A complete classification of hereditarily equivalent plane continua
AA COMPLETE CLASSIFICATION OFHEREDITARILY EQUIVALENT PLANE CONTINUA
L. C. HOEHN AND L. G. OVERSTEEGEN
Abstract.
A continuum is hereditarily equivalent if it is homeomorphic toeach of its non-degenerate sub-continua. We show in this paper that the arcand the pseudo-arc are the only non-degenerate hereditarily equivalent planecontinua. Introduction
By a continuum , we mean a compact connected metric space. A continuum is non-degenerate if it contains more than one point. We refer to the space R , withthe Euclidean topology, as the plane . The Euclidean distance between two points x, y in R (or R ) will be denoted (cid:107) x − y (cid:107) . An arc is a space which is homeomorphicto the interval [0 , map we mean a continuous function.A continuum X is hereditarily equivalent if it is homeomorphic to each of itsnon-degenerate subcontinua. This concept was introduced by Mazurkiewicz, whowas interested in topological characterizations of the arc. In the second volumeof Fundamenta Mathematicae in 1921, Mazurkiewicz [Maz21] asked (Probl`eme 14)whether the arc is the only non-degenerate hereditarily equivalent continuum.A continuum X is decomposable if it is the union of two proper subcontinua, and indecomposable otherwise. X is hereditarily indecomposable if every subcontinuumof X is indecomposable. X is arc-like (respectively, tree-like ) if for every ε > ε -map from X to [0 ,
1] (respectively, to a tree), where f : X → Y is an ε -map if for each y ∈ Y the preimage f − ( y ) has diameter less than ε . Henderson [Hen60] showed that the arc is the only decomposable hereditarilyequivalent continuum. Cook [Coo70] has shown that every hereditarily equivalentcontinuum is tree-like.Probl`eme 14 of Mazurkiewicz was formally answered by Moise [Moi48] in 1948,who constructed another hereditarily equivalent plane continuum which he calledthe “pseudo-arc”, due to this property it has in common with the arc. The pseudo-arc is a one-dimensional fractal-like hereditarily indecomposable arc-like continuum.Such a space was constructed by Knaster [Kna22] in 1922, and another by Bing[Bin48] in 1948 which he proved was topologically homogeneous. Bing [Bin51]proved in 1951 that the pseudo-arc is the only hereditarily indecomposable arc-like continuum. From this characterization it follows that the spaces of Knaster, Date : December 24, 2018.2010
Mathematics Subject Classification.
Primary 57N05; Secondary 54F15, 54F65.
Key words and phrases. plane continua, hereditarily equivalent, pseudo-arc, hereditarilyindecomposable.The first named author was partially supported by NSERC grant RGPIN 435518.The second named author was partially supported by NSF-DMS-1807558. a r X i v : . [ m a t h . GN ] D ec L. C. HOEHN AND L. G. OVERSTEEGEN
Moise, and Bing are all homeomorphic, and also it can immediately be seen thatthe pseudo-arc is hereditarily equivalent.Since Moise’s article, the question has been: What are all hereditarily equivalentcontinua? The main result of this paper is:
Theorem 1. If X is a non-degenerate hereditarily equivalent plane continuum,then X is homeomorphic to the arc or to the pseudo-arc. It remains an open question whether there exists any other hereditarily equiva-lent continuum in R .As part of the sequel we will also give a new characterization (Theorem 7) of thepseudo-arc. 2. Plane strips
If a continuum admits an ε -map to an arc then it can be covered by a chain ofopen sets whose diameters are less than ε (i.e. a set that roughly looks like a tubeof small diameter). The notion of an ε -strip (see Definition 3 below), introducedin [OT82] in a slightly different form, conveys a similar feeling. However, it wasobserved in [OT82, Figure 1] that, for arbitrarily small ε >
0, there exists an ε -stripwhich does not admit a 1-map to an arc. Nevertheless we show in this paper (The-orem 8 below) that if a hereditarily indecomposable plane continuum is containedin an ε -strip for arbitrarily small ε >
0, then it must in fact be homeomorphic tothe pseudo-arc.Given two points x, y in the plane R we denote by xy the straight line segmentjoining them. Given points v , . . . , v n in R , the polygonal arc A with vertices v , . . . , v n is the union of the straight line segments v v , . . . , v n − v n . Denote the vertex set of A by V A = { v , . . . , v n } . If v n = v , then we call A a polygonal closedcurve . We will need the following lemma which was proved in [OT82]. Lemma 2 ([OT82], Lemma 2.1) . Let T be a polygonal closed curve in R withvertex set V T . Given any z ∈ R \ T we say z is odd (respectively, even ) withrespect to T if there exists a polygonal arc A with vertex set V A from z to a point inthe unbounded component of R \ T so that A ∩ V T = ∅ = T ∩ V A and | T ∩ A | is odd(respectively, even). Then this notion of odd/even is well-defined, i.e. independentof the choice of A . A component U of R \ T is called odd (respectively, even ) if each point of U isodd (respectively, even ) with respect to T . Clearly the unbounded complementarydomain of T is even.A map f : [0 , → R is piecewise linear if there are finitely many points0 = t < t < . . . < t n = 1 such that for each i = 1 , . . . , n −
1, as t runs from t i to t i +1 , f ( t ) parameterizes the straight line segment f ( t i ) f ( t i +1 ). If f is a piecewiselinear map, then clearly f ([0 , Definition 3 ( ε -strip) . Suppose that f, g : [0 , → R are two piecewise linearmaps into the plane such that f ([0 , ∩ g ([0 , ∅ and for all t ∈ [0 , (cid:107) f ( t ) − g ( t ) (cid:107) < ε . Let B t = f ( t ) g ( t ), and let T t = B ∪ f ([0 , t ]) ∪ g ([0 , t ] ∪ B t . Wedenote the union of all odd (respectively, even) complementary domains of T t by S − t (respectively, S + t ). If B ∩ B = ∅ , then we say that S − is an ε -strip withdisjoint ends . EREDITARILY EQUIVALENT PLANE CONTINUA 3
Figure 1.
An illustration of an ε -strip, with a generic bridge B t drawn. This strip has one odd domain, which is shaded gray.See Figure 1 for an illustration of a simple ε -strip.We say a continuum X is contained in an ε -strip with disjoint ends if thereexist such f, g as in the above definition such that X ⊂ S − . Observe that in thissituation, X ∩ B = ∅ = X ∩ B .If X is an indecomposable and hereditarily equivalent plane continuum, thenit contains uncountably many pairwise disjoint copies of itself. In particular itcontains a copy of X × C , where C is the Cantor set [vD93]. This is the keyobservation behind the following result. Lemma 4 ([OT84], Theorem 15) . Suppose that X is a non-degenerate, inde-composable and hereditarily equivalent plane continuum. Then there exists a non-degenerate subcontinuum Y such that for each ε > , Y is contained in an ε -stripwith disjoint ends. Separators
In light of Lemma 4 above, to prove Theorem 1 it suffices to show that anyhereditarily indecomposable continuum X contained in arbitrarily small plane stripsis homeomorphic to the pseudo-arc (Theorem 8). Our strategy below is to considera small strip containing the continuum X , and to approximate X by a graph G contained in that strip. If we vary t from 0 to 1, the bridge B t in the strip sweepsacross the graph G . As it does so, it may wander back and forth in G , in a patternwhose essential property is captured in the following result. We will then use thecrookedness of the hereditarily indecomposable continuum X to match with thatpattern (see Theorems 6 and 7 below) to obtain an ε -map to an arc. Lemma 5.
Suppose that a graph G is contained in an ε -strip S − with disjointends. Let C = { ( x, t ) ∈ G × [0 ,
1] : x ∈ B t } . Then C separates G × { } from G × { } in G × [0 , .Proof. Define the function ϕ : G × [0 , → R by ϕ ( x, t ) = + d ( x, B t ) if x ∈ S + t − d ( x, B t ) if x ∈ S − t L. C. HOEHN AND L. G. OVERSTEEGEN where d ( x, B t ) = inf {(cid:107) x − b (cid:107) : b ∈ B t } . Then ϕ is a continuous function (see theproof of Lemma 2.3 in [OT82]). Since S − = ∅ = B ∩ G , ϕ ( x, > x ∈ G .Similarly, since G ⊂ S − , ϕ ( x, < x ∈ G . Hence the set of points C where ϕ ( x, t ) = 0 must separate G × { } from G × { } in G × [0 , (cid:3) In [HO16, Theorem 20], the authors gave a characterization of hereditarily inde-composable continua in terms of sets as in Lemma 5 which separate G × { } from G × { } in the product G × [0 ,
1] of a graph G with [0 , Theorem 6.
A continuum X is hereditarily indecomposable if and only if for anymap f : X → G to a graph G , and for any open set U ⊆ G × (0 , which separates G ×{ } from G ×{ } in G × [0 , , there exists a map h : X → U such that f = π ◦ h (where π : G × [0 , → G is the first coordinate projection).Proof. According to [HO16, Theorem 20], a continuum X is hereditarily indecom-posable if and only if for any map f : X → G to a graph G with metric d , for any set M ⊆ G × (0 ,
1) which separates G × { } from G × { } in G × [0 , U ⊆ G × [0 ,
1] with M ⊆ U , and for any ε >
0, there exists a map h : X → U suchthat d ( f ( x ) , π ◦ h ( x )) < ε for all x ∈ X . The condition in the present theorem isclearly stronger than this condition from [HO16, Theorem 20]. Therefore, to provethe present theorem we need only consider the forward implication.Suppose X is hereditarily indecomposable, let f : X → G be a map to a graph G with metric d , and let U ⊂ G × (0 ,
1) be an open set which separates G × { } from G × { } in G × [0 , § M ⊂ U which also separates G × { } from G × { } in G × [0 , ε > U = { ( g, t ) ∈ G × [0 ,
1] : there exists ( g (cid:48) , t (cid:48) ) ∈ M such that d ( g, g (cid:48) ) < ε and | t − t (cid:48) | < ε } is contained in U . Let U = { ( g, t ) ∈ G × [0 ,
1] : there exists ( g (cid:48) , t (cid:48) ) ∈ M such that d ( g, g (cid:48) ) < ε and | t − t (cid:48) | < ε } , and apply [HO16, Theorem 20] to obtain a map h (cid:48) : X → U such that d ( f ( x ) , π ◦ h (cid:48) ( x )) < ε for all x ∈ X .Define h : X → U by h ( x ) = ( f ( x ) , π ◦ h (cid:48) ( x )), where π : G × [0 , → [0 , h is continuous, and f = π ◦ h . To see that the range of h is really contained in U , let x ∈ X , anddenote h (cid:48) ( x ) = ( g, t ), so that h ( x ) = ( f ( x ) , t ). Because h (cid:48) ( x ) ∈ U , there exists( g (cid:48) , t (cid:48) ) ∈ M such that d ( g, g (cid:48) ) < ε and | t − t (cid:48) | < ε . Moreover, by choice of h (cid:48) wehave d ( f ( x ) , g ) < ε . So by the triangle inequality, we have d ( f ( x ) , g (cid:48) ) < ε , whichmeans h ( x ) ∈ U ⊆ U , as desired. (cid:3) By Bing’s [Bin51] result a hereditarily indecomposable continuum is homeomor-phic to the pseudo-arc if and only if it is arc-like. A new characterization of thepseudo-arc, involving the notion of span zero (see [Lel64]), was obtained in [HO16].It states that a hereditarily indecomposable continuum is a pseudo-arc if and only ifit has span zero. The more technical characterization of the pseudo-arc in Theorem
EREDITARILY EQUIVALENT PLANE CONTINUA 5 X has span zero.In the statement below we assume that all spaces (i.e., X , G , and I ) are containedin Euclidean space R . One could just as well use the Hilbert cube [0 , N , dependingon the intended application. Theorem 7.
Suppose that X ⊂ R is a hereditarily indecomposable continuum.Then the following are equivalent:(1) X is homeomorphic to the pseudo-arc;(2) For each ε > there exist a graph G ⊂ R , a map f : X → G with (cid:107) ( x − f ( x ) (cid:107) < ε for each x ∈ X , and an arc I ⊂ R with endpoints a and b , such that the set U = { ( x, t ) ∈ G × ( I \ { a, b } ) : (cid:107) x − t (cid:107) < ε } separates G × { a } from G × { b } in G × I .Proof. Suppose X is homeomorphic to the pseudo-arc, and fix ε >
0. Note X isarc-like and, hence [Lel64], X has span zero. Therefore, according to Theorem 4 of[HO16], there exists δ > G ⊂ R and arc I ⊂ R bothwithin Hausdroff distance δ from X , the set U = { ( x, t ) ∈ G × I : (cid:107) x − y (cid:107) < ε } separates G × { a } from G × { b } in G × I , where a, b are the endpoints of I . Wemay assume that δ < ε . Since X is arc-like, we may choose an arc G ⊂ R withinHausdorff distance δ of X and a map f : X → G such that (cid:107) x − f ( x ) (cid:107) < ε for all x ∈ X . Choose any arc I within Hausdorff distance δ from X . Then G , f , and I satisfy the conditions of statement (2), as desired.Conversely, suppose statement (2) holds. To prove that X is homeomorphic tothe pseudo-arc, by [Bin51] it suffices to show that for each ε > ε -map from X to an arc. Fix ε >
0. Suppose that G ⊂ R is a graph, f : X → G is a map such that (cid:107) x − f ( x ) (cid:107) < ε for all x ∈ X , I ⊂ R is an arc with endpoints a and b , and U = (cid:110) ( x, t ) ∈ G × ( I \ { a, b } ) : (cid:107) x − t (cid:107) < ε (cid:111) separates G × { a } from G × { b } in G × I . Denote by π : G × I → G the firstcoordinate projection and by π : G × I → I the second coordinate projection. ByTheorem 6 there exists a map h : X → U such that f = π ◦ h . We claim that π ◦ h ( x ) : X → I is an ε -map. To see this suppose that π ◦ h ( x ) = π ◦ h ( x ).Then (cid:107) x − x (cid:107) ≤ (cid:107) x − f ( x ) (cid:107) + (cid:107) π ◦ h ( x ) − π ◦ h ( x ) (cid:107) ++ (cid:107) π ◦ h ( x ) − π ◦ h ( x ) (cid:107) + (cid:107) f ( x ) − x (cid:107) < ε. (cid:3) Proof of main result
We now apply the results established above to prove the following key theorem.
Theorem 8.
Let X ⊂ R be a hereditarily indecomposable plane continuum suchthat for each ε > , there is an ε -strip with disjoint ends containing X . Then X ishomeomorphic to the pseudo-arc. L. C. HOEHN AND L. G. OVERSTEEGEN
Proof.
Let ε >
0, and consider an ε -strip with disjoint ends containing X . That is,consider piecewise linear maps f, g : [0 , → R such that f ([0 , ∩ g ([0 , ∅ , (cid:107) f ( t ) − g ( t ) (cid:107) < ε for each t ∈ [0 , X ⊂ S − . Identify R with R × { } ⊂ R ,and adjust f slightly to obtain a map f (cid:48) : [0 , → R which is one-to-one (so that f (cid:48) ([0 , (cid:107) f ( t ) − f (cid:48) ( t ) (cid:107) < ε for all t ∈ [0 , X is 1-dimensional, so there exists a graph G ⊂ S − and a map h : X → G such that (cid:107) x − h ( x ) (cid:107) < ε for all x ∈ X . By Lemma 5, the set C = { ( x, t ) ∈ G × [0 ,
1] : x ∈ B t } separates G × { } from G × { } in G × [0 , C is contained in U = (cid:110) ( x, t ) ∈ G × (0 ,
1) : (cid:107) x − f ( t ) (cid:107) < ε (cid:111) , and the image of this set U under the homeomorphism id × f (cid:48) : G × [0 , → G × f (cid:48) ([0 , U (cid:48) = { ( x, y ) ∈ G × f (cid:48) ((0 , (cid:107) x − y (cid:107) < ε } . Therefore U (cid:48) separates separates G × { f (cid:48) (0) } from G × { f (cid:48) (1) } in G × f (cid:48) ([0 , X is homeomorphic to the pseudo-arc. (cid:3) We are now ready to prove our main result, Theorem 1.
Proof of Theorem 1.
Let X be a non-degenerate hereditarily equivalent plane con-tinuum. If X is decomposable, then X is an arc by [Hen60]. Suppose then that X is indecomposable, and hence hereditarily indecomposable. By Lemma 4, we mayassume that X is embedded in the plane so that for each ε > X is contained in an ε -strip with disjoint ends. It then follows from Theorem 8 that X is homeomorphicto the pseudo-arc. (cid:3) References [Bin48] R. H. Bing,
A homogeneous indecomposable plane continuum , Duke Math. J. (1948),729–742. MR 0027144 (10,261a)[Bin51] , Concerning hereditarily indecomposable continua , Pacific J. Math. (1951),43–51. MR 0043451 (13,265b)[Coo70] H. Cook, Tree-likeness of hereditarily equivalent continua , Fund. Math. (1970), 203–205. MR 0266164 (42 Proof that every compact decomposable continuum which is topo-logically equivalent to each of its nondegenerate subcontinua is an arc , Ann. of Math. (2) (1960), 421–428. MR 0119183 (22 A complete classification of homogeneous planecontinua , Acta Math. (2016), no. 2, 177–216. MR 3573330[Kna22] B. Knaster,
Un continu dont tout sous-continu est ind´ecomposable , Fund. Math. (1922),247–286.[Kur68] K. Kuratowski, Topology. Vol. II , New edition, revised and augmented. Translated fromthe French by A. Kirkor, Academic Press, New York, 1968.[Lel64] A. Lelek,
Disjoint mappings and the span of spaces , Fund. Math. (1964), 199–214.MR 0179766 (31 Probl`eme 14 , Fund. Math. (1921), 286.[Moi48] Edwin E. Moise, An indecomposable plane continuum which is homeomorphic toeach of its nondegenerate subcontinua , Trans. Amer. Math. Soc. (1948), 581–594.MR 0025733 (10,56i)[OT82] Lex G. Oversteegen and E. D. Tymchatyn, Plane strips and the span of continua. I ,Houston J. Math. (1982), no. 1, 129–142. MR 666153 (84h:54030) EREDITARILY EQUIVALENT PLANE CONTINUA 7 [OT84] ,
Plane strips and the span of continua. II , Houston J. Math. (1984), no. 2,255–266. MR 744910 (86a:54042)[vD93] Eric K. van Douwen, Uncountably many pairwise disjoint copies of one metrizable com-pactum in another , Topology Appl. (1993), no. 2, 87–91. MR 1229705(L. C. Hoehn) Nipissing University, Department of Computer Science & Mathematics,100 College Drive, Box 5002, North Bay, Ontario, Canada, P1B 8L7
E-mail address : [email protected] (L. G. Oversteegen) University of Alabama at Birmingham, Department of Mathemat-ics, Birmingham, AL 35294, USA
E-mail address ::