Topology and its Applications 267 (2019) 106829

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Topology and its Applications

www.elsevier.com/locate/topol

Hyperspaces of generalized continua which are infinite cylinders ✩

Włodzimierz J. Charatonik a, Tomás Fernández-Bayort b,∗, Antonio Quintero c

a Missouri University of Science and Technology, Department of Mathematics and Statistics, 65409 Rolla, MO, USAb Departamento de Matemáticas, Instituto Pablo Neruda, C/ Manuel Rodríguez Navarro 4, 41950 Castilleja de la Cuesta, Sevilla, Spainc Departamento de Geometría y Topología, Facultad de Matemáticas, Universidad de Sevilla, Apartado 1160, 41080 Sevilla, Spain

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

Article history:Received 14 February 2019Received in revised form 24 July 2019Accepted 26 July 2019Available online 3 September 2019

MSC:54B2054F1554C10

Keywords:Hyperspace(Generalized) continuumProper mapKelley propertyConstituantcone = hyperspace property

We deal with an analogue of the cone = hyperspace property for generalized continua. Namely, we study the class Cyl of those generalized continua X for which the hyperspace C(X) is homeomorphic to the infinite cylinder X × R≥0. The class Cyl is characterized by using continuous selections and compactwise Whitney maps, extending a theorem due to Illanes to the non-compact setting. Also it is shown that Cyl contains all 1-dimensional atriodic Kelley generalized continua whose constituants are one-to-one continuous images of the line R.

© 2019 Published by Elsevier B.V.

1. Introduction

A continuum is a compact connected metric space. Given a continuum X, let C(X) denote the hyperspace of its subcontinua endowed with the Hausdorff metric (or, equivalently, the Vietoris topology). It is an elementary fact that for X = [0, 1] and X = S1, the hyperspace C(X) is canonically homeomorphic to the corresponding cones, Cone([0, 1]) ∼= B2 ∼= Cone(S1), which are homeomorphic to the 2-ball.

✩ This work was partially supported by the project MTM2015-65397-P.* Corresponding author.

E-mail addresses: wjcharat@mst.edu (W.J. Charatonik), tfernandez@andaluciajunta.es (T. Fernández-Bayort), quintero@us.es (A. Quintero).

https://doi.org/10.1016/j.topol.2019.1068290166-8641/© 2019 Published by Elsevier B.V.

2 W.J. Charatonik et al. / Topology and its Applications 267 (2019) 106829

In the non-compact setting (i.e., compactness is replaced by local compactness) the cone is no longer well defined. Notwithstanding, for the half-line R≥0 and the line R the map which carries the continuum (i.e., the compact interval) A to the pair (m(A), �(A)), where m(A) is the minimum of A and �(A) is the length of A, defines homeomorphisms C(R≥0) ∼= R≥0 ×R≥0 and C(R) ∼= R × R≥0 onto the corresponding products, which are both homeomorphic to the half-plane {(x, y) ∈ R2; y ≥ 0}.

While the problem of characterizing the continua X for which C(X) is homeomorphic to Cone(X), the so-called “cone = hyperspace property”, has been studied by several authors since the early 1970’s (see [8] for a general reference), the problem of characterizing the generalized continua1 for which C(X) is homeomorphic to the infinite cylinder X ×R≥0 seems to have passed unnoticed in the literature.

This paper is focused on the class Cyl of those generalized continua X admitting a homeomorphism ϕ : C(X) → X ×R≥0 such that ϕ({x}) = (x, 0) for each x ∈ X.

Some of the results given here are essentially extensions of well known results for continua. In partic-ular, the characterization theorem of the cone = hyperspace property due to Illanes in [9] extends to a characterization of the class Cyl in terms of selections (Theorem 4.1).

However the absence of the whole space X in C(X) leads to essential differences with the classical case. While in the compact setting there exist only eight hereditarily decomposable continua whose cones are homeomorphic to their hyperspaces [8, Theorem 7.4], there are uncountably many hereditarily decomposable generalized continua in the class Cyl (Corollary 6.5). This is a consequence of Theorem 6.3 which shows that the class Cyl contains all 1-dimensional atriodic Kelley generalized continua whose constituants are one-to-one continuous images of the line R.

2. Preliminaries

By an admissible space we mean a locally compact separable metrizable space. A generalized continuumis a connected admissible space. In particular ordinary continua are compact generalized continua.

Henceforth all generalized continua will be assumed to be non-compact.Local compactness together with σ-compactness (provided by separability and local compactness) yield

the existence of increasing sequences of compact subsets Xn ⊂ X such that X =⋃∞

n=1 Xn and Xn ⊂intXn+1. Such sequences are termed exhausting sequences.

Recall that a continuous map f : X → Y is said to be proper if f−1(K) is compact for each compact subset K ⊂ Y . Proper maps between admissible spaces are characterized as closed maps for which f−1(y)is compact for all y ∈ Y [5, Theorem 3.7.18]. A proper map α : R≥0 → X is termed a ray in X.

We are mainly concerned with the hyperspace C(X) of subcontinua of an admissible space X; that is, C(X) = {A ⊂ X is a subcontinuum �= ∅}. This hyperspace will be regarded as a subspace of the hyperspace CL(X) = {A ⊂ X; A closed (non-empty) in X} endowed with the Fell topology.

Recall that given any space X, the Fell topology on CL(X) is generated by the subbase

〈X − C〉 = {A ∈ CL(X); A ∩ C = ∅}

〈X,U〉 = {A ∈ CL(X); A ∩ U �= ∅},(2.A)

where C and U range over the (possibly empty) compact sets and non-empty open sets of X, respectively. We refer to [3] for details; see also [4].

It must be pointed out that for Y ⊂ X the Fell topology on C(Y ) is, in general, strictly coarser than the relative topology of C(X); see [4] for an example. Henceforth C(Y ) will be considered with the relative topology of C(X).

1 Terminology will be introduced below.

W.J. Charatonik et al. / Topology and its Applications 267 (2019) 106829 3

We will use in Section 5 below the following two lemmas collecting basic facts of the Fell topology. The proofs are straightforward. Here for any A ⊂ C(X) we write ∪A = ∪{A; A ∈ A}.

Lemma 2.1. Let A a connected set of the admissible space X, then C(A) is a connected set of C(X). Con-versely, if A is a connected set in C(X), then ∪A is a connected set in X. Moreover, if A is a connected component of Z ⊂ X, then C(A) is a component of C(Z).

Lemma 2.2. Let X be an admissible space. If U is an open set in X, then C(U) is an open set in C(X).

For admissible spaces, the topology of C(X) can be described alternatively by the Hausdorff metric Hd

defined by any metric on X, bounded or not. In this way C(X) can be identified with the open set of C(X+), C(X) = {A ∈ C(X+); ∞ /∈ A} where X+ = X ∪ {∞} is the one-point compactification of X. In particular, a sequence {An}n≥1 converges to A in C(X) if and only if it does in the sense of Kuratowski; that is, x ∈ A

if and only if there is a sequence xn ∈ An converging to x in X.It is known that the hyperspace C(X) is an admissible space (a generalized continuum, respectively)

if and only if so is X. Moreover, if {Xn}n≥1 is an exhausting sequence of the admissible space X, then {C(Xn)}n≥1 is an exhausting sequence of C(X).

By the previous observations, the hyperspace operator can be iterated and the union map u : C2(X) =C(C(X)) → C(X) defined by u(A) = ∪A is a well defined proper surjection.

Concerning arcwise connectedness it can be proved that, for an admissible space X, C(X) is arcwise connected if and only if X is continuumwise connected; that is, given any two elements x, x′ ∈ X there exists a subcontinuum A ⊂ X with x, x′ ∈ A.

3. The class Cyl

Recall from the Introduction that Cyl stands for the class of (noncompact) generalized continua X for which there is a homeomorphism C(X) ∼= X ×R≥0 carrying {x} to (x, 0) for each x ∈ X. This is a natural extension to the non-compact setting of the well-known cone = hyperspace property studied in continuum theory.

In this section we analyze the structure of the finite dimensional spaces in Cyl. We start by showing that all finite dimensional spaces in Cyl are 1-dimensional. This is a consequence of the following proposition which is an easy extension of [8, Theorems 72.4 and 7.39].

Proposition 3.1. Let X be a generalized continuum X with dimX ≥ 2, then dim C(X) = ∞.

Proof. Given any exhausting sequence of X, {Xn}n≥1, there exists n0 such that dimXn0 ≥ 2. This follows from [6, Theorem 1.5.3]. Thus, by [6, Theorem 1.9.10] we find a component Y ⊂ Xn0 with dimY ≥ 2. Then we apply [8, Theorems 72.4 and 7.39] to get dim C(X) ≥ dim C(Y ) = ∞. �

Similarly to the compact context ([10]), triods are ruled out as subcontinua of elements of Cyl. Recall that a triod is a continuum T for which there is a subcontinuum Z ⊂ T such that T − Z consists of three mutually separated sets. It is well known that dimC(T ) ≥ 3 ([8, Theorem 70.1]), and so the following proposition is an immediate consequence of Proposition 3.1.

Proposition 3.2. Any 1-dimensional generalized continuum X in Cyl is atriodic.

Regarding the projection π : X × R≥0 → X as a retraction, the following proposition is an immediate consequence of [4, Proposition 11.15].

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Proposition 3.3. Each subcontinuum B of a 1-dimensional generalized continuum X in Cyl is a dendroid.

Recall that an admissible space X is said to be a dendroid if it is arcwise connected and all its subcontinua are unicoherent. In fact Propositions 3.2 and 3.3 can be improved as follows.

Proposition 3.4. Any subcontinuum A of a 1-dimensional generalized continuum X in Cyl is an arc.

Proof. By Proposition 3.3, A is a dendroid with dim C(A) ≤ dim C(X) = 2. So A is a graph by [10, Lemma 2.6]. Moreover, since X and hence A is atriodic it follows that A is an arc. �Proposition 3.5. Let X be a 1-dimensional generalized continuum such that all its subcontinua are arcs. Then each constituant of X is an L-constituant for L = R≥0 or L = R.

Recall from [7] that the constituant of a point x in a generalized continuum X is the union of all subcontinua of X containing x. We define a constituant A ⊂ X to be an R-constituant if there is a continuous bijection f : R → A. Similarly we define an R≥0-constituant.

In the proof of Proposition 3.5 we will use the following lemma; compare with [7, Theorem V.47.VIII.2]. To fix notation henceforth, by an expansive sequence in an admissible space X we mean a sequence {xn}n≥1 ⊂ X

which converges to ∞ in the one-point compactification X+ = X ∪ {∞}; that is, it does not contain convergent subsequences.

Lemma 3.6. Any constituant A of a generalized continuum X is the union of an increasing sequence of continua. Moreover A contains an expansive sequence.

Proof. Let {Xn}≥1 be an exhausting sequence of X. We can assume without loss of generality that X1∩A �=∅. Given p ∈ X1 ∩A we consider for each n ≥ 1 the connected component of p in Xn, say An. Notice that An ⊂ A for all n ≥ 1. We claim that A = ∪n≥1An; indeed, we simply observe that given a ∈ A there is a continuum C ⊂ X with p, a ∈ C. The compactness of C yields C ⊂ An for some n.

Moreover, each An meets FrXn by [7, Theorem V.47.III.2] applied to the one-point compactification of X. Then by choosing xn ∈ An ∩ FrXn for each n ≥ 1 we get an expansive sequence contained in A. �Proof of Proposition 3.5. By Lemma 3.6 any constituant A ⊂ X can be written A = ∪n≥1An for an increasing sequence of continua A1 ⊂ A2 . . . . By hypothesis, each An is an arc and so A is the union of an increasing sequence of arcs. Then, if there exists some n0 for which one of the extremes of An0 is an extreme of all An for n ≥ n0, one can define a continuous bijection ψ : R≥0 → A with ψ([0, k+1]) = An0+k

for k ≥ 0. Otherwise one chooses a subsequence A1 = An1 ⊂ An2 ⊂ . . . such that the extremes of Ankare

distinct of the ones of Ank−1 and one defines a continuous bijection ψ : R → A with ψ([−k, k]) = Ankfor

k ≥ 1. �As usual, a set A ⊂ X is said to be unbounded if its closure A is not compact. We say that the image of

a continuous map f : R → X is doubly unbounded if both f((−∞, 0]) and f([0, ∞)) are unbounded.

Corollary 3.7. Each constituant of a 1-dimensional generalized continuum X in Cyl is either a unbounded R≥0-constituant or a doubly unbounded R-constituant.

Proof. The constituants of X are L-constituants with L = R≥0, R by Propositions 3.4 and 3.5. Furthermore, Lemma 3.6 yields that any R≥0-constituant is unbounded. Assume that the R-constituant A given by the image of a one-to-one continuous map f : R → X is not doubly unbounded and assume that Z = f([0, ∞))is bounded. By Proposition 3.4 the closure Z is an arc. In particular Z ⊂ A and Z − Z reduces to a point,

W.J. Charatonik et al. / Topology and its Applications 267 (2019) 106829 5

Fig. 1. X is a generalized continuum that contains the sin 1x -curve.

say a. Then we easily find a triod in A containing a. This contradicts Proposition 3.4. Thus A is necessarily doubly unbounded. �

As observed in the Introduction, both R≥0 and R are in Cyl. In fact, they are characterized in the following way.

Corollary 3.8. The half-line R≥0 and the line R are the only finite dimensional generalized Peano continua in Cyl.

Proof. Any generalized Peano continuum X admits an exhausting sequence X1 ⊂ X2 ⊂ . . . consisting of Peano subcontinua (see [1, Theorem 2.4]). Moreover X is arcwise connected and it reduces to a single constituant. If in addition X is in Cyl, then by Proposition 3.4 and Corollary 3.7 each Xn is an arc and there is a continuous bijection ψ : L → X for L = R≥0 or R. As {Xn}n≥1 is an exhausting sequence, one readily derives that ψ is, in fact, a homeomorphism. �Remark 3.9. A generalized continuum which is a 1 − 1 continuous image of R containing the sin 1

x -curve is depicted in Fig. 1. Thus the converse of Proposition 3.5 does not hold in general and X is not in Cyl.

The following proposition gives a partial converse of Proposition 3.5. We will use this proposition in Section 6.

Proposition 3.10. Let X be a 1-dimensional generalized continuum such that all its constituants are L-constituants for L = R or R≥0. Assume in addition that all R-constituants of X are doubly unbounded. Then all subcontinua of X are arcs.

Proof. Let B be any continuum in X and let A ⊂ X be the constituant with B ⊂ A. We will prove the proposition when A is an R-constituant. The case when A is an R≥0-constituant is similar and simpler. Let h : R → A be a continuous bijection. We claim that Z = h−1(B), and so B, is an arc.

Firstly, let us check that Z does not contain unbounded components. Indeed, otherwise there exists t such that (−∞, t] ⊂ Z or [t, +∞) ⊂ Z. Assume the latter, and hence the closure K = h([t,+∞)) is a continuum in B.

Let an = h(tn) ∈ A be an expansive sequence in X with tn → +∞ given by the assumption that A is doubly unbounded. Then the sequence {tn}n≥1 is contained eventually in [t, +∞). Hence an ∈ K eventually, which is a contradiction. Analogously if (−∞, t] ⊂ Z.

After the previous observation all components of Z must be compact sets. In particular one finds sequences {xi}i≥0 and {xi}i≤−1 converging to +∞ and −∞, respectively, in R − Z with . . . x−n < · · · < x−1 < x0 <

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x1 < x2 < · · · < xn < . . . . Hence Z turns to be the disjoint union of the compact sets Zi = Z ∩ [xi−1, xi](i ∈ Z). Assume that Zi �= ∅ for infinitely many i. Then Z can be decomposed as a countable union of non-empty compact sets, and so can be B since h is a bijection. This contradicts the well-known Sierpinski theorem [7, Theorem V.47.III.6]. Thus Zi �= ∅ only for finitely many i and so Z is compact and h : Z → B

is a homeomorphism. Therefore Z and B are arcs as claimed since B is a continuum. �Corollary 3.11. Let X be a 1-dimensional generalized continuum having only R≥0-constituants. Then all subcontinua of X are arcs.

4. A proper analogue for the class Cyl of a characterization theorem by Illanes

In this section we extend to non-compact generalized continua a theorem due to Illanes characterizing the cone = hyperspacce property by using selections and Whitney maps ([9, Theorem 2]).

We recall that a selection is a continuous map s : C(X) → X such that s(A) ∈ A for all A ∈ C(X). Also recall that a Whitney map is a continuous map w : C(X) → R≥0 such that w({x}) = 0 for very x ∈ X and w(A) < w(B) whenever A ⊂ B and A �= B. Moreover, a Whitney map w is defined in [4] to be compactwise proper if for each compact set K ⊂ X, the restriction

wK : C(X � K) = {A ∈ C(X);A ∩K �= ∅} → R≥0

is proper. For K = {x} we write C(X|x) instead of C(X � {x}).In [4, Lemma 8.20] it is observed that, for any Heine-Borel metric d on X, the original Whitney map,

w : C(X) → R≥0 constructed by Whitney from d is a compactwise proper Whitney map. Recall that a Heine-Borel metric on X is a metric for which every closed bounded set is compact. It is known that any generalized continuum admits such a metric compatible with its topology (see [12, Theorem 1]).

We are now ready to prove the following characterization of the class Cyl as an extension of [9, Theo-rem 2].

Theorem 4.1. Let X be a 1-dimensional generalized continuum. Then the following statements are equivalent:

(1) X is in Cyl.(2) There exists a continuous selection σ : C(X) → X which restricts to a homeomorphism σ : w−1(t) → X

for each Whitney level of any compactwise proper Whitney map w : C(X) → R≥0.(3) There exist a continuous selection σ : C(X) → X and a compactwise proper Whitney map w : C(X) →

R≥0 such that σ : w−1(t) → X is a homeomorphism for each t ≥ 0.

For the proof of Theorem 4.1 we modify Illanes’s original proof in [9] by using compactwise proper Whitney maps to surmount the lack of compactness of X.

Proof of Theorem 4.1. The implication 2) =⇒ 3) is obvious.To check 3) =⇒ 1), let w : C(X) → R≥0 be a compactwise proper Whitney map and σ : C(X) → X be

a continuous selection such that the restriction σt : w−1(t) → X is a homeomorphism for all t ≥ 0. Then the map h : C(X) → X × [0, ∞) given by h(A) = (σ(A), w(A)) is a homeomorphism. Indeed, h is clearly a continuous bijection. Next we check that the map h is proper and so closed. Given any compact sets K ⊂ X and L ⊂ R≥0, we have the inclusion h−1(K × L) ⊂ C(X � K) ∩ w−1(L) = w−1

K (L). Here we use σ(A) ∈ A ∩K for any A ∈ σ−1(K). As the restriction wK is proper, the closed set h−1(K ×L) is compact.

It remains to show 1) =⇒ 2). Let h : X ×R≥0 → C(X) be a homeomorphism with h(x, 0) = {x} for all x ∈ X, and w : C(X) → R≥0 be any compactwise proper Whitney map.

W.J. Charatonik et al. / Topology and its Applications 267 (2019) 106829 7

As done in the proof in [9, Theorem 2], define the continuous map H : X × R≥0 → C(X) by H(x, t) =⋃r≤t h(x, r). In fact, H is proper because H−1(C(Xn)) ⊂ h−1(C(Xn)). In particular, for each x ∈ X,

H : {x} ×R≥0 → C(X) is a ray. Moreover the map H has the following two properties:

If H(x, t) = H(y, u), then x = y, (4.A)

and

H is a surjection. (4.B)

When X is a continuum, (4.A) and (4.B) are exactly Claims 1 and 2 in the proof of [9, Theorem 2]. In fact the proof of Claim 1 in [9] does not require the compactness of X and it can be repeated here, word by word, to show (4.A).

However, some changes are needed for the proof of (4.B) in the non-compact setting. We will give the details below and now proceed to define the selection σ.

From properties (4.A) and (4.B) we have a well defined selection σ : C(X) → X by setting σ(A) = x

where H(x, t) = A.Moreover, σ is continuous. To show this, assume for a moment that σ is not continuous at A ∈ C(X).

Then there exist a sequence {An}n≥1 ⊂ C(X) that converges to A and ε > 0 such that

d(σ(Ank), σ(A)) ≥ ε (k ≥ 1), (4.C)

for a subsequence {Ank}k≥1. Here d is a metric on X.

Consider A = H(x, t) and Ank= H(xk, tk). We have that {(xk, tk)}k≥1 ⊂ H−1(K) for the compact set

K = {Ak}k≥1 ∪ {A} ⊂ C(X). From the properness of H we can assume that {(xk, tk)}k≥1 converges to (x0, t0) ∈ X ×R≥0. Thus, H(xk, tk) = Ank

converges to H(x0, t0) = A and, by (4.A), x0 = x. Hence σ(An)converges to x0. This contradicts (4.C).

Next we will see that σ restricts to a homeomorphism σt : w−1(t) → X for each t ≥ 0. Consider A, B ∈ w−1(t) with σ(A) = σ(B). By the definition of σ and (4.A), if σ(A) = σ(B) = x, then A = H(x, t)and B = H(x, t′). If t′ < t then B = H(x, t′) ⊂ H(x, t) = A. Since w(A) = w(B) then A = B. Thus σis injective on w−1(t). In order to show the surjectivity of σt, for any x ∈ X we consider the composite

{x} × R≥0H→ C(X|x) ⊂ C(X) w→ R≥0. Since H and the restriction of w to C(X|x) are proper maps, this

composite defines a ray in R≥0 that begins at 0. Thus, for t ∈ R≥0 there exists some r with w(H(x, r)) = t

and hence A = H(x, r) ∈ w−1(t) verifies that σt(A) = x.Finally we will prove that σt is a proper map and so it will be an homeomorphism. Let {Xn}n≥1

be an exhausting sequence of X. The required properness readily follows from the inclusion σ−1t (Xn) ⊂

w−1(t) ∩ C(X � Xn) = (wXn)−1(t) and the properness of wXn

for all n ≥ 1.It remains to check (4.B). Notice that singletons are already in the image of H. Let B be a subcontinuum

of X with more than one point and w(B) = t0.For any x ∈ X the composite H : {x} × R≥0 → C(X|x) ⊂ C(X) w→ R≥0 is a ray in R≥0 containing

w(H(x, 0)) = w({x}) = 0, so its image is the whole half-line. Thus there exists sx ∈ R≥0 with w(H(x, sx))= t0.

Define ϕ : X → C(X) by ϕ(x) = H(x, sx). Notice that ϕ does not depend on the election of sxbecause if w(H(x, r)) = t0 with r < sx then H(x, r) ⊂ H(x, sx) with w(H(x, r)) = w(H(x, sx)), and so H(x, r) = H(x, sx). Similarly if sx < r.

Notice also that {x} = H(x, 0) ⊂ H(x, sx) = ϕ(x). To check the continuity of ϕ, assume on the contrary that there exist a sequence {xn}n≥1 ⊂ X converging to some x and ε > 0 such that

d(ϕ(xnk), ϕ(x)) ≥ ε (k ≥ 1) (4.D)

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for a subsequence {xnk}k≥1. As xnk

∈ ϕ(xnk) then ϕ(xnk

) ∈ C(X � K) with K = {xnk, x}k≥1.

Since wK and H are proper maps, the sequence {(xnk, snk

)}k≥1 with snk= sxnk

has a convergent sub-sequence in H−1w−1

K (t0). For the sake of simplicity we assume that the whole subsequence {(xnk, snk

)}k≥1converges to (x, s) for some s. By the continuity of H, ϕ(xnk

) = H(xnk, snk

) converges to H(x, s) and hence w(H(x, s)) = t0. Then ϕ(x) = H(x, s) by definition and then {ϕ(xnk

)} converges to ϕ(x). This contradicts (4.D).

Now the checking of (4.B) ends as in the proof of [9, Theorem 2]. We give the details for the sake of completeness.

As B is an arc by Proposition 3.4, let p �= q be its endpoints and consider the closed sets P = B ∩ϕ−1(C(X|p)) and Q = B ∩ ϕ−1(C(X|q)). As p ∈ P and q ∈ Q we have P, Q �= ∅. Since x ∈ B ∩ ϕ(x), we have that ϕ(x) ∪ B is a subcontinuum of X and so an arc by Proposition 3.4 for all x ∈ B. If ϕ(x) ⊂ B

for some x ∈ B, then w(ϕ(x)) = w(B) = t0 and so ϕ(x) = H(x, sx) = B, whence B is in the image of H. Otherwise, p ∈ ϕ(x) or q ∈ ϕ(x) for all x ∈ B, that is, B = P ∪ Q. By connectedness there exists x0 ∈ P ∩ Q. Thus the arc ϕ(x0) contains both p and q and so B ⊂ ϕ(x0). As w(B) = w(ϕ(x0)) we get B = ϕ(x0) = H(x0, sx0). This shows (4.B), and the proof is complete. �5. Obstructions to being in the class Cyl

As in the compact setting ([9, Corollary 4]), the existence of a continuous selection σ : C(X) → X for a generalized continuum X implies that X is not of type N at any of its subcontinua. Namely, the same proof of [9, Corollary 4] shows the following proposition.

Proposition 5.1. Let X be a generalized continuum and σ : C(X) → X be a continuous selection. Then Xis not of type N at any of its subcontinua.

Recall that a generalized continuum X is said to be of type N at the subcontinuum A ⊂ X (A �= X) if there exist four sequences of subcontinua converging to A, {Ak

n}n≥1 (k ∈ {1, 2, 3, 4}) such that A1n ∩ A2

n = {pn}and A3

n∩A4n = {qn} for all n ≥ 1, and the sequences {pn}n≥1 and {qn}n≥1 converge, respectively, to p, q ∈ A

with p �= q.As an immediate consequence of Theorem 4.1 and Proposition 5.1, we get

Corollary 5.2. Any generalized continuum in the class Cyl is not of type N at any of its subcontinua.

Example 5.3. By Corollary 5.2, the generalized continuum X depicted below does not belong to the class Cyl. This shows that the converse of Corollary 3.7 does not hold with full generality. (See Fig. 2.)

Another obstruction to being in the class Cyl appears with the existence of R3-points. Recall that a set A ⊂ X is an R3-set if there exists an open neighbourhood of A, U , and a sequence of components of U , {Cn}n≥1, such that lim inf Cn = A. A point p ∈ X is said to be an R3-point if {p} is an R3-set. We have the following crucial fact which generalizes [2, Theorem 3.7] to the non-compact setting.

Proposition 5.4. If a generalized continuum X contains an R3-set L, then C(X) contains an R3-set L such that the equality ∪L = L holds. In particular, if p is an R3-point in X, then {p} is an R3-point in C(X).

Proof. Assume L = lim inf Cn where {Cn}n≥1 are components of an open neighbourhood of L, U . Then by Lemmas 2.1 and 2.2, C(Cn) are components of the open set C(U).

Moreover, for the canonical embedding s : X → C(X), s(x) = {x}, we have s(L) ⊂ lim inf C(Cn) and so L = lim inf C(Cn) �= ∅. Indeed, let W = 〈X −K, V1, . . . , Vs〉 be an open neighbourhood of {x} in C(X)

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Fig. 2. The generalized continuum X is of type N at the subcontinuum A.

with x ∈ L. Thus Vx = (X −K) ∩ (∩si=1Vi) is an open neighbourhood of x and so Vx ∩ Cn �= ∅ eventually.

Take pn ∈ Vx ∩ Cn. Then {pn} ∈ C(Cn) ∩W, whence {x} ∈ L. So s(L) ⊂ L and L ⊂ ∪L.To check the inclusion ∪L ⊂ L, let q ∈ ∪L and choose A ∈ L with q ∈ A. Let V be any open

neighbourhood of q in X. As A ∩ V �= ∅, V = 〈X, V 〉 is a basic open neighbourhood of A in C(X). Thus there exists Zn ∈ C(Cn) ∩ V if n ≥ n0 for some n0. Hence ∅ �= Zn ∩ V ⊂ Cn ∩ V and so q ∈ L.

Finally, as ∪L = L ⊂ U , we get A ⊂ U for all A ∈ L and so L ⊂ C(U). �Corollary 5.5. Let X be a generalized continuum containing an R3-point. Then X does not belong to the class Cyl.

Proof. According to Proposition 5.4, C(X) contains an R3-point. Therefore, it will suffice to check that X × R≥0 does not contain R3-points. Indeed, assume on the contrary that {(x, t)} = lim inf Cn where Cn

are components of an open neighbourhood U of (x, t). We can find ε > 0 and an open neighbourhood of xin X, V0, with compact closure and such that V = V 0 × (t − ε, t + ε) ⊂ U . Hence the components Cn meet V whenever n ≥ n0 for some n0. Then for each n ≥ n0 there is a component Dn of V such that Dn ⊂ Cn

and Dn = D′n× (t − ε, t + ε) where D′

n is a component of V0. Let (xn, t + ε2 ) ∈ Dn ⊂ Cn. By compactness we

can assume that the sequence {xn}n≥n0 converges to some x0 ∈ V0. Thus (x0, t + ε2 ) ∈ lim inf Cn = {(x, t)}.

This contradiction shows that there are no R3-points in X ×R≥0. �6. Sufficient conditions for a generalized continuum to be in the class Cyl

In this section we give two results involving the well-known Kelley property which provide sufficient conditions for a generalized continuum to be in the class Cyl. Recall that a generalized continuum X is said to have the property of Kelley provided that for each point x ∈ X, for each continuum C ⊂ X containing x and for each sequence {xn}n≥1 converging to x in X, there exists a sequence of continua Cn ⊂ X with xn ∈ Cn converging to C in C(X).

We first state and prove the following partial converse of Corollary 3.7. Recall that all generalized continua are assumed to be non-compact.

Theorem 6.1. Let X be a 1-dimensional generalized continuum with the Kelley property. Assume that all constituants of X are doubly unbounded R-constituants. Then X is in Cyl.

Let us start with the following lemma.

Lemma 6.2. [4, Lemma 8.21] Let X be a generalized continuum with the Kelley property. Given a compact-wise proper Whitney map w : C(X) → R≥0, the map H : X × R≥0 → C(X) given by H(x, s) = ∪{A ∈C(X); x ∈ A and w(A) = s} is a proper map.

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Proof of Theorem 6.1. We already know from Proposition 3.10 that all subcontinua of X are arcs. This allows us to replicate the proof of [8, Theorem 80.3]. We give the details for the sake of completeness and also to point out the differences with the compact case.

Let w : C(X) → R≥0 be a compactwise proper Whitney map. Then the map H : X × R≥0 → C(X)given by H(x, s) = {A ∈ C(X); x ∈ A and w(A) = s} is proper by Lemma 6.2. We proceed to show that His a bijection and so a homeomorphism. Analogously to the proof of [8, Theorem 80.3], this will be derived from the following alternative description of the map H.

Let {Xn}n≥1 be an exhausting sequence of X. For x ∈ X and s ∈ R≥0, let Z be the R-constituant of x and h : R → Z be a continuous bijection with h(tx) = x. As X is doubly unbounded, there is a sequence in R, tx = t0 < t1 < · · · < tn < . . . , converging to +∞ such that h(tn) /∈ Xn. Let Γ0 = {x} and Γn = h([tn−1, tn]) (n ≥ 1). Let η : R≥0 → C(X) be the ray obtained by the juxtaposition of ordered arcs from Γn to Γn+1 for all n ≥ 0.

Notice that η(0) = Γ0 = {x} and so x ∈ η(t) for all t ≥ 0. Then the composite w ◦ η : R≥0 → C(X|x) →R≥0 is a proper map fixing 0, and so it is onto. In particular there exists some vx ∈ R≥0 with vx ≥ tx and w(h([tx, vx]) = s. Similarly, by using a decreasing sequence t0 > t′1 > · · · > t′n > . . . with t′n /∈ Xn, we find ux ≤ tx such that w(h([ux, tx])) = s. In particular, the arcs h([ux, tx]) and h([tx, vx]) are contained in H(x, s) and so h([ux, vx]) = h([ux, tx]) ∪ h([tx, vx]) ⊂ H(x, s).

We claim that H(x, s) = h([ux, vx]). Indeed, let B ∈ C(X|x) with w(B) = s. Then B is an arc and so is h−1(B) (see the proof of Proposition 3.10). Here we use again that the constituants of X are doubly unbounded. Thus B = h([a, b]) for some a < b. Since w(h([a, tx])) ≤ w(B) = s = w(h([ux, tx])), we have ux ≤ a. Similarly b ≤ vx, and so B ⊂ h([ux, vx]).

We can now check that H is one-to-one as done in the proof of [8, Theorem 80.3].Suppose H(x, s) = H(y, s′). Then h([ux, vx]) = h([uy, vy]), and so ux = uy and vx = vy since h is

one-to-one. If tx < ty (similarly, ty < tx), then s = w(h([ux, tx]) < w(h([uy, ty]) = s′ = w(h([ty, vy]) <w([tx, vx]) = s. This contradiction shows that tx = ty. Thus, x = y and s = s′, and H is injective.

In order to check that H is onto, let C be any non-degenerate subcontinuum of X, and Z be the constituant containing C. Given a continuous bijection h : R → Z, we have h([a, b]) = C for some a ≤ b

since h−1(C) is an arc. Then the continuous map g : [a, b] → R defined by g(t) = w(h([a, t])) − w(h([t, b]))satisfies g(a) < 0 < g(b), and so there exists t0 with g(t0) = 0. Hence, with the alternative definition of H, we have for x = h(t0) the equalities: tx = t0, ux = a and vx = b. Thus H(h(t0), w(h[a, t0])) = h([a, b]) = C. �Theorem 6.3. Let X be a 1-dimensional atriodic generalized continuum with the Kelley property and all constituants of X are R-constituants. Then X is in Cyl.

Proof. According to Theorem 6.1, it will suffice to check that all R-constituants are doubly unbounded. For this, assume that some R-constituant Z ⊂ X fails to be doubly unbounded. Let h : R → Z be a continuous bijection.

By Lemma 3.6, given an exhausting sequence of X, {Xn}≥1, there exists at least one unbounded sequence {tn}n≥1 ⊂ R such that h(tn) /∈ Xn for all n ≥ 1. Assume t1 > t2 > · · · > tn > . . . .

As Z is not doubly unbounded, the image Y = h([t1, +∞)), and so its closure Y , is contained in some Xn0 . Notice that Y is a continuum and Y ⊂ Z.

Let L ⊂ Y denote the set of limit points of images of increasing unbounded sequences in [t1, ∞). The compactness of Y yields that L is not empty. Moreover, it is readily checked that L = ∩t≥t1f([t,∞)), and so L is a continuum.

If L reduces to a point p = h(s), then for t large enough A1 = h([t, ∞)) ∪ {p} is an arc hitting the arc A2 = h[s − 1, s + 1]) exactly in p and A1 ∪A2 is then a triod in X. This is ruled out by the hypothesis, and we necessarily have #L ≥ 2.

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Choose a = h(x) ∈ L and t0 = tn from the sequence above with t0 < x and h(t0) /∈ Xn0 . Let B be the arc h([t0, x]), so that h : [t0, x] → B is a homeomorphism. Clearly B ∩ L �= ∅ is compact, and so the compact set h−1(L ∩B) has a minimum, say x0.

Next we consider the subarc B0 = h([t0, x0]) for which L ∩B0 = {h(x0)} and pick an increasing unbounded sequence {sn}n≥1 ⊂ [t1, ∞) such that the images h(sn) converge to h(x0).

The Kelley property provides a sequence of continua in X, {Bn}n≥1, converging to B0 with h(sn) ∈ Bn. In particular, Bn ⊂ Z = h(R) for all n ≥ 1.

We claim that Cn = h−1(Bn), and hence Bn, are arcs eventually. Indeed, if Cnkis not an arc for a

subsequence {nk}k≥1, then Cnkcontains an unbounded component (see the proof of Proposition 3.10).

Clearly Cnkdoes not contains an interval of the form (−∞, yk] since the Bn’s are eventually contained in

any compact neighbourhood of B0 and the h((−∞, yk])’s do not. On the other hand, if Cnkcontains an

interval [yk, ∞), we can assume yk ≥ t1. Then L ⊂ h([yk,∞) ⊂ Bnkand so L ⊂ B0, whence L reduces to

{h(x0}. This contradicts #L ≥ 2.For the sake of simplicity, let us assume that the whole sequence {Bn}n≥1 are arcs which can be written

as images of intervals Zn = [pn, qn].As #L ≥ 2, choose b �= h(x0) in L and let {s′n}n≥1 be an increasing unbounded sequence in [t1, ∞) such

that {h(s′n)}n≥1 converges to b. There is no loss of generality in assuming in addition s1 < s′1 < s2 < s′2 < . . .

for the sequence {sn}n≥1 above.The convergence of the Bn’s to B0 = h([t0, x0]) yields a sequence of points bn = h(xn) ∈ Bn converging

to h(t0) /∈ Xn0 . Moreover, the points bn are outside the compact set Xn0 eventually. We can assume without loss of generality bn /∈ Xn0 for all n ≥ 1, and so xn ≤ t1 for all n ≥ 1. In particular, xn < si < s′i < si+1

for all n ≥ 1 and i ≥ 1. As sn, xn ∈ [pn, qn], we get s′n−1 ∈ [xn, sn] ⊂ [pn, qn]. Hence h(s′n−1) ∈ Bn for all n ≥ 2, and so b ∈ B0 ∩ L = {h(x0)}. This contradiction completes the proof. �

An immediate consequence of Theorem 6.3 is the following result.

Corollary 6.4. Let X be a 1-dimensional generalized continuum with the Kelley property and such that all of its constituants are homeomorphic to the line R. Then X is in Cyl.

A well-known result due to S.B. Nadler shows that there are only eight hereditarily decomposable continua X for which C(X) is homeomorphic to Cone(X); see [8, Theorem 7.4].

Recall that a generalized continuum X is said to be decomposable if X = X1 ∪ X2 is the union of two generalized subcontinua X1, X2 �= X. A generalized continuum is said to be hereditarily decomposableprovided that all of its generalized subcontinua (�= {∗}) are decomposable.

With regard to the class Cyl we can show by using Corollary 6.4 the following result in sharp contrast with Nadler’s theorem.

Corollary 6.5. There exist uncountably many hereditarily decomposable generalized continua in Cyl.

Proof. Let X ⊂ R2 be the generalized continuum X = (∪m≥0Rm) ∪ (∪n≥1Σn) depicted in Fig. 3.For each n ≥ 1, let Σ+

n be a disjoint copy of Σn. Given any set J of non-negative integers, let XJ =X ∪s∈J Σ+

s ; see Fig. 4.Clearly each XJ is hereditarily decomposable. Moreover, XJ is in Cyl by Corollary 6.4. We claim that

XJ is homeomorphic to XJ ′ if and only if J = J ′, and so the family of the XJ ’s is uncountable.In order to prove the claim, for l ≥ k ≥ 0, let

Ak,lJ = (∪k≤m≤kRm) ∪ (∪k+1≤n≤lΣn) ∪

(∪s∈J,k+1≤s≤lΣ+

s

).

12 W.J. Charatonik et al. / Topology and its Applications 267 (2019) 106829

Fig. 3. X is a generalized continuum contained in R2.

Fig. 4. XJ for J the set of all odd integers.

Fig. 5. X is a generalized continuum with no continua of type N and the point p is an R3-point.

Similarly we define Ak,lJ ′ . It is obvious that Rm, Σn and Σ+

s are the constituants of XJ . Furthermore, any homeomorphism h : XJ

∼= XJ ′ carries necessarily each Rm to some Rm′ since they are closed sets. Moreover h(R0) = R0; indeed if h(R0) = Rm with m ≥ 1 then h(A0,1

J ) = Am−1,m+1J ′ but the closure of the former

contains the constituants R0 and R1 while the closure of the latter contains the constituants Rm−1, Rm

and Rm+1.Thus h(R0) = R0 and so h(A0,1

J ) = A0,1J ′ , whence h(R1) = R1 and h(A1,2

J ) = A1,2J ′ . By proceeding

inductively we get J = J ′ as claimed. �7. Final remarks and open questions

It is natural to ask for the role of the Kelley property in Theorem 6.3. Namely,

Open Question 7.1. Can the Kelley property be replaced by a weaker condition in Theorem 6.3?

The generalized continuum depicted in Fig. 5 rules out the no existence of continua of type N as a possible candidate to replace the Kelley property; see Corollary 5.5.

W.J. Charatonik et al. / Topology and its Applications 267 (2019) 106829 13

Fig. 6. Topological model of Cone(X+).

Fig. 7. The generalized continuum X.

With regard to Corollary 6.5, an alternative uncountable family of hereditarily decomposable generalized continua can be constructed from any of the six hereditarily decomposable continua X �= [0, 1], S1 in Nadler’s theorem [8, Theorem 7.4]. Namely, choose any 0-dimensional compact set K ⊂ X such that the open set YK = X − K remains connected and all constituants (= arcwise components) of YK are R-constituants. We require in addition that, if X is the Warsaw circle, K contains the non Kelley point of X. Then, as an immediate consequence of Corollary 6.4, YK is a hereditarily decomposable generalized continuum in Cylsuch that YK � YL if K � L.

Also the following question is left open.

Open Question 7.2. Under which conditions can Theorem 6.3 be extended to atriodic generalized continua with R≥0-constituants?

Examples, with and without the Kelley property, of generalized continua in Cyl with R≥0-constituants are easily found. For instance, let X be the generalized continuum which is the complement of an extreme (termed here ∞) of the limit arc of the sin 1

x -curve.A homeomorphism C(X) ∼= X × R≥0 follows from the analysis of the identification of C(X+) with the

topological model of Cone(X+) in Fig. 6, taken from Figure 4.15 in [11], where the vertex of the cone is the arc R+ = [b, ∞] and [x, R+] ⊂ X+ denotes the irreducible subcontinuum between x ∈ X −R+ and R+.

This way the subcontinua of X+ containing ∞ correspond to the points of the bold arcs Γ and Ω in Fig. 6. As C(X) ∼= {A ∈ C(X+); ∞ /∈ A}, the homeomorphism C(X+) ∼= Cone(X+) restricts to a homeomorphism C(X) ∼= X ×R≥0.

The following variation of the previous example gives us another example of generalized continuum in Cyl with R≥0-constituants, this time without the Kelley property.

Consider the half-plane R × R≥0 as the union of the three regions A− = {(x, y); 0 ≤ y ≤ −x, x ≤ 0}, A+ = {(x, y); 0 ≤ y ≤ x, x ≥ 0} and A0 = {(x, y); −y ≤ x ≤ y; y ≥ 0}.

It is not hard to check that any homeomorphism ψ− : A− ∼= R≤0×R≥0 → C(R≤0) satisfies ψ−(A−∩A0) =C(R≤0|0). Similarly, for any homeomorphism ψ+ : A+ ∼= R≥0 × R≥0 → C(R≥0) we have ψ+(A+ ∩ A0) =

14 W.J. Charatonik et al. / Topology and its Applications 267 (2019) 106829

C(R≥0|0). Thus, given ψ− and ψ+ we get a homeomorphism ψ : R × R≥0 → C(R) by extending ψ− and ψ+ on A0 by setting ψ(x, y) = [ay, byy (x + y)] if ψ−(−y, y) = [ay, 0] and ψ+(y, y) = [0, by], for each y ≥ 0and −y ≤ x ≤ 0. Analogously, for each y ≥ 0 and 0 ≤ x ≤ y, we define ψ(x, y) = [ay − ay

y x, by].Then by using the homeomorphism ψ one easily extends the homeomorphism C(X) ∼= X ×R≥0 derived

from Fig. 6 above to a homeomorphism C(Y ) ∼= Y ×R≥0 for Y = X ∪ ({0} ×R≤0); see Fig. 7. Notice that Y does not have the Kelley property.

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