Topology and its Applications 160 (2013) 63–73

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Topology and its Applications

www.elsevier.com/locate/topol

Towards the complete classification of generalized tent maps inverselimits ✩

Iztok Banic a,c, Matevž Crepnjak a,b,∗, Matej Merhar a, Uroš Milutinovic a,c

a Faculty of Natural Sciences and Mathematics, University of Maribor, Koroška 160, Maribor 2000, Sloveniab Faculty of Chemistry and Chemical Engineering, University of Maribor, Smetanova 17, Maribor 2000, Sloveniac Institute of Mathematics, Physics and Mechanics, Jadranska 19, Ljubljana 1000, Slovenia

a r t i c l e i n f o a b s t r a c t

Article history:Received 11 October 2011Received in revised form 21 September2012Accepted 25 September 2012

MSC:54C6054B10

Keywords:ContinuaInverse limitsGeneralized tent mapsKnaster continua

We study generalized tent maps inverse limits, i.e. inverse limits of inverse sequencesof unit segments I with a generalized tent map being the only bonding function. Asthe main result we identify an infinite family of curves in I2 such that if top pointsof graphs of generalized tent maps belong to the same curve, the corresponding inverselimits are homeomorphic, and if they belong to different curves, the inverse limits arenon-homeomorphic. The inverse limits corresponding to certain families of top points areexplicitly determined, and certain properties of the inverse limit are proved in the case of(0,1) as the top point.

© 2012 Elsevier B.V. All rights reserved.

1. Introduction

Continua as inverse limits have been studied for a long time. One reason for such intense research in this area is the factthat inverse sequences with very simple spaces and simple bonding maps can give extremely complicated continua as theirinverse limits. The inverse limits may be both complicated and useful even in the case, when all the spaces are unit intervals[0,1] and all the bonding functions are the same. Such inverse sequences and their inverse limits play an important role inthe continuum theory as well as in the theory of topological dynamical systems. They also appear in applications in suchdiverse areas as economy, mechanics of fluids, physics and more; see [37,39–41,44,45].

Even such a simple case when the graph of the bonding function is the union of two segments may be highly non-trivial.Such functions are called generalized tent maps and they are the main object of our study. In Definition 1.1 we introducethe basic notation related to them, which we use later in the paper.

Definition 1.1. For any a,b ∈ [0,1], the generalized tent function f(a,b) : [0,1] → [0,1] is defined as the set-valued func-tion with the graph Γ ( f(a,b)) being the union of the segment (possibly degenerate) from (0,0) to (a,b) and the segment

✩ This work was supported in part by the Slovenian Research Agency, under Grants L2-7207 (C) and J1-6150 (B).

* Corresponding author at: Faculty of Natural Sciences and Mathematics, University of Maribor, Koroška 160, Maribor 2000, Slovenia.E-mail addresses: iztok.banic@uni-mb.si (I. Banic), matevz.crepnjak@um.si (M. Crepnjak), matej.merhar@uni-mb.si (M. Merhar),

uros.milutinovic@uni-mb.si (U. Milutinovic).

0166-8641/$ – see front matter © 2012 Elsevier B.V. All rights reserved.http://dx.doi.org/10.1016/j.topol.2012.09.017

64 I. Banic et al. / Topology and its Applications 160 (2013) 63–73

(possibly degenerate) from (a,b) to (1,0). The point (a,b) is called the top point of the graph Γ ( f(a,b)). The inverse limitobtained from the inverse sequence of closed unit intervals [0,1] and the bonding function f(a,b) is denoted by

K(a,b) = lim←−{[0,1], f(a,b)

}∞n=1.

Note that f(a,b) is single-valued if and only if a /∈ {0,1} or (a,b) ∈ {(0,0), (1,0)}.The first and the most famous example of continua K(a,b) is K

( 12 ,1)

, called the Brouwer–Janiszewski–Knaster continuum

or sometimes just the Knaster continuum.The whole family of continua K

( 12 ,b)

, 12 < b � 1, has been called Knaster continua and the famous Ingram conjecture,

stated in 1992, claimed that all of them are pairwise non-homeomorphic. It generated a large number of articles, suchas Barge, Brucks, Diamond [6], Barge, Diamond [8,9], Barge, Jacklitch, Vago [11], Barge, Martin [12–14], Block, Jakimovik,Kailhofer, Keesling [15], Block, Keesling, Raines, Štimac [16], Brucks, Bruin [17], Brucks, Diamond [18], Bruin [19–21], Collet,Eckmann [23], Good, Knight, Raines [27], Good, Raines [28], Kailhofer [35,36], Raines [52], Raines, Štimac [53,54], Štimac[55–58], Swanson, Volkmer [59], and others, in which certain special cases of the conjecture were proved. Finally, theconjecture was proved in 2009 by M. Barge, H. Bruin and S. Štimac [7]. In this paper the authors exploited results on inverselimit spaces of inverse sequences of closed intervals with bonding maps belonging to the logistic family fa(x) = ax(1 − x),a ∈ [0,4]; among other sources it uses the highly relevant paper [10].

In spite of such great effort and many obtained results the complete classification of all inverse limits K(a,b) is still anopen problem. In this paper we continue the study of inverse limits of generalized tent maps f(a,b) and their classification.More specifically we explicitly determine the inverse limits corresponding to the top points (a,b) for a, b satisfying 0 <

a < b � 1 − a. Together with cases covered by [5,7] the inverse limits are thus recognized as already known continua for alltop points (a,b) except for those where 1

2 �= max{a,1 − a} < b < 1 or (a,b) = (0,1). In the present paper we also introducecertain curves Ct , subsets of the region {(a,b) ∈ [0,1] × [0,1] | max{a,1 − a} < b < 1}, and prove that the inverse limits oftwo generalized tent maps are homeomorphic if their top points belong to the same curve Cn , for some positive integer n.Also such an inverse limit space can be recognized as one appearing in the Ingram conjecture. Finally, we prove certain newproperties of the inverse limit in the case when (0,1) is the top point.

Note that in some cases the generalized tent maps are not single-valued and therefore the concept of inverse limits ofinverse sequences with upper semicontinuous set-valued bonding functions is needed. Such a generalization of the conceptof inverse limits was introduced in [34,42] by W.T. Ingram and W.S. Mahavier. They gave conditions under which the inverselimit of an inverse sequence of Hausdorff spaces with upper semicontinuous set-valued bonding functions is a Hausdorffcontinuum, provided some interesting examples of such inverse limits, and discussed their dimension. The concept of thesegeneralized inverse limits has become very popular since its introduction and has been studied by many authors and manypapers appeared; for examples see [1–5,22,24,32–34,38,42,50,51,60].

A similar problem was studied in [30,31] by W.T. Ingram. There he studied inverse limits of inverse sequences of [0,1]and piecewise linear unimodal bonding map with the graph of the map being the union of the segment from (0,b) to (c,1)

and the segment from (c,1) to (1,0), where (b, c) ∈ [0,1] × (0,1). Considering different values of the parameters b and c,he gives a partial classification and many useful properties of such inverse limits.

2. Definitions and notation

Our definitions and notation mostly follow [34] and [48].A continuum is a nonempty, compact and connected metric space.Let W = {(x, sin 1

x ) ∈ R2 | 0 < x � 1}. Any continuum homeomorphic to Cl(W ) is called a sin 1

x -continuum.A harmonic fan is any continuum, homeomorphic to the continuum, defined as the union (

⋃∞n=1 Kn) ∪ K , where for

each n, Kn is the segment in the plane from (0,0) to (1, 1n ), and K is the segment from (0,0) to (1,0).

Let (Xn,dn) be a sequence of metric spaces, where all metrics are bounded by 1. Then

D(x, y) = supn∈{1,2,3,...}

{dn(xn, yn)

n

},

where x = (x1, x2, x3, . . .), y = (y1, y2, y3, . . .), will be used for the metric on the product space∏∞

n=1 Xn (it is well knownthat the metric D induces the product topology [25, p. 190], [47, p. 123]).

If (X,d) is a compact metric space, then 2X denotes the set of all nonempty closed subsets of X . Let for each ε > 0 andeach A ∈ 2X

Nd(ε, A) = {x ∈ X

∣∣ d(x,a) < ε for some a ∈ A}.

We will always equip the set 2X with the Hausdorff metric Hd , which is defined as

Hd(H, K ) = inf{ε > 0

∣∣ H ⊆ Nd(ε, K ), K ⊆ Nd(ε, H)},

for H, K ∈ 2X . Then (2X , Hd) is a metric space, called the hyperspace of the space (X,d). For more details see [29,48].

I. Banic et al. / Topology and its Applications 160 (2013) 63–73 65

Let X and Y be compact metric spaces. A single-valued function f : X → 2Y is also called a set-valued function f :X → Y . A set-valued function f : X → Y is upper semicontinuous (abbreviated u.s.c.) if for each open set V ⊆ Y the set{x ∈ X | f (x) ⊆ V } is an open set in X .

The graph Γ ( f ) of an u.s.c. set-valued function f : X → Y is the set of all points (x, y) ∈ X × Y such that y ∈ f (x).Ingram and Mahavier gave the following characterization of u.s.c. functions [34, p. 120]:

Theorem 2.1. Let X and Y be compact metric spaces and f : X → Y a set-valued function. Then f is u.s.c. if and only if its graph Γ ( f )is closed in X × Y .

In this paper we deal with inverse sequences {Xn, fn}∞n=1, where Xn are compact metric spaces and fn : Xn+1 → Xn areu.s.c. set-valued functions.

The inverse limit of an inverse sequence {Xn, fn}∞n=1 is defined to be the subspace of the product space∏∞

n=1 Xn of allx = (x1, x2, x3, . . .) ∈ ∏∞

n=1 Xn , such that xn ∈ fn(xn+1) for each n. The inverse limit is denoted by lim←−{Xn, fn}∞n=1. The notionof the inverse limit of an inverse sequence with u.s.c. bonding functions was introduced by W.S. Mahavier in [42] andW.T. Ingram and W.S. Mahavier in [34].

For any compact metric space X we use dim(X) for the topological (covering) dimension of X (for the definition see [26,p. 385] or [49, p. 10]).

For the reader’s convenience we list the following well-known results that will be used later:

Theorem 2.2. ([26, p. 393]) Let X and Y be compact metric spaces such that dim(X) = 0. Then

dim(X × Y ) = dim(Y ).

Theorem 2.3. ([49, p. 15]) Let {Xn}∞n=1 be a sequence of compact subspaces of a metric space and let k be a nonnegative integer, suchthat dim(Xn) � k for all n. Then

dim

( ∞⋃n=1

Xn

)� k.

3. Main results

In this section we formulate and prove the main results.First we introduce basic notation and facts.Since cases when b � a or b = 1 or a = 0 were already covered in [5], in this section we restrict our attention to the

remaining case when a ∈ (0,1), b ∈ (0,1) and b > a. Although the case (a,b) = (0,1) has already been studied in [5,22],we include a short section (Section 4) dealing exclusively with K(0,1) , adding some new insight into its very complicatedstructure.

We see that in this section according to Definition 1.1 for each (a,b) satisfying a ∈ (0,1), b ∈ (0,1) and b > a,

f(a,b)(t) ={

ba t; if t ∈ [0,a],

ba−1 (t − 1); if t ∈ [a,1].

Note that f(a,b) in this case is single-valued.

The point e = f(a,b)(b) = b(b−1)a−1 plays an important role in the study of K(a,b) . The restrictions of f(a,b) mapping [0,a]

onto [0,b], and [a,b] onto [e,b], respectively, are bijections. Therefore they have the inverse functions and we denote themby L : [0,b] → [0,a] and R : [e,b] → [a,b], respectively. It is easy to see that L(t) = a

b t and R(t) = a−1b t + 1.

For any point (x1, x2, x3, . . .) ∈ K(a,b) and any positive integer n, xn = f(a,b)(xn+1). If xn+1 < a then xn+1 = L(xn), if xn+1 > athen xn+1 = R(xn), and finally if xn+1 = a then xn = b and xn+1 = L(xn) = R(xn). Note that if xn+1 � a then xn+1 ∈ [a,b],because xn+1 = f(a,b)(xn+2) � b. Therefore in that case xn ∈ [e,b]. This fact was the main reason for the introduction of eand our choice of the restriction of f(a,b) in the definition of R .

That means that xn+1 = R(xn) is possible only for xn � e, but note that for xn+1 = L(xn) there are no restrictions.We continue with the following lemma that gives a convenient description of an arc that we will use in the proof of

Theorem 3.2.

Lemma 3.1. Let X be a compact metric space and A its subspace, and for each positive integer n, let An be an arc in X from an to an+1such that

(1) for each positive integer n, An ∩ An+1 = {an+1},(2) Ai ∩ A j �= ∅ if and only if |i − j|� 1,

66 I. Banic et al. / Topology and its Applications 160 (2013) 63–73

(3) there is a point z ∈ X \ (⋃∞

n=1 An) such that limn→∞ An = {z} in 2X ,(4) A = (

⋃∞n=1 An) ∪ {z}.

Then A is an arc.

Proof. Let for each positive integer n, fn : [n−1n , n

n+1 ] → An be a homeomorphism such that fn(n−1n ) = an and fn( n

n+1 ) =an+1. Next, let f : [0,1] → X be the function, defined by f (t) = fn(t) if t ∈ [n−1

n , nn+1 ] for some positive integer n, and

f (1) = z. Since fn( nn+1 ) = an+1 = fn+1(

nn+1 ) for each positive integer n, it follows that f is continuous on [0,1) [25, p. 83].

Let {ti}∞i=1 be a sequence in [0,1] such that limi→∞ ti = 1. Then it follows from (3) that limi→∞ f (ti) = z and therefore f iscontinuous at t = 1. It also follows from (4) that f is surjective and from (1), (2), and (3) that f is injective. Therefore f isa homeomorphism from [0,1] onto A. �

In the following theorem we prove that for certain values of the parameters a and b the inverse limit K(a,b) is an arc. Inthe proof a point z and certain subspaces An of K(a,b) satisfying conditions of Lemma 3.1 are recognized. The fact that thesubspaces An are arcs is proved by direct explicit construction of homeomorphisms.

Theorem 3.2. Let a,b ∈ (0,1), a < b and b < 1 − a. Then K(a,b) is an arc.

Proof. In order to simplify notation denote f(a,b) by f . Let e = f (b) = b(b−1)a−1 . From a < b < 1 and b < 1 − a it follows that

| ba−1 | < 1 and hence e − a = (b−a)(1−a−b)

1−a > 0. Therefore e > a, implying that f ([a,b]) = [e,b] ⊆ [a,b].Using the language and techniques of itineraries (as defined and used in [35,36,15]) one can see that the only backward

itineraries that appear in this case are R∞ and Rk L∞ , where k � 0. This leads naturally to a family of subspaces An of K(a,b)

that have the desired properties of Lemma 3.1 and then the claim of the present theorem follows by Lemma 3.1.But we will present a different proof that will give explicit homeomorphisms from closed intervals onto a similar fam-

ily An (which mostly coincide with the subspaces obtained by the itinerary approach). We choose this approach in order tomake our result more easily understood by those readers that are not familiar to the itinerary methods.

Define A1 to be the set of all points (x1, x2, x3, . . .) ∈ K(a,b) such that xi+1 = L(xi) for each positive integer i. It followsthat A1 is an arc from

a1 = (0,0,0, . . .) to a2 = (b,a, L(a), L2(a), . . .

),

since one easily proves that t �→ (t, L(t), L2(t), . . .) is a homeomorphism from [0,b] onto A1.Let A2 be the set of all points (x1, x2, x3, . . .) ∈ K(a,b) such that x2 = R(x1) and xi+1 = L(xi) for each positive integer i � 2.

It follows that A2 is an arc from

a2 = (b,a, L(a), L2(a), . . .

)to a3 = (

e,b,a, L(a), L2(a), . . .),

since t �→ (t, R(t), L(R(t)), L2(R(t)), . . .) is a homeomorphism from [e,b] onto A2.Also, for each positive integer n � 3 define the set An to be the set of all points (x1, x2, x3, . . .) ∈ K(a,b) such that xn =

R(xn−1) and xi+1 = L(xi) for each positive integer i � n. It follows that An is an arc from

an = (f n−2(e), . . . , f 2(e), f (e), e,b,a, L(a), L2(a), L3(a), . . .

)to

an+1 = (f n−3(e), . . . , f 2(e), f (e), e,b,a, L(a), L2(a), L3(a), . . .

),

since t �→ ( f n−2(t), . . . , f 2(t), f (t), t, R(t), L(R(t)), L2(R(t)), . . .) is a homeomorphism from [e,b] onto An .Since R is an expansive map the only point of the form (t, R(t), R2(t), . . .) is obtained in the case when t = R(t). One

easily checks that t0 = b1+b−a is the only solution of the equation t = R(t). Let z = (t0, t0, t0, . . .).

Since a < e and L(b) = a, it follows that L(t) � L(b) = a < e for each t ∈ [e,b], and therefore R(L(t)) is not defined. Itfollows that K(a,b) = (

⋃∞n=1 An) ∪ {z}.

One can easily check that An ∩ An+1 = {an+1} for each positive integer n and that for positive integers m and n such that|m − n| > 1 it holds that Am ∩ An = ∅.

We have shown properties (1), (2) and (4) of Lemma 3.1 and now need only to show (3).The function f is a contraction mapping on [e,b] with the contraction factor M = b

1−a < 1. Using this we show that

limn→∞ An = {z} in 2K(a,b) . Take any ε > 0. Let n0 be a positive integer such that Mn < εb−e for each n � n0. Then for each

n � n0 and for each t ∈ [e,b],∣∣ f n(t) − t0∣∣ = ∣∣ f n(t) − f n(t0)

∣∣ < Mn|t − t0|� Mn(b − e) < ε.

I. Banic et al. / Topology and its Applications 160 (2013) 63–73 67

Fig. 1. Ct for t = 1,2,3,4.

Choose a positive integer k, such that 1k < ε. Next we prove that for each n � n0 + k + 1 and for each x ∈ An it holds that

D(x, z) < ε. Take any

x = (x1, x2, . . . , xk, . . .)

= (f n−2(t), . . . , f n−k−1(t)︸ ︷︷ ︸

k

, . . . , f 2(t), f (t), t, R(t), L(

R(t)), L2(R(t)

), . . .

)in An . Take arbitrary positive integer m. If m � k then

d(xm, t0)

m� 1

m� 1

k< ε.

If m < k then n − m − 1 > n − k − 1 � n0, hence

d(xm, t0)

m= | f n−m−1(t) − t0|

m< ε.

It follows that D(x, z) < ε for each x ∈ An and therefore H D(An, {z}) < ε. Using Lemma 3.1 it follows that K(a,b) is an arc. �Definition 3.3. For any t ∈ [1,∞) let

Ct = {(x, y) ∈ [0,1] × [0,1] ∣∣ xt+1 − xt = yt+1 − yt, 0 < x < y

}.

See Fig. 1.

It is easy to see that C1 is the graph of the function f : (0, 12 ) → ( 1

2 ,1), f (x) = 1 − x.In the next theorem we prove that the inverse limits K(a,b) and K(c,d) are homeomorphic if (a,b) and (c,d) belong to

the same set Cn , where n is a positive integer.

Theorem 3.4. Let n be a positive integer and (a,b), (c,d) ∈ Cn. Then K(a,b) and K(c,d) are homeomorphic.

Proof. Let e1 = f(a,b)(b) = b(b−1)a−1 and e2 = f(c,d)(d) = d(d−1)

c−1 .

From (a,b) ∈ Cn it follows that ( ba )n−1e1 = ( b

a )n−1 b(b−1)a−1 = a, ( b

a )ne1 = ( ba )n b(b−1)

a−1 = b. Note that as ab < 1 then e1 < a.

Similarly for (c,d) one gets ( d )n−1e2 = c, ( d )ne2 = d, and e2 < c.

c c

68 I. Banic et al. / Topology and its Applications 160 (2013) 63–73

Moreover

0 < e1 < f(a,b)(e1) < f 2(a,b)(e1) < · · · < f n−1

(a,b)(e1) = a < f n

(a,b)(e1) = b,

and f k(a,b)

(e1) = ( ba )ke1 for each k = 0,1, . . . ,n. Similarly

0 < e2 < f(c,d)(e2) < f 2(c,d)(e2) < · · · < f n−1

(c,d)(e2) = c < f n

(c,d)(e2) = d,

and f k(c,d)

(e2) = ( dc )ke2 for each k = 0,1, . . . ,n.

Let I0 = [0, e1],

Ik = [f k−1(a,b)

(e1), f k(a,b)(e1)

] =[(

b

a

)k−1

e1,

(b

a

)k

e1

]for each k = 1, . . . ,n, and In+1 = [b,1].

Analogously, we define J0 = [0, e2],

Jk = [f k−1(c,d)

(e2), f k(c,d)(e2)

] =[(

d

c

)k−1

e2,

(d

c

)k

e2

]for each k = 1, . . . ,n, and Jn+1 = [d,1].

Now define the continuous piecewise linear increasing function ϕ : [0,1] → [0,1], which maps each interval Ik affinelyonto Jk , for k = 0,1,2, . . . ,n,n + 1, in the obvious way.

It is easily checked that f(c,d) ◦ϕ = ϕ ◦ f(a,b) by noticing that values of both compositions at each endpoint of any intervalIk coincide and that their restrictions to Ik are linear functions. Since ϕ is a homeomorphism, that means that f(a,b) andf(c,d) are conjugate and by a well-known result it follows that the corresponding inverse limit spaces K(a,b) and K(c,d) arehomeomorphic. �

One can easily see that for each t ∈ (1,∞), Ct is a subset of [0,1] × [0,1] containing ( 12 , y) for exactly one y ∈ ( 1

2 ,1).

Corollary 3.5. If (a,b) ∈ Cm and (c,d) ∈ Cn for some positive integers m,n � 2, m �= n, then the continua K(a,b) and K(c,d) are nothomeomorphic.

Proof. Let t1, t2 ∈ ( 12 ,1) such that ( 1

2 , t1) ∈ Cm , ( 12 , t2) ∈ Cn . It follows from Theorem 3.4 that K(a,b) is homeomorphic to

K( 1

2 ,t1)and K(c,d) is homeomorphic to K

( 12 ,t2)

. Since K( 1

2 ,t1)and K

( 12 ,t2)

are not homeomorphic, by the positive solution of

the Ingram conjecture [7], it follows that K(a,b) and K(c,d) are not homeomorphic. �We have proved in Theorem 3.4 that K(a,b) and K(c,d) are homeomorphic for any (a,b), (c,d) ∈ C1 (recall that C1 is the

open interval from (0,1) to ( 12 , 1

2 )), but we are able to give more precise information about these continua, as shown in thefollowing theorem. The proof consists of an explicit construction of a homeomorphism from K(a,b) to a conveniently chosencopy of the sin 1

x -continuum for any (a,b) ∈ C1.

Theorem 3.6. If (a,b) ∈ C1 , then K(a,b) is a sin 1x -continuum.

Proof. Note that in this case e = a and R(t) = 1 − t for each t ∈ [a,b] and that (a,b) ∈ C1 if and only if 1 > b > a andb = 1 − a. General remarks about the structure of elements of K(a,b) from the beginning of this section therefore in thepresent special case yield the following subspaces of K(a,b):

A0 = {(t,1 − t, t,1 − t, . . .)

∣∣ t ∈ [a,b]},A1 = {(

t, L(t), L2(t), L3(t), . . .) ∣∣ t ∈ [0,b]},

and for each positive integer n,

A2n = {(t,1 − t, t,1 − t, . . . , t,1 − t︸ ︷︷ ︸

2n

, L(1 − t), L2(1 − t), . . .) ∣∣ t ∈ [a,b]},

A2n+1 = {(t,1 − t, t,1 − t, . . . , t,1 − t, t︸ ︷︷ ︸

2n+1

, L(t), L2(t), . . .) ∣∣ t ∈ [a,b]},

where L has the usual meaning (L : [0,b] → [0,a], L(t) = ab t for any t).

Obviously K(a,b) = ⋃∞n=0 An .

We will begin by defining a convenient sin 1 -continuum and then show that K(a,b) is homeomorphic to it.

x

I. Banic et al. / Topology and its Applications 160 (2013) 63–73 69

Let 1 = x0 > x1 > x2 > x3 > · · · be a sequence in [0,1] converging to 0.Also let B0 = 0 × [a,b], and let B1 be the straight line segment from (x0,0) to (x1,b), and for each positive integer n, let

B2n be the straight line segment from (x2n−1,b) to (x2n,a) and B2n+1 = the straight line segment from (x2n,a) to (x2n+1,b).Obviously

⋃∞n=0 Bn is a sin 1

x -continuum.Now for each nonnegative integer n we define ϕn : An → Bn by

ϕn(t, . . .) = (z(t), t

),

where (t, . . .) ∈ An and (z(t), t) ∈ Bn . Later we will have to work with functions ϕn for different n simultaneously and insuch cases we will write ϕn(t, . . .) = (zn(t), t). Note that each ϕn is a homeomorphism. Note also that for each t z0(t) = 0and xn � zn(t) � xn−1 for each positive integer n.

Next, let ϕ : ⋃∞n=0 An → ⋃∞

n=0 Bn be a function, defined by

ϕ(x) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

ϕ0(x); if x ∈ A0,

ϕ1(x); if x ∈ A1,

ϕ2(x); if x ∈ A2,

...

ϕn(x); if x ∈ An,

...

It is easy to see that ϕ is a bijection and that ϕ is continuous on⋃∞

n=1 An . To see that ϕ is continuous on A0, take arbitraryu0 = (t0,1 − t0, t0,1 − t0, . . .) ∈ A0 and let {un}∞n=1 be a sequence of points in

⋃∞n=0 An such that limn→∞ un = u0. We will

show that limn→∞ ϕ(un) = (0, t0) = ϕ(u0).

(a) Suppose there is a positive integer n0 such that un ∈ A0 for all n � n0. Then for each n � n0 fix the unique tn ∈ [a,b],such that

un = (tn,1 − tn, tn,1 − tn, . . .).

From limn→∞ un = u0 it follows that limn→∞ tn = t0. Obviously, for each positive integer n � n0, ϕ(un) = ϕ0(un) ∈ B0and hence

limn→∞ϕ(un) = lim

n→∞ϕ0(tn,1 − tn, tn,1 − tn, . . .)

= limn→∞(0, tn)

= (0, t0)

= ϕ(u0).

(b) Suppose there is a positive integer n0, such that for each n � n0, un /∈ A0. Then for each n � n0 there is a positiveinteger kn such that un ∈ Akn . Obviously limn→∞ kn = ∞, since otherwise there would be infinitely many un ’s in someAm and from compactness of Am it would follow that the sequence un would have an accumulation point in Am

and hence would not converge to u0 ∈ A0. Therefore limn→∞ xkn = 0 and from xkn � zkn (tn) � xkn−1 it follows thatlimn→∞ zkn (tn) = 0. Also note that from limn→∞ un = u0 it follows that limn→∞ tn = t0. Therefore

limn→∞ϕ(un) = lim

n→∞ϕkn (un)

= limn→∞

(zkn (tn), tn

)= (0, t0)

= ϕ(u0).

(c) If both un ∈ A0 and un /∈ A0 hold true for infinitely many n, we apply (a) and (b) respectively for the two subsequencessatisfying each of the above conditions. Clearly in each of these cases we get limn→∞ ϕ(un) = ϕ(u0).

We have shown that ϕ is a continuous bijection from the compact space⋃∞

n=0 An onto the metric space⋃∞

n=0 Bn andtherefore ϕ is a homeomorphism. �

Again, itineraries may be used in an alternative proof. The allowed backward itineraries in this case are again R∞ andRk L∞ , where k � 0. The difference from the first case is that the itinerary R∞ gives an arc instead of a point.

70 I. Banic et al. / Topology and its Applications 160 (2013) 63–73

4. K(0,1)

K(0,1) turns out to be a very complicated continuum. In this section we give a detailed description of K(0,1) , which helpsus to recognize some of its subcontinua as certain familiar continua. The continuum has already been studied in [5,22].

Let x = (x1, x2, x3, . . .) ∈ K(0,1) . Suppose there is an integer n such that xn ∈ {0,1}. If xn = 0, then xn+1 ∈ {0,1}. If xn = 1,then xn+1 = 0. In the case where xn = t ∈ (0,1), one can easily see that xn+1 ∈ {0,1 − t}.

Let

A = {(x1, x2, x3, x4, . . .) ∈ K(0,1)

∣∣ x1 = 0} ⊆ K(0,1)

and

B = {(x1, x2, x3, x4, . . .) ∈ K(0,1)

∣∣ x1 = 1} ⊆ K(0,1).

If x = (x1, x2, x3, . . .) ∈ K(0,1) , then exactly one of the following is possible.

• x ∈ A.• x ∈ B .• There are an odd positive integer n and a ∈ A such that

x ∈ An(a) = {( t,1 − t, t,1 − t, . . . , t︸ ︷︷ ︸

n

,a)∣∣ t ∈ (0,1)

}.

• There are an even positive integer n and a ∈ A such that

x ∈ An(a) = {( t,1 − t, t,1 − t, . . . , t,1 − t︸ ︷︷ ︸

n

,a)∣∣ t ∈ (0,1)

}.

• x ∈ A∞ = {(t,1 − t, t,1 − t, . . .) | t ∈ (0,1)}.

One can easily see that Cl(A∞) = {(t,1 − t, t,1 − t, . . .) | t ∈ [0,1]} is the arc from (0,1,0,1, . . .) ∈ A to (1,0,1,0, . . .) ∈ Bin K(0,1) . Here and in the rest of this section by Cl we denote the closure operator in the Hilbert cube.

For each n, Cl(An(a)) is the arc {( t,1 − t, t,1 − t, . . . , t︸ ︷︷ ︸n

,a) | t ∈ [0,1]} from

(0,1,0,1, . . . ,0︸ ︷︷ ︸n

,a) ∈ A to (1,0,1,0, . . . ,1︸ ︷︷ ︸n

,a) ∈ B

if n is odd, and the arc {( t,1 − t, t,1 − t, . . . , t,1 − t︸ ︷︷ ︸n

,a) | t ∈ [0,1]} from

(0,1,0,1, . . . ,1︸ ︷︷ ︸n

,a) ∈ A to (1,0,1,0, . . . ,0︸ ︷︷ ︸n

,a) ∈ B

if n is even. We will show that

K(0,1) =( ∞⋃

n=1

( ⋃a∈A

Cl(

An(a))))

∪ Cl(A∞). (1)

For each a ∈ A there are two possibilities: either a = (0,0, . . .) or a = (0,1, . . .). In the first case there is an a0 ∈ A such thata = (0,a0) and hence a ∈ Cl(A1(a0)). In the second case a is of the form a = (0,1,0, . . .) and therefore there is an a0 ∈ Asuch that a = (0,1,a0) and hence a ∈ Cl(A2(a0)). That proves

A ⊆( ∞⋃

n=1

( ⋃a∈A

Cl(

An(a))))

∪ Cl(A∞).

For each b ∈ B there is a0 ∈ A such that b = (1,a0). Therefore b ∈ Cl(A1(a0)) and hence

B ⊆( ∞⋃

n=1

( ⋃a∈A

Cl(

An(a))))

∪ Cl(A∞),

and (1) follows.

I. Banic et al. / Topology and its Applications 160 (2013) 63–73 71

Using (1) we are able to prove some additional properties of K(0,1) as follows.

(a) K(0,1) contains sin 1x -continua. For example, a proof similar to the proof of Theorem 3.6 can be obtained in order to

prove that for a = (0,0,0, . . .) ∈ A( ∞⋃n=1

Cl(

An(a))) ∪ Cl(A∞)

is a sin 1x -continuum. See also [5].

(b) K(0,1) contains harmonic fans. For example,

F1 =( ∞⋃

n=1

Cl(

A2n−1(a))) ∪ Cl(A∞),

and

F2 =( ∞⋃

n=1

Cl(

A2n(a))) ∪ Cl(A∞),

where a = (0,1,0,1,0,1, . . .) ∈ A, are harmonic fans in K(0,1) .(c) K(0,1) is one-dimensional. It is easy to see that A is a Cantor set and that for each positive integer n,⋃

a∈A

Cl(

An(a))

is homeomorphic to the product A × [0,1]. Since dim(A) = 0, it follows from Theorem 2.2 that

dim

( ⋃a∈A

Cl(

An(a))) = 1.

A countable union of one-dimensional compacta is a one-dimensional compactum, see Theorem 2.3, therefore

dim(K(0,1)) = dim

(( ∞⋃n=1

( ⋃a∈A

Cl(

An(a))))

∪ Cl(A∞)

)= 1.

See also [22].(d) It has been proved in [22] that K(0,1) has trivial shape and is therefore tree-like.

5. A question

Unfortunately the techniques that were used in the proof of Theorem 3.4 do not work in general, i.e. using them onecannot prove that K(a,b) is homeomorphic to K(c,d) if (a,b), (c,d) ∈ Ct for arbitrary t ∈ [1,∞). Initially we conjectured sucha result, but many unsuccessful attempts to prove it provided us with evidence of a very complicated behavior, and we arenot so confident anymore.

Therefore we only pose the following question.

Question 5.1. Is it true that for any a,b, c,d ∈ (0,1), from (a,b), (c,d) ∈ Ct , for some t ∈ [1,∞), it follows that K(a,b) andK(c,d) are homeomorphic?

Let us also note that [43,46] studied the kneading sequences of skew tent maps

Fλ,μ(x) ={

1 + λx; x � 0,

1 − μx; x � 0,

and obtained results about coinciding kneading sequences for (λ,μ) belonging to certain algebraic curves.The results of [46] can be understood as certain reparametrizations of ours, but since their skew tent maps map R to

R this does not apply globally (but it applies to the cores [ f(a,b)(b),b]). In our case, each inverse limit is a continuumconsisting of the core of the inverse limit and a ray with the endpoint (0,0,0, . . .). In [46] the corresponding inverse limitcontains an additional ray with the endpoint (0,0,0, . . .) which destroys the compactness of the inverse limit.

In [46] the authors prove that there is a family of curves that fill up the whole region

D ={(λ,μ)

∣∣∣ λ � 1, μ > 1,1 + 1 � 1

},

λ μ

72 I. Banic et al. / Topology and its Applications 160 (2013) 63–73

and have the property that the functions Fλ,μ for which (λ,μ) belong to the same curve have the same kneading sequences.With a proper reparametrization, certain restrictions of the functions Fλ,μ can be transformed to the functions f(a,b) andthe family of the curves from [46] transforms to a family of curves in {(x, y) ∈ [0,1] × [0,1] | y � x, y > 1 − x} such that ifthe top points of the graphs of the generalized tent maps belong to the same curve, the corresponding kneading sequencesare the same. However, in [46] the authors do not give the explicit formulas for the curves.

In [43] the authors give similar results for the kneading sequences of skew tent maps Fλ,μ , for λ � 1,μ > 1. Even more,they give the explicit formulas for the curves in region

D1 = {(λ,μ)

∣∣ λ � 1, μ > 1}

with the same property. In this case there is no closed interval J = [α,β], with α < 0 < β , that is mapped by Fλ,μ to J ,such that Fλ,μ(α) = Fλ,μ(β) = α. Therefore there is no reparametrization as above that translates a restriction of thefunction Fλ,μ , (λ,μ) ∈ D1, to a generalized tent function f(a,b) , but this only makes the single ray containing the point(0,0,0, . . .) non-compact, as explained above. Other than this, the inverse limit spaces are homeomorphic.

Acknowledgements

We wish to thank to the anonymous referees for constructive and helpful remarks. We are also very grateful to HenkBruin for several useful remarks and comments on the earlier draft of the paper.

This work was supported in part by the Slovenian Research Agency, under Grants P1-0285 and P1-0297.

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