Hairy Cantor sets
Davoud Cheraghi, Mohammad Pedramfar

TL;DR
The paper introduces hairy Cantor sets, a new topological object with universal features, and proves their ambient homeomorphism in the plane, linking complex dynamics and renormalisation theory.
Contribution
It provides an axiomatic characterization of hairy Cantor sets and establishes their topological equivalence in the plane, connecting dynamics of holomorphic maps with arithmetic conditions.
Findings
Hairy Cantor sets share universal topological features.
Any two hairy Cantor sets in the plane are ambiently homeomorphic.
They link polynomial-like renormalisation with circle diffeomorphism dynamics.
Abstract
We introduce a topological object, called hairy Cantor set, which in many ways enjoys the universal features of objects like Jordan curve, Cantor set, Cantor bouquet, hairy Jordan curve, etc. We give an axiomatic characterisation of hairy Cantor sets, and prove that any two such objects in the plane are ambiently homeomorphic. Hairy Cantor sets appear in the study of the dynamics of holomorphic maps with infinitely many renormalisation structures. They are employed to link the fundamental concepts of polynomial-like renormalisation by Douady-Hubbard with the arithmetic conditions obtained by Herman-Yoccoz in the study of the dynamics of analytic circle diffeomorphisms.
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Taxonomy
TopicsMathematical Dynamics and Fractals
Hairy Cantor sets
Davoud Cheraghi
Department of Mathematics, Imperial College London, London, SW7 2AZ, UK
and
Mohammad Pedramfar
Department of Mathematics, Imperial College London, London, SW7 2AZ, UK
Abstract.
We introduce a topological object, called hairy Cantor set, which in many ways enjoys the universal features of objects like Jordan curve, Cantor set, Cantor bouquet, hairy Jordan curve, etc. We give an axiomatic characterisation of hairy Cantor sets, and prove that any two such objects in the plane are ambiently homeomorphic.
Hairy Cantor sets appear in the study of the dynamics of holomorphic maps with infinitely many renormalisation structures. They are employed to link the fundamental concepts of polynomial-like renormalisation by Douady-Hubbard with the arithmetic conditions obtained by Herman-Yoccoz in the study of the dynamics of analytic circle diffeomorphisms.
2010 Mathematics Subject Classification:
54F65 (Primary), 57R30, 37F10 (Secondary)
1. Introduction
We introduce a topological object which enjoys a similar level of universal features as objects like Jordan curve, Cantor set, Cantor bouquet, hairy Jordan curve, Lelek fan, etc [Jor93, MK19, Lel61, BO90, Nad92, HO16]. These are topological objects uniquely determined, up to ambient homeomorphisms, by some simple axioms. These objects frequently appear in dynamical systems, in particular, as the Julia sets or the attractors of holomorphic maps on complex spaces. The object presented here has a delicate fine-scale structure, for instance, it is not locally connected. However, due to the way these emerge in iterations of holomorphic mappings, they turn out ubiquitous.
We start by presenting simple examples of our favourite object.
Definition 1.1**.**
A set in the plane is called a straight hairy Cantor set, if there are a Cantor set of points and a function such that
[TABLE]
and the function satisfies the following properties:
- (i)
the set of with is dense in ;
- (ii)
if is an end point 111 is called an end point, if there is such that either or ., then
[TABLE]
- (iii)
if is not an end point, then
[TABLE]
By property (ii) in the above definition, on a dense subset of . Also, by properties (ii) and (iii), any straight hairy Cantor set is compact. In particular, is finite. On the other hand, as every arc in is accumulated from both sides by arcs, any straight hairy Cantor set is not locally connected.
It is not immediately clear, but true, that for any open interval with , the closure of the set forms a connected interval in . Due to such features, straight hairy Cantor sets in the plane are topologically unique.
Theorem 1.2**.**
All straight hairy Cantor sets in the plane are ambiently homeomorphic.
We are pursuing topological objects in the plane which are ambiently homeomorphic to a straight hairy Cantor set. However, for practical reasons, one requires an axiomatic description of such objects. Let , and consider the following axioms:
- ()
any connected component of is either a single point or a Jordan arc;
- ()
the closure of the set of point components of is a Cantor set of points, say ;
- ()
any arc component of meets at one of its end points;
- ()
whenever within , the unique arc in connecting to converges to the unique arc in connecting to , with the convergence in the Hausdorff topology;
- ()
for every arc component of , and every in minus the end points of , is not accessible from ;
- ()
is dense in ;
- ()
the set of the end points of the arc components of is dense in ;
All of the above axioms, except , are well-defined for any metric space.
Theorem 1.3**.**
Let be a compact metric space. If satisfies axioms to , and , then is homeomorphic to a straight hairy Cantor set.
Theorem 1.4**.**
Let be a compact set. If satisfies axioms to , then is ambiently homeomorphic to a straight hairy Cantor set.
Definition 1.5**.**
A compact set is called a hairy Cantor set, if it satisfies the axioms to .
Combining Theorems 1.2 and 1.4 we conclude that all hairy Cantor sets in the plain are ambiently homeomorphic.
The proof of Theorem 1.2 is fairly elementary and is based on a careful analysis of bump-functions associated to straight hairy sets. In contrast, the proof of Theorem 1.4 is more involved. The main idea is to identify a one dimensional topological foliation of the plane, equivalent to the foliation of by vertical lines, and each component of lies in a single leaf of the foliation. The (extra portions of) leaves come from hyperbolic geodesics in . The topology of the leaves/geodesics are studied in the framework of Caratheodory’s prime ends [Car13], and using classical complex analysis, in paticular, Gehring-Hayman theorem [GH62]. The main technical difficulty is to build a global cross section for the foliation. We form a uniformisation of the hairy Cantor set to an straight hairy Cantor set using hyper-spaces and Whitney maps [Whi33].
Hairy Cantor sets have emerged in the study of the dynamics of holomorphic maps. In the counterpart paper [CP19] we explain the appearance of hairy Cantor sets as the attractors of a wide class of holomorphic maps with infinitely many renormalisation structures. Roughly speaking, successive perturbations of holomorphic maps with parabolic cycles leads to infinite renormalisation structures discovered by Douady and Hubbard in early 80s [DH85]. Although such systems were the subject of profound studies in the 80s and 90s, recent significant advances on the topic fuels interest in these objects. In [CP19], hairy Cantor sets are employed to make a striking connection between the satellite renormalisation structures, and the arithmetic conditions which appeared in the study of the linearisation of analytic circle diffeomorphisms by Herman-Yoccoz [Her79, Yoc02]. Also, our results shed light on the remarkable examples of positive area Julia sets by Buff-Chéritat [BC12], the non-locally connected Julia sets by Sorensen [Sr00] and Levin [Lev11], as well as the infinitely renormalisable maps without a priori bounds in [CS15].
Conjecturally, for a dense set of maps on the bifurcation locus of any non-trivial family of rational functions, the attractor of the map contains a hairy Cantor set. One may compare this to the appearance of Cantor bouquets in the dynamics of exponential maps [DK84, AO93, Six18], and hairy Jordan curves in the attractors of holomorphic maps with an irrationally indifferent fixed/periodic point [Che17].
2. Straight hairy Cantor sets
The main purpose in this section is to prove Theorem 1.2. Before we embark on the proof of that theorem, we present a straight hairy Cantor set, which essentially represents the general form of a straight hairy Cantor set.
Lemma 2.1**.**
There exists a straight hairy Cantor set in the plane.
Proof.
For , consider the set of multi-indexes such that for each , . We aim to define a Cantor set
[TABLE]
where each is a closed interval in , and , for and , whenever and the first entries of are the same as the first entries of . For , we let . Assume that for some , is defined for all . In order to define , for , we divide each , , into intervals with equal lengths. Then, starting from the left most interval, we alternate discarding one interval, and keeping the next interval, until the last interval. In other words, with the order on the real line, we keep the odd numbered intervals. Thus, we are left with closed intervals in . We label these intervals as , such that the first entries of are the same as , and use the last entry to label these intervals in an order preserving fashion. This completes the definition of , for . With this construction, every has a unique address , where
[TABLE]
By an inductive process, we define a sequence of continuous functions as follows. For , let for all . Now assume that is defined for some . For , we let
[TABLE]
For we let , for we let , and on any interval in the complement of with and in , we use a linear interpolation of and . This gives a continuous function on .
Note that for all , , which implies . Therefore, , for all and . In particular, we may define
[TABLE]
For , we have
[TABLE]
Since each is continuous, the set
[TABLE]
is closed in . Therefore,
[TABLE]
is also closed in . This implies that is upper semi-continuous, that is, for all ,
[TABLE]
We claim that the set
[TABLE]
is a straight hairy Cantor set.
It follows from the product formula in Equation (1) that for any with , for large enough , we have . Evidently, the set of such points is dense in . This gives us property (i) in Definition 1.1.
If is an end point of , we must have either or , for large enough . Using the product formula for , those imply that . By Equation (2), . Thus, property (ii) in Definition 1.1 holds.
Now assume that is not an end point. If , by Equation (2), we must have and . If , by Equation (1), there must be such that for all we have . For we may consider the unique point whose address is
[TABLE]
Evidently, is a strictly increasing sequence that tends to . Moreover,
[TABLE]
Therefore, . Similarly, we may define a strictly decreasing sequence in converging to such that . This implies part (iii) of Definition 1.1. ∎
Below we introduce some basic definitions and notations which will be used for the proof of Theorem 1.2.
Given a straight hairy Cantor set , the corresponding function determining in Definition 1.1 is unique. We refer to as the length function of , and often denote it by . Also, we refer to the Cantor set as the base Cantor set of , or equivalently, say that is based on the Cantor set . Note that is the unique Cantor set of points in which contains all point components of .
Let be a Cantor set of points. A set is called a Cantor partition for , if is the union of a finite number of closed intervals in with and . A nest of Cantor partitions shrinking to , by definition, is a nest of Cantor partitions for such that . When we write a Cantor partition for as , we mean that each is a non-empty closed interval in and , for . Also, we assume that the intervals are labelled in a way that if and only if .
Let be a straight hairy Cantor set based on the Cantor set of points . For any Cantor partition for , by property (ii) in Definition 1.1, , for all .
Let be a Cantor set, and let be an upper semi-continuous function. For any closed interval with , we use the notation
[TABLE]
By the upper semi-continuity of , there is with
[TABLE]
Clearly, with this property might not be unique. For any such choice of , we define
[TABLE]
as
[TABLE]
We refer to as the bump function of on the interval associated to .
Assume that is a Cantor partition for . We may consider a bump function on each interval , say , and combine them to define a bump function for associated to the partition ,
[TABLE]
according to
[TABLE]
The following lemma is fairy easy, but will be used several times in the upcoming arguments.
Lemma 2.2**.**
Let be a Cantor set, and be an upper semi-continuous function. Also assume that be a sequence of closed intervals in shrinking to a point . Then
[TABLE]
Proof.
For all , we have , which gives
[TABLE]
On the other hand, since is upper semi-continuous, we have
[TABLE]
The key property of bump functions is stated in the following lemma.
Lemma 2.3**.**
Let be a straight hairy Cantor set, based on a Cantor set , and with length function . Assume that is a Cantor partition for . Any bump function is piecewise monotone and continuous.
Proof.
Recall that for any , . Therefore, it is enough to show that each is piecewise monotone and continuous. Fix an arbitrary , and let and . Assume that is associated to some satisfying .
By definition, is increasing on the interval and is decreasing on .
If , then and there is nothing to prove. When , by property (i) in Definition 1.1, is positive at some point in , and therefore, .
By the increasing property of on , and properties (ii) and (iii) in Definition 1.1, we note that
[TABLE]
This implies the continuity of at . Similarly, one obtains the continuity of at .
Since , by property (ii), is not an end point of , and by property (iii), we have
[TABLE]
Hence,
[TABLE]
This implies the continuity of at .
For an arbitrary , if , then is constant near , and in particular, it is continuous at . If , then
[TABLE]
This implies the continuity of at . Similarly, one proves the continuity of on . ∎
The proof of Theorem 1.2 will be based on the following two propositions.
Proposition 2.4**.**
Let and be straight hairy Cantor sets based on the same Cantor set and with length functions and . There are nests of Cantor partitions and , for , shrinking to , such that for all we have
- (i)
for all we have \big{|}J^{X}_{n,i}\big{|}<2^{-n}|C| and \big{|}J^{Y}_{n,i}\big{|}<2^{-n}|C|; 222We use the notation to denote the Euclidean diameter of a given .
- (ii)
for all and , if and only if ;
- (iii)
whenever for some and , then
[TABLE]
Proof.
We shall use the notations
[TABLE]
with
[TABLE]
and
[TABLE]
By an inductive argument, we shall simultaneously define the partitions and . For , we set and . That is, each of and consists of one closed interval. Property (i) in the proposition holds, and there is nothing to verify for items (ii) and (iii).
Now assume that the partitions and are defined for some and satisfy the properties in items (i)-(iii) in the proposition. Below we defined and .
Let us assume that is even (the odd case is mentioned at the end of the proof). Let be an arbitrary Cantor partition of such that is a subset of and
[TABLE]
We shall use to construct . To that end, it is enough to identify all the intervals which are contained in , for each with . Before identifying those intervals, we note that the number of such intervals for each must be equal to the number of the intervals which are contained in . This will clearly guarantee property (ii). To ensure property (iii) we need more detailed analysis.
Fix an arbitrary with . Recall that
[TABLE]
There are and such that
[TABLE]
Since there are such that
[TABLE]
We set
[TABLE]
If , then we are done. Below we assume that . We need to identify and , for .
There is a unique integer with such that
[TABLE]
We shall determine and recursively by going from to , if there are any such , and independently going from to , if there are any such , so that
[TABLE]
We only explain the process for , the other case being similar.
Let us assume that for some , we have already defined
[TABLE]
Since
[TABLE]
and , we have
[TABLE]
Therefore, we may use to define a bump function on the set . By Lemma 2.3, as we move from to , the value of continuously increases from [math] to . Therefore, since
[TABLE]
there must be such that
[TABLE]
Now, by slightly moving to the right, so that it becomes an end point of , we may find an interval such that
[TABLE]
and
[TABLE]
Here, and . Also, we have used that the map depends continuously on . This completes the induction step to identify and . Note that at the end of this construction, , thus,
[TABLE]
We assumed at the induction step that is even. For odd , we interchange the role of and in the above process, that is, we start with an arbitrary partition for which is contained in and satisfies Equation (3), and build in the same fashion. This guarantees that for all and we have \big{|}J^{X}_{n,i}\big{|}\leq 2^{-n}|C| and \big{|}J^{Y}_{n,i}\big{|}\leq 2^{-n}|C|. ∎
Proposition 2.5**.**
Let and be straight hairy Cantor sets based on the same Cantor set of points and with length functions and . Let , for , be a nest of Cantor partitions shrinking to , with disjoint non-empty closed intervals , for each . Assume that whenever for some integers , , and , we have
[TABLE]
Then, and are ambiently homeomorphic.333Recall that two subsets of the plane are ambiently homeomorphic if there is a homeomorphism of the plain which maps one bijectively onto the other.
Let us denote the Euclidean ball of radius about with . In the same fashion, given , let
[TABLE]
Proof.
Without loss of generality we may assume that , with [math] and in . This may be achieved by applying a translation and then a linear rescaling. Also, by applying two rescalings in the second coordinate in , we may assume that
[TABLE]
These changes do not alter the ratios of the maximums in the hypotheses of the proposition.
For , we define the function as follows. If , we identify the unique integers and with , and set
[TABLE]
For , we set . Clearly, is constant on each closed interval of the partition .
By the assumption in the proposition, for all and all , we have
[TABLE]
By an inductive argument, we aim to build homeomorphisms , for , such that each is a straight hairy Cantor set based on and its length function satisfies the relation
[TABLE]
whenever for some .
For , we let be the identity map on . Assume that for some , is defined and satisfies Equation (4). Below we define .
Let be a bump function for associated to the Cantor partition . Since is constant on each interval of , we have
[TABLE]
In particular, for all ,
[TABLE]
Define as the identity map outside of , and for , let
[TABLE]
Evidently, is identity outside , and also on the real slice . For , maps the line segment onto itself in a piecewise linear fashion, with
[TABLE]
[TABLE]
In particular,
[TABLE]
It follows from the continuity of in Lemma 2.3 that is continuous, and hence, a homeomorphism of the plane. We define .
The set is a straight hairy Cantor set. That is because, the homeomorphism sends each vertical line into itself, and is the identity on the horizontal line .
To prove that satisfies Equation (4), choose , and note that implies
[TABLE]
This completes the process of defining the maps , for .
Next we show that the maps converge uniformly to a homeomorphism . To prove this, first note that when ,
[TABLE]
and when ,
[TABLE]
These imply that
[TABLE]
Hence, the sequence is uniformly Cauchy. In particular, the limiting map exists and is continuous.
We claim that is injective. To see this, fix . For any , and all and in , we have
[TABLE]
For and in , we have
[TABLE]
Finally, since is a homeomorphism of , for , we have
[TABLE]
Combining the above inequalities, we conclude that for all , , and in , as well as all , we have
[TABLE]
We note that . By virtue of the above inequality, is injective.
The map is surjective. To see this, we note that is compact. Therefore, if there is , then there must be such that the disk of radius around , , does not meet . Choose large enough so that . Then, we have , which is a contradiction.
Every (and therefore every ) sends vertical lines to vertical lines and is identity on the boundary of . Therefore also has this property. The convergence of to implies that for all , we have
[TABLE]
We claim that . To prove this, it is enough to show that , for all .
Fix . For each , choose with . By Lemma 2.2,
[TABLE]
Therefore, if , using Equation (4), we obtain
[TABLE]
On the other hand, from the hypothesis, and using for , we note that for all , we have
[TABLE]
Similarly, using for , we obtain
[TABLE]
Thus, if and only if . However, since is a homeomorphism, if and only if . ∎
Proof of Theorem 1.2.
First we note that it is enough to show that any two straight hairy Cantor sets based on the same Cantor set are ambiently homeomorphic. That is because, for any two Cantor sets and in , there is a homeomorphism with . The map may be extended to a homeomorphism of through . Given a straight hairy Cantor set based on , is a straight hairy Cantor set based on .
Now consider two straight hairy Cantor sets and based on the same Cantor set and with length functions and , respectively. By Proposition 2.4 there are nests of Cantor partitions and shrinking to , which enjoy the three properties in that proposition. In particular, by properties (i) and (ii) in that proposition, there is a homeomorphism such that for all and all ,
[TABLE]
Evidently, maps onto . We extend to a homeomorphism of through
[TABLE]
It follows that is a straight hairy Cantor set based on . Moreover, by part (iii) of Proposition 2.4, the straight hairy Cantor sets and , as well as the nest of Cantor partitions satisfy the hypothesis of Proposition 2.5. Therefore, we obtain a homeomorphism of which maps to . This completes the proof. ∎
The following proposition is the main technical step towards the proof of Theorem 1.2.
Proposition 2.6**.**
Let be a Cantor set, and let be a function such that the set
[TABLE]
satisfies the following properties:
- (i)
* is compact;*
- (ii)
* is dense in .*
Then, is homeomorphic to a straight hairy Cantor set.
Proof.
We aim to modify , by successively applying homeomorphisms close to the identity, so that in the limit we obtain a straight hairy Cantor set. That is, we build a chain of homeomorphisms as in
[TABLE]
Each is a piecewise translation, shuffling the arcs in so that any bump function of has jump discontinuities of sizes at most . We present the details below.
Since all Cantor sets in the real line are homeomorphic, without loss of generality, we may assume that is the middle-third Cantor set. Let , and for , recursively define444Given and , we define and .
[TABLE]
Then, is a shrinking sequence of Cantor partitions for . For , let , where each is a connected component of . We may label the subscripts so that , whenever . For , and , consider the set
[TABLE]
We have
[TABLE]
We divide the remaining argument into three steps.
Step 1. There are a sequence of integers , homeomorphisms , and functions , for , such that for all , and all we have
- (i)
[TABLE]
- (ii)
for all , , and ;
- (iii)
for every ,
[TABLE]
- (iv)
for all with ,
[TABLE]
Let , be the identity map, and . Fix an arbitrary . Assume that , , and are define for all , and we aim to define , , and .
Consider the set
[TABLE]
It follows that is a homeomorphism from onto . Since is compact, must be also compact. This implies that is upper semi-continuous.
By virtue of the homeomorphism from onto , and the hypothesis of the proposition, the set is dense in . Therefore, for every with , there is a finite set , such that
[TABLE]
By Lemma 2.2, for every we have
[TABLE]
Therefore, we may find an integer such that for all with , we have
[TABLE]
We define on each , for , so that it induces a homeomorphism of . Fix such an integer . By virtue of Equations (5) and (6), there is a permutation
[TABLE]
such that
- •
for , we have
[TABLE]
- •
for all integers with , we have
[TABLE]
For instance, to identify such , one may first find a permutation of such that , whenever in . Then, compose with the following permutation of ,
[TABLE]
Once we have , there is a unique homeomorphism such that for all in
[TABLE]
and each is a translation by a constant on each of .
Carrying out the above process for all , we obtain a homeomorphism . Then, we define
[TABLE]
This completes the proof of Step 1.
Consider the homeomorphisms , defined according to
[TABLE]
Step 2. The sequence , for , uniformly converges to a homeomorphism
[TABLE]
By property (i) in Step 1, for all and all , we have
[TABLE]
Hence,
[TABLE]
Therefore, forms a Cauchy sequence, and converges to a continuous map .
To prove that is surjective, note that is compact. Therefore, if there is in , then there is such that
[TABLE]
Choose so that . Then, we have , which is a contradiction.
To prove that is injective, fix arbitrary points in . There are and such that and . By the above construction, there are distinct integers and in such that and . On the other hand, for all , we have
[TABLE]
Therefore, and similarly . In particular, .
Step 3. The set is a straight hairy Cantor set.
Consider the function , and the set
[TABLE]
By Step 2, is a homeomorphism.
We claim that for all and , we have
[TABLE]
Recall that . To prove the above property, it is sufficient to show that , or equivalently, . However, by property (i) in Step 1, for all , . Using , we conclude that preserves .
Since is a homeomorphism, the set is dense in . This implies property (i) in Definition 1.1.
Let be an end point of . There are and such that . Then, for all , with . By property (iii) in Step 1, we must have
[TABLE]
Then, by Lemma 2.2 we have . This completes the proof of property (ii) in Definition 1.1.
Fix an arbitrary with , and let be arbitrary. Assume that are the sequence of integers with , for all . Using Lemma 2.2, there exists such that , and
[TABLE]
On the other hand, since ,
[TABLE]
Therefore,
[TABLE]
and by property (iv) in Step 1,
[TABLE]
[TABLE]
Let be a point in with . Then, ,
[TABLE]
and
[TABLE]
Similarly, we may find such that
[TABLE]
This completes the proof of property (iii) in Definition 1.1. ∎
We will need the following Lemma for the proof of Theorem 1.4.
Lemma 2.7**.**
Let be a Cantor set, , and
[TABLE]
Then, is a straight hairy Cantor set if and only if the following properties hold:
- (i)
* is compact;*
- (ii)
the set is dense in ;
- (iii)
any point in is not accessible from .
Proof.
Assume that is a straight hairy Cantor set. By the uniqueness of the length function of , , we must have . Thus, properties (i), (ii), and (iii) in the above lemma hold.
Assume that satisfies the three properties in the lemma. We get property (i) in Definition 1.1 for free. To see property (ii) in Definition 1.1, let be an end point. If , any with will be accessible from . This contradiction with property (iii) in the lemma shows that we must have .
Now assume that is not an end point. Define
[TABLE]
By the compactness of , and are finite values. Moreover, we must have and . If , then any with will be accessible from the right hand side. Similarly, if , then any with will be accessible from the left hand side. These contradict property (iii) in the lemma, so we must have . Thus, we also have property (iii) in Definition 1.1. ∎
Proposition 2.8**.**
Assume that
[TABLE]
is a straight hairy Cantor set, where is a Cantor set, and is the length functions of . Then, we have
- (i)
the set is dense in the interval ;
- (ii)
the set is dense in .
Proof.
As is compact, the value is realised and is finite. Assume in the contrary that there is an open interval such that for all , . Let . Since is compact, must be compact.
By property (ii) in Definition 1.1 every is not an end point of . Thus, by property (iii) in that definition, for every there is such that and . Since , and , we must have . Thus, . In other words, for every , there is with . This is not possible, since is a compact set.
To prove item (ii), let be an arbitrary point in , and fix an arbitrary . There are and in such that is a Cantor set in . Let . From Definition 1.1, one may see that the set
[TABLE]
is a straight hairy Cantor set. As belongs to the above set, . Applying item (i) to the above straight hairy Cantor set, we conclude that there is with . Therefore,
[TABLE]
3. Height functions and base curves
In this section we establish some topological features of hairy Cantor sets. These will be used in Section 4 to prove Theorem 1.4.
Recall that a curve lands at , if exists and is equal to . A point is called accessible from , if there is a curve which lands at .
Given a set satisfying axioms to in the introduction, the base of is the unique Cantor set which contains all point components of . We shall often say that is based on the Cantor set .
Any arc 555A Jordan arc, or simply an arc, in is the image of a continuous and injective map . The end points of this Jordan arc are and . component of has two distinct end points, one of which belongs to . The end point of which belongs to is called the base of . The other end point of , which does not belong to , is called the peak of . If is a point component of , we may either refer to as the base of or as the peak of . For , the base of is defined as the base of the component of containing , and similarly, the peak of is defined as the peak of that component. We may define the base map and the peak map
[TABLE]
where is the base of and is the peak of .
Slightly abusing the notation, we say that is a base point, if , and we say that is a peak point if . Since and are the identity map on , any element of is both a base point, and a peak point. It follows from axiom that is continuous on . However, is far from continuous.
The base points of and the peak points of are defined solely using the topology of . It follows that any homeomorphism of a hairy Cantor set must send base points to base points, and peak points to peak points.
Assume that satisfies axioms to , and is the base Cantor set of . A function is called a height function on , if the following properties hold:
- (i)
h is continuous on ;
- (ii)
;
- (iii)
is injective on any connected component of .
For instance, for a straight hairy Cantor set , is a height function on . However, there are many other height functions on . For the proof of Theorem 1.4 we require a height function on a given hairy Cantor set. To this end we employ a general theory of Whitney maps, which we explain below.
Assume that is a compact metric space, and let denote the set of all non-empty compact subsets of . The space may be equipped with the Hausdorff topology induced from the Hausdorff metric. That is, given non-empty compact sets and in , the Hausdorff distance between and is defined as the infimum of all such that neighbourhood of contains and neighbourhood of contains . A function is called a Whitney map for , if
- (i)
is continuous on ;
- (ii)
for every ;
- (iii)
for every and in with , we have .
In [Whi33], Whitney proves the existence of maps satisfying the above properties. The existence of a Whitney map for a compact subset of an Euclidean space is a classical result in topology. One may refer to the general reference [Nad92, Excercise 4.33], or the extensive monograph on the hyperspaces [IN99].
Proposition 3.1**.**
Any compact metric space which satisfies axioms to admits a height function.
Proof.
Fix , and assume that denotes the base Cantor set of . Let be a Whitney map for , that is, enjoys the above properties. Consider the function , define as
[TABLE]
where denotes the unique arc in which connects to the base point .
To see the continuity of on , let within . By axiom , , in the Hausdorff topology. Thus, by the continuity of on , we conclude that .
By property (ii) of Whitney maps,
[TABLE]
By property (iii) of Whitney maps, is injective on every connected component of . ∎
The existence of the height function in the above proposition is sufficient to prove Theorem 1.3. A reader interested only in that theorem may directly go to the proof of Theorem 1.3 at the end of Section 4. The remaining statements in this section are required for the proof of Theorem 1.4. So, from now on we assume that is a subset of .
Let be a hairy Cantor set with base Cantor set . A Jordan curve is called a base curve for , if , and the bounded connected component of does not intersect . In particular, the closure of the unbounded component of contains .
We aim to prove that any hairy Cantor set admits a base curve. To this end we need to introduce some basic notions and prove some technical lemmas.
By a topological disk in the plane we mean a connected and simply connected open subset of the plane. Let be a hairy Cantor set with base Cantor set . We say that an open set is admissible for , if the following properties hold:
- (i)
consists of a finite number of pairwise disjoint topological disks, and ;
- (ii)
for all , ;
- (iii)
for all , is accessible from both of and ;
- (iv)
if and are distinct connected components of , then .
Evidently, any finite union of pairwise disjoint Jordan domains which contains is an admissible set for . However, it is not immediately clear that nontrivial admissible sets for exist. Over the next few pages we aim to show that any hairy Cantor set admits a nest of admissible sets shrinking to its base Cantor set. This will be employed to build a base curve for .
Lemma 3.2**.**
Let be a hairy Cantor set with base Cantor set . Assume that is an open set which satisfies properties (i)-(iii) in the definition of admissible sets for , and let , for , be the connected components of . There are topological disks , for , such that is an admissible set for .
Proof.
If , we let , and there is nothing to prove. Below we assume that . Let . We claim that for distinct and in ,
[TABLE]
To prove this, it is enough to show that for every , if then . Fix an arbitrary . There is a sequence in , for , which converges to . By property (ii) of admissible sets, , and hence . Since , we must have .
In two steps, we shrink each component of so that the closures of distinct components become disjoint.
Fix an arbitrary . If for all , we let . Otherwise, first note that the sets and are compact, and by the above paragraph, are disjoint. Let be a finite union of Jordan domains such that contains , but does not meet . Then, define,
[TABLE]
Repeating the above process, we obtain topological disks for all . Note that since for all , , we have . With these modifications, for distinct values of and in , we have
[TABLE]
That is because, the intersection must lie on , but and .
Fix an arbitrary . If does not meet for all , we let . If there is such that , we further modify as follows. By the above paragraph, the sets and are disjoint. Then, let be a finite union of Jordan domains such that contains , but does not meet . Then, define
[TABLE]
It follows that is connected and simply connected. Repeating the above process for all we obtain topological disks for all .
By construction, , for distinct and . Moreover, since , we must have . This implies that contains and satisfies property (ii) of admissibility. As satisfies property (iii), and each of and is a finite union of Jordan domains, also satisfies property (iii). ∎
Lemma 3.3**.**
Let be a hairy Cantor set with base Cantor set . For any , there is an admissible set for which is contained in .
Proof.
Fix an arbitrary . Let be a finite union of Jordan domains such that , , and the closures of the connected components of are pairwise disjoint. Below we modify so that it becomes an admissible set for .
By Proposition 3.1 there is a height function . Since is open, , and , there must be such that . Let us define
[TABLE]
This is a closed set in with . Define
[TABLE]
Let denote the union of and the bounded components of . Let denote the components of which meet . Then, ( and hence) is contained in , and every component of is simply connected. Also, since covers the compact set , has a finite number of connected components.
Fix an arbitrary . First assume that . Then , and hence . In particular, since is connected and meets , is contained in . Now assume that . Note that , , and for every , . This implies that there is such that . Now, , and since , the connected set must be contained in the bounded component of . Thus, .
Note that . As has no interior points, we have . On the other hand, , which implies . Combining these we obtain
[TABLE]
The set is a finite union of Jordan curves, so it is locally connected. It follows that every point in is accessible from both and . By the above equation, . This implies that every point in enjoys the same property. Since consists of a finite number of the components of , we conclude that every point in enjoys that property.
So far we have shown that satisfies properties (i)–(iii) in the definition of admissibility for . If necessary, we employ Lemma 3.2 to shrink each component of so that it becomes an admissible set for . ∎
We need the following general feature of admissible sets.
Lemma 3.4**.**
Let be an admissible set for a hairy Cantor set . Then, for any connected component of , is connected.
We shall use a criterion of Borsuk [ES52, Chapter XI, Theorem 3.7] in the proof of the above lemma. That is, for a closed set in the Riemann sphere , the set is connected if and only if every continuous map from to the unit circle is null-homotopic. Recall that a continuous map is called null-homotopic, if is homotopic to a constant map. For instance, this criterion may be used to show that the complement of a Cantor set in the plane is connected. To see this, let be a Cantor set in the plane, and be a continuous map. As is compact, is uniformly continuous. Therefore, we may decompose into a finite number of disjoint Cantor sets such that the image of on each is contained in an arc of of length at most . It follows that on each , is homotopic to a constant map. Combining these homotopies, one obtains a homotopy from to a constant map on all of .
Proof.
Fix an arbitrary hairy Cantor set . Let be an admissible set for , and be a connected component of . As is compact, without loss of generality we may assume that is bounded in the plane, that is, is compact. Define as the union of the connected components of which intersect both and . Every connected component of is a Jordan arc. Define as the union of the connected components of which are contained in . Note that . We shall first show that is connected, and simply connected, and then show that is connected. Mostly due to perhaps complicated topological structure of relative the argument is rather long. We shall deal with in Steps 1–4 below, and then deal with in Steps 5–7. The main idea in both cases is to work in a quotient space where the boundary of is identified to a single point.
Let be the base Cantor set of . By Proposition 3.1, we may choose a height function on , which by rescaling, if necessary, we may assume that . Let be a homeomorphism from to a Cantor set in . We may extend onto , using the height function , according to
[TABLE]
By the continuity of the base map and properties of height functions, is continuous and injective on . As is compact, is a homeomorphism from to . This provides us with a rather simple coordinate system on , which we shall use to build homotopies of maps on subsets of .
Step 1: The set is closed in .
Let , for , be a sequence in that converges to some . We need to show that and . Since each component of meets , there is with , for each . As is compact, passing to a subsequence if necessary, we may assume that converges to some . By the continuity of , we have
[TABLE]
In particular, as , . However, by properties of admissible sets, , which implies that . Hence, we must have , and therefore . The above equation also implies that , thus, . This completes Step 1.
Let
[TABLE]
This is a closed set in . By property (iv) of admissible sets, for any connected component of , there is such that . It follows that there is a function such that
[TABLE]
Consider the equivalence relation on defined according to if and only if , or both and belong to . The quotient space is the one-point compactification of the connected and simply connected open set . Hence, it is a topological sphere, which may be identified with . Let be the quotient map, that is,
[TABLE]
Define
[TABLE]
Then is a single point in , and is the corresponding one-point compactification of , and is a compact set in .
Consider the map ,
[TABLE]
Step 2. The map is continuous, and is a homeomorphism.
Let and . By Step 1, is closed, which implies that is closed in . We claim that is also closed in . To see this, first note that is closed in , and is a homeomorphism on , which imply that is closed. Since , this set must be also closed, and hence compact. Using the compactness of as well, we note that
[TABLE]
Hence, , and therefore, is closed in .
The map is continuous on both and . That is because, is continuous on , and is continuous on . On th e set , is a constant map only taking value . It follows that is continuous on .
As is a homeomorphism, is a homeomorphism on , which takes values in . The map is also a homeomorphism on . Therefore, is a homeomorphism from onto .
Step 3. Any continuous map is null-homotopic.
Consider , defined as
[TABLE]
This is a continuous map with and . Let be a continuous map. Define the map as
[TABLE]
For all , and . We need to show that is continuous.
By Step 2, is continuous on . On the set , is constant, and hence continuous. To prove that is continuous, it is sufficient to show that for any convergent sequence in with , we have .
Let , for . Since , by Step 2, the sequence may not have a limit point in . Hence, 666Given compact sets and in , we define
[TABLE]
As is compact, for every , there is in with
[TABLE]
Recall that
[TABLE]
In particular, implies that . Therefore, since is not expanding distances,
[TABLE]
Therefore,
[TABLE]
By the continuity of and , the above equation implies that . This completes Step 3.
Step 4. The set is connected, and simply connected.
As is injective and continuous on , and is open, is a homeomorphism. It follows that is connected if and only if is connected. By Borsuk’s criterion, discussed before the proof, and Step 3, is connected. Thus, is connected.
Let . If is not simply connected, there is a closed curve in which is not null-homotopic in . Note that . Since is not null-homotpic, both connected components of must intersect . However, this is not possible since is connected. This completes Step 4.
Consider the equivalence relation on where if and only if , or both and belong to . By Step 4, is a topological disk. Thus, is the one-point compactification of the topological disk , which is a topological sphere. Hence, we may identify with . Let be the quotient map,
[TABLE]
Define
[TABLE]
Then, is a single point, while is a compact set. Note that may or may not be compact. Indeed, if is compactly contained in , then is compact, and is away from . However, if is not compactly contained in , then is a limit point of .
Define
[TABLE]
This is a continuous map on . Moreover, since is continuous and injective on , and , induces a homeomorphism from to .
Step 5 : Any continuous map is homotopic to a map which is constant on each connected component of .
Consider , defined as
[TABLE]
This a continuous map, with , and . Also, for each , maps into .
Let be a continuous map. Consider defined as
[TABLE]
We have . On the other hand, since maps each connected component of to a vertical line segment, is constant on each connected component of . There remains to show that is continuous.
Evidently, is continuous on , and is constant on . If is away from , then we are done. Otherwise, assume that a sequence in converges to , and converges to . Below we show that .
Note that . That is because, if is a sequence in converging to some , then . However, , which implies that . Thus, , and hence . Now recall that , for , and is a homeomorphism. Since , the sequence may not have a limit point in . Therefore, any limit point of this sequence belongs to , that is,
[TABLE]
which implies
[TABLE]
For , let . As is compact, for every , there is in such that
[TABLE]
Since , for all , and is non-expanding,
[TABLE]
Thus, converges to . Recall that . Using the continuity of and , we conclude that . This completes Step 5.
Step 6. The map is null-homotopic.
Define
[TABLE]
We have , and is closed and totally disconnected. The only point in which might be isolated is . Hence is either a Cantor set, or the union of an isolated point and a Cantor set. By the discussion preceding the proof of this lemma, is null-homotopic. Hence, there is a constant , and a continuous map such that
[TABLE]
Now consider the map defined as
[TABLE]
For , , and therefore, is well-defined. Since is constant on each component of , for ,
[TABLE]
For , . Thus, . Also, . Below we show that is continuous.
Let , for , be a convergent sequence in . To prove that is continuous, it is sufficient to consider sequences with which converge to . Let . We claim that . If not, let be a limit point of the sequence . Then is a limit point of the sequence . Therefore, has a limit point in . Hence, has a limit point in , which is a contradiction. Thus, , which implies that tends to , as . Therefore, is continuous.
Step 7 : The set is connected.
The map is injective and continuous on . Hence, is a homeomorphism from to . This implies that is connected if and only if the set is connected. By Borsuk’s criterion discussed before the proof, and Steps 5–6, the latter set is connected. Thus, is connected. ∎
Let be a hairy Cantor set, and be an admissible set for . By a marking for we mean a collection of points in such that for every connected component of there is a unique such that . We refer to the pair as a marked admissible set for .
Let and be marked admissible sets for with . Let be a connected component of which is contained in a component of . We say that is a free component of in , if there is a curve
[TABLE]
such that
[TABLE]
Let us say that is a free admissible set for within , if is an admissible set for , , and every component of is a free component of in .
Lemma 3.5**.**
Let be a hairy Cantor set, be a marked admissible set for , and be an admissible set for with . There exists a marked admissible set for such that and is free in .
Proof.
Let be a component of , which is contained in a component of . Let us say that is an accessible component of in , if there is a curve
[TABLE]
such that
[TABLE]
If every component of is an accessible component of in , we define , and be the collection of over all components of . If has non-accessible components within , we modify the components of so that all components of become accessible. We explain this process below.
Let be a non-accessible component of . Let be the component of which contains , and let be the unique element in . Let , for , denote the components of which lie in . Rearranging the indexes if necessary, we may assume that . Recall from Lemma 3.4 that is connected. As is also open, it must be path-connected. Also, recall that is accessible from . These imply that there is a curve
[TABLE]
such that and .
For , let denote the union of the connected components of which intersect . We define
[TABLE]
Since each is a topological disk, the connected components of are topological disks. As , for each , , and . It follows that satisfies properties (i)–(iii) in the definition of admissible sets.
At this stage, the set might not satisfy property (iv) in the definition of admissible sets. However, we may employ Lemma 3.2 to shrink the components of so that property (iv) holds as well. Let denote this modified admissible set obtain from employing that lemma. Note that the number of components of is the same as the number of components of .
If meets some , for , then every component of is an accessible component of in (via a portion of the curve ). Also, if some , for , is an accessible component of within , and , then is an accessible component of within . Moreover, as , is an accessible component of in , via the curve .
By the above paragraph, the number of non-accessible components of within is strictly less than the number of non-accessible components of within . On the other hand, if is a component of which is contained in a component of , and is an accessible component of within , then is an accessible component of within . Therefore, the number of non-accessible components of within is strictly less than the number of non-accessible components of within .
Recall that has a finite number of connected components. By repeating the above process, we may reduce the number of non-accessible components of until all its components become accessible. ∎
Lemma 3.6**.**
Let be a hairy Cantor set with base Cantor set . There are marked admissible sets , for , such that the following hold:
- (i)
;
- (ii)
for , is compactly contained in ;
- (iii)
for , is a free admissible set in .
Proof.
For , let be a Jordan domain containing . Let consist of an arbitrary point on . Evidently, is a marked admissible set for .
Assume that is defined for all , for some , and satisfy properties (ii) and (iii) listed in the lemma. Below, we build .
Let . By Lemma 3.3, there is an admissible set for such that . It follows from the choice of that is compactly contained in . By the induction hypothesis, is a marked admissible set for . Therefore, by Lemma 3.5, there is a marked admissible set for such that and is free within .
Note that for , and . Therefore, . ∎
Proposition 3.7**.**
Every hairy Cantor set admits a base curve.
Proof.
Let be a hairy Cantor set, and let be the base Cantor set of . By Lemma 3.6, there is a nest of marked admissible sets , for , such that is free in , and shrinks to . Recall that is a Jordan domain containing . To simplify the forthcoming presentation, let us choose a Jordan domain which contains , and let contain a single point. For each , let , for , denote the connected components of . We have . Also, let denote the unique point in . For and , define as the set of integers with and .
For each , is free in . Therefore, there are
[TABLE]
such that
[TABLE]
for all and . We may extend onto by setting and . If necessary, we may modify these curves such that , for distinct values of and .
For , let denote the union of the curves , for all , , and . Each forms a finite tree embedded in the plane. Let
[TABLE]
Since , is a compact set in the plane. Moreover, .
We aim to turn into a base curve for . The curve will be the limit of a sequence of curves , where is obtained from slightly “thickening” the tree . We present the details of this construction below. See Figure 3.
Note that the curves are pairwise disjoint, for distinct values of the triple . Hence, there are tubular neighbourhoods of these curves, which are pairwise disjoint, and do not meet . In other words, there are orientation preserving, continuous and injective maps
[TABLE]
such that for all , we have
[TABLE]
and for distinct triples and we have
[TABLE]
We may extend onto by setting and , for all . Consider
[TABLE]
This is a compact set in , with .
Fix and . There are curves on which land at . These are the curves and , for , as well as and , for some with . Let , , …, denote the elements of , labelled in such a way that the curves , for , followed by and then land at in a clockwise fashion.
Let be small enough, and consider pairwise disjoint curves
[TABLE]
for , such that
[TABLE]
Similarly, let and be such that
[TABLE]
Let be a decreasing sequence of positive numbers tending to [math] such that , for all and . We may choose the above maps so that the union of the curves , for , , , for , , and forms a closed loop which encloses the mark point and is contained in .
Recall that denotes the circle of radius about . We may choose closed and connected sets , for and , satisfying the following properties:
- •
is contained in the interior of if and only if ;
- •
if , then ;
- •
the Euclidean length of is bounded by .
For , let and define
[TABLE]
It follows that is a Cantor set of points on . We aim to build a Jordan curve such that , and if and only if . The map will be defined as the limit of a sequence of maps , for . We define inductively as follows.
For , we define such that
[TABLE]
The choice of the parameterisation of this curve is not important; any choice will work.
Now assume that the curve is defined for some . We define as follows. On we let . On each , we define in such a way that for all
[TABLE]
and becomes
[TABLE]
Again, the specific parametrisation of the curve is not important. This completes the definition of for all . Note that for all and we have
[TABLE]
Now we show that , for , forms a Cauchy sequence. Fix , and assume that and are bigger than . If , we have . If , for some , by Equation (7), . It follows from the construction that and also belong to . Therefore, for all , we have
[TABLE]
However, as , tends to zero as tends to infinity. Hence, the sequence of maps converges uniformly on to a map . In particular, is continuous.
If , there are and such that . This implies that for all , . Therefore, . If , there are an arbitrarily large and with . By Equation (7), . Since , we conclude that . On the other hand, for every there is with . To see this, choose such that . There is with . If is a convergent subsequence of , converging to some , we obtain . It follows from these statements that .
To see that is injective, let and be distinct elements of . If and belong to , then there must be and such that and . By Equation (7) and the decreasing property of , for all , and . Since , we must have . If and do not belong to , then there is such that and . Then, , by the definition of , and the injectivity of . If exactly one of and belongs to , say and , by the above paragraph, and . Thus, .
There remains to show that does not meet the bounded connected component of . To see this, note that may be continuously deformed into the tree in the complement of . This implies that does not meet the bounded component of . Since converges to uniformly on , one infers that does not meet the bounded component of . ∎
Proposition 3.8**.**
Let be a hairy Cantor set, and be a base curve for . Then, the set has two connected components, where the bounded component is simply connected, and the unbounded component is doubly connected.
Proof.
Let denote the bounded component of , and . Since is a base curve for , does not meet . Therefore, is a bounded component of . Evidently, is simply connected. Let . We aim to show that is connected and doubly connected. To this end, we use Borsuk’s criterion (see the paragraph after Lemma 3.4), and show that is connected and simply connected in the Riemann sphere .
By Borsuk’s criterion, in order to prove that is connected, it is sufficient to show that any continuous map is null-homotopic. Below we construct a homotopy between a given such map and a constant map.
By Proposition 3.1, there is a height function on , which after rescaling we may assume . Let be a homeomorphism from onto a Cantor set in the line . We extend the map to , according to
[TABLE]
As is continuous and injective on a compact set, it must be a homeomorphism onto its image.
Consider the family of map , defined as
[TABLE]
Define the family of maps , defined as
[TABLE]
One may see that is continuous. Moreover, , for all . Hence, is a homotopy between and the map
[TABLE]
Note that , for all , and , for all .
Now we construct a homotopy between and a constant map. The set is homeomorphic to the closed unit disk . Hence, the map is null-homotopic. Therefore, there is a continuous map and a constant such that and . Now define as . Evidently, is continuous, and for all we have and . Hence, is null-homotopic and therefore, is connected.
The complement of in , which is , is connected. This implies that must be simply connected. ∎
4. Uniformisation of hairy Cantor sets
In this section we prove Theorem 1.4. Let us fix a hairy Cantor set , with base Cantor set , and a base curve . By Proposition 3.8, divides the plane into two connected components. Let denote the bounded connected component of . We have . We fix the notations
[TABLE]
Then, is an open, connected, and simply connected subset of .
We aim to complete into a one dimensional (topological) foliation of the plane so that each component of is contained in a leaf of that foliation. To that end, we identify a single chart (“a box neighbourhood”) containing so that within that chart may be completed into a foliation which is homeomorphic to the foliation of the unit square by vertical lines. The required portions of leaves will come from hyperbolic geodesics in and . This analysis will be carried out in the framework of Carathéodory’s theory of prime ends, which relates the geometry of a plane domain to the topology of its boundary. For the general theory of prime ends one may see the classic references [Car13] and [Koe15], or the modern treatment [Pom75]. However, below we briefly recap the basic notions and statements of the theory which are employed here.
We use the notations and for the distance and the diameter, respectively, with respect to the spherical metric on .
Let be an open, connected, and simply connected set whose boundary contains at least two points. A fundamental chain in is a nest of open sets in satisfying the following properties:
- •
for , is a simple curve in , which is homeomorphic to and its closure contains two distinct points in ;
- •
, for ;
- •
for , ;
- •
, as .
Note that by the Jordan curve theorem, divides into two connected components. Equivalently, some authors define the notion of fundamental chains using sequences of Jordan arcs in with end points in .
Two fundamental chains and in are called equivalent if every contains some and every contains some . Any two fundamental chains in are either equivalent or eventually disjoint, that is, for large enough and . An equivalence class of fundamental chains in is called a prime end of .
The impression of a prime end of , represented by a fundamental chain , is defined as . This is a non-empty compact and connected subset of . Evidently, the impression of a prime end is independent of the choice of the fundamental chain . Any point on is contained in the impression of a prime end of .
We say that is the principal point of a fundamental chain in , if . In general, equivalent fundamental chains might have different principal points. The set of principal points of a prime end is defined as the set consisting of all principal points of the fundamental chains in the class of that prime end. Every prime end of has at least one principal point. By the theory of prime ends, is accessible from , if and only if, there is a prime end of whose impression contains , and is the only point in the set of principal points of that prime end. Abusing the terminology, we say a prime end is accessible if it has a unique principal point.
Lemma 4.1**.**
We have .
Proof.
Let . Since and are open in , and , may not be in . As , . Therefore, .
Let and . Since is a Jordan domain, is a non-empty open set. As has empty interior, may not be contained in , and hence, must meet . This implies that . Thus, . ∎
Recall the base map and the peak map defined at the beginning of Section 3. We may extend these maps to
[TABLE]
by setting , for all . Also, for , we set . As is continuous on both and , and and are closed sets, is continuous on .
We present the properties of the prime ends of in the following lemma.
Lemma 4.2**.**
The following properties hold:
- (i)
For any prime end of there is such that the impression of is equal to . On the other hand, for any , there is a prime end of whose impression is equal to .
- (ii)
Every prime end of is accessible. Moreover, is accessible from if and only if for some .
- (iii)
Distinct prime ends of have disjoint impressions.
Proof.
(i): By Lemma 4.1, . Therefore, any belongs to the impression of a prime end of . That is, there is a fundamental chain in whose impression contains . We aim to show that .
For , is a Jordan arc connecting two points on . Let and denote those points. Since is connected, for all , is a closed arc on , which is bounded by and . As belongs to , we must have , with respect to a fixed cyclic order on . On the other hand, as , , which implies that . By the uniform continuity of , we conclude that
[TABLE]
By the definition of hairy Cantor sets, the only accessible points of from belong to . It follows that and belong to . Using , for , one concludes that for all ,
[TABLE]
In particular, , for all , and hence, . On the other hand, for all , , for all . By the above paragraph, and , which leads to . Therefore, . Combining the two relations, we obtain .
(ii): Let be a prime end in , and let be a fundamental chain in in the class of which has a principal point, say . We aim to show that .
Assume in the contrary that . In particular, , and is a Jordan arc. We may extend into so that we obtain a Jordan arc in with of the end points of in . There is a Jordan domain such that , , consists of two points, and has two connected components.
Let and denote the landing points of on . Since is the principal point of , we must have and , as . Moreover, since belongs to the impression of , as in Equation (9), we have , for all . By the choice of , this implies that for large enough , and belong to distinct components of . On the other hand, and belong to , with and . This implies that for large enough , both and belong to the same component of . This contradiction shows that we must have .
Let be an arbitrary prime end of . By part (i), the impression of is of the form , for some . By the above argument, if is a principal point of , we must have . Since , and there is a single point in with , we conclude that has a unique principle point. Thus, is accessible. On the other hand, by Part (i), for any , there is a prime end of whose impression is equal to . By the above argument, is the unique principal point of that prime end. Therefore, the set of principal points of the prime ends of is equal to .
(iii): Assume in the contrary that there are distinct prime ends and in whose impressions intersect. By part (i), the impressions of and must be equal to , for some . Let and be fundamental chains in the classes of and , respectively. Since and are not equivalent, there is such that . For , choose . We have .
Let and denote the landing points of on . We may relabel these points so that , with respect to a fixed cyclic order on . The open set
[TABLE]
is a Jordan domain, which is bounded by
[TABLE]
In particular, . Since , as , there must be such that , and hence . Therefore,
[TABLE]
which is a contradiction. ∎
Let be a connected and simply connected domain in whose boundary contains at least two points. Let be a Riemann map, that is, a one-to-one and onto holomorphic map. By the general theory of prime ends, for any fundamental chain in , is a fundamental chain in . This correspondence induces a bijection between the set of prime ends of and the set of prime ends of .
The prime ends of are easy to understand. Any prime end of is accessible, and distinct prime ends of have distinct principal points. The map which sends a prime end of to its unique principal point induces a bijection between the set of prime ends of and . Combining with the above paragraph, induces a one-to-one correspondence between and the set of the prime ends of . Let be a prime end of which is the image of under this bijection and has a unique principal point in . Then, the radial limit exists and is equal to the principal point of . See Corollary 2.17 in [Pom92].
By Proposition 3.8, the set is connected, and simply connected. We may consider a Riemann map
[TABLE]
with . By the above paragraphs, induces a bijection between the set and the set of the prime ends of . By Lemma 4.2, every prime end in is accessible, and has a unique principal point which belongs to . Thus, there is a bijection between and . For every , the radial limit
[TABLE]
exists and belongs to . Moreover, for in , .
We extend the base map to
[TABLE]
as follows. For an arbitrary we let , and define
[TABLE]
For each , there is a unique such that . Then, for all on the curve , . This implies that for each , the set in is a Jordan arc consisting of and . This curve meets only at .
Proposition 4.3**.**
The map is continuous.
Proof.
We aim to prove that is closed in , for every closed set . Since is a Jordan curve, it is sufficient to prove that is closed whenever is either a point or a Jordan arc. If is a single point, then is a Jordan arc and therefore is closed. Let be a Jordan arc and let and be its end-points. We know that and are disjoint Jordan arcs which land at and on . Therefore, divides into two disjoint Jordan domains, denoted by and . The sets and are Jordan arcs in with end points at and . Hence, one of these sets, say , is equal to . On the other hand, is a disjoint union of , for . Therefore, is a disjoint union of such arcs. This implies that
[TABLE]
For , is a Jordan arc in which lands at a single point on . For and in , we define to be the unique Jordan arc in which ends at and . When , we have . Note that if and belong to , this notation is consistent with our earlier notation .
Proposition 4.4**.**
For any convergent sequence in with , we have
[TABLE]
in the Hausdorff topology.
When , for all , we already have the convergence of to from the definition of hairy Cantor sets. When , is the union of and , where the latter set is a hyperbolic geodesic in . For the above proposition, we need to show that such a sequence of hyperbolic geodesics may not have pathological behaviour. We control the location of those hyperbolic geodesic by employing a theorem of Gehring and Hayman, see [GH62] or [Pom92, Page 88]. That is, let be a simply connected domain whose boundary contains at least two points, and let and be distinct points in . Assume that is an arbitrary curve in connecting to , and is a geodesic in , with respect to the hyperbolic metric on , connecting to . By Gehring-Hayman theorem,
[TABLE]
where is a universal constant independent of , , , and .
For , let
[TABLE]
This is a compact annulus in .
Proof of Proposition 4.4.
For any , is a Jordan arc which is homeomorphic to . So there is a linear order on each , where is the smallest point.
Let us fix such that for all , , where is the compact annulus defined by Equation (11). Then, for all , . The set of non-empty compact subsets of with respect to the Hausdorff topology is compact. Therefore, to prove the proposition, it is sufficient to prove that if is a convergent sequence in the Hausdorff topology, then .
Assume that converges in the Hausdorff topology, and let denote the limit of this sequence. Then,
[TABLE]
By Proposition 4.3, is continuous. This implies that is contained in . On the other hand, since each is connected, must be also connected. Therefore, , for some . Moreover, since , we have , and hence , with respect to the linear order on . We need to show that . Assume in the contrary that . We shall derive a contraction by considering two cases.
Case (i): Assume that and . Define
[TABLE]
Consider the compact annulus defined by Equation (11). Since , and , there is an integer such that for all , . In particular, for , is contained in . As is compact, the Hausdorff limit of , that is , must be contained in . But, and , since . This is a contradiction.
Case (ii): Assume that and . By passing to a subsequence, we may assume that is monotone on , that is, for all , , with respect to a fixed cyclic order on . Define
[TABLE]
By virtue of the continuity of in Proposition 4.3, , as , with respect to the Hausdorff topology.
The Jordan arcs and divide into two Jordan domains. Let
[TABLE]
Since in this case, we may choose two points and in such that . There are Jordan arcs and such that
[TABLE]
and
[TABLE]
Each of the curves and divides into two connected components.
Let us fix , for , such that as . Let denote the unique point in . Then, , for all .
The curve divides into two connected components. As , , and , there must be such that for all , and lie in distinct components of . In particular, for every , both of the curves and meet the curve . It follows that for , the sets and are non-empty compact sets in . Since , there must be a Jordan arc
[TABLE]
such that
[TABLE]
Note that the last property in the above equation implies that . Then, the union of the Jordan arcs and forms a Jordan curve in the plane. Let denote the bounded component of the complement of that Jordan curve. Thus, we have
[TABLE]
and
[TABLE]
The closure of the Jordan domain is homeomorphic to . Moreover, as belongs to , contains and . Since and lie on distinct components of , the curve must divide into at least two connected components. It follows that there is a Jordan arc
[TABLE]
such that
[TABLE]
and divides into two components, one of which contains and the other one contains .
We claim that for every , the Jordan domain enjoys the following property:
[TABLE]
To see this, fix an arbitrary . By Equation (14), , but . Therefore, the Jordan arc must cross . However, . That is because, and , while and are either disjoint or identical. But, belongs to , and by Equation (13)), (since ). This completes the proof of the claim.
Since all the points , , , and lie on , we may compare them using the linear order on . We have
[TABLE]
Since and belong to , we must have
[TABLE]
Similarly, as and belong to , we must have
[TABLE]
In particular, we have
[TABLE]
Moreover, since ,
[TABLE]
Now note that at least one of the following three possibilities occurs:
- (a)
there are infinitely many with ,
- (b)
there are infinitely many with ,
- (c)
there are infinitely many with .
Below we show that each of the above scenarios leads to a contradiction.
If (a) occurs, let be an increasing sequence such that . In particular, the points and do not belong to . Then, the Jordan arc does not meet , and hence, it is a hyperbolic geodesic in . Therefore, according to the Gehring-Hayman theorem discussed before the proof, for , we must have
[TABLE]
This contradicts the limiting behaviours in Equations (19) and (20).
If (b) occurs, let be an increasing sequence with . By Equation (16), we must have . Therefore, by property (v) in the definition of hairy Cantor sets, and Equation (18), converges to . On the other hand, by Equation (16), we have , and by Equation (17), , But , which is a contradiction.
If (c) occurs, let be an increasing sequence with . Let . By Equation (15), . Let . As , from Equation (17), we conclude that as . On the other hand, since , by property (v) in the definition of hairy Cantor sets, we have . But, and . This is a contradiction. ∎
Proof of Theorem 1.4.
By virtue of Theorem 1.2, it is sufficient to prove that every hairy Cantor set in the plane is ambiently homeomorphic to a straight hairy Cantor set in the plane. Let be an arbitrary hairy Cantor set, with base Cantor set , that is, is the closure of the set of point components of . By Proposition 3.7, admits a base curve, that is, a Jordan curve such that and the bounded component of does not meet . Then we define the sets in Equation (8).
Let denote the closed annulus defined by Equation (11). The base map extends to the map ; see Equation (10). By Proposition 4.3, is continuous. On the other hand, following the discussions in Section 3, we may consider a Whitney map for the set . Then, we define
[TABLE]
By virtue of Proposition 4.4 and properties of Whitney maps, is continuous. Moreover, if and for some and in , by the properties of height functions, we have , which implies .
Let be a Jordan arc such that . Consider a homeomorphism
[TABLE]
Then, we may define the map
[TABLE]
according to
[TABLE]
By the above paragraph, is continuous and injective. As is a compact set, must be a homeomorphism from its domain onto its image. Since the domain of is a Jordan curve, its image must be also a Jordan curve. Therefore, one may extend to a homeomorphism of the plane.
There remains to show that is a straight hairy Cantor set. To prove that, we employ Lemma 2.7. Let . Since is a homeomorphism, and is a Cantor set of points, must be a Cantor set of points. Define according to
[TABLE]
As is compact, is well-defined and non-negative for all . We have
[TABLE]
We need to verify the three conditions in Lemma 2.7. Evidently, is compact, so we have item (i). By property (iii) in the definition of hairy Cantor sets, is dense in . In particular, the closure of contains . This implies that the closure of contains . In particular, we have item (ii) in Lemma 2.7. Property (iv) in the definition of hairy Cantor sets directly implies item (iii) in Lemma 2.7. Therefore, by Lemma 2.7, is a straight hairy Cantor set. ∎
Proof of Theorem 1.3.
Fix a compact metric space which satisfies axioms to , and . Let denote the base Cantor set of , and its base map.
By Proposition 3.1, there is a height function , which is continuous on , injective on any connected component of , and .
Let be a Cantor set, and let be an arbitrary homeomorphism. Define the map as
[TABLE]
The map is injective. That is because, if and , for some and in , then and belong to the same connected component of , and we have . This implies that .
The map is continuous and injective on a compact set. Therefore, it is a homeomorphism onto its image. We claim that is homeomorphic to a straight hairy Cantor set. To show that, we apply Proposition 2.6 to the set .
Consider the function such that
[TABLE]
As is a homeomorphism, and is compact, must be compact.
Define the sets
[TABLE]
By axiom , is dense in . For any ,
[TABLE]
which implies that
[TABLE]
Therefore,
[TABLE]
which implies that
[TABLE]
Therefore, is dense in . This completes the proof of the theorem. ∎
Axiom is stronger than axiom . But it turns out that any compact set which satisfies axioms to , also satisfies axiom . We state the following corollary for future application.
Corollary 4.5**.**
For any hairy Cantor set the set of peak points of is dense in .
Proof.
By Theorem 1.4, is ambiently homeomorphic to a straight hairy Cantor set. By Proposition 2.8, for every straight hairy Cantor set, the set of peak points is dense in that straight hairy Cantor set. Therefore, the set of peak points of must be dense in . ∎
In [DR14], the authors propose a topological model for the attractor (post-critical set) of a especial class of infinitely satellite renormalisable maps. The main object in that paper is built as the intersection of a nest of plain domains. However, they do not study the topological features of the model presented in the paper. One may see that the topological model presented in that paper is a Hairy Cantor set.
Acknowledgements.
The first author acknowledges funding from EPSRC(UK) - grant No. EP/M01746X/1 - rigidity and small divisors in holomorphic dynamics. The second author acknowledges funding from ERC advanced grant No 339523 – Rigidity and global deformations in dynamics.
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