fn(D).We now show that f(C) ⊂ f(D). Let y ∈ f(C). Let c ∈ C be suchthat f(c) = y. Then {c} ∈ C and therefore {y} ∈ fn(C) ⊂ fn(D).Let A ∈ D be such that fn(A) = {y}. Taking x ∈ A, we havethat x ∈ N = D and f(x) = y. Hence y = f(x) ∈ f(D). Thusf(C) ⊂ f(D). Therefore f is semi-confluent. �3.4. Light Mappings.

Definition 3.22. A topological space X is said to be totally dis-connected if every component of X is degenerate.

It is well known that in compact metric spaces the componentsand quasi-components coincide. The next corollaries follow directlyfrom this fact.

Corollary 3.23. If X is a metric compact space, then X is totallydisconnected if and only if for every pair of different points x andy there exists an open and closed subset U of X such that x ∈ Uand y /∈ U .

Corollary 3.24. If X is a metric compact space, then X is totallydisconnected if and only if for every finite subset A of X and everypoint y ∈ X \A there exists an open and closed subset U of X suchthat A ⊂ U and y /∈ U .


Definition 3.25. A mapping between continua f : X → Y is lightif f−1(y) is totally disconnected for every y in Y .

Theorem 3.26. The following three statements for a mappingf : X → Y are equivalent:

a) f is light,b) fn is light for some n ∈ N,c) fn is light for every n ∈ N.

Proof. b) ⇒ a). Suppose that fn is light for some n ∈ N. Giveny ∈ Y , f−1n ({y}) is totally disconnected. Notice that, x ∈ f−1(y)if and only if {x} ∈ f−1n ({y}). Thus f−1(y) is homeomorphic tof−1n ({y})∩F1(X). Therefore, f−1(y) is totally disconnected and fis light.

a) ⇒ c). Let n ∈ N. Suppose that f is light. Let B ={y1, . . . , yk} ∈ Fn(Y ). Then f−1(yi) is totally disconnected foreach i ∈ {1, . . . , k}.

It is easy to check that f−1n (B) =⟨f−1(y1), . . . , f

−1(yk)⟩n.

Let A1 = {x1, . . . , xm} and A2 = {u1, . . . , us} be two differentelements of f−1n (B). We may assume that x1 /∈ A2 and that x1 ∈f−1(y1). By hypothesis f−1(y1) is totally disconnected. Since A2∩f−1(y1) is finite, by Corollary 3.24, there exists an open and closedsubset K of f−1(y1) such that x1 ∈ K and

(A2 ∩ f−1(y1)

)∩K = ∅.

Let L = f−1(y1) \K. Then K and L are closed in X, K ∩ L = ∅,f−1(y1) = K ∪ L, x1 ∈ K and A2 ∩ f−1(y1) ⊂ L.

Let K ={A ∈ f−1n (B) : A ∩K = ∅

}= f−1n (B) ∩ ⟨K,X⟩n and

let L =⟨L, f−1(y2), . . . , f

−1(yk)⟩n. Clearly K and L are closed

subsets of f−1n (B), A1 ∈ K and A2 ∈ L. Given A ∈ f−1n (B), ifA ∩ K = ∅, then A ∈ K, and if A ∩ K = ∅, since f−1n (B) =⟨f−1(y1), . . . , f

−1(yk)⟩n, then ∅ = A ∩ f−1(y1) ⊂ L, so A ∈ L.

This shows that f−1n (B) = K ∪ L.If there is an element A ∈ K ∩ L, then there exists x ∈ A ∩ K

and A ⊂ L ∪ f−1(y2) ∪ · · · ∪ f−1(yk). Since K ⊂ f−1(y1), x ∈A ∩ f−1(y1) ⊂ L and then x ∈ K ∩ L, a contradiction. ThereforeK ∩ L = ∅.

This shows that K is an open and closed (relative to f−1n (B))subset of f−1n (B), A1 ∈ K and A2 /∈ K. Hence f−1n (B) is totallydisconnected. Therefore fn is light. �


3.5. Universal Mappings.

Definition 3.27. A mapping f : X → Y is universal if for everycontinuous function g : X → Y there exists p ∈ X such that f(p) =g(p).

In this definition, a particular interesting case is when f is theidentity. Notice that the identity id : X → X is universal if andonly if for every continuous function g : X → X there exists p ∈ Xsuch that g(p) = p. In other words, the identity is universal if andonly if X has the fixed point property.

Then, a first step to see if the universality of a mapping is pre-served by the induced functions is to see if this happens when thismapping is the simplest of all, the identity.

Example 3.28 (J. Oledzki, [28]). There exists a continuumX withthe fixed point property such that the hyperspace F2(X) does nothave the fixed point property.

Example 3.28 implies the following.

Example 3.29 ([28]). There exists a continuum X and the map-ping id: X → X such that id is universal, but id2 : F2(X) → F2(X)is not universal.

The other implication does not hold either. The authors havegiven an example in [15].

Example 3.30 ([15]). There exists a continuumX such that F2(X)has the fixed point property but X does not have the fixed pointproperty.

Thus, we obtain the following.

Example 3.31. There exists a continuum X and the mappingid: X → X such that id2 : F2(X) → F2(X) is universal but id isnot universal.

3.6. Atomic Mappings.

Definition 3.32. A mapping between continua f : X → Y isatomic if for every subcontinuum K of X we have that f(K) isdegenerate or f−1(f(K)) = K.


Theorem 3.33. If fn : Fn(X) → Fn(Y ) is atomic for some n ≥ 2and Y is nondegenerate then f is a homeomorphism.

Proof. Let d be a metric for Y . It suffices to show that f isone-to-one. Suppose on the contrary that there exist x1 = x2in X such that f(x1) = f(x2). Since Y is nondegenerate and fis surjective there exists x3 ∈ X such that f(x1) = f(x3). Letd0 = d(f(x1), f(x3)). By Theorem 2.4 there exists an order arc αfrom {x3} to X. Let F : C(X) → C(Y ) be the induced mappingby f . Then F ◦ α : [0, 1] → C(Y ) is a path from {f(x3)} to Y .

Consider the function g : C(Y ) → [0,∞) given by g(K) =d(f(x1),K) = min{d(f(x1), y) : y ∈ K}. Clearly g is continuous.So the function β = g◦ F ◦ α : [0, 1] → [0,∞) is continuous. Thusβ([0, 1]) is a connected set. We have that β(1) = d(f(x1), Y ) = 0and that β(0) = d(f(x1), {f(x3)}) = d0. Since β([0, 1]) is con-

nected, we can choose a number t ∈ β−1(d02 ).

Notice that F (α(t)) is a subcontinuum of Y containing f(x3)and with distance to f(x1) less than d0, so F (α(t)) is nondegener-ate. And the distance between F (α(t)) and f(x1) is greater thanzero, so f(x1) /∈ F (α(t)). Therefore A = α(t) is a subcontinuumof X that does not intersect f−1(f(x1)) and f(A) = F (α(t)) isnondegenerate.

Consider K = {{x1, v} : v ∈ A}, then K is homeomorphic to Aand, hence, is a subcontinuum of Fn(X). Thus fn(K) = {{f(x1), w}: w ∈ f(A)} is nondegenerate. Also, fn({x2, x3}) = {f(x2), f(x3)}= {f(x1), f(x3)} ∈ fn(K) and {x2, x3} /∈ K, so f−1n (fn(K)) = K.Therefore fn is not an atomic mapping. �

3.7. Linking Mappings.

Definition 3.34. A mapping between continua f : X → Y is link-ing if for every subcontinuum B of Y and every pair of componentsC and D of f−1(B) we have that f(D) ∩ f(C) = ∅.Theorem 3.35. If fn : Fn(X) → Fn(Y ) is a linking mapping forsome n ∈ N, then f : X → Y is a linking mapping.

Proof. Let B be a subcontinuum of Y and let C and D be two com-ponents of f−1(B). Let C and D be the components of f−1n (F1(B))that contain F1(C) and F1(D), respectively. Let M=

∪{E :E∈ C}

and N =∪

{E : E ∈ D}. Proceeding as in the proof of Theorem3.19, it follows that C = M and D = N .


Since fn is linking, we have that fn(C) ∩ fn(D) = ∅. Let A1 ∈ Cand A2 ∈ D be such that fn(A1) = fn(A2). Let a1 ∈ A1 anda2 ∈ A2 be such that f(a1) = f(a2). Since a1 ∈ A1 ∈ C, we havethat a1 ∈ M = C. Analogously, a2 ∈ D and then f(D)∩ f(C) = ∅.Therefore f is linking. �Example 3.36. There exist a continuumX and a mapping f : X →X such that f is linking and its induced mapping f2 : F2(X) →F2(X) is not linking.

Proof. Let X = [0, 1] and f : [0, 1] → [0, 1] be given by:

f(x) =

12 + 2x, if x ∈

[0, 14

],

32 − 2x, if x ∈ [14 ,

34 ],

2x− 32 , if x ∈

[34 , 1

].

Clearly f is a linking mapping.

To see that f2 is not linking, consider the following set: A ={{3

4 , y} : y ∈ [14 ,34 ]} ∪ {{y, 14} : y ∈ [14 ,

34 ]}. Notice that A is a

subcontinuum of F2([0, 1]) and

f−1n (A) =

{{18 , x} : x ∈ [0,18 ]} ∪ {{1

8 , x} : x ∈ [38 ,58 ]} ∪ {{1

8 , x} : x ∈ [78 , 1]}∪{{3

8 , x} : x ∈ [0,18 ]} ∪ {{38 , x} : x ∈ [38 ,

58 ]} ∪ {{3

8 , x} : x ∈ [78 , 1]}∪{{5

8 , x} : x ∈ [0,18 ]} ∪ {{58 , x} : x ∈ [38 ,

58 ]} ∪ {{5

8 , x} : x ∈ [78 , 1]}∪{{7

8 , x} : x ∈ [0,18 ]} ∪ {{78 , x} : x ∈ [38 ,

58 ]} ∪ {{7

8 , x} : x ∈ [78 , 1]}.

Also notice that the sets D = {{18 , x} : x ∈ [0, 18 ]} and C =

{{78 , x} : x ∈ [78 , 1]} are closed, disjoint to each other and disjoint

to all the other sets whose union is f−1n (A). Hence D and C arecomponents of f−1n (A). Notice that f2(D) = {{3

4 , y} : y ∈ [12 ,34 ]}

and f2(C) = {{14 , y} : y ∈ [14 ,

12 ]} do not meet. It follows that f2 is

not a linking mapping. �3.8. Hereditary Properties.

Definition 3.37. Given a property R defined for mappings. Wesay that a maping f : X → Y is hereditarily R if for every subcon-tinuum A of X we have that f |A : A → f(A) has property R.

Theorem 3.38. If fn : Fn(X) → Fn(Y ) is hereditarily R, for someproperty R and some n ∈ N, then f : X → Y is hereditarily R.


Proof. Notice that for every subcontinuum A of X, the mappingfn|F1(A) : F1(A) → F1(f(A)) behaves topologically identical tof |A : A → f(A). So since F1(A) is a subcontinuum of Fn(X) andfn is hereditarily R, we have that fn|F1(A), and therefore f |A, hasthe property R. �

Some of the most commonly studied hereditary properties arehereditarily monotone, hereditarily confluent, hereditarily weaklyconfluent, etc. For these properties we have the following results.

Theorem 3.39. If fn : Fn(X) → Fn(Y ) is hereditarily monotonefor some n ≥ 2 and Y is nondegenerate, then f : X → Y is ahomeomorphism.

Proof. It suffices to show that f is one-to-one. Supose that thereexist two different points a, b ∈ X such that f(a) = f(b) = y.Since Y is nondegenerate, there exists v ∈ Y \ {y}. Fix x0 ∈X \ (f−1(y) ∪ f−1(v)).

LetA={{a, x} : x ∈ X}, C={{b, x} : x ∈ X} andD = {{x0, x} :x ∈ X}. Then A, C and D are homeomorphic to X and thereforethey are continua. Notice that A ∩ C = {a, b} and C ∩ D = {b, x0},so G = A ∪ C ∪ D is a subcontinuum of Fn(X).

We analyze the map fn|G : G → fn(G). Since f(x0) /∈ {v, y},(fn|G)−1({y, v}) ⊂ A ∪ C, and in fact, (fn|G)−1({y, v}) = {{a, x} :x ∈ f−1(v)} ∪ {{b, x} : x ∈ f−1(v)}, which are closed, disjointand nonempty. Therefore (fn|G)−1({y, v}) is not connected andfn is not hereditarily monotone. This concludes the proof of thetheorem. �

Theorem 3.40. If fn : Fn(X) → Fn(Y ) is hereditarily weakly con-fluent for some n ≥ 3 (and Y is nondegenerate), then f : X → Yis a homeomorphism.

Proof. Suppose that there exist two different points a, b ∈ X suchthat f(a) = f(b) = y0. Using order arcs we can constuct twonondegenerate and disjoint subcontinua K and L of Y which donot contain y0. Let e ∈ f−1(L) and c ∈ f−1(K). Let A ={{a, b, x} : x∈X}, B={{a, e, x} : x∈X} and C={{b, c, x} : x∈X}.Then A, B and C are subcontinua of Fn(X).


Since {a, b, e} ∈ A ∩ B and {a, b, c} ∈ A ∩ C, M = A ∪ B ∪ Cis a subcontinuum of Fn(X). Let K = {{y0, f(e), w} : w ∈ K} andL = {{y0, f(c), w} : w ∈ L}. Since {y0, f(c), f(e)} ∈ K ∩ L, R =K∪L is a subcontinuum of Fn(Y ). Since K ⊂ fn(B) and L ⊂ fn(C),R ⊂ fn(M). Since f(e), y0 /∈ K, K ⊂ F3(X)\F2(X). Analogously,L ⊂ F3(X) \ F2(X) and then R ⊂ F3(X) \ F2(X).

We will show that fn|M : M → fn(M) is not weakly confluent.

Note that (fn|M)−1(R)=(f−1n (R)∩A)∪(f−1n (R)∩B)∪(f−1n (R)∩C).Notice that, for every x ∈ X, we have that fn({a, b, x}) =

{y0, f(x)} ∈ F2(Y ). Hence f−1n (R) ∩ A = ∅ and (fn|M)−1(R) =(f−1n (R) ∩ B) ∪ (f−1n (R) ∩ C).

It is easy to prove that f−1n (R) ∩ B = {{a, e, x} : x ∈ f−1(K)}and f−1n (R) ∩ C =

{{b, c, x} : x ∈ f−1(L)

}.

This implies that fn(f−1n (R)∩B) = K and fn(f

−1n (R)∩ C) = L.

Let N be a component of (fn|M)−1(R), notice that f−1n (R) ∩ Band f−1n (R) ∩ C are closed and disjoint, so N ⊂ f−1n (R) ∩ B orN ⊂ f−1n (R) ∩ C. Since K and L are nondegenerate, K R andL R, so fn(f

−1n (R) ∩ B) R and fn(f

−1n (R) ∩ C) R. Thus

fn(N ) R. Hence fn|M : M → fn(M) is not weakly confluent,so fn is not hereditarily confluent. �Definition 3.41. A subcontinuum A of X is terminal in X if forevery subcontinuum B of X, we have that B ⊂ A or A ⊂ B orA ∩B = ∅.Theorem 3.42. If f2 : F2(X) → F2(Y ) is hereditarily confluent,then f : X → Y is monotone and f−1(y) is a terminal subcontin-uum of X for each y ∈ Y .

Proof. The proof is based in the following claim.

Claim 1. If there exist y ∈ Y and a subcontinuum A of Xsuch that A ∩ f−1(y) = ∅, f−1(y) * A and A * f−1(y), thenf2 : F2(X) → F2(Y ) is not hereditarily confluent.

Proof of Claim 1. Let a ∈ A∩f−1(y), b ∈ f−1(y)\A and c ∈ A\f−1(y). Notice that f(a), f(c) ∈ f(A) and f(a) = y = f(c), so f(A)is nondegenerate. Let A = {{a, x} : x ∈ X}, B = {{b, x} : x ∈ X},C = {{c, x} : x ∈ A} and M = A ∪ B ∪ C. Notice that A, B and Care closed, connected, {a, b} ∈ A ∩ B and {a, c} ∈ A ∩ C, hence Mis a subcontinuum of F2(X).


Let K = {{f(c), z} : z ∈ f(A)}, then K is a nondegenerate sub-continuum of F2(Y ) such that f2(C) = K, so K is a subcontinuumof f2(M). We show that f2|M : M → f2(M) is not confluent.

Since (f2|M)−1 (K) = (f−12 (K) ∩A) ∪ (f−12 (K) ∩ B) ∪ (f−12 (K) ∩C) = {{a, x} : x ∈ f−1(f(c))}∪{{b, x} : x ∈ f−1(f(c))}∪C and thatthe sets {{b, x} : x ∈ f−1(f(c))} and

{{a, x} : x ∈ f−1(f(c))

}∪ C

are closed, disjoint and nonempty, there exists a component D of(f2|M)−1 (K) that is contained in

{{b, x} : x ∈ f−1(f(c))

}. But

then f2(D) ⊂ f2({{b, x} : x ∈ f−1(f(c))}

)= {{y, f(c)}} K. It

follows that f2|M : M → f2(M) is not confluent and therefore f2is not hereditarily confluent. This finishes the proof of Claim 1.

Suppose that there exists y ∈ Y such that f−1(y) is not con-nected. Then there exist two closed, disjoint and nonempty subsetsK and L of X such that f−1(y) = K ∪ L. Let C be a componentof f−1(y) contained in K. By Theorem 2.4 there exists an or-der arc α : [0, 1] → C(X) from C to X. Since α(0) = C ⊂ K,α(0) ⊂ X \ L, there exists s > 0 such that α(s) ⊂ X \ L andα(s) = α(0). Let A = α(s). Then A is a subcontinuum of X suchthat ∅ = L ⊂ f−1(y) \ A, ∅ = C ⊂ f−1(y) ∩ A. If A ⊂ f−1(y),then A must be contained in a component of f−1(y), so C = A.This is absurd, so A * f−1(y). We have obtained a subcontinuumA of X such that f−1(y) ∩A = ∅, f−1(y) * A and A * f−1(y). Itfollows from Claim 1 that f2 is not hereditarily confluent; that is acontradiction. This shows that f−1(y) must be connected for eachy ∈ Y and therefore f is monotone.

Finally, Claim 1 also implies that f−1(y) is a terminal subcon-tinuum of X for every y ∈ Y . �

Definition 3.43. A continuumK is decomposable if it is the unionof two proper subcontinua.

Theorem 3.44. If f2 : F2(X) → F2(Y ) is hereditarily confluentthen for every decomposable subcontinuum K of Y and for everyy ∈ Y \K the set f−1(y) is degenerate.

Proof. Let K ⊂ Y be a decomposable subcontinuum. Let A and Bbe proper subcontinua of K such that K = A∪B and let y ∈ Y \K.By Theorem 3.42, f−1(y), f−1(A) and f−1(B) are subcontinua ofX.


Suppose that f−1(y) contains two points x1 = x2. Let b ∈f−1(B) \ f−1(A). Let A =

{{x1, x} : x ∈ f−1(K)

}, B = {{b, x} :

x∈f−1(y)}, C =

{{x2, x} : x ∈ f−1 (B)

}and M = A ∪ B ∪ C.

Since {x1, b} ∈ A ∩ B and {b, x2} ∈ B ∩ C, we have M is a subcon-tinuum of F2(X).

We will show that f2|M : M → f2(M) is not confluent. Let K ={{y, a} : a ∈ A}. Notice that K ⊂ f2(A), so K is a subcontinuumof f2(M). Since b /∈ f−1(A) ∪ f−1(y), f−12 (K) ∩ B = ∅. Then

(f2|M)−1 (K) = (f−12 (K)∩A)∪(f−12 (K)∩C). and the sets f−12 (K)∩Aand f−12 (K)∩C are closed and disjoint. Clearly f−12 (K)∩A = ∅ and

since K is a continuum, A∩B = ∅. This implies that f−12 (K)∩C =

∅. Then there exists a component D of (f2|M)−1 (K) such thatD ⊂ f−12 (K) ∩ C. Thus

f2(D) ⊂ f2(f−12 (K) ∩ C) = f2(C) ∩ K = {{y, x} : x ∈ B} ∩ K

= {{y, x} : x ∈ A ∩B} Kand therefore f2|M : M → f2(M) is not confluent. This contradic-tion proves that f−1(y) is a one-point set. �

Problem 3.45. Suppose that f2 : F2(X) → F2(Y ) is hereditarilyconfluent (weakly confluent) and Y is nondegenerate, must f be ahomeomorphism?

3.9. Refinable Mappings.

Definition 3.46. LetX and Y be continua and let d be a metric forY . Given continuous functions f, g : X → Y , we define dsup(f, g) =max {d(f(x), g(x)) : x ∈ X}.

The following lemma is immediate.

Lemma 3.47. Let f, g : X → Y be a pair of continuous functionsbetween continua, then for every n ∈ N, dsup(f, g) < ε if and onlyif dsup(fn, gn) < ε.

Definition 3.48. A mapping f : X → Y is refinable if for everyε > 0, there exists a mapping g : X → Y such that dsup(f, g) < εand diam(g−1(y)) < ε for every y ∈ Y .

Theorem 3.49. If f : X → Y is refinable, then fn : Fn(X) →Fn(Y ) is refinable for every n ∈ N.


Proof. Let ε > 0. Since f is refinable, there exists a mappingg : X → Y such that dsup(f, g) < ε and diam(g−1(y)) < ε for everyy ∈ Y . By Lemma 3.47, dsup(fn, gn) < ε. Let A = {y1, . . . , ym} ∈Fn(Y ) and take B,C ∈ g−1n (A). Let c ∈ C and i ∈ {1, . . . ,m} besuch that g(c) = yi. Since gn(B) = A, there exists b ∈ B suchthat g(b) = yi. Since diam(g−1(yi)) < ε, d(c, b) < ε. ThereforeC ⊂ Nε(B), analogously B ⊂ Nε(C), then H(B,C) < ε. Thisshows that diam(g−1n (A)) < ε. Therefore, fn is refinable. �Problem 3.50. Is it true that if fn : Fn(X) → Fn(Y ) is refinable forsome n ≥ 2, then f : X → Y is refinable?

4. Dynamical Properties

In this section all the functions are of the form f : X → X andthey are continuous but not necessarily surjective, also, X alwaysdenotes a continuum. For each k ∈ {0, 1, . . .} we define, inductively,fk as f0 = idX and fk = f ◦ fk−1.

4.1. Transitivity, Mixing properties and Chaos.

Definition 4.1. A continuous function f : X → X is:

(a) transitive if for every pair of nonempty open subsets U andV of X, there exists k ∈ N such that fk(U) ∩ V = ∅;

(b) mixing if for every pair of nonempty open subsets U and Vof X there exists N ∈ N such that fk(U)∩ V = ∅ for everyk ≥ N ;

(c) weakly mixing if for all nonempty open subsets U1, U2, V1

and V2 of X there exists k ∈ N such that fk(Ui) ∩ Vi = ∅for each i ∈ {1, 2}.

Lemma 4.2. The family {⟨U1, . . . , Un⟩n ⊂ Fn(X) : U1, . . . , Un areopen in X} is a basis for the topology of Fn(X).

Proof. We know that the family B = {⟨U1, . . . , Um⟩n ⊂ Fn(X) :m ∈ N and U1, . . . , Um are open in X} is a basis for Fn(X). Thenit is enough to show that for every ⟨U1, . . . , Um⟩n in that family, andA ∈ ⟨U1, . . . , Um⟩n, there exist n open subsets of X, say V1, . . . , Vn,such that A ∈ ⟨V1, . . . , Vn⟩n ⊂ ⟨U1, . . . , Um⟩n.

Case 1. m ≤ n. In this case, we take Vi = Ui for eachi ∈ {1, . . . ,m} and Vi = Um for each i ∈ {m+ 1, . . . , n}. Then⟨V1, . . . , Vn⟩n = ⟨U1, . . . , Um⟩n and we are done.


Case 2. m > n. Let A = {x1, . . . , xk}. For each i ∈ {1, . . . , k},let Wi =

∩{Uj : xi ∈ Uj}. It is easy to prove that A ∈

⟨W1, . . . ,Wk⟩n ⊂ ⟨U1, . . . , Um⟩n. Clearly xi ∈ Wi, for each i ∈{1, . . . , k}, so A ∈ ⟨W1, . . . ,Wk⟩n.

Since A ∈ Fn(X), we know that k ≤ n and then it followsfrom Case 1 that there exist open subsets V1, . . . , Vn of X suchthat A ∈ ⟨V1, . . . , Vn⟩n ⊂ ⟨W1, . . . ,Wk⟩n ⊂ ⟨U1, . . . , Um⟩n. Thisconcludes the proof of the lemma. �Theorem 4.3. The following statements for a continuous functionf : X → X are equivalent:

a) f is mixing,b) fn is mixing for some n ∈ N,c) fn is mixing for every n ∈ N.

Proof. a) ⇒ c). Suppose that f is mixing and let n ∈ N. Let⟨U1, . . . , Un⟩n and ⟨V1, . . . , Vn⟩n be two basic nonempty open sub-sets of Fn(X) (Lemma 4.2). For each i ∈ {1, . . . , n} there ex-ists Ni ∈ N such that fk(Ui) ∩ Vi = ∅ for each k ≥ Ni. TakeN = max {N1, . . . , Nn}, then fk(Ui) ∩ Vi = ∅ for every k ≥ Nand every i ∈ {1, . . . , n}. Given k ≥ N and i ∈ {1, . . . , n},we can pick a point xi ∈ Ui ∩ f−k(Vi). Let Ak = {x1, . . . , xn}.Clearly Ak ∈ ⟨U1, . . . , Un⟩n and (fn)

k(Ak) ∈ ⟨V1, . . . , Vn⟩n. Hence

(fn)k(⟨U1, . . . , Un⟩n)∩⟨V1, . . . , Vn⟩n = ∅ for every k ≥ N . Therefore

fn is mixing.

b) ⇒ a). Suppose that fn is mixing for some n ∈ N. Let U andV be two nonempty open subsets of X. Then ⟨U⟩n and ⟨V ⟩n aretwo nonempty open subsets of Fn(X). Since fn is mixing, thereexists N ∈ N such that (fn)

k(⟨U⟩n) ∩ ⟨V ⟩n = ∅ for every k ≥ N .

For each k ≥ N , let Ak ∈ ⟨U⟩n be such that (fn)k(Ak) ∈ ⟨V ⟩n and

let xk ∈ Ak ⊂ U . Since fk(xk) ∈ (fn)k(Ak) ∈ ⟨V ⟩n, fk(xk) ∈ V .

Therefore fk(U) ∩ V = ∅ for each k ≥ N. This shows that f ismixing. �

The following theorem is well known (see for example [1]). Theequivalence a) ⇐⇒ b) is due to H. Furstenberg.

Theorem 4.4. The following statements for a continuous functionf : X → X are equivalent:

a) f is weakly mixing.


b) For each m≥2 and every nonempty open subsets U1, . . . ,Um,V1, . . . , Vm of X, there exists k ∈ N such that fk(Ui)∩Vi = ∅for each i ∈ {1, . . . ,m}.

c) For every nonempty open subsets U, V1, V2 of X, there existsk ∈ N such that fk(U) ∩ Vi = ∅ for each i ∈ {1, 2}.

Theorem 4.5 (compare with Theorem 2 of [2]). The followingstatements for a continuous function f : X → X are equivalent:

a) f is weakly mixing.b) fn is weakly mixing for each n ∈ N.c) fn is transitive for each n ∈ N.d) fn is weakly mixing for some n ≥ 2.e) fn is transitive for some n ≥ 2.

Proof. a) ⇒ b). Suppose f is weakly mixing and let n ∈ N. LetU1 =

⟨U11 , . . . , U

1n

⟩n, U2 =

⟨U21 , . . . , U

2n

⟩n, V1 =

⟨V 11 , . . . , V

1n

⟩n

and V2 =⟨V 21 , . . . , V

2n

⟩nbe nonempty basic open subsets of Fn(X)

(Lema 4.2). Since f is weakly mixing, by Theorem 4.4 there exists

k ∈ N such that fk(U ji ) ∩ V j

i = ∅ for every i ∈ {1, . . . , n} andj ∈ {1, 2}. Given i ∈ {1, . . . , n} and j ∈ {1, 2} we choose a point

xji ∈ f−k(V ji )∩U j

i . Let A1 ={x11, . . . , x

1n

}and A2 =

{x21, . . . , x

2n

}.

Clearly Aj ∈ Uj and (fn)k(Aj) ∈ Vj , for each j ∈ {1, 2}. Hence

(fn)k(Uj)∩Vj = ∅ for each j ∈ {1, 2}. This shows that fn is weakly

mixing.

The implications c) ⇒ e) and b) ⇒ d) are obvious; b) ⇒ c) andd) ⇒ e) follow directly from the definitions.

e) ⇒ a). We use Theorem 4.4. Let U, V1 and V2 be nonemptyopen subsets of X. Let U = ⟨U⟩n and V = ⟨V1, V2⟩n. Since n ≥2, V = ∅. Since fn is transitive, there exists k ∈ N such that(fn)

k(U) ∩ V = ∅. Then there exists A ∈ U such that (fn)k(A) ∈ V,

then fk(A) ∩ V1 and fk(A) ∩ V1 are both nonempty. Since A ⊂ Uit follows that fk(U) ∩ Vi = ∅ for each i ∈ {1, 2}. This shows thatf is weakly mixing and finishes the proof of the theorem. �Notation 4.6. Given a continuous function f : X → X, per(f)denotes the set of periodic points of f .

Lemma 4.7 (compare with Lemma 1 of [2]). For each continuousfunction f : X → X and each n ∈ N, the set per(fn) is dense inFn(X) if and only if per(f) is dense in X.


Proof. (Necessity). Let n ∈ N. Suppose that per(fn) is dense inFn(X). Let x ∈ X and ε > 0. Since per(fn) is dense in Fn(X),there exists A ∈ per(fn) ∩ BH(ε, {x}). Since A ∈ per(fn), thereexists k ∈ N such that (fn)

k(A) = A. This means that fk|A : A →A is a permutation of the elements of A. Since A is finite andpermutations of finite sets have finite order, there exists r ∈ N suchthat

(fk

)r |A = idA. Then A ⊂ per(f). Since A ∈ BH(ε, {x}), wehave that A ⊂ B(ε, x), this shows that B(ε, x) ∩ per(f) = ∅ andper(f) is dense in X.

(Sufficiency). Suppose per(f) is dense in X. Let n ∈ N, A ={x1, . . . , xm} ∈ Fn(X) and ε > 0. Since per(f) is dense, for eachi ∈ {1, . . . ,m}, there exist pi ∈ per(f) ∩ B(ε, xi) and ni ∈ N suchthat fni(pi) = pi. Take k = n1 · · ·nm, then fk(pi) = pi for each i ∈{1, . . . ,m}. Let B = {p1, . . . , pm} . By construction, (fn)

k (B) ={fk(p1), . . . , f

k(pm)}

= B so B ∈ per(fn). Since for every i ∈{1, . . . ,m}, pi ∈ B(ε, xi), we have B ∈ BH(ε,A). This shows thatB ∈ per(fn)∩BH(ε,A). Therefore, per(fn) is dense in Fn(X). �Definition 4.8. A continuous function f : X → X is chaotic if itis transitive and per(f) is dense in X.

Theorem 4.9. The following statements for a continuous functionf : X → X are equivalent:

a) f is chaotic and weakly mixing.b) fn is chaotic for some n ≥ 2.c) fn is chaotic for each n ≥ 2.

Proof. It follows directly from Lemma 4.7 and Theorem 4.5. �Example 4.10. There exists a continuum X and a continuouschaotic function f : X → X such that fn : Fn(X) → Fn(X) is notchaotic for any n ≥ 2.

Let X = [0, 1] and f : [0, 1] → [0, 1] be defined as follows:

f(x) =

2x+ 1

2 , if x ∈[0, 14

],

32 − 2x, if x ∈

[14 ,

12

],

1− x, if x ∈[12 , 1

].

By Theorem 4.9 it is enough to show that f is chaotic but notweakly mixing.

Claim 1. f is not weakly mixing.


In order to prove Claim 1, let U1 = U2 = V1 = (0, 12) and V2 =

(12 , 1). Let W1 = [0, 12 ] and W2 = [12 , 1]. Clearly, f(W1) = W2 andf(W2) = W1. Thus

fk(W1) =

{W1, if k is even,W2, if k is odd.

Suppose that there exists k ∈ N such that fk(U1) ∩ V1 = ∅and fk(U2) ∩ V2 = ∅. Take points x ∈ U1 and y ∈ U2 such thatfk(x) ∈ V1 and fk(y) ∈ V2. If k is odd, then fk(x) ∈ fk(W1) = W2,so fk(x) ∈ W2 ∩ V1 = ∅, a contradiction. If k is even, fk(y) ∈fk(W1) = W1, so fk(y) ∈ W1 ∩ V2 = ∅, a contradiction. Thisproves that f is not weakly mixing.

Claim 2. f is chaotic.In order to show Claim 2, lets take a look at f2

f2(x) =

12 − 2x, if x ∈

[0, 14

],

2x− 12 , if x ∈

[14 ,

34

],

52 − 2x, if x ∈

[34 , 1

].

Note that we can consider f2|[0, 12 ] :[0, 12

]→

[0, 12

]and f2|[ 12 ,1] :[

12 , 1

]→

[12 , 1

]as independent dynamical systems. It is easy to see

that both of this mappings are topologically conjugate ([12]) to themap known as the“tent map” which was described in Example 3.13in the previous section. It is known that this mapping is chaotic([12]). It easily follows from this that f is chaotic.

4.2. Specification and the P property.

Definition 4.11. A continuous function f : X → X has speci-fication if for every ε > 0 there exists Mε such that for eachk ≥ 2, for every x1, . . . , xk ∈ X and for every nonnegative inte-gers a1 ≤ b1 < a2 ≤ b2 < · · · < ak ≤ bk such that ai − bi−1 ≥ Mε,there exists z ∈ X such that for every i ∈ {1, . . . , k} and everym ∈ {ai, . . . , bi} we have that d(fm(z), fm(xi)) < ε.

Definition 4.12. A continuous function f : X → X has the Pproperty if for every pair of nonempy open subsets U0, U1 ofX thereis anN ∈ N such that for each k ≥ 2 and each s = (s(1), . . . , s(k)) ∈{0, 1}k there exists x ∈ X such that x ∈ Us(1), f

N (x) ∈ Us(2), f2N (x)

∈ Us(3), . . . , f(k−1)N (x) ∈ Us(k).


The relation between a mapping having one of these propertiesand the induced mapping to some hyperspaces having it was studiedby Rongbao Gu and Wenjing Guo in [14]. The proofs of Theorems4.1 and 4.2 of [14] can be very easily adapted to prove the followingtheorems.

Theorem 4.13. The following statements for a continuous func-tion f : X → X are equivalent:

a) f has specification,b) fn has specification for some n ≥ 2,c) fn has specification for each n ≥ 2.

Theorem 4.14. The following statements for a continuous func-tion f : X → X are equivalent

a) f has the P property,b) fn has the P property for some n ≥ 2,c) fn has the P property for each n ≥ 2.

4.3. Expansive Homeomorphisms. Given a homeomorphismf : X → X we can define its negative iterations as f−n =

(f−1

)nfor each n ∈ N, where f−1 is the inverse of f .

Definition 4.15. A homeomorphism f : X → X is expansive ifthere exists c > 0 such that for every two differents points x, y ∈ Xthere is k ∈ Z such that d(fk(x), fk(y)) > c. In such case we call can expansion constant for f .

Theorem 4.16. If fn : Fn(X) → Fn(X) is an expansive homeo-morphism for some n ∈ N, then f : X → X is an expansive home-omorphism.

Proof. Let n ∈ N be such that fn is an expansive homeomorphism.By Theorem 3.1, f is a homeomorphism. Let c > 0 be an expansionconstant for fn. We will show that c is also an expansion constantfor f .

Given x = y ∈ X we know that {x} = {y} ∈ Fn(X) and thenthere exists a k ∈ Z such that

c < H((fn)k ({x}), (fn)k ({y})) = d(fk(x), fk(y)).

Thus c is an expansion constant for f . Therefore f is an expan-sive homeomorphism. �


4.3.1. Inverse Limits and Shift Homeomorphisms.

Definition 4.17. An inverse sequence is a sequence of pairs{(Xk, fk)}∞k=1, where for every k ∈ N, Xk is a continuum andfk : Xk+1 → Xk is a continuous function.

Definition 4.18. Given an inverse sequence {(Xk, fk)}∞k=1 we de-fine its inverse limit as:

lim←−

{(Xk, fk)} = {(xk)∞k=1∈∞∏k=1

Xk : fk(xk+1) = xk for every k∈N}

The metric in∞∏k=1

Xk is given by ρ((x1, x2, . . .), (y1, y2, . . .)) =

∞∑k=1

dk(xk,yk)2k

, where dk is a metric for Xk bounded by a number

M . It is known that the inverse limit of each inverse sequence{(Xk, fk)}∞k=1, where each Xk is a continuum, is a continuum.

Notation 4.19. Given a continuous function f : X → X one canconsider the inverse sequence {Xk, fk}∞k=1, where Xk = X and fk =f for every k ∈ N. The inverse limit of this particular inversesequence is simply denoted by lim

←−(X, f).

Definition 4.20. Given Y = lim←−

(X, f) we define a function f : Y →

Y as f((x1, x2, . . .)) = (f(x1), f(x2), . . .) = (f(x1), x1, x2, . . .).

Notice that (f(x1), x1, x2, . . .) belongs to Y , so f is well defined.

The following theorem is immediate.

Theorem 4.21. Given a continuous function f : X → X, if Y =

lim←−

(X, f), then the function f : Y → Y is a homeomorphism, called

the shift homeomorphism, and the inverse of f is given by (f)−1

((x1, x2, . . .)) = ((x2, x3, . . .)).

Example 4.22. There exists a continuum X and an expansivehomeomorphism g : X → X such that gn : Fn(X) → Fn(X) is notan expansive homeomorphism for any n ≥ 2.

Let n ≥ 2. Consider the unit circle S1, centered at the ori-gin in the complex plane C. We give to S1 the metric of theshorter arc, that is, given z, w ∈ S1 we define their distance tobe d(z, w) = the length of the shortest arc containing z and w.


This is clearly a metric which is compatible with the norm metricinherited from C. Let f : S1 → S1 be defined as f(z) = z2.

Let X = lim←

(S1, f). It is known that f : X → X is an ex-

pansive homeomorphism (see for example [31]). We show that

fn : Fn(X) → Fn(X) is not an expansive homeomorphism for anyn ≥ 2.

Claim 1. Given z, w ∈ S1 there exist u, v ∈ S1 such that f(u) =

z, f(v) = w and d(u, v) = d(z,w)2

To prove Claim 1, we consider two cases.

Case 1. The complex number 1 is not in the interior of theshortest arc of S1 joining z and w.

In this case, we write z = eiα and w = eiβ with α, β ∈ [0, 2π).

Then d(z, w) = |α− β|. In this case, define u = eiα2 and v = ei

β2 .

Case 2. The complex number 1 is in the interior of the shortestarc of S1 joining z and w.

In this case, the complex number −1 is not in the shortest arcjoining z and w. Then we write z = eiα and w = eiβ with α, β ∈(−π, π). In this case, let u = ei

α2 and v = ei

β2 .

The following claim is easy to prove.

Claim 2. Let z, w ∈ S1 be such that d(z, w) < π2 then d(f(z),

f(w)) = 2d(z, w).

Claim 3. The homeomorphism fn : Fn(X) → Fn(X) is notexpansive.

Proof of Claim 3. Let 0 < c < π4 . We will show that c is not an

expansion constant for fn. Let p ∈ N and z1, w1 ∈ S1 be such thatπ2p < c and 0 < d(z1, w1) < c

2p . Applying Claim 1, succesively, it

is possible to construct two sequences {zk}∞k=1 and {wk}∞k=1 in S1

such that f(zk+1) = zk, f(wk+1) = wk and d(zk+1, wk+1) =d(z1,w1)

2k

for every k ∈ N.

Notice that the points z = (z1, z2, . . .) and w = (w1, w2, . . .)belong to X.

First, we will show that ρ(fm(z), fm(w)) < c for every m ∈{. . . , p− 1, p}.


By the definition of (f)−1 we have that, for each m ∈ {0, 1, . . .},

ρ(f−m(z), f−m(w)) = ρ((zm+1, zm+2, . . .), (wm+1, wm+2, . . .))

=∞∑k=1

d(zm+k, wm+k)

2k=∞∑k=1

d(z1,w1)2m+k

2k

=

∞∑k=1

d(z1, w1)

22k+m≤∞∑k=1

d(z1, w1)

2k

= d(z1, w1) <c

2p< c .

Hence, we have shown that ρ(fm(z), fm(w)) < c for each m ∈{. . . ,−2,−1, 0}.

Next, by induction, we will show that for each m ∈ {0, 1, . . . , p},d(fm(z1), f

m(w1)) = 2md(z1, w1). The case m = 0 is immediate.Suppose thatm ∈ {0, 1, . . . , p− 1} is such that d(fm(z1), f

m(w1))=2md(z1, w1). Sincem<p and 2pd(z1, w1)<c< π

4 , d(fm(z1), f

m(w1))

< π4 . Claim 2 implies that d(fm+1(z1), f

m+1(w1)) = 2d(fm(z1),

fm(w1)) = 2m+1d(z1, w1). This concludes the induction.

Finally, again by induction, we will prove that ρ(fm(z), fm(w)) ≤2md(z1, w1) for each m ∈ {0, 1, . . . , p}. The case m = 0 wasshowed in the chain of inequalities above. Therefore, suppose that

if m ∈ {1, . . . , p}, then ρ(fm−1(z), fm−1(w)) ≤ 2m−1d(z1, w1). By

the definitions of ρ and f we have

ρ(fm(z), fm(w)) =ρ((fm(z1), . . . , f

0(z1), z2, . . . ), (fm(w1), . . . f

0(w1), w2, . . .))

=m−1∑k=0

d(fm−k(z1),fm−k(w1))2k+1 +

∞∑k=1

d(zk,wk)2m+k

= d(fm(z1),fm(w1))2 +

m−1∑k=1

d(fm−k(z1),fm−k(w1))2k+1 +

∞∑k=1

d(zk,wk)2m+k

= d(fm(z1),fm(w1))2 + 1

2

m−2∑k=0

d(f (m−1)−k(z1),f (m−1)−k(w1))2k+1 + 1

2

∞∑k=1

d(zk,wk)

2(m−1)+k

= d(fm(z1),fm(w1))2 + ρ(fm−1(z),fm−1(w))

2 ≤ 2md(z1,w1)2 + 2m−1d(z1,w1)

2

≤ 2md(z1, w1).


We have shown that ρ(fm(z), fm(w)) ≤ 2md(z1, w1). This con-

cludes the induction and shows that ρ(fm(z), fm(w)) ≤ 2md(z1, w1)for all m ∈ {0, 1, . . . , p}. By the choice of p, 2pd(z1, w1) < c, it fol-

lows that ρ(fm(z), fm(w)) < c for each m ∈ {0, 1, . . . , p}.

This completes the proof that ρ(fm(z), fm(w)) < c for all m ∈{. . . , p− 1, p}.

Let z′=(eπiz1, eπ2iz2, e

π4iz3, . . .) and w′=(eπiw1, e

π2iw2, e

π4iw3, . . .).

Clearly z′, w′ ∈ X. It is clear that rotations preserve the given

metric on S1. Therefore, ρ(fm(z′), fm(w′)) < c for each m ∈{. . . ,−1, 0, 1, . . . , p}.

Let A = {z, w′} , B = {w, z′} ∈ Fn(Y ). We are going to show

that, for each m ∈ Z, H((fn)m(A), (fn)

m(B)) < c. We have seen

that ρ(fm(z), fm(w)) and ρ(fm(z′), fm(w′)) < c, for each m ∈{. . . , p − 1, p}. This implies that H((fn)

m(A), (fn)m(B)) < c, for

each m ∈ {. . . , p− 1, p}.

Let m ∈ {p+ 1, p+ 2, . . .}. We will show that H((fn)m(A),

(fn)m(B)) < c. Since f(eπiz1) = e2πi(z1)

2 = (z1)2 = f(z1), we

have that the first m coordinates of fm(z) coincide with the first

m coordinates of fm(z′). By definition, the metric d is bounded byπ and by the choice of p, π

2p < c. So

ρ(fm(z), fm(z′)) =

∞∑k=m+1

d(zk−m, eπ

2k−m−1 izk−m)

2k

≤∞∑

k=m+1

π

2k=

π

2m<

π

2p< c.

Similarly, ρ(fm(w), fm(w′)) < c. Therefore, H((fn)m(A),

(fn)m(B)) < c.

This completes the proof that H((fn)m(A), (fn)

m(B)) < c, for

each m ∈ Z and that c is not an expansion constant for fn. Since

c was chosen arbitrarily, we have shown that fn has no expansion

constant. Therefore, fn is not an expansive homeomorphism. �


Definition 4.23. A homeomorphism f : X → X is continuum-wise expansive if there exists a constant c > 0 such that for eachnondegenerate subcontinuum Y of X, there exists k ∈ Z such thatdiam(fk(Y )) > c. In such case c is a continuum-wise expansionconstant for f .

Lemma 4.24. Let K be a nondegenerate subcontinuum of Fn(X)and let D be a component of the set

∪{E : E ∈ K}. Then diam(K)

≥ diam(D)2n .

Proof. Fix A = {a1, . . . , am} ∈ K and let r < diam(D)2n . For every

i ∈ {1, . . . ,m} let Ui = B(r, ai)∩D. Then Ui = ∅ or diam(Ui) ≤ 2r.

We show that U1 ∪ · · · ∪ Um D. Suppose, to the contrary,that D = U1 ∪ · · · ∪ Um. Let x, y ∈ D be such that d(x, y) =diam(D). Let i0, j0 ∈ {1, . . . ,m} be such that x ∈ Ui0 and y ∈Uj0 . Since D is connected and every Ui is open in D, there exists{i0, i1, . . . , ik} ⊂ {1, . . . ,m} such that ik = j0, ip = iq for everyp = q and Uip ∩ Uip+1 = ∅ for each p ∈ {0, 1, . . . , k − 1}. This is achain of k+1 open subsets from x to y and every link has diameterat most 2r. Then, d(x, y) ≤ 2r(k + 1) ≤ 2rm ≤ 2rn < diam(D);this is a contradiction. Therefore U1 ∪ · · · ∪ Um D.

Let b ∈ D \ (U1 ∪ · · · ∪ Um) = D \∪{B(r, ai) : i ∈ {1, . . . ,m}}.

This implies that ai /∈ B(r, b), for any i ∈ {1, . . . ,m}. Let B ∈ Kbe such that b ∈ B. Notice that H(A,B) ≥ r. Then diam(K) ≥ r.

Since r was an arbitrary number less than diam(D)2n , we conclude

that diam(K) ≥ diam(D)2n . �

Theorem 4.25. The following statements for a continuous func-tion f : X → X are equivalent:

a) f is a continuum-wise expansive homeomorphism,b) fn is a continuum-wise expansive homeomorphism for some

n ≥ 2,c) fn is a continuum-wise expansive homeomorphism for each

n ≥ 2.

Proof. b) ⇒ a). Suppose that fn is a continuum-wise expansivehomeomorphism and let c > 0 be a continuum-wise expansion con-stant for fn. We claim that c is an continuum-wise expansionconstant for f . Let K be a nondegenerate subcontinuum of X.


Then F1(K) is a nondegenerate subcontinuum of Fn(X). There-

fore there exists k ∈ Z such that diam((fn)k (F1(K))) > c. Clearly

diam(fk(K)) = diam((fn)k (F1(K))). Therefore c is a continuum-

wise expansion constant for f .

a) ⇒ c). Let n ∈ N. Suppose that f is a continuum-wise ex-pansive homeomorphism and let c > 0 be a continuum-wise ex-pansion constant for f . We claim that c

2n is a continuum-wiseexpansion constant for fn. Let K be a nondegenerate subcontin-uum of Fn(X). Lemma 2.1 implies that

∪{E : E ∈ K} has at most

n components, but K has infinitely many elements, so one of thosecomponents must be nondegenerate. Let D be one nondegeneratecomponent of

∪{E : E ∈ K}. Since c is a continuum-wise expan-

sion constant for f , there exists k ∈ Z such that diam(fk(D)) >c. Given a ∈ fk(D) there exists x ∈ D such that fk(x) = aand since D ⊂

∪{E : E ∈ K} there is A ∈ K such that x ∈

A. Then a = fk(x) ∈ (fn)k (A) ∈ (fn)

k (K). This shows that

fk(D) ⊂∪{

E : E ∈ (fn)k (K)

}. Let E0 be the component of∪{

E : E ∈ (fn)k (K)

}that contains fk(D). Then diam(E0) ≥

diam(fk(D)) > c. Lemma 4.24 implies that diam((fn)k (K)) ≥

diam(E0)2n and therefore diam((fn)

k (K)) > c2n . This shows that c

2nis a continuum-wise expansion constant for fn. Therefore, fn is acontinuum-wise expansive homeomorphism. �

Acknowledgment

The authors wish to thank Leonardo Espinosa for his technicalhelp during the preparation of this paper.

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Instituto de Matematicas, Universidad Nacional Autonoma de Me-xico, Circuito Exterior, Cd. Universitaria, Mexico 04510, D.F.Mexico

E-mail address: [email protected] address: [email protected]

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