A fixed point theorem for Whitney blocks

Topology and its Applications 125 (2002) 315–321

A fixed point theorem for Whitney blocks

Jorge Bustamante∗, Raúl Escobedo, Fernando Macías-Romero

Facultad de Ciencias Físico Matemáticas, BUAP, Av. San Claudio y Río Verde, Ciudad Universitaria,San Manuel, 72570 Puebla, Mexico

Received 17 January 2001; received in revised form 25 October 2001

Abstract

General theorems concernings-connectedness and hyperspaces are first obtained. These results areapplied to prove that: for a continuumX having zero surjective semispan, (1) each Whitney blockin the hyperspace of subcontinua ofX, C(X), has the fixed point property and (2) iff :Y → X ismap from a continuumY ontoX, then the induced mapf :C(Y ) → C(X) is universal. Both resultsprovide new proofs to some theorems for arc-like continua. The first one answers a question askedby Nadler. 2001 Elsevier Science B.V. All rights reserved.

MSC:54B20; 54F15; 54H25

Keywords:Fixed point property; Hyperspaces; Induced maps;s-connectedness; Semispan;Universal maps; Whitney blocks

1. Introduction

A continuumis a nonempty compact connected metric space. For a continuumX, letC(X) (respectively 2X) denote the hyperspace of all continua (respectively nonemptycompacta) inX with the Hausdorff metric, see [4] or [11]. Amapis a continuous function.A Whitney mapfor C(X) is a mapµ :C(X) → [0,1] such that (a)µ({x}) = 0 for everyx ∈ X, (b) if A,B ∈ C(X) and A ⊂ B �= A, thenµ(A) < µ(B), and (c)µ(X) = 1.A Whitney blockin C(X), respectively aWhitney levelin C(X), is a set of the formµ−1([s, t]), respectivelyµ−1(t), where 0� s < t � 1. These concepts are defined for 2X

similarly. It is known that Whitney maps, for bothC(X) and 2X, always exist [4, Theo-rem 13.4]. Moreover, Whitney blocks and Whitney levels inC(X) are continua [4, The-orem 19.9]. However, for 2X this is not true [4, Theorem 24.2]. Given a mapf :Y → X

* Corresponding author.E-mail addresses:[email protected] (J. Bustamante), [email protected] (R. Escobedo),[email protected] (F. Macías-Romero).

0166-8641/01/$ – see front matter 2001 Elsevier Science B.V. All rights reserved.PII: S0166-8641(01)00284-X

316 J. Bustamante et al. / Topology and its Applications 125 (2002) 315–321

between continuaY andX, we let f :C(Y ) → C(X) denote the correspondinginducedmapdefined byf (A)= f (A) for everyA ∈ C(Y ).

For a continuumX, let π1 denote the first projection ofX × X onto X. FollowingLelek [8],X haszero surjective semispanif wheneverZ is a continuum inX×X such thatπ1(Z) = X, thenZ intersects the diagonal,∆X , in X × X, where∆X = {(x, x): x ∈ X}.It is known that each arc-like continuum has zero surjective semispan [7,8]. The converseis an open problem [1, Houston Problem Book, Problem 82].

A mapf :Y →X is universal[2] providedf has a coincidence with every map fromYintoX, this means for any mapg :Y →X there exists a pointp ∈ Y such thatf (p) = g(p).A continuumX is said to have thefixed point propertyprovided that every mapf fromX into X has a fixed point, i.e., a pointx ∈ X such thatf (x) = x. It is easy to verify thatif f :Y → X is a universal map, thenf is surjective andX has the fixed point property.Also it can be proved that a continuumX has zero surjective semispan if and only if eachmap from any continuum ontoX is universal. In particular, continua having zero surjectivesemispan have the fixed point property.

In this paper we prove: For a continuumX with zero surjective semispan, Theorem 3.1,each Whitney block inC(X) has the fixed point property and, Theorem 4.1, iff :Y → X

is a map from a continuumY ontoX, then the induced mapf :C(Y )→C(X) is universal.In a conference [5] Nadler asked whether Whitney blocks inC(X) have the fixed point

property for an arc-like continuumX, Theorem 3.1 led to the answer to this question.Segal [19, Theorem 3] has shown thatC(X) has the fixed point property for an arc-likecontinuumX. Some easier proofs of this result are in [6,12,18]. We obtain Segal’s fixedpoint theorem as an immediate corollary of each of Theorems 3.1 and 4.1. Also we mentionthat Theorem 4.1 simultaneously strengthens results of Nadler [12] and Oversteegen andTymchatyn [15].

Our approach follows some ideas of Marsh [10] related withs-connectedness. We recallthis notion. LetA andB be disjoint closed subsets of a connected spaceX. A closed setH ⊂X cuts weakly betweenA andB in X provided that wheneverC is a closed connectedset inX that intersects each ofA andB, thenC intersectsH . We say thatX is s-connectedbetweenA andB provided that wheneverH is a closed set inX that cuts weakly betweenA andB, then some componentK of H cuts weakly betweenA andB. A connected spaceX is said to bes-connectedprovided that wheneverA andB are disjoint closed connectedsubsets ofX, thenX is s-connected betweenA andB.

If f :X → Y is a map andA ⊂X, thenf |A denotes the restriction off to A.

2. s-connectedness and hyperspaces

We obtain some general results concernings-connectedness and hyperspaces, which weuse later.

Theorem 2.1. For any continuumX, C(X) is s-connected.

J. Bustamante et al. / Topology and its Applications 125 (2002) 315–321 317

Proof. Let X be a continuum. It follows from [4, Theorem 19.2] thatC(X) is an inverselimit of absolute retracts. On the other hand, in [10, Theorem 3] is proved that an inverselimit of absolute retracts iss-connected. ThereforeC(X) is s-connected. ✷Remark 2.2. It is known that, for any continuumX, C(X) is unicoherent [4, Theo-rem 19.8]. Sinces-connectedness implies unicoherence [10, Lemma 1], Theorem 2.1 pro-vides a new proof to this result.

The arguments used in the proof of Theorem 2.1 also work to show that a Whitney blockof the formµ−1([t,1]) is s-connected, see [4, Exercise 19.14].

The proofs of Theorems 2.4 and 3.1 use the following result.

Theorem 2.3. Let X be a continuum,µ be a Whitney map forC(X) and0 � s < t � 1.If H is a closed set that cuts weakly betweenµ−1(s) andµ−1(t) in the Whitney blockµ−1([s, t]), thenH cuts weakly betweenµ−1(s) andµ−1(t) in C(X).

Proof. Let C be a continuum inC(X) which intersects each ofµ−1(s) andµ−1(t). Weclaim that there is a component,D, of C ∩ µ−1([s, t]) such thatD intersects bothµ−1(s)

andµ−1(t). Next we prove this claim.If no component ofC ∩ µ−1([s, t]) intersects each ofµ−1(s) andµ−1(t) then, by the

cut wire theorem, [13, Theorem 5.2],

C ∩µ−1([s, t]) =Z1 ∪Z2

whereZ1 andZ2 are disjoint closed subsets ofC ∩µ−1([s, t]) with(C ∩µ−1(s)

) ⊂Z1 and(C ∩µ−1(t)

) ⊂Z2.

Notice that

C = (Z1 ∪ {

A ∈ C: µ(A)� s}) ∪ (

Z2 ∪ {A ∈ C: µ(A)� t

}),

thus, we have a contradiction to connectedness ofC.Then, we have thatD is a continuum in the blockµ−1([s, t]) which intersects each of

µ−1(s) andµ−1(t), and so, by hypothesis,D ∩H �= ∅. Therefore,C ∩H �= ∅. ✷It is easy to see that there exist Whitney blocks which are nots-connected, for example

consideringC(S1) whereS1 is the unit circle in the plane. However, as we show in nexttheorem, for any continuumX, every Whitney block inC(X) is s-connected between itslids.

Theorem 2.4. LetX be a continuum. Ifµ is a Whitney map forC(X) and0 � s < t � 1,then the Whitney blockµ−1([s, t]) is s-connected betweenµ−1(s) andµ−1(t).

Proof. Let H be a closed set that cuts weakly betweenµ−1(s) andµ−1(t) in the blockµ−1([s, t]). By Theorem 2.3, we have thatH cuts weakly betweenµ−1(s) andµ−1(t)

in the hyperspaceC(X). Now, by Theorem 2.1, there is a componentK of H which cuts


weakly betweenµ−1(s) andµ−1(t) in C(X). It is clear thatK cuts weakly betweenµ−1(s)

andµ−1(t) in the blockµ−1([s, t]) also. ✷Remark 2.5. Theorems 2.1 and 2.3 also holds for 2X. Theorem 2.4 can be proved for aWhitney mapµ for 2X if, in addition,µ is monotone, i.e., the Whitney levelsµ−1(t) areconnected sets.

3. A fixed point theorem

In [3, Corollary 3.3] Hosokawa proved that the property of having zero surjectivesemispan is a Whitney property. It means, ifX is a continuum with zero surjectivesemispan, then each Whitney level inC(X) has zero surjective semispan. Therefore, forsuch a continuum the Whitney levels have the fixed point property. In next theorem weinclude this fact.

Theorem 3.1. LetX be a continuum with zero surjective semispan. Ifµ is a Whitney mapfor C(X) and0� s � t � 1, thenµ−1([s, t]) has the fixed point property.

Proof. We only consider the cases < t (because the comment above the theorem). Letf :µ−1([s, t]) →µ−1([s, t]) be a map and set

H = {A ∈ µ−1([s, t]): µ(A)= µ

(f (A)

)}.

Sinceµ|µ−1([s,t ]) :µ−1([s, t])→ [s, t] is a universal map, it has a coincidence with the mapµ ◦ f :µ−1([s, t]) → [s, t], thenH is a nonempty closed subset ofµ−1([s, t])).

We claim thatH cuts weakly betweenµ−1(s) andµ−1(t) in the blockµ−1([s, t]).To prove this fact, letC be a continuum inµ−1([s, t]) which intersects bothµ−1(s) andµ−1(t). Notice thatµ(C) = [s, t]. Thusµ|C :C → [s, t] is a universal map. Then, byconsidering the mapµ ◦ f |C :C → [s, t], we obtain that there exists an elementA ∈ Csuch thatµ(A)= µ(f (A)). That is,H ∩ C �= ∅.

From Theorem 2.3, we have thatH cuts weakly betweenµ−1(s) andµ−1(t) in thehyperspaceC(X). Then, by Theorem 2.1, we can fix a componentK of H which cutsweakly betweenµ−1(s) andµ−1(t) in C(X). Next we prove that

X =⋃

{A: A ∈ K}. (1)

Takea ∈ X. Let Γ be an order arc inC(X) from the singleton{a} to X (see [4, p. 110]).We have thatΓ is a continuum inC(X) which intersects each ofµ−1(s) andµ−1(t).Therefore,Γ ∩ K �= ∅. Let A ∈ Γ ∩ K. Since each element ofΓ contains the pointa, itfollows thata ∈ A. This proves (1).

We assert that there existsA ∈ K such that eitherA ⊂ f (A) or f (A)⊂ A. Let us proveit. If the assertion is false, by an argument of compactness, there existsε > 0 such that, foreveryA ∈ K

f (A) �⊂N(ε,A) and A �⊂N(ε,f (A)

),


whereN(ε,A) = {x ∈ X: d(x, a) < ε for somea ∈ A} andd denotes the metric onX.Now, for eachA ∈ K set

[A] = (A× f (A)

) \ (N(ε,f (A)×N(ε,A)

).

By Lemma 3.1 of [3], we have that[A] is a subcontinuum ofX ×X such that

π1([A]) = A and [A] ∩∆X = ∅. (2)

SetM = ⋃{[A]: A ∈ K}. It is easy to prove thatM is a continuum inX × X. From (1)and the first equation in (2), it follows thatπ1(M) =X. By the second equation in (2), weobtainM ∩∆X = ∅. SinceX has zero surjective semispan, we have a contradiction. Thisproves our assertion.

Now, we can fixA ∈ K such thatA ⊂ f (A) or f (A)⊂ A. SinceK ⊂H, A ∈ H. Thus,µ(A)= µ(f (A)). It follows from the properties of the Whitney map thatA = f (A). ✷

Since the whole hyperspaceC(X) is a Whitney block, we obtain the following corollary.

Corollary 3.2. If X is a continuum with zero surjective semispan, thenC(X) has the fixedpoint property.

It is known that an arc-like continuum has zero surjective semispan [7] or [8]. Then, asa direct consequence of Theorem 3.1 and Corollary 3.2, we obtain the following results,respectively.

Corollary 3.3. For an arc-like continuumX, each Whitney block inC(X) has the fixedpoint property.

Corollary 3.4 (Segal [19, Theorem 3]).If X is an arc-like continuum, thenC(X) has thefixed point property.

4. A universal mapping theorem

Our next theorem simultaneously strengthens results of Nadler [12] and Oversteegenand Tymchatyn [15], see Corollaries 4.2 and 4.3 below.

Theorem 4.1. LetX be a continuum with zero surjective semispan. Iff :Y → X is a mapfrom a continuumY ontoX, then the induced mapf :C(Y )→C(X) is universal.

Proof. Let g :C(Y )→C(X) be a map. Fix a Whitney mapµ for C(X). Denote

H = {A ∈ C(Y ): µ

(f (A)

) = µ(g(A)

)}.

It is easy to see thatµ ◦ f mapsC(Y ) onto the unit interval[0,1]. Hence,µ ◦ f is auniversal map. Then, with similar arguments to the ones given in the first part of the proofof Theorem 3.1, we can prove thatH is a nonempty closed set which cuts weakly between


F1(Y ) andY in C(Y ), whereF1(Y ) = {{y}: y ∈ Y }. Now, by Theorem 2.1, we have thatC(Y ) is s-connected, and thus there is a componentK of H which cuts weakly betweenF1(Y ) andY in C(Y ).

Next we claim that there existsA ∈ K such that eitherf (A) ⊂ g(A) andg(A) ⊂ f (A).In other case we chooseε > 0 such that, for eachA ∈ K, f (A) �⊂ N(ε,g(A)) andg(A) �⊂N(ε, f (A). Then we put

[A] = (f (A)× g(A)

) \ (N

(ε, g(A)

) ×N(ε, f (A)

))

andM = ⋃{[A]: A ∈ K}. The rest goes as in the proof of Theorem 3.1.✷Corollary 4.2 (Nadler [12, Theorem 2.11]).LetX be an arc-like continuum. Iff :Y → X

is a map from a continuumY ontoX, then the induced mapf :C(Y )→C(X) is universal.

Recall that a mapf :Y → X is weakly confluentprovided that every subcontinuumof X is the image underf of a subcontinuum ofY . A continuumX is said to be inclass(W) provided that every map from any continuum ontoX is weakly confluent. Itis easy to see that a mapf :Y → X is weakly confluent if and only if the induced mapf :C(Y )→C(X) is surjective. Then, since any universal map is surjective, we obtain nextresult as a consequence of Theorem 4.1.

Corollary 4.3 (Oversteegen and Tymchatyn [15, Theorem 8]).If X is a continuum withzero surjective semispan, thenX is in class(W).

Remark 4.4. Theorem 4.1 also implies Segal’s fixed point theorem that we stated asCorollary 3.4. Statement in Corollary 4.3 was first obtained by Read [16, Theorem 4] forarc-like continua.

In [9,14] Marsh and Nadler, respectively, have defined a continuumX to be inclass(U)

(class(U )) if for every mapf of a continuumY onto X, f (the induced mapf ) isuniversal. Considering these notions we have the following result.

Theorem 4.5. class(U) is a subclass of class(U ), and class(U ) is a proper subclass ofclass(W).

Proof. It is clear that a continuumX has zero surjective semispan if and only ifX is inclass(U), then the first part follows from Theorem 4.1. The argument to state Corollary 4.3actually prove thatclass(U ) is a subclass ofclass(W). LetZ be the circle-with-a-spiral [4,Example 7.1]. It can be proved thatZ is in class(W). It is known thatC(Z) does not havethe fixed point property [17]. Thus, there is not a universal map ontoC(Z). Therefore,Zis not inclass(U ). ✷Question 4.6. Is class(U) = class(U )?


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A fixed point theorem for Whitney blocks

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