@
Appl. Gen. Topol. 23, no. 2 (2022), 235-242

doi:10.4995/agt.2022.17492

© AGT, UPV, 2022

Dynamics of induced mappings on symmetric

products, some answers

Alejandro Illanes and Verónica Mart́ınez-de-la-Vega

Instituto de Matemáticas, Universidad Nacional Autónoma de México, Circuito Exterior, Cd. Uni-

versitaria, México 04510, Ciudad de México. (illanes@matem.unam.mx, vmvm@matem.unam.mx)

Communicated by A. Linero

Abstract

Let X be a metric continuum and n ∈ N. Let Fn(X) be the hyperspace
of nonempty subsets of X with at most n points. If 1 ≤ m < n, we con-
sider the quotient space F nm(X) = Fn(X)/Fm(X). Given a mapping
f : X → X, we consider the induced mappings fn : Fn(X) → Fn(X)
and fnm : F

n
m(X) → F nm(X). In this paper we study relations among the

dynamics of the mappings f, fn and f
n
m and we answer some questions,

by F. Barragán, A. Santiago-Santos and J. Tenorio, related to the prop-
erties: minimality, irreducibility, strong transitivity and turbulence.

2020 MSC: 54F16; 37B02; 54C05; 54F15.

Keywords: continuum; dynamical system; induced mapping; irreducibility;
symmetric product; turbulence.

1. Introduction

A continuum is a compact connected metric space with more than one point.
Given a nonempty compact metric space X and integers 1 ≤ m < n we con-
sider the following hyperspaces of X:

2X = {A ⊂ X : A is a nonempty closed subset of X},

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

Received 08 April 2022 – Accepted 18 May 2022

http://dx.doi.org/10.4995/agt.2022.17492
https://orcid.org/0000-0002-7109-4038
https://orcid.org/0000-0002-1694-6947


A. Illanes and V. Mart́ınez-de-la-Vega

and the quotient space F nm(X) = Fn(X)/Fm(X).

The hyperspace 2X is considered with the Hausdorff metric [13, Theorem
2.2]. Given subsets U1, . . . , Uk of X, let

〈U1, . . . , Uk〉 = {A ∈ Fn(X) : A ⊂ U1 ∪·· ·∪Uk
and A∩Ui 6= ∅ for each i ∈{1, . . . , k}}.

Then the family of sets of the form 〈U1, . . . , Uk〉, where the sets Ui are open
subsets of X, is a base of the topology in Fn(X) [13, Theorem 3.1]. The
hyperspace Fn(X) is called the n

th-symmetric product of X. We denote the
quotient mapping by qm : Fn(X) → F nm(X) (or qnm, if necessary) and we denote
by F mX the element in F

n
m(X) such that qm(Fm(X)) = {F mX }. A mapping is

a continuous function. Given a mapping f : X → X, the induced mapping
2f : 2X → 2X is defined by 2f (A) = f(A) (the image of A under f). The
induced mapping fn : Fn(X) → Fn(X) (also denoted in some papers by Fn(f))
is the restriction of 2f to Fn(X). The induced mapping f

n
m : F

n
m(X) → F nm(X)

(also denoted by SF nm(f)) is the mapping that makes commutative the following
diagram [8, Theorem 4.3, Chapter VI].

Fn(X)
fn //

qm

��

Fn(X)

qm

��
F nm(X)

fnm

// F nm(X)

A dynamical system is a pair (X, f), where X is a non-degenerate compact
metric space and f : X → X is a mapping. Given a point p ∈ X, the orbit of p
under f is the set orb(p, f) = {fk(p) ∈ X : k ∈ N∪{0}}. A dynamical system
(X, f) induces the dynamical systems (2X, 2f ), (Fn(X), fn) and (F

m
n (X), f

n
m).

H. Hosokawa [12] was the first author that studied induced mappings to
hyperspaces. Since then, this topic has been widely studied. The most common
problem studied in this area is the following. Given a class of mappings M,
determine whether one of the following statements implies another:

(a) f ∈M,
(b) 2f ∈M,
(c) fn ∈M,
(d) fnm ∈M.
Of course, this problem has also been considered for other hyperspaces. Dy-
namical properties of induced mappings on symmetric products have been con-
sidered in [2], [3], [4], [5], [6], [9], [10], [11] and [14]. In particular, in [4] and
[5], the properties of being: exact, mixing, weakly mixing, transitive, totally
transitive, strongly transitive, chaotic, minimal, irreducible, feebly open and
turbulent were studied.

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 236



Dynamics of induced mappings on symmetric products, some answers

The aim of this paper is to solve most of the problems posed by F. Barragán,
A. Santiago-Santos and J. Tenorio in [4] and [5], related to the properties:
minimality, irreducibility, strong transitivity and turbulence.

Throughout this paper the word space means a non-degenerate compact
metric space.

We are aware that some of our proofs can be copied to obtain results with
less restrictions either on the spaces or in the functions, however we consider
that, point out the more general setting under each result holds, is worthless
and breaks the continuity of the paper.

2. Minimality

Let X be a space. A mapping f : X → X is minimal [1, p.7] if there is no
nonempty proper closed subset M of X which is invariant under f (invariance
of M means that f(M) ⊂ M); equivalently, if the orbit of every point of X is
dense in X. The mapping f is totally minimal if fs is minimal for each s ∈ N.

Given n ∈ N, in this section we consider the following statements.
(1) f is minimal,
(2) fn is minimal, and
(3) fn1 is minimal.

In [4, Theorem 4.18], it was proved that (2) implies (3), (3) implies (1), (2)
implies (1), (1) does not imply (2) and (1) does not imply (3), for the case
that X is a continuum. Moreover, in [4, Question 4.2] it was asked whether (3)
implies (2). The following theorem solves this question and even when it has
a very simple proof, it shows that the question and several results on minimal
induced mappings are irrelevant.

Theorem 2.1. Let X be a space, f : X → X a mapping and 1 ≤ m < n.
Then:

(a) fn(F1(X)) ⊂ F1(X),
(b) fnm(F

m
X ) = F

m
X ,

(c) for each A ∈ Fm(X), orb(A, fn) ⊂ Fm(X). Thus, orb(A, fn) is not dense
in Fn(X) and fn is not minimal, and

(d) orb(F mX , f
n
m) = {F mX }. Thus, orb(F

m
X , f

n
m) is not dense in F

n
m(X) and f

n
m

is not minimal.

Proof. Take a point p ∈ X. Then fn({p}) = f({p}) = {f(p)} ∈ F1(X).
Moreover, fnm(F

m
X ) = f

n
m(qm({p})) = qm(fn({p})) = qm({f(p)}) = F mX . This

proves (a), (b) and (d). The proof of (c) is similar. �

Theorem 2.1 (b) implies that the mappings fn and f
n
m are never minimal or

totally minimal. Then proved results in which minimality or total minimality
of fn or f

n
m is either assumed or concluded become irrelevant or partially ir-

relevant, such is the case of the following results by Barragán, Santiago-Santos
and Tenorio: Theorem 4.18, Corollary 4.19, Corollary 4.20 and Corollary 4.21
of [4]; Corollary 5.13 and Corollary 5.17 of [6].

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 237



A. Illanes and V. Mart́ınez-de-la-Vega

3. Irreducibility

Let X be a space. A mapping f : X → X is irreducible [1, p.171] if the only
closed subset A of X for which f(A) = X is A = X;

Given n ∈ N, in this section we consider the following statements.
(1) f is irreducible,
(2) fn is irreducible,
(3) fn1 is irreducible, and
(4) fnm is irreducible.

Using [4, Theorem 5.1], in [5, Theorem 4.1] it was shown that each one of
the statements (2), (3) and (4) implies (1). The authors of [4] and [5] supposed
that the spaces are continua, however, it is easy see that the proofs for these
results are valid for infinite compact metric spaces without isolated points. The
rest of the implications among (1), (2), (3) and (4) are left as questions in [4,
Questions 5.5] and [5, Question 4.2]. The purpose of this section is to complete
the proof that, in fact, all the statements (1)-(4) are equivalent.

Theorem 3.1. Let X be a space without isolated points, f : X → X a mapping
and 1 ≤ m < n. If f is irreducible, then fn is irreducible.

Proof. Suppose that f is irreducible.
Claim 1. If U is a nonempty open subset of X, then there exists p ∈ U

such that f(p) /∈ f(X \U).
In order to prove Claim 1, let A = X\U. Then A is a proper closed subset of

X. Since f is irreducible, f is onto. Thus there exist q ∈ X such that q /∈ f(A)
and p ∈ X such that f(p) = q. Observe that p ∈ U. This finishes the proof of
Claim 1.

Claim 2. If U is a nonempty open subset of Fn(X), then there exists B ∈U
such that B ∈ Fn(X) \Fn−1(X) and f(B) /∈ fn(Fn(X) \U).

We prove Claim 2. Since X does not have isolated points, Fn(X)\Fn−1(X) is
dense in Fn(X). Then there exists D = {p1, . . . , pn}∈ (Fn(X)\Fn−1(X))∩U.
Then there exist pairwise disjoint open subsets U1, . . . , Un of X such that for
each i ∈ {1, . . . , n}, pi ∈ Ui and D ∈ 〈U1 . . . , Un〉 ⊂ U. By Claim 1, for each
i ∈{1, . . . , n}, there exists ui ∈ Ui such that f(ui) /∈ f(X \Ui).

Define B = {u1, . . . , un}. Clearly, B ∈ Fn(X) \ Fn−1(X). Suppose that
there exists E ∈ Fn(X) \ U such that f(E) = f(B). Given i ∈ {1, . . . , n},
let ei ∈ E be such that f(ei) = f(ui). By the choice of ui, ei ∈ Ui. Thus
E ∈ 〈U1 . . . , Un〉⊂U, a contradiction. This proves that f(B) /∈ fn(Fn(X)\U).
This finishes the proof of Claim 2.

We are ready to prove that fn is irreducible. Let A be a proper closed
subset of Fn(X) and U = Fn(X) \A. By Claim 2, there exists B ∈ U such
that B ∈ Fn(X) \Fn−1(X) and f(B) /∈ fn(A). Therefore fn(A) 6= Fn(X) and
fn is irreducible. �

Theorem 3.2. Let X be an space without isolated points, f : X → X a map-
ping and 1 ≤ m < n. If fn is irreducible, then fnm is irreducible.

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 238



Dynamics of induced mappings on symmetric products, some answers

Proof. Suppose that fnm is not irreducible. We will prove that fn is not ir-
reducible. Then there exists a proper closed subset A of F nm(X) such that
fnm(A) = F nm(X). Let B = q−1m (A ∪ {F mX }). Then B is a closed subset of
Fn(X).

We check that B 6= Fn(X). Set U = F nm(X) \A. Then U is a nonempty
open subset of F nm(X). This implies that q

−1
m (U) is a nonempty open subset

of Fn(X). Since X does not have isolated points, Fn(X) \ Fn−1(X) is dense
in Fn(X). So, there exists G ∈ (Fn(X) \ Fn−1(X)) ∩ q−1m (U). Thus qm(G) ∈
U \{F mX } = F

n
m(X) \ (A∪{F mX }). Hence G /∈B. Therefore B 6= Fn(X).

Now, we prove that fn(B) = Fn(X). Since fn is surjective, we have that
f is surjective [4, Theorem 3.2]. Take E ∈ Fn(X). In the case that E =
{q1, . . . , qk}∈ Fm(X), with k ≤ m. Since f is surjective, for each i ∈{1, . . . , k}
there exists pi ∈ X such that f(pi) = qi. Thus {p1, . . . , pk} ∈ Fm(X) =
q−1m (F

m
X ) ⊂B. Therefore E = fn({p1, . . . , pk}) ∈ fn(B). Now we suppose that

E /∈ Fm(X). Let A ∈ A be such that fnm(A) = qm(E). Let B ∈ Fn(X) be
such that A = qm(B). Then B ∈ B. Since qm(E) = fnm(A) = fnm(qm(B)) =
qm(fn(B)) and E /∈ Fm(X), we have that {E} = q−1m (qm(E)) = fn(B). There-
fore E ∈ fn(B). We have shown that fn(B) = Fn(X). Therefore fn is not
irreducible. Therefore, we have shown that if fnm is not irreducible, then fn is
not irreducible. �

Corollary 3.3. Let X be a space without isolated points, 1 ≤ m < n and
f : X → X a mapping. Then the following statements are equivalent.

(1) f is irreducible,
(2) fn is irreducible, and
(3) fnm is irreducible.

4. Strong transitivity

Let X be a space. A mapping f : X → X is strongly transitive [15, p.369]
if for each nonempty open subset U of X, there exists r ∈ N such that⋃r

i=0 f
i(U) = X.

Given 1 ≤ m < n, in this section we consider the following statements.

(1) f is strongly transitive,
(2) fn is strongly transitive,
(3) fn1 is strongly transitive, and
(4) fnm is strongly transitive.

Using [4, Theorem 4.13], in [5, Theorem 3.17] it was shown that (2) implies
(1), (2) implies (3), (2) implies (4), (3) implies (1), (4) implies (1), (1) does not
imply (2), (1) does not imply (3) and (1) does not imply (4). The authors of
[4] and [5] supposed that the spaces are continua, however it is easy to see that
the proofs for these results are valid for non-degenerate compact metric spaces
without isolated points.

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 239



A. Illanes and V. Mart́ınez-de-la-Vega

The questions whether the rest of the implications hold are contained in [4,
Question 4.1] and [5, Question 3.18]. With the following theorem we show that
all these implications hold.

Theorem 4.1. Let X be a space without isolated points, f : X → X a mapping
and 1 ≤ m < n. If fnm is strongly transitive, then fn is is strongly transitive.

Proof. Let U be a nonempty open subset of Fn(X).
Fix an element A = {a1, . . . , ak}∈U, where k ≤ n and the cardinality of A

is k. Let W ′1, . . . W
′
k be pairwise disjoint open subsets of X such that for each

i ∈ {1, . . . , k}, ai ∈ W ′i and W
′ = 〈W ′1, . . . , W ′k〉 ⊂ U. For each i ∈ {1, . . . , k},

choose an open subset Wi of X such that ai ∈ Wi ⊂ clX (Wi) ⊂ W ′i . Let
W = 〈W1, . . . , Wk〉. Since X does not have isolated points, Fn(X) \ Fm(X)
is dense in Fn(X) and the set V = W \ Fm(X) is a nonempty open subset
of Fn(X). Observe that qm(V) is a nonempty open subset of F nm(X). By
hypothesis there exists r ∈ N such that

⋃r
i=0(f

n
m)

i(qm(V)) = F nm(X).
We claim that

⋃r
i=0(fn)

i(W′) = Fn(X). Take an element B ∈ Fn(X). Let
{Bs}∞s=1 be a sequence in Fn(X) \ Fn−1(X) such that lims→∞Bs = B. Given
s ∈ N, there exist Ds ∈ V and is ∈ {0, 1, . . . , r} such that (fnm)is (qm(Ds)) =
qm(Bs). This implies that qm(f

is (Ds)) = qm(Bs).
Since Fn(X) is compact, we may suppose that the sequence {Ds}∞s=1 con-

verges to an element D ∈ Fn(X) and there exists j ∈ {0, 1, . . . , r} such that
for each s ∈ N, is = j.

Given s ∈ N, qm(fj(Ds)) = qm(Bs). Since Bs /∈ Fm(X), we obtain that
fj(Ds) = Bs. By the continuity of f

j, fj(D) = B. Since Ds ∈ V ⊂ W ⊂
clFn(X)(W), we conclude that D ∈ clFn(X)(W) ⊂ 〈clX (W1), . . . , clX (Wk)〉 ⊂
〈W ′1, . . . , W ′k〉 = W

′. Therefore B ∈ (fn)j(W′). This finishes the proof that⋃r
i=0(fn)

i(W′) = Fn(X), so,
⋃r

i=0(fn)
i(U) = Fn(X) and completes the proof

of the theorem. �

5. Turbulence

Let X be a space. A mapping f : X → X is turbulent [7, p.588] if there are
compact non-degenerate subsets K and L of X such that K ∩ L has at most
one point and K ∪L ⊂ f(K) ∩f(L).

Given 1 ≤ m < n, in this section we consider the following statements.
(1) f is turbulent,
(2) fn is turbulent,
(3) fn1 is turbulent, and
(4) fnm is turbulent.

Using [4, Theorem 5.6] in [5, Theorem 4.5] it follows that (1) implies (2),
(3) and (4). In [5, Questions 4.6], it was asked whether one of the rest of the
possible implications holds, when X is a continuum.

The following example shows that (2) and (3) does not imply (1), when X
is a compact metric space.

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 240



Dynamics of induced mappings on symmetric products, some answers

Problem 5.1. Does one of the statements (2), (3) or (4) implies another for
a compact metric space?

Example 5.2. There exist a non-degenerate compact metric space X and a
mapping f : X → X such that f2 and f21 are turbulent but f is not turbulent.

Define X = {0}∪{1
n

: n ∈ N}. For each m ∈ N, let am = 13m−2 , bm =
1

3m−1
and cm =

1
3m

. Then

X = {0}∪{am : m ∈ N}∪{bm : m ∈ N}∪{cm : m ∈ N}.
Define f : X → X by

f(p) =




0, if p = 0,

ck, if p = a2k−1,

bk, if p = a2k,

ak, if p ∈{b2k, c2k, b2k−1, c2k−1}.
Clearly, f is an onto mapping.
Suppose to the contrary that f is turbulent. Then there are compact non-

degenerate subsets K and L of X such that K ∩ L has at most one point and
K ∪L ⊂ f(K) ∩f(L).

If there exists k ≥ 2 such that ck ∈ K ∪ L, since f−1(ck) = {a2k−1}, we
have that a2k−1 ∈ K ∩ L. Since f−1(a2k−1) = {b4k−2, c4k−2, b4k−3, c4k−3},
there is p ∈ {b4k−2, c4k−2, b4k−3, c4k−3} ∩ K such that f(p) = a2k−1. Since
f−1(p) = {ai} for some i > 4k − 3 > 2k − 1, we have that ai ∈ K ∩ L. Thus
{ai, a2k−1} ⊂ K ∩ L, a contradiction. Thus (K ∪ L) ∩ {ck : k ≥ 2} = ∅.
Similarly, (K ∪ L) ∩{bk : k ≥ 2} = ∅. Therefore K ∪ L ⊂ {ak : k ∈ N}∪
{b1, c1}∪{0}.

If there exists k ≥ 2 such that ak ∈ K ∪ L, then there exists k′ > 2 such
that {bk′, ck′}∪(K ∪L) 6= ∅. This contradicts what we proved in the previous
paragraph. Thus K∪L ⊂{a1, b1, c1}∪{0}. Since ({a1, b1, c1}∪{0})∩f−1(b1) =
∅, we have that b1 /∈ K ∪ L. Hence K ∪ L ⊂ {a1, c1}∪{0}. Since ({a1, c1}∪
{0}) ∩f−1(a1) = {c1} and ({a1, c1}∪{0}) ∩f−1(c1) = {a1}, we obtain that if
{a1, c1}∩ (K ∪ L) 6= ∅, then {a1, c1} ⊂ K ∩ L, a contradiction. This proves
that K ∪ L ⊂ {0}, a contradiction. This completes the proof that f is not
turbulent.

Now, we check that f2 is turbulent. Define

K = {{am, bm}∈ F2(X) : m ∈ N}∪{{0}}, and
L = {{am, cm}∈ F2(X) : m ∈ N}∪{{0}}.

Then K and L are compact non-degenerate subsets of F2(X) and K∩L =
{{0}}.

Given m ∈ N, {am, bm} = {f(c2m), f(a2m)} = f2({c2m, a2m}) ∈ f2(L).
Moreover, {am, bm} = {f(b2m), f(a2m)} = f2({b2m, a2m}) ∈ f2(K). Since
{0} = {f(0)} = f({0}}) = f2({0}) ∈ f2(K) ∩ f2(L). We have shown that
K⊂ f2(K) ∩f2(L). Similarly, L⊂ f2(K) ∩f2(L). Therefore, f2 is turbulent.

© AGT, UPV, 2022 Appl. Gen. Topol. 23, no. 2 241



A. Illanes and V. Mart́ınez-de-la-Vega

Using K0 = q1(K) and L0 = q1(L), it can be proved that f21 is turbulent.

Acknowledgements. This paper was partially supported by the projects
“Teoŕıa de Continuos, Hiperespacios y Sistemas Dinámicos III”, (IN 106319)
of PAPIIT, DGAPA, UNAM; and “Teoŕıa de Continuos e Hiperespacios, dos”
(AI-S-15492) of CONACYT. The authors wish to thank Leonardo Espinosa for
his technical help during the preparation of this paper.

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