Julia s two faces. The dichotomy theorem on Julia sets of quadratic polynomials. Rasila Hoek July 1, Bachelor thesis Supervisor: dr.

Transcription

1 Julia s two faces The dichotomy theorem on Julia sets of quadratic polynomials Rasila Hoek July 1, 2013 Bachelor thesis Supervisor: dr. Han Peters Korteweg-de Vries Instituut voor Wiskunde Faculteit der Natuurwetenschappen, Wiskunde en Informatica Universiteit van Amsterdam

2 Abstract In this thesis, the behavior of points under iteration of complex-valued, quadratic polynomials is studied. Any quadratic polynomial can, under an affine transformation, be written as f c (z) = z 2 + c, where c, z C. We define the set K c, consisting of all points that have a bounded orbit under iteration of f c. The Julia set J c is defined as the boundary of K c. The aim of this thesis is to prove the dichotomy theorem, which consists of the following two equivalences: 1. 0 K c J c is connected / K c J c is totally disconnected. The first step in the proof is showing that there is some disk V c so that, if an orbit contains points outside of V c, it will converge to infinity. Only orbits that lie completely in V c have starting points that lie in K c. To find these points, we take repeated pre-images of V c. We prove that the limit of the intersection of these pre-images equals K c. If 0 lies in K c, f(0) = c lies in all of the pre-images. We prove that the pre-image of a set that contains c, consists of 1 connected component. This will result in a K c that consists of only one connected component. We prove that this K c is even simply connected. Finally we prove that the boundary of K c, the Julia set, is connected, using the facts that K c is simply connected and bounded. If 0 does not lie in K c, f(0) = c will not lie in all of the pre-images of V c. We prove that the pre-image of a set that does not contain c, consists of 2 components. This will result in a K c that consists of uncountably many connected components. We note that in this case, the Julia set is equal to K c. We show in the last chapter that each of these components consists of exactly one point. Title: Julia s two faces Author: Rasila Hoek, rasila.hoek@science.uva.nl, Supervisor: dr. Han Peters Second grader: prof. Jan Wiegerinck Final date: July 1, 2013 Korteweg-de Vries Instituut voor Wiskunde Universiteit van Amsterdam Science Park 904, 1098 XH Amsterdam 2

3 Contents 1 Introduction 4 2 The dichotomy theorem on Julia sets Three important sets The dichotomy theorem Julia s two faces A forbidden circle The pre-image of an open set A connected Julia Set A disconnected Julia Set The pig snout intersection 22 Popular Summary 25 Bibliography 29 3

4 1 Introduction The subject of this thesis lies in the field of complex dynamical analysis. Complex, because the functions that we will study live on the complex plane. Dynamical, because we will look at the iteration of these functions. The functions that we will study are polynomials, where we are interested in the points that have a bounded orbit under iteration. The boundary of this set of points gives us the Julia set. When studying polynomials, a logical first step would be to start with linear polynomials. Their dynamical behavior however, though interesting, is after some finite and even short time rather predictable. When making the step towards quadratic polynomials, the complexity increases surprisingly much, just as the visual beauty of the corresponding Julia sets. On the appearance of those Julia sets we will formulate our main theorem: the dichotomy theorem. It will tell us about the two different looks that such a Julia set can have: Julia s two faces. Proving this theorem is the main objective of this thesis. For the sake of simplicity, we will work with quadratic polynomials written as f(z) = z 2 + c. We can justify this choice by noting that any quadratic polynomial can, under an affine transformation, be written in this form. The dichotomy theorem has been proved for the first time by Gaston Julia and Pierre Fatou in In that sense, nothing new will be brought up in this thesis. On the other hand, no existing proof of the dichotomy theorem was used in writing this thesis. The idea of the proof that will be given here has been provided by my supervisor Dr. Han Peters. The details I developed myself or together with Han in our weekly meetings. I have been striving not to use (advanced) literature: all proofs and definitions ought to be understandable for anyone with some background in elementary topology and complex analysis. At the few points in chapter 4 where something more advanced is necessary, references are given. The process that lead to this thesis has been valuable to me in a number of ways. Where the regular bachelor courses were, in my experience, marked by the attempt to get familiar with a big amount of mathematical objects, the bachelor project felt like actually using these objects for the first time. The benefits of writing this thesis can be described as an improvement of skill rather than an enlargement of knowledge. My ambition has been to write a correct, precisely formulated and clear proof, critical in its assumptions, rather than to deliver surprising results or explore new areas. I would like to thank Han Peters, both for his patience, and for his willingness to discuss also the most elementary theory in a critical way. The weekly meetings with him, and 4

5 our correspondence by during his absence, have been of great help. Furthermore I would like to thank Jeroen, for his permanent availability for LateX advice, and Freek, whose faith in my mathematical talents is indestructible. Rasila Hoek 5

6 2 The dichotomy theorem on Julia sets In this chapter, we will introduce three sets that are very important in this thesis, as well as the main theorem that we will prove. Throughout this thesis, we will write f c for the function f : C C defined by z z 2 + c, where c, z C. 2.1 Three important sets We will study the behavior of points in C under iteration of f c. In order to define the Julia set of a point z, we will have to distinguish between points with a bounded orbit and points with an unbounded orbit. Firstly we will define a safe set K c, consisting of points that do not move too far away under iteration of f c. Definition 2.1. The safe set K c is the set of all values z C for which {f n c (z) n N} is bounded. Of course, K c depends on the choice of c. The simplest example is the case c = 0, when f(z) = z 2. Iterating f 0 geometrically means for a point in C that its modulus is squared and its angle is doubled. A point that lies outside of the unit disk will move away from the origin when its modulus is repeatedly squared. A point that lies on the unit disk will not move away from the origin when its modulus is squared. In this simplest case, the safe set K 0 is exactly the unit disk. The boundary of the safe set forms the second set that is important to us. Definition 2.2. The Julia set of c C, denoted by J c, is the boundary of the safe set K c. In other words, the Julia set of a point c consists of all the points z for which any neighborhood of z contains both points with a bounded orbit, and points with an unbounded orbit. In the case c = 0, as explained above, the safe set K 0 is the unit disk, so the corresponding Julia set J 0 is the unit circle. We will be studying the properties of J c for different values of c. It will turn out that there are certain values of c for which the Julia set is connected. In figure 2.1 two examples of connected Julia sets can be seen. Those values of c that result in a connected Julia set make up the third set discussed in this section: the Mandelbrot set. 6

7 Figure 2.1: The shape of K c for two different values of c. In these cases, J c = δk c is connected. Figure 2.2: The Mandelbrot set M in C Definition 2.3. The Mandelbrot set, denoted by M, is the set of all values c C for which J c is connected. The Mandelbrot set is shown in figure 2.2. It can be seen that the point 0 lies in the Mandelbrot set. Note that indeed, 0 should be a point in M, since its Julia set, the unit circle, is connected. 2.2 The dichotomy theorem If c M, the Julia Set J c of c is connected. If c / M, the Julia set is disconnected. It will be proved in this thesis that, in the latter case, the Julia set is totally disconnected. In other words, the Julia set of a point c either consists of one component, or it consists of infinitely many components, where each of the components consists of exactly one point. 7

8 On top of that, it will turn out that there is a relative simple criterion to check which of the two faces J c will have, just by looking at the behavior of 0 under iteration of f c. This is summarized in the following theorem. Theorem 2.4. (dichotomy theorem) Let c C. Then the following equivalences hold: 1. 0 K c J c is connected (c M) / K c J c is totally disconnected (c / M). The proof of this theorem is the main aim of this thesis, and will be developed step by step in the following two chapters. 8

9 3 Julia s two faces In this chapter, we will prove the biggest part of the dichotomy theorem. The origin of Julia s two faces will become clear. 3.1 A forbidden circle Important to our proof of the dichotomy theorem is the fact that we can, at a certain moment, be sure that an orbit is unbounded. More precisely, we have some disk V centred at the origin, so that the following implication holds: An orbit contains a point outside of V = This orbit is unbounded. Such a disk we can find for our function f c, but more generally we can prove that such a disk exists for all holomorphic complex-valued polynomials. This is stated in the following lemma. Lemma 3.1. (a forbidden circle) Let g(z) = a d z d + a d 1 z d 1 + a d 2 z d a 0 be a complex-valued polynomial. Then there is some disk V R with radius R so that: g m (z) V R for some integer m = {g n (z) n N} is unbounded. Proof. Using an adjusted version of the triangle inequality, namely x + y x y, we can conclude that: g(z) a d z d a d 1 z d a 0 It is clear that z will have an unbounded orbit under iteration of a d z d when z > 1. If we can find an R bigger than 1, so that 1 2 a dz d a d 1 z d a 0 holds, then the proof is finished, since that would imply that g(z) 1 2 a dz d. 9

10 Now, 1 2 a dz d a d 1 z d a 0 z d 2 a d a d 1 z d a 0 z 2 a d a d 1 + a d 2 1 z a 0 1 z d 1 For the last part we can use the triangle equality again, this time in its usual form x + y x + y, together with our assumption that z > 1, which gives 2 a d a d 1 + a d 2 1 z a 0 2 a d a d a d a d z a d 2 a d ( a d 1 + a d a 0 ) 1 z d 1 a 0 z d+1 So, to complete the proof, we take R = max{1, 2 a d ( a d 1 + a d a 0 )}. For this R, V R satisfies the conditions. Focusing again on our specific polynomial f c, the corresponding disk V c we call the forbidden circle. Its radius can be different for different values of c. For reasons that will become clear later on, it is important that V c is open. So let us make V c slightly bigger, and remove its boundary. Note that our new V c still satisfies the conditions stated in the above lemma. It is useful to look at the pre-image of V c. The points that lie outside this pre-image, will end up outside the forbidden circle after one iteration step, and therefore they can not belong to K c. If we take the pre-image of the pre-image of the forbidden circle, we exclude the points that will lie outside the forbidden circle after two iteration steps. Continuing to take pre-images, we will enclose the safe set K c. In other words: K c = f n (V c ) n=0 Since we are interested in the Julia set, which is the boundary of the set K c, we need to know what the pre-image of a disk in C looks like. We will develop some more generally formulated theory on this in the next section. 3.2 The pre-image of an open set In this section, we will look at the pre-image under f c of an open set in C. It will turn out that there are two options for this pre-image, which will be very important in 10

11 understanding the origin of Julia s two faces. We begin with the definition of a topological covering map ([3]). Definition 3.2. A function f : V U is a topological covering map if there exists an open covering U = {U α } α A of U so that for all α in A, the pre-image f 1 (U α ) can be written as a disjoint union of V β s so that for any β, the function f Vβ : V β U α is a homeomorphism. Our function f c is an example of such a map. This is our next lemma, and its proof might clarify this abstract definition. Lemma 3.3. The function f : C \ {0} C \ {c} given by f(z) = z 2 + c is a topological covering map. Proof. Without loss of generality (since z z + c is an affine transformation), we can consider f 0 (z) = z 2, from C \ {0} to C \ {0}. Now take a point z 0 in C \ {0}, with polar coordinates (r 0, θ 0 ). Since C \ {0} is open, we can take an open neighborhood U z0 given by: { r (r 0 ε, r 0 + ε) U z0 = θ (θ 0 ε, θ 0 + ε) for some sufficiently small ε. Now we look at f 1 (U z0 ), which is a union of V z0,1 and V z0,2, given by: { r ( r 0 ε, r 0 + ε) V z0,1 = θ ( 1(θ 2 0 ε), 1(θ ε)) { r ( r 0 ε, r 0 + ε) V z0,2 = θ ( 1(θ 2 0 ε) + π, 1(θ ε) + π) See figure 3.1. The sets V z0,1 and V z0,2 are clearly disjoint and f Vz0,1 : V z0,1 U z0 given by z z 2 f Vz0,2 : V z0,2 U z0 given by z z 2 are homeomorphisms. So if we take U z for all z C \ {0}, this gives us the desired open covering. Knowing now that f c is a covering map, we can use this in our search for the pre-image of an open set. To say something about the connectedness of such a pre-image, we need the following lemma. 11

12 Figure 3.1: A sketch of the proof of lemma 3.3 Figure 3.2: the lifting lemma for fc 12

13 Figure 3.3: to the proof of lemma 3.4 Lemma 3.4. (lifting lemma) Let f : V U be a topological covering map. If γ : [0, 1] U is a continuous path with γ(0) = a and we have a point x0 V so that f (x0 ) = a, then there exists a unique path γ in V so that γ (0) = x0 and that for all t in [0, 1], we have γ = f (γ (t)). See figure 3.2. Proof. Let γ : [0, 1] U be a continuous path, with γ(0) = a, and let x0 V so that f (x0 ) = a. Since γ is continuous, and [0, 1] is compact, we know that γ([0, 1]) is compact. Since f is a topological covering map, there is a covering U0 = {Uα }α A0 of U so that for all α A0, the pre-image f 1 (U ) can be written as a disjoint union of Vβ s that are mapped isomorphically onto Uα. Since γ([0, 1]) is compact, we have a finite subcovering U = {Uα }α A of U0 that covers γ([0, 1]). Since a lies in γ([0, 1]), a must lie in Uα for some α A. This α we will call α1. We assumed that f (x0 ) = a, so x0 lies in f 1 (Uα1 ). By the definition of a covering map, x0 must, for exactly one β1, lie in Vβ. Now take the maximal ε so that γ([0, ε)) Uα1. This ε we will call ε1. The sets Uα1 and Vβ1 are isomorphic, so there must be a continuous path γ 1 : [0, ε1 ) Vβ1 so that f (γ 1 ) = γt for all t [0, ε1 ) (see figure 3.3). We know that γ(ε1 ) is a point in γ([0, 1]), so for some α2 A, γ(ε1 ) lies in Uα2. Again, Uα2 is isomorphic to some Vβ2. Now let ε2 be the maximal ε so that γ([ε1, ε)) Uα2. We then find a second part of our path in V : we have a γ 2 : [ε1, ε2 ) Vβ2 so that f (γ 2 ) = γt for all t [ε1, ε2 ). 13

14 Repeating this procedure n times, we end up with a path γ in V, defined as follows: γ 1 (t) if 0 t < ε 1 γ 2 (t) if ε 1 t < ε 2. γ(t) =... γ n (t) if ε n 1 t 1 ε n will equal 1, since U αn must contain the endpoint of γ. Now we created a path γ in V, in such a way that γ = f( γ(t)) for all t [0, 1]. Next we will prove the uniqueness of γ. Let us assume that there are two continuous paths γ 1 and γ 2, so that γ 1 (0) = x 0 and γ 2 (0) = x 0, and for both paths, the image under f is equal to γ. Furthermore, we assume that there is some t [0, 1], such that γ 1 (t) γ 2 (t). We know that γ 1 (0) = γ 2 (0). Now let t 0 be the supremum of all t for which γ 1 (t) = γ 2 (t). By assumption, f( γ 1 (t 0 )) = f( γ 2 (t 0 )) := y 0. Since y 0 γ([0, 1]), and f is a covering map, we have an open neighbourhood U y0 of y 0 so that f 1 (U y0 ) is a disjoint union of V β s with every V β isomorphic to U y0. Now we consider two cases: If γ 1 (t 0 ) γ 2 (t 0 ): Say that γ 1 (t 0 ) V β1 and γ 2 (t 0 ) V β2.then, by continuity of γ 1 and γ 2, we must have a δ such that γ 1 (t 0 δ) V β1 and γ 2 (t 0 δ) V β2. Remember that t 0 was the supremum of the values of t for which γ 1 and γ 2 were equal, so γ 1 (t 0 δ) = γ 2 (t 0 δ). But V β1 and V β2 are disjoint, so this leads to a contradiction. So γ 1 equals γ 2 for every t in [0, 1]. If γ 1 (t 0 ) = γ 2 (t 0 ), we say that γ 1 (t 0 ) and γ 2 (t 0 ) both lie in V β. Since V β is isomorphic to U y0, and V β is open, we must have some δ so that γ 1 (t 0 + δ) = γ 2 (t 0 + δ). But t 0 was the supremum of all t for which γ 1 (t) and γ 2 (t) are equal. So again, we have a contradiction, and γ 1 equals γ 2 for every t in [0, 1]. We will use the lifting lemma to prove the next lemma, which tells us something about the possible looks of the pre-image of an open set under f c. Depending on the location of c, this pre-image consists of 1 or 2 components. This will be crucial in our proof of the dichotomy theorem. Lemma 3.5. (the pre-image of an open set) Let U be an open, simply connected subset of C. Then the following implications hold: 1. c U = f 1 (U) is connected. 14

15 2. c / U = f 1 (U) consists of two disconnected components. Proof. We now prove the first part. 1. Let U C be an open, connected set and let c U. Let w, z f 1 (U). Our aim is to show that there is a continuous path from w to z in f 1 (U). Since U is connected, and c, f(z) U, there is a continuous path in U between f(z) and c, which we will call γ z. Also, there is a path between f(w) and c, which we will call γ w. γ z : [0, 1] U, where γ z (0) = f(z) and γ z (1) = c γ w : [0, 1] U, where γ w (0) = f(w) and γ w (1) = c We would like to use lemma 3.4 here. However, f is not covering in 0. Therefore we want to restrict γ z and γ w in such a way that they do not reach c, the image of 0 under f. U is open, so we can take an open environment B(c, ε) of 0 so that B(c, ε) U. Now choose a point a B(c, ε), a 0, that lies on γ z, and a point b B(c, ε), b 0, that lies on γ w. We define the following restrictions of γ z and γ w : γ r z : [0, 1] U, with γ z (0) = f(z) and γ z (1) = a γ r w : [0, 1] U, with γ w (0) = f(w) and γ w (1) = b Since f is a covering map on C \ {0}, there must, by lemma 3.4, be a continuous path γ z r : [0, 1] f 1 (U) with γ z r(0) = z and so that for all t [0, 1], we have γz r = f( γ z). r We know that the inverse values of a B(c, ε) must lie in B(0, ε). So γz(1) r = f( γ z (1)) implies that γ z (1) lies in B(0, ε). Analogously, there must be a continuous path γ w r : [0, 1] f 1 (U) with γ w(0) r = w and so that for all t [0, 1], we have r = f( γ w ). Again, γw(1) r = f( γ w(1)) r implies that γw(1) r must lie in B(0, ε) We know that B(0, ε) is connected. So a and b can be connected by a continuous path γ a,b. The three continuous paths γ r z, γ a,b, and γ r w connect w to z. So f 1 (U) is connected. Remark. We did not deal with the case that w is 0. However, since 0 lies in f 1 (U), and we proved that any point in an open neighbourhood is connected to z, 0 must be connected to z as well. 2. Now we proceed to the second part. Let U C be an open, simply connected set and let c / U. Firstly we will show that f 1 (U) has at most 2 connected components. Let z 1 f 1 (U). We say f(z 1 ) = b, b U. Let z 2 be the other point that is mapped onto b. 15

16 Let w f 1 (U) so that w z 1 and w z 2. We say f(w) = a, a U. Since U is connected, there is a continuous path γ : [0, 1] U so that γ(0) = a and γ(1) = b. We know that f is a covering map, so by lemma 3.4 there is a continous path γ : [0, 1] f 1 (U) so that γ(0) = w and so that for all t [0, 1], we have γ(t) = f( γ(t)). Since z 1 and z 2 are the only inverse values of b, γ(1) = b = f( γ(1)) implies that γ(1) is either z 1 or z 2. In other words, γ is either a continuous path between w and z 1, or between w and z 2. So w either lies in the same component as z 1, or in the same component as z 2. That means that f 1 (U) has at most two connected components. Next we will show that f 1 (U) has at least 2 connected components. Let a U. The pre-image of a consists of two points, we call them b and d. We assume that there is a continuous path γ from b to d in f 1 (U), hoping to end up with a contradiction. Since f(b) = f(d) = a, we know that f(γ) is a closed path with starting and ending point a. Since U is simply connected, we know that f(γ) is contractible, that is, homotopy-equivalent to the constant path in a. That means we have a homotopy so that for all t [0, 1] we have and for all y [0, 1], we have (See figure 3.4). η : [0, 1] [0, 1] U η(0, t) = f(γ(t)) η(1, t) = a η(y, 0) = η(y, 1) = a Since f and η are both continuous functions, a function γ y (t) : [0, 1] [0, 1] f 1 (U) such that f(γ y (t)) = η y (t), for all y, t [0, 1], must be continuous in y and in t. We know that γ 0 (t) is either b or d, say it is b. (Otherwise the proof works with the roles of b and d swapped). By the uniqueness in the lifting lemma, f(γ 1 (t)) = η 1 (t) and f(b) = a implies that γ 1 (t) = γ(t). Knowing that γ y (0) is continuous in y, and the fact that, for all y [0, 1], we have f(γ y (0)) = η y (0) = η y (1) = a, gives us: γ 0 (0) = γ(0) = b = γ y (0) = b, for all y [0, 1] Intuitively, we used that γ y (0) can not jump from b to d at once. We proved above that γ 1 (t) = γ(t), for all t in [0, 1]. So γ(t) would be the constant path in b. This is a contradiction. We can conclude that there is no continuous path between b 16

17 Figure 3.4: Contracting f (γ) and d in f 1 (U ), which therefore consists of at least two components. We now proved that f 1 (U ) has at least two connected components, as well as at most two connected components. We can conclude that f 1 (U ) has exactly two connected components. 3.3 A connected Julia Set In this section we will prove the first equivalence in the dichotomy theorem, the case that c lies in the Mandelbrot set: 0 Kc Jc is connected Let 0 Kc, or in other words, let {f n (0)}n N be bounded. That means that the orbit of 0 lies within the forbidden circle Vc. Remember from section 3.1, that: Kc = \ f n (Vc ) n=0 Now {f (0)}n N is bounded, which means that 0 lies in f n (Vc ) for every n N. Since f (0) = c, this implies that c lies in f n (Vc ) for every n N as well. Noting that all of the f n (Vc ) are open, and applying lemma 3.5, we conclude that f n (Vc ) must be n 17

18 connected for every n N. In order to show that this also implies that the intersection of these inverse images is connected, we first need to see that, for all n, f n (V c ) f n+1 (V c ) f 1 (V c ) V c. This is easily shown by induction. Let V c be the forbidden circle of c. If there would be a point z 0 in f 1 (V c ) that does not lie in V c, that would imply that f(z 0 ) lies in V c. However, since z 0 lies outside of V c, its orbit will not re-enter V c (see the proof of lemma 3.1). So f ( z 0 ) can not lie in V c. From this we can conclude that f 1 (V c ) V c. Now assume that for some n, we know that f n (V c ) f n+1 (V c ) f 1 (V c ) V c. Then f n 1 (V c ) = f 1 (f n (V c )) f 1 (f n+1 (V c )) = f n (V c ). So by induction, the claim is proved. Remark. Remember our construction of V c : we made a disk that satisfied the conditions, and then we made it slightly bigger and open. Therefore we can even conclude that the closure of f 1 (V c ) in C is a subset of V c. So f 1 (V c ) is relatively compact in V c, which we denote by f 1 (V c ) V c. This we will use later on. Now the next two lemmas will tell us that in this case, K c is connected. Lemma 3.6. Let A 1, A 2, A 3,... be compact, non-empty subsets of C, and A 1 A 2 A 3..., that is, A n is a nested sequence of sets. Then n=1 A n is non-empty. Proof. Let (z i ) i N be a sequence of points so that z i lies in A i. Then all of these points lie in A 1, and A 1 is compact. This means that there is a convergent subsequence z i1, z i2,... and we say lim j z ij = z. Since A 1 is closed, and z i1, z i2,... all lie in A 1, we know that z must lie in A 1. More generally we know that, for every n, there must be some j so that all z ij, z i,j+1,... lie in A n. Since all of the A n are closed, this implies that z lies in all of the A n, that is, z n=1 A n. So n=1 A n is non-empty. Lemma 3.7. Let B 1, B 2, B 3,... be connected, compact and non-empty subsets of C, and B 1 B 2 B 3..., that is, B n is a nested sequence of sets. Then B := n=1 B n is connected. Proof. We assume that B is not connected. Then we have two open subsets U and V in C so that 1. B U V 2. B U and B V 3. U V = Since B n is a nested sequence of sets, we have B B n for all n, so B n U and B n V for all n. By assumption, all B n are connected, so U V = implies that B n U V. In other words, A n := B n (U V ) c is non-empty for all n. Now B 1 B 2 B 3... implies that also A n A n 1 A 1, so by lemma 3.6 this means that the intersection of the A n is non-empty. But this intersection is exactly B (U V ) c, which should be empty by the connectedness of B. This is a contradiction, so B is connected. 18

19 By this lemma, K c is connected. We are now close to proving that J c, the boundary of the safe set K c, is connected too. However, it does not follow immediately. We need a stronger property of K c, and the next lemma tells us that it is sufficient if the safe set is simply connected. Lemma 3.8. If A, a subset of C, is simply connected and bounded, then δa is connected. Proof. Let A be a bounded, simply connected set. Assume that δa is not connected. Then we have two open subsets of δa, R 1 and R 2, so that δa = R 1 R 2 and R 1 R 2 is empty. Since R 1 and R 2 are complements of each other in δa, we have that R 1 and R 2 are both closed in δa as well. By definition δa is closed itself, so R 1 and R 2 are closed, bounded sets in C, which means that they are compact. Now take a point x R 1. Since the intersection of R 1 and R 2 is empty, x does not lie in R 2, so we have some ε x > 0 so that B(x, ε x ) R 2 =. This makes {B(x, ε x ) x R 1 } an open covering of R 1. Since R 1 is compact, we can take a finite subcovering {B(x i, ε xi ) i = 1,..., n} of R 1. Now if we take ε = min{ε xi i = 1,...n} we have an ε so that for all x R 1 and all y R 2, d(x, y) > ε. We know that R 1 is bounded since A is bounded. Let us look at a curve γ that has lies on a distance of ε around R 2 1 (or R 2 ) so that γ lies in A completely (note that such a curve exists, because R 1 and R 2 form the boundary of A, and A is connected). This γ is not contractable, since it lies around R 1 (or R 2 ). This contradicts the fact that A is simply connected. So δa is connected. The last thing we need to do now for Julia s connected face, is proving that K c is simply connected indeed. Lemma 3.9. The safe set K c is simply connected. Proof. Let us define the set I c := C \ K c, consisting of all points with an unbounded orbit. We proved in the last section that K c is connected. Furthermore we know that K c is bounded by the forbidden circle V c. If K c is not simply connected, then I c must be disconnected, with some bounded connected component. Let us call this component U. We know that U is open, since I c is the complement of the closed set K c. Remark. K c is closed, since it can be written as the infinite intersection of closed preimages of a closed disk (recall our construction of V c, which started as a closed disk). So the orbits of points in U converge to infinity, and the orbits of points on the boundary of U are bounded. Let z 0 be a point in I c. Then we have some n N so that f n (z 0 ) > R. But f n (w) R for all w K c, and in particular for w δu. This contradicts the maximum principle (see [5]). From this we can conclude that I c does not have a bounded connected component, and K c is simply connected. 19

20 Figure 3.5: Pre-images of the n-th pre-image of Vc. The numbering gives information on the location of the set Ux (the first numbers indicate the smallest pre-image in which Ux is included. The last number indicates if Ux is mapped onto the set in which it is included (1) or into the other set of the pair (2)). Now lemma 3.8 tells us that the Julia set Jc, the boundary of Kc, is connected. This completes Julia s first face, the connected one. 3.4 A disconnected Julia Set In this section we will prove the second equivalence in the dichotomy theorem, the case that c does not lie in the Mandelbrot Set: 0 / Kc Jc is totally disconnected If the orbit of 0 is unbounded, the orbit of f (0) = c is unbounded too, so we have some smallest n so that f n (c) lies outside of Vc. In other words, we know that c does not lie in f n (Vc ), whereas c does lie in f n+1 (Vc ). So this n-th pre-image of the forbidden circle, which we will call U0, is an open subset of C \ {c}. Since c lies in the image of U0 under f, U0 is connected by lemma 3.5, part 1. To U0 we can hence apply lemma 3.5, part 2, which says that f 1 (U0 ) consists of two connected components. These components we will call U1 and U2. These components are both included, and even relatively compact, in U1, as proved by induction in the previous section. When we continue taking pre-images, we include Kc, which will clearly consist of lots of connected components. The numbering of these components is showed in figure 3.5. To see that this will result in a Kc that consists of uncountably many connected components, we consider a point z in Kc. 20

21 As discussed in section 3.1, K c = f n (V c ) n=0 which now means that we have some sequence x i, where x i {1, 2}, so that z lies in all of the sets U 0, U x1, U x1 x 2, U x1 x 2 x 3,... In other words, we can find K c by taking the intersections of the open sets corresponding to all of these possible sequences. So the number of components of K c equals the number of infinite sequences we can make with only 1 s and 2 s. Hence K c consists of uncountably many connected components. We would like to show that any of these intersections consists of exactly one point. We will use the last chapter to prove this last claim. 21

22 4 The pig snout intersection We are very close to completing our proof of the dichotomy theorem. Julia s connected face has become completely clear to us. It is her totally disconnected face that remains somewhat incomplete. What part do we miss? So far we proved that Julia s second face is a disconnected one. We even showed that it consists of uncountably many connected components. We have got some idea of the looks of these components: they can be enclosed by taking infinitely many inverse images of V c. This has given us an image of infinitely many pig snouts, included in each other, each level resembling the level above. In this chapter we will follow one path through an infinite number of pig snouts, and look at the intersection of all of these sets. We will need a surprising amount of sub-results to conclude, in the end, that the intersection consists of only one point indeed. Now let us formulate the theorem that we would like to prove in this chapter: Theorem 4.1. U x1...x n = {p}, k N where x i {1, 2} and {p} is denoting any set of one point. The first lemma of this chapter links a pig snout minus one of the nostrils, for instance U 0 \ U 1, to an annulus. This leads to a definition that gives information about the size ratio of the nostril and the snout. Lemma 4.2. Let U be a simply connected, open, bounded subset of C, and V U. Then we have a conformal map ϕ : U \ V Ann(r 1, r 2 ) := {z C r 1 < z < r 2 } The ratio r 2 r 1 is independent of the choice of ϕ (see [4], theorem 14.22). Proof. This is one of the two lemmas that we will not prove. For a proof, see [1], section 6.5, theorem 10. We define the modulus of U \ V, using the above theorem. Definition 4.3. Mod (U \ V ) = 1 2π log r 1 r 2 Why do we want to use this definition? It is our goal to prove that U consists of one point. The trick we use is looking at U 0 \U, and using the modulus of this set to conclude that U must consist of one point. In order to do this, we will use a lemma that tells us how to make computations with moduli. 22

23 Lemma 4.4. Let U, V, W be simply connected, open and bounded subsets of C, and W V U. Then Mod (U \ V ) + Mod (V \ W ) Mod (U \ W ) Proof. This is (a special case of) Grötsch inequality. See [2], proposition 9. We would like to apply this to U 0 \ U. Using the above lemma, we deduce: So Mod (U 0 \ U 1 ) + Mod (U 1 \ U 11 ) Mod (U 0 \ U 11 ) Mod (U 0 \ U 1 ) + Mod (U 1 \ U 11 ) + Mod (U 11 \ U 111 ) Mod (U 0 \ U 11 ) + Mod (U 11 \ U 111 ) Mod (U 0 \ U 111 ) and continuing this, we find Mod (U 0 \ U 1 ) + Mod (U 1 \ U 11 ) + Mod (U 11 \ U 111 ) + Mod (U 0 \ U). We are interested in Mod (U 0 \ U), and therefore we would like to know the limit of the sum of the sequence: Mod (U 0 \ U 1 ), Mod (U 1 \ U 11 ), Mod (U 11 \ U 111 ),... Now let ϕ be a conformal mapping between U 0 \ U 1 and an annulus, Ann(r 1, r 2 ). As defined, Mod (U 0 \ U 1 ) = 1 log r 2 2π r 1. We know that U 11 is a pre-image of U 1 under f c, just as U 1 is a pre-image of U 0. So U 11 \ U 1 can be mapped onto Ann(r 1, r 2 ) by ϕ f c. Note that this composition is a conformal mapping as well, so we can conclude that Mod (U 1 \ U 11 ) = 1 log r 2 2π r 1. In an analogous way, using ϕ fc 2, we see that Mod (U 11 \ U 111 ) = 1 log r 2 2π r 1. Generalizing this, denoting a sequence of n ones by x n, we get: Mod(U xn, U xn+1 ) = Mod(U xn+1, U xn+2 ) = Mod(U 0 \ U 1 ) for all n N On an intuitive level, this means that, when zooming in, we will see the same pig snout on every level. We can conclude that every term in our sequence has the same value. Since U xn+1 U xn for all n N, the ratio r 2 r 1 is strictly positive, and therefore the modulus of every set U xn \ U xn+1 is strictly positive. So we have a sequence of infinitely many positive terms, all equal, which means that the sum converges to infinity. So Mod (U 0 \ U) must be infinite as well. Remark. At this point, we can justify our choice to look at the intersection U 1 U 11 U instead of considering all components of the intersection n=0 f n (V c ). Recall that every component of the intersection can be written as k N U x 1 x 2...x k, where x i 23

24 {1, 2}. Note that U x1 x 2...x k will be mapped onto U 0 after k iterations. After the same number of iterations, U x1 x 2...x k+1 will be mapped either onto U 1 or onto U 2. In the same way as above, we can compose the conformal mapping ϕ i, from U 0 \ U i to an annulus, with fc k (i is either 1 or 2). The family of conformal mappings ϕ i fc k maps each pig snout (minus one nostril) onto one of two possible annuli. So since the choice of a nested sequence of sets always results in an infinite sequence of moduli that can have two possible values (that is, a sequence with a sum converging to infinity), it is without loss of generality that we use the intersection given by a sequence of only ones. So we proved: Mod (U 1 \ U) = What does this mean for U? The next lemma will enable us to conclude that U is a set of 1 point. Lemma 4.5. Let U, V be simply connected, open, bounded subsets of C, and V U. Then Mod (U \ V ) = = V = {p}. Proof. If Mod (U \ V ) =, then r 1 = 0. This means that Ann (r 1, r 2 ) is isomorphic to a punctured disk D. So we can reformulate our conformal mapping ϕ as ϕ : U \ V D Since ϕ is a conformal mapping, we have an inverse function ϕ 1 from D to U \ V. This inverse function is not defined in 0. We recall that this singularity can be of three different types ([5], definition 7.5.1): An essential singularity A pole A removable singularity Since U \ V is bounded, 0 must be a removable singularity ([5], theorem 7.5.2). That is, ϕ 1 is holomorphically extendable to D, the non-punctured disk. D is simply connected, so the image of D under ϕ 1 (the holomorphic extension of ϕ 1 ) is simply connected too. So the point 0, that we added to the domain, must be mapped onto V. This is the only way to add a point to U \ V so that the resulting set is simply connected. So V consists of one point, and V must consist of one point. Since we showed that Mod (U 0 \ U) =, we can conclude that U = U 1 U 11 U consists of one point. As explained in the above remark, we can conclude hereby that every intersection k N U x 1 x 2...x k, where x i {1, 2}, consists of exactly one point. So K c = n=0 f n (V c ) consists of uncountably many separated points. Now J c must equal K c. This completes the last part of Julia s second face. 24

25 Popular Summary Imagine a planet called Origo, lying in the origin of the complex plane C. The inhabitants of Origo leave a satellite in space, that is supposed to circle around Origo. The satellite is placed on a distance 1 from Origo and starts to move. Some clever mathematicians living on Origo, understand that the movement of this satellite can be described by the iteration of a simple function that they call f 0 : f 0 : C C given by z z 2 In other words, if the satellite starts in point z, it will be in the point z 2 after one unit of time. After another unit of time, the satellite will be in the point (z 2 ) 2 = z 4. What does this look like? Remember that squaring in the complex plane geometrically means that the angle is doubled, and the modulus is squared. When the satellite starts in a position on a distance of 1, the modulus will not change, and the angle will be doubled in every time step. The satellite will move on the circle around the origin of radius one. See figure 4.1. Now imagine a cosmic wind blowing around Origo, that will blow the satellite a little out of place. This can be represented by adding a complex-valued constant c to our function, the value of c depending on the strength and the direction of the cosmic wind. f c : C C given by z z 2 + c For example, when we choose c = 1, this is a wind blowing from the west, shifting the satellite 1 step to the east in every time step. The choice c = i is a southern wind with the same strength, and c = 1 + i represents a somewhat stronger wind that is directed to the north-east. See figure 4.2. The inhabitants of Origo are worried, and decide to investigate what happens to their satellite on the long run. Of course, this depends on the strength and direction of the wind and on the starting position of the satellite. For every wind type c, the mathematicians make a map that shows which starting positions are safe (see figure 4.3): if the satellite starts in these positions, it will not be lost in space. It strikes the people of Origo that some of the maps show a connected area of safe starting positions (like the maps in figure 4.3), whereas other maps rather consist of a set of loose points. The mathematicians make one other map (figure 4.4), which shows the wind types, that is, the values of c, for which the area of safe starting positions is 25

26 connected. The set of points in figure 4.4, is a famous set in complex dynamical systems: the Mandelbrot set, written as M. If a point c belongs to the Mandelbrot set, the area of safe starting positions corresponding to wind type c, is connected. In this thesis, we will focus on the boundary of this safe area. This boundary is called the Julia set, written as J c. The conclusions of the inhabitants of Origo, can be extended as follows to a theorem: 1. When 0 belongs to the safe set, J c is connected and c belongs to the Mandelbrot set. 2. When 0 does not lie in the safe set, J c is disconnected and consists of infinitely many points, and c does not belong to the Mandelbrot set In this thesis, this set of claims, called the dichotomy theorem, is proved. 26

27 Figure 4.1: Origo, a planet in the origin of the complex plane, with a satellite circling around it. Figure 4.2: Different types of wind, represented by the complex-valued constant c. 27

28 Figure 4.3: Two maps of safe starting positions Figure 4.4: Safe types of cosmic winds 28

29 Bibliography [1] L. Ahlfors, Complex Analysis: an Introduction to the Theory of Analytic Functions of One Complex Variable, edition 2, 1966 [2] B. Branner, Puzzles and Para-puzzles of Quadratic and Cubic Polynomials, from: Complex Dynamical Systems: the Mathematics Behind the Mandelbrot and Julia Sets, 1994 [3] B.J.J. Moonen, Topologie, syllabus, 2008 [4] W. Rudin, Real and Complex Analysis, edition 3, 2007 [5] J. Wiegerinck and P.J.I.M de Paepe, Analyse 3: functietheorie, syllabus,