s4 continua in the sense of michael are dendrites

Topology and its Applications 129 (2003) 19–27

www.elsevier.com/locate/topol

S4 continua in the sense of Michael are dendrites

Francis Jordan1

Department of Mathematics, University of Mississippi, University, MS 38677, USA

Received 18 January 2001; received in revised form 2 May 2002

Abstract

We show thatS4 spaces in the sense of Michael are dendrites. The proof involves functions whosegraphs are connected and monotone maps onto the arc. 2002 Elsevier Science B.V. All rights reserved.

MSC:primary 54C65, 54E40, 54C08; secondary 26A15

Keywords:Connectivity functions; Continuous selections; Monotone maps;S4 spaces; Irreducible continua

1. Introduction

We let 2X denote the space of all non-empty, compact subsets of a metric spaceX; 2X

has the Vietoris topology or, equivalently, the Hausdorff metric topology [9].LetF ⊆ 2X . A continuous selection forF is a continuous functionσ :F →X such that

σ(F ) ∈ F for eachF ∈ F .Michael [8, p. 178] defined aS4 space to be a spaceX such that there is a continuous

selection for everyF ⊆ 2X such that the members ofF are mutually disjoint and⋃F = X. The question of which spaces areS4 spaces is asked in [8, p. 155], and some

partial answers are given in [8, pp. 178, 179]. In particular, it is mentioned that noS4 spacecan contain a simple closed curve. A dendrite is a continuum that is locally connected andcontains no simple closed curve. The question of whetherS4 continua are dendrites is dueto Gail S. Young [12].

In [6] X is defined to be a weakS4 space provided that there is a continuous selectionfor everyF ⊆ F2(X), whereF2(X) denotes the 2-fold symmetric product ofX [9], such

E-mail address:[email protected] (F. Jordan).1 The results in this paper were found while the author was visiting Loyola University, New Orleans, LA

between 2000 and 2001.

0166-8641/02/$ – see front matter 2002 Elsevier Science B.V. All rights reserved.PII: S0166-8641(02)00135-9

20 F. Jordan / Topology and its Applications 129 (2003) 19–27

that the members ofF are mutually disjoint and⋃

F = X. Clearly, everyS4 space is aweakS4 space. Also, no simple closed curve is a weakS4 space. Both theS4 and weakS4 property are hereditary with respect to subsets [6].

We use the ideas developed in [6], namely maps with connected graphs. To prove thefollowing theorem:

Theorem. If X is a weakS4 continuum, thenX is an dendrite.

It is still unknown if the converse of this theorem is true.I would like to thank the referee for many helpful comments regarding the presentation

of this work, particularly, his or her comments on Lemma 3.5 and Corollary 3.1.

2. Terminology

We denote the cardinality ofR by c and the first uncountable cardinal byω1. By ω wedenote simultaneously the non-negative integers and the cardinality of the non-negativeintegers. For an arbitrary setS we denote its cardinality by|S|. The closed unit interval[0,1] will be denoted byI .

A continuum is a non-empty compact connected metric space. We useH(A,B) todenote the Hausdorff distance between two setsA and B. Given setsA,B ⊆ X anda metric d on X we let d(A,B) = inf{d(x, y): x ∈ A and y ∈ B} and diam(X) =sup{d(x, y): x, y ∈ X}. Given a functionf we denote the set of its discontinuity pointsby D(f ). We use cl to denote closure.

If X is a topological space andA ⊆ X is such thatX \ A is separated (i.e., notconnected), we sayA is a separator ofX. For a disconnected spaceX we writeX = A|Bto denote a partition ofX into disjoint closed subsets.

Given a functionf :X→ Y we denote the graph off by Γ (f ). A functionf :X→ Y

is called a connectivity function provided thatΓ (f |C) is connected for every connectedsubsetC of X. Connectivity functions have been studied extensively (e.g., [1,4,5]).

A continuumX is said to be irreducible between two pointsp andq provided that noproper subcontinuum ofX contains bothp andq . By 4.34 of [10], it is easy to see thatgiven any two distinct pointsp andq in a continuum there is a minimal continuumKcontainingp andq . Clearly,K must be irreducible betweenp andq .

Let X be a continuum. We sayf :X → I is aT map provided thatf is a monotonecontinuous surjection such thatf−1(w) is nowhere dense inX for everyw ∈ I . We useT map to denote these maps because of the following proposition due to E.S. Thomas[11, Theorem 10]:

Proposition 2.1. If X is hereditarily decomposable and irreducible between two points,then there is aT map fromX ontoI .

For f :X → I and 0< w < 1 defineAfw = cl(f−1([0,w))) ∩ cl(f−1((w,1])). Forw ∈ {0,1} we defineAfw = f−1(w). Observe that iff is a T map, thenAfw = ∅ forall w ∈ I .

F. Jordan / Topology and its Applications 129 (2003) 19–27 21

Given a compact metric spaceX andf :X → I continuous, we definef ∗ : I → 2X

to bef ∗(w) = f−1(w). Recall that a functionf :X → Y is said to be of Borel Class 1provided thatf−1(U) is a countable union of closed sets for every open setU ⊆ Y . WesayA⊆X is meager or of the first category provided thatA is contained in the countableunion of closed nowhere dense sets.

We state some well-known and useful facts aboutf ∗.

Proposition 2.2. The functionf ∗ has the following properties:

(i) f ∗ is in Borel Class1,(ii) D(f ∗) is a meager set, and(iii) if K ⊆X is closed, then{w ∈ I : f ∗(w)∩K = ∅} is closed.

Proof. We outline a proof. First one may verify (iii) using the fact thatf [K] is closed bycontinuity and thatf [K] = {w ∈ I : f ∗(w) ∩K = ∅}. Next (i) can be verified by lookingat the basic open sets of 2X and using the fact that every open set inX is the countableunion of closed sets and (iii). Finally (ii) follows from (i) and the well-known fact, see,for example [7], that the points of discontinuity of a Borel Class 1 function form a meagerset. ✷

Now, we introduce a simple version of the Cantor–Bendixson Theorem which we statebelow. We will only use the result forI , but we state it more generally. We will use thisproposition in the proofs of Lemmas 3.1 and 3.6. LetX be a topological space andK ⊆X.Inductively define sequences{Kα}0<α<ω1 and{Wα}α∈ω1 so that

(a) K0 =K,(b) Wβ = {x ∈Kβ : x is isolated inKβ},(c) if α has the formβ + 1, thenKα =Kβ \Wβ , and(d) if α is a limit cardinal, thenKα =⋂

β<α Kβ .

Proposition 2.3. If X is a complete metric space,K ⊆X is closed, and we define{Kα}α∈ω1

and{Wα}α<ω1 as above, thenKα is closed for allα ∈ ω1 and there is anζ ∈ ω1 such that

(i) Kζ =Kβ for β � ζ and(ii) Kζ = ∅ if and only ifK is countable if and only ifK =⋃

α∈ω1Wα .

3. Proof of theorem

Lemma 3.1. LetX be a continuum andf :X→ I be aT map. IfC is a closed separatorofX, then|f [C]| = c or there is aw ∈ (0,1) such thatAfw ⊆ C.

Proof. LetK = f [C]. By way of contradiction, assume that|K| = c andAfw \C = ∅ forall w ∈ I . SinceK is closed,|K| � ω. LettingK0 =K, define sequences of sets{Kα}α∈ω1

as we did in the discussion prior to the statement of Proposition 2.3. By Proposition 2.3(ii),


there is anζ ∈ ω1 such thatKζ = ∅. Notice thatC must be nowhere dense inX, sincef isaT map andf [C] is countable.

We claim that for everyβ � ζ if J is a complementary interval ofKβ , thenf−1(J ) \Cis connected. Forβ = 0 the claim follows immediately from the monotonicity off . Soassume we have shown the claim for allξ < β . We now establish the claim forβ . LetJ bea complementary interval ofKβ . Consider the following collection of open intervals:

C = {L⊆ J : f−1(L) \C is connected

}.

Notice that C = ∅ since K0 is countable and closed. It is easily checked that any⊆-increasing chain inC has an upper bound inC, so by the Hausdorff Maximal Principle,there is a⊆-maximal elementM of C. We show thatM = J which will prove the claimfor β . LetM = (a, b) andJ = (c, d). By way of contradiction, assume thatb = d . Sinceb /∈ Kβ , there is a maximalξ < β such thatb ∈ Kξ . Sinceb /∈ Kξ+1, we have thatb isisolated inKξ . So, there is ane ∈ J ∪ {d} such that(b, e) is a complementary intervalof Kξ . Now f−1((a, b)) \ C is connected and, by inductive hypothesis,f−1((b, e)) \ Cis connected. SinceC is nowhere dense, cl(f−1((b, e)) \ C) = cl(f−1((b, e))). So,(f−1((b, e)) \ C) ∪ H is connected for anyH ⊆ A

f

b . Similarly, (f−1((a, b)) \ C) ∪ His connected for anyH ⊆A

fb . By assumption,Afb \C = ∅. Thus, lettingH =A

fb \C, we

havef−1((a, e))\C = ((f−1((b, e))\C)∪H)∪((f −1((a, b))\C)∪H) connected whichcontradicts maximality ofM. Thus,b = d . Similarly, we can show thata = c. Hence,J =M. So, the claim holds forβ . Therefore, the claim holds for allβ � ζ .

SinceKζ = ∅, we haveX \ C = f−1(I) \ C which is connected by the claim. Thiscontradicts thatC is a separator ofX. ✷Lemma 3.2. If X is compact andf :X→ I is a function, thenf−1(w)= A

fw for all but

countably manyw ∈ I .

Proof. Suppose the lemma is false. Without loss of generality, we may assume that there isan uncountableW ⊆ I and numbersε, δ > 0 such thatd(xw,f−1((w,w+ δ))) > ε wherefor eachw ∈ W we havexw ∈ f−1(w). SinceW is uncountable, there is a decreasingconvergentsequence{wn}∞n=1 such thatwn ∈W for all n. There is an integerk large enoughthat|wn −wm|< δ for all m,n� k. By our choice ofδ we must haved(xwn, xwm) > ε forall n,m� k, this however contradicts the compactness ofX. Thus, we have the lemma.✷

Given a functionf :X → Y we sayg :Y → X is an inverse off provided thatf (g(y))= y for all y ∈ f [X] andg(f (x))= x for all x ∈ g[Y ].

Lemma 3.3. Let X be a continuum andf :X → I be aT map. There is a connectivityfunctiong : I → X such thatg is an inverse off and g[I ] is dense inX. Moreover, ifp ∈ f−1(0) andq ∈ f−1(1) we may assumeg(0)= p andg(1)= q .

Proof. Let {Cα}α∈c be an enumeration of the closed separatorsC of X with the propertythat |f [C]| = c. LetR = {w: f−1(w) = A

fw} ∪ {0,1}, by Lemma 3.2,R is countable. By


transfinite induction we can construct a sequence{xα}α∈c such that for everyα ∈ c wehave,

xα ∈ f [Cα] \ ({xξ : ξ < α} ∪R).Notice such choices can be made since|f [Cα]| = c and|{xξ : ξ < α} ∪ R| < c for everyα ∈ c. For eachα ∈ c defineg(xα) so thatg(xα) ∈ f−1(xα) ∩ Cα . Forx ∈ I \ {xα: α ∈ c}defineg(x) so thatg(x) ∈ Afx , in particular we may defineg(0) = p andg(1) = q . It isimportant to notice at this point thatg(x) ∈ Afx for all x ∈ I . In particular,g is an inverseof f sinceg(x) ∈Afx ⊆ f−1(x).

We claim thatg[I ] is connected. By way of contradiction, assume thatg[I ] is notconnected. Sinceg[I ] ⊆X andX is connected, there is, by [7, Theorem 3, p. 155], a closedsetC ⊆ X such thatC separatesX andg[I ] ∩ C = ∅. We must now consider two cases.First suppose|f [C]| = c. Pickα ∈ c such thatC = Cα . By the way we definedg, we haveg(xα) ∈ Cα which contradicts thatg[I ] ∩C = ∅. We now consider the case whenf [C] iscountable. SinceC ⊆ X, andf is aT map, Lemma 3.1 tells us that there is aw ∈ (0,1)such thatAfw ⊆ C. Nowg(w) ∈Afw ⊆ C which contradicts thatC∩g[I ] = ∅. Thus,g[I ] isconnected.

Let f1 :X→ I ×X be the continuous injection defined byf1(x)= 〈f (x), x〉. Clearly,f1[g[I ]] is connected. Ifx ∈ g[I ], then

f1(x)=⟨f (x), x

⟩= ⟨f (x), g

(f (x)

)⟩ ∈ Γ (g).Thus,f1[g[I ]] ⊆ Γ (g). If 〈w,y〉 ∈ Γ (g), then sinceg is an inverse off and f is asurjection, we have

〈w,y〉 = ⟨w,g(w)

⟩= ⟨f(g(w)

), g(w)

⟩= f1(g(w)

).

Thus,Γ (g) ⊆ f1[g[I ]]. Therefore,Γ (g)= f1[g[I ]] is connected. Since the domain ofgis I , the fact thatΓ (g) is connected implies thatg is connectivity.

To see thatg[I ] is dense inX it is enough, by [6, Lemma 4.1] to notice thatg[I ]intersects each closed separator ofX. ✷

I have been informed by the referee that the construction in the following lemma is whatis known as a pullback. In particular, the construction we use here is very similar to the oneused in [3, Theorem 3.9].

Lemma 3.4. Let X and Y be continua. Iff :X → I and g :Y → I are T maps andD(f ∗) ∩ D(g∗)= ∅, then

R = {〈x, y〉: f (x)= g(y)}

is a continuum andh :R→ I defined byh(〈x, y〉)= f (x)= g(y) is aT map.

Proof. First note that the function(f, g) :X × Y → I2 defined by (f, g)(〈x, y〉) =〈f (x), g(y)〉 is continuous. It is easily checked thatR = (f, g)−1(diag(I × I)) so R iscompact.


We now show thath :R→ I is monotone. Letw ∈ I . By way of contradiction, assumethat h−1(w) is not connected. Leth−1(w) = A|B. Sinceg−1(w) is connected, we havethat{x} × g−1(w)⊆ A or {x} × g−1(w)⊆ B for all x ∈ f−1(w). We now have

f−1(w) = {x ∈ f−1(w):

{x} × g−1(w)⊆A} ∣∣ {x ∈ f−1(w): {x} × g−1(w)⊆ B

},

which contradicts thatf−1(w) is connected. Thus,h is monotone. SinceI is connected,we also have thatR is a continuum.

We now show thath−1(w) is nowhere dense for allw ∈ I . Let 〈x, y〉 ∈ h−1(w). SinceD(f ∗)∩ D(g∗)= ∅, we may assume thatg∗ is continuous atw. Sincef is aT map, thereexists a sequence{xn ∈X \ f−1(w)}n∈ω converging tox such that{f (xn)}n∈ω convergesto f (x)=w. Sinceg∗ is continuous atw andf (xn)→ w, we haveg∗(f (xn))→ g∗(w).So, we may pick a sequence{yn ∈ Y }n∈ω converging toy such thatg(yn)= f (xn). Now{〈xn, yn〉}n∈ω converges to〈x, y〉 and {h(〈xn, yn〉)}n∈ω converges tow and 〈xn, yn〉 /∈h−1(w) for all n ∈ ω. Thus,h−1(w) is nowhere dense.✷Lemma 3.5. LetX andY be continua,f :X→ I andg :Y → I beT maps such thatf ∗ iscontinuous at0 andg∗ is continuous at0. There exist connected dense subsetsN ⊆X andM ⊆ Y and a connectivity injectionh :N → M such that fora ∈ f−1(0), b ∈ f−1(1),c ∈ g−1(0), andd ∈ g−1(1), we haveh(a)= d andh(b)= c.

Proof. Sincef andg are continuous, we have, (using Proposition 2.2(ii)) thatf ∗ andg∗,are continuous outside of meager setsMf andMg , respectively. By [2] and the factthat 0 /∈ Mf and 0 /∈ Mg there is a decreasing homeomorphisml : I → I such thatl[Mf ] ∩Mg = ∅. Let f1 = l ◦ f . Notice thatD(f ∗

1 )= l[Mf ]. Sincef1, andg areT mapsandD(f ∗

1 )∩ D(g∗)= ∅, Lemma 3.4 yields that

R = {〈x, y〉 ∈X× Y : f1(x)= g(y)}

is a continuum andk :R → I defined byk(〈x, y〉) = f1(x) = g(y) is a T map. Noticethat 〈a, d〉 ∈ k−1(1) and〈b, c〉 ∈ k−1(0), sincel is decreasing. By Lemma 3.3, there is aconnectivity inversej : I → R of k with the property thatj (1)= 〈a, d〉 andj (0)= 〈b, c〉andj [I ] is dense inR.

Let N = {x ∈ X: there is ayx ∈ Y such that〈x, yx〉 ∈ j [I ]}. Sincej [I ] is dense andconnected inR, it follows thatN is dense and connected inX.

Defineh :N → Y by h(x)= yx . To see thath is well defined takex ∈ N and supposethat 〈x, yx〉, 〈x, y∗

x 〉 ∈ j [I ]. Sincej [I ] ⊆ R, we havek(〈x, yx〉) = f1(x) = k(〈x, y∗x 〉).

Since j is an inverse ofk, 〈x, yx〉 = j (k(〈x, yx〉)) = j (k(〈x, y∗x 〉)) = 〈x, y∗

x 〉. Hence,yx = y∗

x . Therefore,h is well defined.The same argument that was applied to the first coordinate can be used to show thath is

injective.We show thath is connectivity. LetP be a connected subset of the domainN of h.

We must show thatΓ (h|P ) is connected. SinceP is connected,f1 is continuous andj is connectivity;j [f1[P ]] is connected. So, it is enough for us to show thatΓ (h|P ) =j [f1[P ]].


If 〈x, y〉 ∈ Γ (h|P ), thenx ∈ P and there is aw ∈ I such thatj (w)= 〈x, y〉. It followsthatw = k(〈x, y〉)= f1(x). So,〈x, y〉 ∈ j [f1[P ]].

Suppose〈x, y〉 ∈ j [f1[P ]]. Let x∗ ∈ P be such thatj (f1(x∗)) = 〈x, y〉. Sincex∗ ∈

P ⊆ N , we have〈x∗, h(x∗)〉 ∈ j [I ] ⊆ R. So, k(〈x∗, h(x∗)〉) = f1(x∗). It follows that,

〈x, y〉 = j (f1(x∗))= j (k(〈x∗, h(x∗)〉))= 〈x∗, h(x∗)〉. Hence,〈x, y〉 ∈ Γ (h).

LetM = h[N]. SinceΓ (h|N)= j [I ] is dense and connected inR, it follows thatM isdense and connected inY . We also have thath(a)= d andh(b)= c. ✷Lemma 3.6. SupposeX is a continuum andf :X→ I is a T map. Ifdiam(X) > ε, thenthere existp,q ∈X such thatd(p, q) > ε/2, f (p) = f (q), f−1(f (q))=A

f

f (q), andf ∗ iscontinuous atf (p).

Proof. Let p ∈ X be such thatf ∗ is continuous atf (p). Since diam(X) > ε, there is aq1 ∈X such thatd(p, q1) > ε/2.

Let 0< δ < (d(p, q1) − ε/2)/4. Let K = {w ∈ I : d(q1, f∗(w)) � δ}. By Proposi-

tion 2.2,K is closed. Starting withK0 = K, define sequences〈Wα〉α∈ω1 and 〈Kα〉α∈ω1

as we did before the statement of Proposition 2.3. LetW =⋃α∈ω1

Wα .We claim that for everyw ∈W we haved(f ∗(w), q1)= δ. First supposew ∈W0. Since

w is isolated inK0 = K, there is an open setU such thatw ∈ U and for allx ∈ U \ {w}we haved(f ∗(x), q1) > δ. Sincef−1(w) is nowhere dense inX, we must have thatd(f ∗(w), q1) = δ. Suppose now thatα ∈ ω1 and we have shown thatd(f ∗(w), q1) = δ

for all w ∈ ⋃β<α Wβ . Let w ∈ Wα . Sincew is isolated inKα , there is an open setUsuch thatw ∈ U and for everyx ∈ U \ {w} we havex ∈⋃β<α Wβ or x /∈ K. It follows

thatd(f ∗(x), q1)� δ for all x ∈ U \ {w}. Sincef−1(w) is nowhere dense inX, we haved(f ∗(w), q1)= δ. Thus,d(f ∗(w), q1)= δ for everyw ∈W .

If K is countable then, by Proposition 2.3(ii),K = W . It follows, then, thatd(q1, f

∗(w))= δ for all w ∈K, which contradicts thatd(q1, f∗(f (q1)))= 0. Thus,K is

uncountable.SinceK \W is uncountable, there is by Lemma 3.1 aw ∈K \ (W ∪ {f (p)}) such that

f−1(w)= Afw. Pickq ∈ f−1(w) such thatd(q, q1)� δ. Clearly,f−1(f (q))= A

f

f (q) andf (p) = f (q). By our choice ofδ we haved(p, q) > ε/2. ✷

An immediate corollary to Lemma 3.6 is:

Corollary 3.1. If X is a continuumf :X→ I is aT map anddiam(X) > ε, then there isa subcontinuumY of X and aT mapf1 :Y → I such that there exist pointsp ∈ f−1

1 (0)andq ∈ f−1

1 (1) with d(p, q) > ε/2 andf ∗1 is continuous at0.

Proof. To see it letp,q ∈ X be as in Lemma 3.6. Without loss of generality, we mayassume thatf (p) < f (q). Let Y = f−1([f (p),f (q)]). Clearly,p,q ∈ Y . We check thatf |Y is a T map. Since(f |Y )−1(w) = f−1(w) for all w ∈ [p,q], we know thatf |Y ismonotone. It remains to check thatf−1(w) is nowhere dense inY for all w ∈ [p,q]. Thisis easily seen to be the case forw ∈ (p, q). For p andq notice thatf−1(p) = A

f

f (p) ⊆cl((p, q]) andf−1(q)=A

f

f (q)⊆ cl([p,q)). So,f |Y is aT map.


We may get the desiredf1 by lettingh : [f (p),f (q)] → I be an increasing homeomor-phism and settingf1(x)= h(f (x)) for all x ∈ Y . ✷

We are now ready to prove our theorem.

Proof of theorem. Let X be a weakS4 continuum. SinceX contains no simple closedcurve, it is enough for us to show thatX is locally connected. We also note that, byCorollary 7.2 of [6],X is hereditarily decomposable.

By way of contradiction, assume thatX is not locally connected. By Exercise 5.22 andTheorem 5.12 of [10] there exist mutually disjoint subcontinuaK and{Kn}n∈ω in X suchthat limn→∞Kn =K andK is non-degenerate. SinceK is non-degenerate, we may alsoassume that there is anε > 0 such that diam(Kn) > ε for everyn ∈ ω.

For eachKn pick points an, bn ∈ Kn such thatd(an, bn) > ε. For eachn ∈ ω letLn ⊆ Kn be a continuum irreducible betweenan and bn. By Proposition 2.1, there isa T mapgn :Ln → I . By Corollary 3.1, we may find a continuumYn ⊆ Ln, a T mapfn :Yn → I , andpn, qn ∈ Yn such thatd(pn, qn) > ε/2, fn(pn) = 0, fn(qn) = 1, andf ∗n is continuous at 0. Picking convergent subsequences if necessary, we may assume

that there existp,q ∈ K such that the limn→∞ pn = p, and limn→∞ qn = q . Notice thatd(p, q) � ε/2. By Lemma 3.5, for everyn ∈ ω there are connected dense setsNn ⊆ Y2nandMn ⊆ Y2n+1 and a connectivity injectionhn :Nn → Mn such thathn(p2n) = q2n+1,andhn(q2n)= p2n+1.

Consider the collection of mutually disjoint doubletons

S =( ∞⋃n=0

{{x,hn(x)

}: x ∈Nn

})∪ {{p,q}}.We show there is no continuous selector forS. By way of contradiction, supposeg :S →X

is a continuous selector.Fix n ∈ ω. Assumeg({p2n,hn(p2n)})= p2n. Since{{x,hn(x)}: x ∈ Nn} is connected

and K2n and K2n+1 are disjoint, we haveg({q2n,hn(q2n)}) = q2n by continuity. Bya similar argument, ifg({p2n,hn(p2n)}) = hn(p2n) = p2n+1, then g({q2n,hn(q2n)}) =hn(q2n)= q2n+1.

Now suppose thatg({p2n,hn(p2n)})= p2n for everyn in some infinite subsetA of ω.Since limn→∞{p2n,hn(p2n)} = {p,q}, we haveg({p,q})= p. However, it is also the casethatg({q2n,hn(q2n)})= q2n for all n ∈A. Since limn→∞{q2n,hn(q2n)} = {p,q}, we musthaveg({p,q})= q . Sincep = q , we have a contradiction. A similar argument shows thatif g({p2n,hn(p2n)}) = hn(p2n) = q2n+1 for infinitely manyn in ω, then one also gets acontradiction. Therefore,X is locally connected and thus a dendrite.✷

References

[1] R.G. Gibson, T. Natkaniec, Darboux-like functions, Real Anal. Exchange 22 (2) (1996–97) 492–533.[2] W. Gorman Jr, The homeomorphic transformation ofc-sets intod-sets, Proc. Amer. Math. Soc. 17 (1966)

825–830.[3] J. Grispolakis, S.B. Nadler Jr, E.D. Tymchatyn, Some properties of hyperspaces with application to continua

theory, Canad. J. Math. 31 (1) (1979) 197–210.


[4] M. Hagan, Equivalence of connectivity maps and peripherally continuous transformations, Proc. Amer.Math. Soc. 17 (1966) 175–177.

[5] O.H. Hamilton, Fixed points for certain noncontinuous transformations, Proc. Amer. Math. Soc. 8 (1957)750–756.

[6] F. Jordan, S.B. Nadler Jr, A result about a selection problem of Michael, Proc. Amer. Math. Soc. 129 (4)(2001) 1219–1228.

[7] K. Kuratowski, Topology, Vol. 2, Academic Press and Polish Scientific Publishers, 1998.[8] E. Michael, Topologies on spaces of subsets, Trans. Amer. Math. Soc. 71 (1951) 152–182.[9] S.B. Nadler Jr, Hyperspaces of Sets, Marcel Dekker, New York, 1978.

[10] S.B. Nadler Jr, Continuum Theory, Marcel Dekker, New York, 1992.[11] E.S. Thomas Jr, Monotone decompositions of irreducible continua, Dissertationes Math. 50 (1966).[12] G.S. Young, in: Lecture Notes in Math., Vol. 171, Springer, New York, 1969.

s4 continua in the sense of michael are dendrites

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