A fixed-point-free map of a tree-like continuum induced by bounded valence maps on trees
AA FIXED-POINT-FREE MAP OF A TREE-LIKE CONTINUUM INDUCED BYBOUNDED VALENCE MAPS ON TREES
RODRIGO HERN ´ANDEZ-GUTI´ERREZ AND L. C. HOEHN
Abstract.
Towards attaining a better working understanding of fixed points of maps of tree-like continua,Oversteegen and Rogers constructed a tree-like continuum with a fixed-point-free self-map, described ex-plicitly in terms of inverse limits. Specifically, they developed a sequence of trees T n , n ∈ N and maps f n and g n from T n +1 to T n for each n , such that the g n maps induce a fixed-point-free self-map of the inverselimit space lim ←− ( T n , f n ).The complexity of the trees and the valences of the maps in their example all grow exponentially with n ,making it difficult to visualize and compute with their space and map. We construct another such example,in which the maps f n and g n have uniformly bounded valence, and the trees T n have a simpler structure. Introduction
The study of fixed points of maps has been a substantial theme in continuum theory for several decades(see e.g. the surveys [2] and [4]). A space X has the fixed point property if for every continuous function f : X → X , there is a point x ∈ X such that f ( x ) = x . By a continuum , we mean a compact connectedmetric space.One of the oldest and most important open problems in continuum theory is the plane fixed point prob-lem (see e.g. [2] Question 3): Does every continuum X ⊂ R which does not separate R have the fixedpoint property? An interesting special case of this problem arises when one assumes further that X is 1-dimensional. It is well known that a 1-dimensional continuum in the plane is non-separating if and only ifit is tree-like (see e.g. [5]). Thus the 1-dimensional special case of the plane fixed point problem is: Problem.
Does every tree-like continuum X ⊂ R have the fixed point property?David Bellamy constructed the first example of a tree-like continuum which does not have the fixed pointproperty in [1]. Fearnley and Wright [3] gave an analytic geometric description of Bellamy’s example in R . Oversteegen and Rogers [12, 13] produced another example similar to that of Bellamy with an explicitinverse limit construction (see Section 1.1 below). Minc has produced a number of variants of Bellamy’sexample having a variety of additional properties in [6], [7], [8], [9], and [10].Each of the above examples is rather complex, making it difficult to ascertain whether they could beembedded in the plane, and to visualize and manipulate them. The purpose of this paper is to provideanother example of a tree-like continuum with a fixed-point-free self-map, which is simpler than previouslygiven examples and lends itself to more manageable visualizations and computations. We consider this tobe an important step towards attacking the plane fixed point problem (for tree-like continua). Date : November 20, 2018.2010
Mathematics Subject Classification.
Primary 54F15; Secondary 54H25.
Key words and phrases. tree-like, continuum, fixed point, fixed point property, coincidence set.The second named author was partially supported by NSERC grant RGPIN 435518, and by the Mary Ellen Rudin YoungResearcher Award. a r X i v : . [ m a t h . GN ] A ug RODRIGO HERN´ANDEZ-GUTI´ERREZ AND L. C. HOEHN
Our example is related to that of Oversteegen and Rogers [12, 13]. To describe its properties, we outlinethe scheme of their construction and ours in the following section.1.1.
Induced maps on inverse limits.
Let T , T , T , . . . be a sequence of trees, and for each n ≥ f n : T n +1 → T n and g n : T n +1 → T n be maps such that f n ◦ g n +1 = g n ◦ f n +1 for all n ≥
0. Then onthe inverse limit space X = lim ←− ( T n , f n ), the maps g n induce a self-map g : X → X defined by the formula (cid:104) x n (cid:105) ∞ n =0 (cid:55)→ (cid:104) g n ( x n +1 ) (cid:105) ∞ n =0 . This latter point belongs to X because if (cid:104) x n (cid:105) ∞ n =0 ∈ X , then for each n ≥ f n +1 ( x n +2 ) = x n +1 so f n ( g n +1 ( x n +2 )) = g n ( f n +1 ( x n +2 )) = g n ( x n +1 ).If f and g do not have a coincidence point , i.e. if there is no point x ∈ T such that f ( x ) = g ( x ), thenthis induced map g is fixed-point-free. This is because for each point (cid:104) x n (cid:105) ∞ n =0 ∈ X , the first coordinate ofthis point is x = f ( x ), and this by assumption is different from g ( x ), which is the first coordinate of theimage of this point under g .We remark that if f and g have no coincidence point, then in fact f n and g n have no coincidence pointfor all n ≥
0. This can be proved by induction as follows: suppose f n and g n have no coincidence point forsome n ≥
0. If f n +1 ( x ) = g n +1 ( x ) for some x ∈ T n +2 , then the equality f n ( g n +1 ( x )) = g n ( f n +1 ( x )) impliesthat the point f n +1 ( x ) = g n +1 ( x ) in T n +1 is a coincidence point for f n and g n , contradicting the assumptionthat f n and g n do not have a coincidence point.Because of the symmetry between the maps f n and g n in this type of construction, each such sequenceof pairs of maps also gives rise to a “dual” example of a tree-like continuum and fixed-point-free self-map,obtained by using the maps g n as bonding maps, and inducing a self-map using the maps f n .According to a well-known result of Mioduszewski [11], every map between inverse limit spaces is inducedby a sequence of maps between some of their factor spaces, but in general these maps form diagrams with thebonding maps which are merely “almost” commuting, instead of exactly commuting as in our construction.Oversteegen and Rogers constructed a sequence of trees T n and pairs of maps f n , g n as above in [12, 13].Their maps f n and g n have valence which is increasing in n . Here by valence of a map f : X → Y we meanthe maximum, taken over all y ∈ Y , of the number of components of f − ( y ). In practice, this increasingvalence makes it very difficult to draw pictures which clearly and accurately depict their example. In thispaper, we produce a new example using the same scheme, in which our maps f n all have valence 6 and g n all have valence 12. This is a substantial simplification, as it enables one to draw pictures which clearly andaccurately describe these maps, such as those in Figures 2, 3, and 4 below.2. Notation
By an arc , we mean a space which is homeomorphic to the interval [0 , F denotes an arc, we will usethe same symbol F to represent a fixed homeomorphism from [0 ,
1] to the set F . Hence if F is an arc, thenthe points of F are of the form F ( t ), for t ∈ [0 , F are F (0) and F (1).A tree is a continuum which is the union of finitely many arcs having pairwise finite intersections, andwhich contains no circles.The n -fold tent map on the unit interval is the piecewise linear function τ n : [0 , → [0 ,
1] defined by τ n ( x ) = (cid:40) nx − i if x ∈ [ in , i +1 n ] for some even i ∈ { , , . . . , n − } i + 1 − nx if x ∈ [ in , i +1 n ] for some odd i ∈ { , , . . . , n − } . In this paper we will make use of τ , τ , and τ .Given a set A ⊂ [0 ,
1] and an integer m ≥
1, we write τ − m A = { x ∈ [0 ,
1] : m times (cid:122) (cid:125)(cid:124) (cid:123) τ ◦ τ ◦ · · · ◦ τ ( x ) ∈ A } .Further, if m = 0 then put τ − m A = A . It will be useful to keep in mind that for any y ∈ [0 , τ − { y } = { y , − y } (and these two points are different unless y = 1). OUNDED VALENCE MAPS ON TREES 3 (a)
The tree T . The sizes of the attached triods indicate the smallest value of m ≥ T m . (b) The tree T , represented as the quotient of a union of disjoint arcs, where the arcs are identifiedat the indicated points to form the space T pictured in part (a). Note that the arcs are arrangedso that the larger triods from the picture in part (a) appear further to the left than the smallerones. This represntation enables us to draw pictures of T × T as in Figure 3. Figure 1.
Pictures of the tree T .Given a pair of function f, g : X → Z , the coincidence set of g and f is the set [ g, f ] = { ( x, y ) ∈ X × X : g ( x ) = f ( y ) } . Observe that f and g have a coincidence point if and only if [ g, f ] ∩ ∆ X (cid:54) = ∅ , where∆ X = { ( x, x ) : x ∈ X } . If ˆ f , ˆ g : W → X are two other functions, then the equality f ◦ ˆ g = g ◦ ˆ f holds if andonly if (cid:16) ˆ f ( w ) , ˆ g ( w ) (cid:17) ∈ [ g, f ] for all w ∈ W .A map f : X → Z is monotone if f − ( z ) is connected for all z ∈ Z . For maps between arcs, this isequivalent to being non-increasing or non-decreasing. An arc A in the product X × X is monotone if both A ∩ ( { x } × X ) and A ∩ ( X × { x } ) are connected (or empty) for all x ∈ X .3. Construction of the example
For each n = 0 , , , . . . , let T n be the tree consisting of the interval [0 , p ∈ { , } ∪ (cid:83) n − m =0 τ − m { , } , T n contains three arcs F pi , for i = 0 , ,
2, such that F pi (0) is identified with p , and these arcs are otherwise disjoint from each other and from [0 , F p ∪ F p ∪ F p as a triod attached to [0 ,
1] at p . See Figure 1 for an illustration of the tree T .Below we will define the sequences of functions f n and g n from T n +1 to T n , for n = 0 , , , . . . . To ensurethe commutativity condition f n − ◦ g n = g n − ◦ f n , we will construct these maps recursively, defining f n and g n in terms of f n − and g n − . We will identify a certain set Γ n ⊂ [ g n − , f n − ] ⊂ T n × T n , then define f n and g n so that ( f n ( x ) , g n ( x )) ∈ Γ n for all x ∈ T n +1 , thus achieving the commutativity condition.In a nutshell, on the arc [0 , ⊂ T n +1 , f n will resemble the map τ , and g n will resemble τ . Observethat τ and τ have two coincidence points, 0 and , both of which are mapped to 0. For this reason, g n (0)and g n ( ) are pulled away from 0 to some point ε n >
0, to avoid a coincidence point. Moreover, the triodsattached at 0 and enable f n and g n to branch in different directions there, so that both are onto but theydon’t have a coincidence point. It turns out that additional triods need to be attached to the trees T n as n increases, to enable us to complete the commutative diagrams. Near points x ∈ [0 ,
1] where τ ( x ) equalsone of these branch points, f n takes a detour up and down one of the legs F τ ( x ) i ; likewise for g n near points RODRIGO HERN´ANDEZ-GUTI´ERREZ AND L. C. HOEHN x ∈ [0 ,
1] where τ ( x ) equals one of these branch points. The choice of which leg to detour up and down willbe dictated by the function j defined below.For each n = 0 , , , . . . , let ε n = · n . Let Γ n be a subset of T n × T n with Γ n ∩ ∆ T n = ∅ , consisting ofelements as follows:( C1 ) The set A = Γ n ∩ ([0 , × [0 , , ε n ) to ( , ,
1) to ( , + ε n ), and one from ( , − ε n ) to (1 , C2 ) A passes through all the points of the form ( p, τ ( p )), where p ∈ τ − m { , } for some m = 0 , , . . . , n .( C3 ) For each i = 0 , ,
2, Γ n ∩ (cid:0) F i × [0 , (cid:1) is a monotone arc from (cid:0) F i (0) , ε n (cid:1) = (0 , ε n ) to ( F i ( ) , i = 0 , ,
2, Γ n ∩ (cid:16) F i × [0 , (cid:17) is the union of two monotone arcs, one from (cid:16) F i (0) , + ε n (cid:17) =( , + ε n ) to (cid:16) F i ( ) , (cid:17) , and one from (cid:16) F i (0) , − ε n (cid:17) = ( , − ε n ) to (cid:16) F i ( ) , (cid:17) .( C4 ) For each p ∈ { , } and each i = 0 , ,
2, Γ n ∩ (cid:0) F pi × F pi +1 (cid:1) is a monotone arc from (cid:0) F pi ( ) , F pi +1 (0) (cid:1) = (cid:0) F pi ( ) , p (cid:1) to (cid:0) F pi (1) , F pi +1 (1) (cid:1) . Here i + 1 is reduced modulo 3.( C5 ) For each p ∈ τ − m { , } , where m = 0 , , . . . , n −
1, and each i = 0 , ,
2, Γ n ∩ (cid:16) F pi × F τ ( p ) i (cid:17) is amonotone arc from (cid:16) F pi (0) , F τ ( p ) i (0) (cid:17) = ( p, τ ( p )) to (cid:16) F pi (1) , F τ ( p ) i (1) (cid:17) .( C6 ) For each p ∈ τ − n { , } and each i = 0 , ,
2, Γ n contains a straight (vertical) arc from (cid:16) p, F τ ( p ) i (0) (cid:17) =( p, τ ( p )) to (cid:16) p, F τ ( p ) i (1) (cid:17) .On the set { , , } ∪ (cid:83) ∞ m =0 τ − m { , , } , define the function j as follows: • j ( ) = 0, j ( ) = 1, j ( ) = 2. Observe that if x ∈ { , , } , then j ( τ ( x )) = j ( x ) + 1 (modulo 3). • Given x ∈ τ − m { , , } for some m = 0 , , , . . . , we have τ m +12 ( x ) ∈ { , , } . Define j ( x ) = j ( τ m +12 ( x )). Observe that this implies j ( x ) = j ( τ ( x )) in this case.We now proceed to define maps f n , g n : T n +1 → T n with ( f n ( x ) , g n ( x )) ∈ Γ n for all x ∈ T n +1 . Note thatsince Γ n ∩ ∆ T n = ∅ , it follows that f n and g n do not have any coincidence point, hence [ g n , f n ] ∩ ∆ T n +1 = ∅ .To define f n , g n : T n +1 → T n , we specify below the pair ( f n ( x ) , g n ( x )) for all x in a particular finite subset R of T n +1 with the property that T n +1 (cid:114) R is a union of finitely many disjoint open arcs. Call two points of R adjacent if they are the endpoints of the closure of an arc in T n +1 (cid:114) R . We will ensure that if x , x ∈ R areadjacent, then there will be a unique monotone arc in Γ n from ( f n ( x ) , g n ( x )) to ( f n ( x ) , g n ( x )). We thenextend f n , g n to all of T n +1 by asserting that for x ∈ [ x , x ], ( f n ( x ) , g n ( x )) parameterizes this monotone arcin Γ n .The set R is the union of the following sets:(a) { , } (b) { ε n +1 , ± ε n +1 } (c) (cid:83) n +1 m =0 τ − m { , } (d) { , , } (e) (cid:83) nm =0 τ − m { , , } (f) { F i ( ) , F i (1) , F i ( ) , F i (1) : i = 0 , , } (g) { F pi (1) : p ∈ (cid:83) nm =0 τ − m { , } , i = 0 , , } We remark that ε n +1 is defined to be small enough so that 0 and ε n +1 are adjacent, as are and ± ε n +1 . OUNDED VALENCE MAPS ON TREES 5
Figure 2.
A picture of a set Γ ⊂ T × T , drawn in heavy black lines.Define f n , g n on each of these points in R as follows: x ∈ R ( f n ( x ) , g n ( x )) ∈ Γ n (a) x = 0 or (0 , ε n )This point is in Γ n by condition ( C1 ).(b) x = ε n +1 or ± ε n +1 ( ε n , t ), where ε n < t < is such that ( ε n , t ) ∈ Γ n (c) x ∈ τ − m { , } for some m = 0 , , . . . , n ( τ ( x ) , τ ( x ))Note τ ( x ) ∈ τ − m { } , so this point is in Γ n bycondition ( C2 ). x ∈ τ − ( n +1)2 { , } ( t, τ ( x )), where t ∈ [0 ,
1] is such that ( t, τ ( x )) ∈ Γ n and t < if and only if τ ( x ) < (d) x = , , or (cid:16) F j ( x ) (1) , F j ( x )+1 (1) (cid:17) This point is in Γ n by condition ( C4 ). RODRIGO HERN´ANDEZ-GUTI´ERREZ AND L. C. HOEHN
Figure 3.
A picture of a set Γ ⊂ T × T , drawn in heavy black lines.(e) x ∈ τ − m { , , } for some m = 0 , , . . . , n − (cid:16) F τ ( x ) j ( x ) (1) , F τ ( x ) j ( x ) (1) (cid:17) Note τ ( x ) ∈ τ − m { } , so this point is in Γ n bycondition ( C5 ). x ∈ τ − n { , , } (cid:16) τ ( x ) , F τ ( x ) j ( x ) (1) (cid:17) Note τ ( x ) ∈ τ − n { } , so this point is in Γ n bycondition ( C6 ). OUNDED VALENCE MAPS ON TREES 7
Figure 4.
The images of points in T under f (above) and g (below). For brevity, inthese pictures we write F pi in place of F pi (1).This completes the definition of f n and g n on [0 , ⊂ T n +1 . We pause here to point out that f n “resembles” τ on [0 ,
1] in the following sense: first, f n (0) = f n ( ) = 0 and f n ( ) = f n (1) = 1. Further, let r : T n → [0 , T n onto [0 , ⊂ T n , which is defined by r ( x ) = x if x ∈ [0 , r ( F pi ( t )) = p for any p in [0 ,
1] where a simple triod is attached and each i ∈ { , , } and t ∈ [0 , r ◦ f n is monotone from 0 to , from to , and from to 1. Similarly, g n “resembles” τ on [0 ,
1] in the sense that g n (0) = g n ( ) = ε n , g n ( ) = g n (1) = 0, g n ( ) = g n ( ) = g n ( ) = 1, and r ◦ g n is monotone between each pair of these points.On the arcs F pi , we define f n and g n as follows: x ∈ R ( f n ( x ) , g n ( x )) ∈ Γ n (f) x = F i ( ) or F i ( ) for i = 0 , , (cid:0) F i ( ) , (cid:1) x = F i (1) or F i (1) for i = 0 , , (cid:0) F i (1) , F i +1 (1) (cid:1) Here i + 1 is reduced modulo 3. These points arein Γ n by condition ( C4 ).(g) x = F pi (1), where p ∈ τ − m { , } for some m =0 , , . . . , n −
1, and i = 0 , , (cid:16) F τ ( p ) i (1) , F τ ( p ) i (1) (cid:17) Note τ ( x ) ∈ τ − m { } , so this point is in Γ n bycondition ( C5 ). x = F pi (1), where p ∈ τ − n { , } and i = 0 , , (cid:16) τ ( p ) , F τ ( p ) i (1) (cid:17) Note τ ( x ) ∈ τ − n { } , so this point is in Γ n bycondition ( C6 ).See Figure 4 for an illustration of the maps f and g .It remains to verify that the coincidence set [ g n , f n ] contains a set Γ n +1 with the above features ( C1 )through ( C6 ) (with n replaced by n + 1). The reader may find it helpful to verify that all the parts of theset Γ depicted in Figure 3 are present in the coincidence set [ g , f ], by refering to Figure 4. RODRIGO HERN´ANDEZ-GUTI´ERREZ AND L. C. HOEHN
The following straightforward observation reduces the task to checking equality of the maps at endpointsof arcs: if g n is monotone on the arc [ x , x ] ⊂ T n +1 and f n is monotone on the arc [ x , x ] ⊂ T n +1 , and if g n ( x ) = f n ( x ) and g n ( x ) = f n ( x ), then [ g n , f n ] contains a monotone arc from ( x , x ) to ( x , x ).We will treat conditions ( C1 ), ( C2 ), and ( C5 ) for the set Γ n +1 , and leave the rest to the reader. Verification of (C1) and (C2).
We first point out that (0 , ε n +1 ) ∈ [ g n , f n ] and ( , ± ε n +1 ) ∈ [ g n , f n ].This is because g n (0) = g n ( ) = ε n by (a), and f n ( ε n +1 ) = f n ( ± ε n +1 ) = ε n by (b). We next show (incases below) that for each other x ∈ R ∩ [0 , x, τ ( x )) ∈ [ g n , f n ]. This includes ( ,
1) and (1 , R ). • Let x ∈ { , } . Then τ ( x ) ∈ { , } . So g n ( x ) = τ ( x ) = 0 by (c) part 1, and f n ( τ ( x )) = 0 by (a). • Let x ∈ τ − m { , } for some m = 1 , , . . . , n + 1. Then τ ( x ) ∈ τ − ( m − { , } . So g n ( x ) = τ ( x ) by(c), and f n ( τ ( x )) = τ ( τ ( x )) = τ ( x ) by (c) part 1. • Let x ∈ { , , } . Then τ ( x ) ∈ { , , } as well. So g n ( x ) = F j ( x )+1 (1) by (d), and f n ( τ ( x )) = F j ( τ ( x )) (1) = F j ( x )+1 (1) by (d) and the property of j . • Let x ∈ { , , } . Then τ ( x ) ∈ { , , } . So g n ( x ) = F τ ( x ) j ( x ) (1) = F j ( x ) (1) by (e) part 1, and f n ( τ ( x )) = F j ( τ ( x )) (1) = F j ( x ) (1) by (d) and the property of j . • Let x ∈ τ − m { , , } for some m = 1 , , . . . , n . Then τ ( x ) ∈ τ − ( m − { , , } . So g n ( x ) = F τ ( x ) j ( x ) (1)by (e), and f n ( τ ( x )) = F τ ( τ ( x )) j ( τ ( x )) (1) = F τ ( x ) j ( x ) (1) by (e) part 1 and the property of j .For any adjacent points x , x ∈ R ∩ [0 , g n is monotone on [ x , x ] and f n is monotone on [ τ ( x ) , τ ( x )].Therefore, the above suffices to demonstrate that [ g n , f n ] contains the set A specified in condition ( C1 ), madeup of monotone arcs between pairs ( x , τ ( x )) and ( x , τ ( x )) where x , x ∈ R ∩ [0 ,
1] are adjacent and areeither both less or equal or both greater or equal . Moreover, according to the first and second bulletsabove, this set A contains all the points specified in condition ( C2 ). Verification of (C5).
Let p ∈ τ − m { , } for some m = 0 , , . . . , n , and let i ∈ { , , } . To prove that[ g n , f n ] contains a monotone arc from (cid:16) F pi (0) , F τ ( p ) i (0) (cid:17) to (cid:16) F pi (1) , F τ ( p ) i (1) (cid:17) , we show that g n ( F pi (0)) = f n ( F τ ( p ) i (0)) = τ ( p ) and g n ( F pi (1)) = f n ( F τ ( p ) i (1)) = F τ ( p ) i (1): • g n ( F pi (0)) = g n ( p )) = τ ( p ) by (c) part 1, and g n ( F pi (1)) = F τ ( p ) i (1) by (g). • If m = 0, then p ∈ { , } , so τ ( p ) ∈ { , } and τ ( p ) = 0. In this case, f n ( F τ ( p ) i (0)) = f n ( τ ( p )) =0 = τ ( p ) by (a), and f n ( F τ ( p ) i (1)) = F i (1) = F τ ( p ) i (1) by (f) part 2.If m >
0, then τ ( p ) ∈ τ − m { , } where m ∈ { , , . . . , n − } . In this case, f n ( F τ ( p ) i (0)) = f n ( τ ( p )) = τ ( τ ( p )) = τ ( p ) by (c) part 1, and f n ( F τ ( p ) i (1)) = F τ ( τ (0)) i (1) = F τ ( p ) i (1) by (g) part1.Thus (cid:16) F pi (0) , F τ ( p ) i (0) (cid:17) ∈ [ g n , f n ] and (cid:16) F pi (1) , F τ ( p ) i (1) (cid:17) ∈ [ g n , f n ]. Now since g n is monotone from F pi (0) to F pi (1) and f n is monotone from F τ ( p ) i (0) to F τ ( p ) i (1), it follows that [ g n , f n ] contains a monotonearc from (cid:16) F pi (0) , F τ ( p ) i (0) (cid:17) to (cid:16) F pi (1) , F τ ( p ) i (1) (cid:17) .The conditions ( C3 ), ( C4 ) and ( C6 ) can be verified similarly. OUNDED VALENCE MAPS ON TREES 9
This completes the proof that if Γ n ⊂ ( T n × T n ) (cid:114) ∆ T n is a set satisfying conditions ( C1 ) to ( C6 ) above,then there exist maps f n , g n : T n +1 → T n such that ( f n ( x ) , g n ( x )) ∈ Γ n for all x ∈ T n +1 , and the coincidenceset [ g n , f n ] ⊂ ( T n +1 × T n +1 ) (cid:114) ∆ T n +1 contains a set Γ n +1 satisfying conditions ( C1 ) to ( C6 ) with n replacedby n + 1.We now construct the sequences of mappings (cid:104) f n (cid:105) ∞ n =0 and (cid:104) g n (cid:105) ∞ n =0 by recursion together with sets Γ n , n = 0 , , , . . . . Begin with a set Γ ⊂ ( T × T ) (cid:114) ∆ T satisfying conditions ( C1 ) to ( C6 ) above for n = 0.For the recursive step, having defined Γ n ⊂ T n × T n , let f n , g n : T n +1 → T n be mappings as above such that( f n ( x ) , g n ( x )) ∈ Γ n for all x ∈ T n +1 , and the coincidence set [ g n , f n ] contains a set Γ n +1 satisfying conditions( C1 ) to ( C6 ) with n replaced by n + 1.This construction ensures that for each n >
0, ( f n ( x ) , g n ( x )) ∈ Γ n ⊂ [ g n − , f n − ] for all x ∈ T n +1 , hence f n − ◦ g n = g n − ◦ f n . Because Γ ∩ ∆ T = ∅ , we have that f and g have no coincidence point.It can be seen from the definitions above that f n and g n attain their highest valence at the points p ∈ τ − m { } for m = 0 , , . . . , n . For such points p , we have | f − n ( p ) | = 6 and | g − n ( p ) | = 12. For all otherpoints x ∈ T n +1 , f − n ( x ) has at most 3 components, and g − n ( x ) has at most 6 components.4. Discussion and questions
For a geometric description of the example of this paper, roughly speaking one should visualize thesequence of trees T n embedded in R in such a way that for each n and each x ∈ T n +1 , the points x and f n ( x ) are very close together. Then the inverse limit X = lim ←− ( T n , f n ) is equal to the Hausdorff limit of thetrees T n so embedded in R .The tree-like continuum described in this paper is obtained from the 3-fold Knaster type continuumlim ←− ([0 , , τ ) by first attaching to it a null sequence of simple triods; and second, taking each arc in theKnaster continuum near one of these simple triods and pulling it out to follow close to one leg of the simpletriod. See Figure 5 for an illustration.It is evident that this space cannot be embedded in the plane. We remark that this space contains noCantor fan, which is a characteristic feature of most previously known examples of tree-like continua withoutthe fixed point property.Although the valences of the maps f n and g n are uniformly bounded in our example, the number of branchpoints in the trees T n increases exponentially with n . It is unknown whether this quantity could be boundedas well: Question 1.
Is there a coincidence-point-free commuting system as in Section 1.1 such that. . .(1) . . . the number of branch points in the trees T n is bounded?(2) . . . each T n is equal to the same tree? the simple triod?Moreover, it is not yet known whether the maps f n could in fact be all the same, and likewise for themaps g n . This is the content of the following question, which we also pose for dendrites: Question 2 (See [4], Problem 4) . Does there exist a pair of disjoint commuting maps on a tree? on thesimple triod? on a dendrite?
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A picture of the 3-fold Knaster type continuum with dots indicating some of theplaces where modifications are made to obtain the continuum X = lim ←− ( T n , f n ). The blackdots indicate where the largest several simple triods are attached. At each grey dot, thearcs are pulled out to follow close to one leg of a nearby triod.
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