Plane Topology and Dynamical Systems Boris Kolev

To cite this version:

Boris Kolev. Plane Topology and Dynamical Systems. Ecole´ th´ematique. Summer School ”Syst`emesDynamiques et Topologie en Petites Dimensions”, Grenoble, France, 1994, pp.35.

HAL Id: cel-00719540 https://cel.archives-ouvertes.fr/cel-00719540 Submitted on 20 Jul 2012

HAL is a multi-disciplinary open access L’archive ouverte pluridisciplinaire HAL, est archive for the deposit and dissemination of sci- destinéeau dépôtet àla diffusion de documents entific research documents, whether they are pub- scientifiques de niveau recherche, publiésou non, lished or not. The documents may come from émanant des établissements d’enseignement et de teaching and research institutions in France or recherche fran¸caisou étrangers,des laboratoires abroad, or from public or private research centers. publics ou privés. Plane Topology and Dynamical Systems

Boris KOLEV

CNRS & Aix-Marseille University

Grenoble, France, June-July 1994

Summary. — These notes have been written for a Summer School, Syst`emes Dynamiques et Topologie en Petites Dimensions, which took place at the Institut Fourier, in June-July 1994. The goal was to provide simple proofs for the Jordan and Schoenﬂies theorems and to give a short introduction to the theory of locally connected continua and indecomposable continua, with applications in Dynamical Systems and the theory of attractors.

E-mail : [email protected] Homepage : http://www.cmi.univ-mrs.fr/~kolev/

○c This work is licensed under a Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported License. Chapter 1

The Jordan Curve Theorem

A homeomorphic image of a closed interval [푎, 푏](푎 < 푏) is called an arc and a homeomorphic image of a circle is called a simple closed curve or a Jordan curve. To begin with, we recall ﬁrst two simple facts about the plane. 2 2 1. If 퐹 is a closed set in R , any component of R − 퐹 is open and arcwise connected. We will call these components, the complementary domains of 퐹 .

2 2 2. If 퐾 is a compact set in the plane R , then R −퐾 has exactly one unbounded component. We will refer to it as the unbounded or exterior component of 퐾. 2 Assertion (1) follows from the local arcwise-connectedness of R and (2) from the boundedness of 퐾. 2 Theorem 1.1 (Jordan Curve Theorem). The complement in the plane R of a simple closed curve 퐽 consists of two components, each of which has 퐽 as its boundary. Furthermore, if 퐽 has complementary domains 푖푛푡(퐽) (the bounded, or interior domain) and 푒푥푡(퐽) (the unbounded, or exterior domain), then 퐼푛푑(푥, 퐽) = 0 if 푥 ∈ 푒푥푡(퐽) and 퐼푛푑(푥, 퐽) = +1 if 푥 ∈ 푖푛푡(퐽). Remark 1.2. Obviously, it follows from this statement that a simple closed curve divides the 2-sphere into exactly two domains, each of which it is the common boundary. The proof we give here is due to Maehara [13] and uses as main ingredient the following well-known theorem of Brouwer. Theorem 1.3 (Brouwer’s Fixed Point Theorem). Every continuous map of the closed unit disc 퐷2 into itself has a fixed point. Proof. We identify here the plane with the complex plane and let 2 퐷 = {푧 ∈ C ; |푧| ≤ 1} . Suppose that 푓 : 퐷2 → 퐷2 has no ﬁxed point. Then 푓(푧) ̸= 푧 for all 푧 ∈ 퐷2 and the degree1 of each map 1 푓푡(푧) = 푡푧 − 푓(푡푧), 푧 ∈ 푆 , 푡 ∈ [0, 1], is well deﬁned and all of them are equal. Since 푓0 is a constant map we have 푑(푓0) = 0 and therefore 푑(푓푡) = 0 for all 푡 ∈ [0, 1]. But since |푧 − 푓(푧)| is strictly positive on the compact set 푆1, it is bounded below by a positive constant 푚 and we get after an easy computation (︀ 푚2 푅푒 (푧 − 푓(푧))¯푧 > > 0, 2 1 for all 푧 ∈ 푆 so that 푑(푓1) = 푑(퐼푑푆1 ) = 1 which leads to a contradiction.

1 Each continuous map 푓 of the circle with values in C lift to a map 푓˜ : R → C (angle determination) which satisfies 푓˜(휃 + 2휋) = 푓˜(휃) + 2푑휋. The integer 푑 is called the degree of the map 푓. It is easily shown to not depend on the particular lift of 푓 and to be a homotopy invariant.

1 Definition 1.4. Let 푋 be any metric space and 퐴 a subspace of 푋. A continuous map 푟 : 푋 → 퐴 such that 푟 = 퐼푑 on 퐴 is called a retraction of the space 푋 on 퐴.

As a corollary of theorem 1.3 (these two statements are in fact equivalent) we have the following.

Theorem 1.5 (No-Retraction Theorem). There is no retraction of the unit disc 퐷2 onto its boundary 푆1.

Proof. Suppose there exists a continuous map 푟 : 퐷2 → 푆1 such that 푟(푧) = 푧 for all 푧 ∈ 푆1 and let 푠 : 푆1 → 푆1 deﬁned by 푠(푧) = −푧. Then the map 푠 ∘ 푟 : 퐷2 → 푆1 ⊂ 퐷2 will be a continuous map without ﬁxed points.

Let 퐸 denote the square [−1, 1] × [−1, 1] and Γ = 퐹 푟(퐸) its boundary. A path 훾 in 퐸 is a continuous map 훾 :[−1, 1] → 퐸.

Lemma 1.6. Let 훾1 and 훾2 be two paths in 퐸 such that 훾1 (resp. 훾2) joins the two opposite vertical (resp. horizontal) sides of 퐸. Then 훾1 and 훾2 have a common point.

Proof. Let 훾1(푠) = (푥1(푠), 푦1(푠)) and 훾2(푠) = (푥2(푠), 푦2(푠)) so that

푥1(−1) = −1, 푥1(1) = 1, 푦2(−1) = −1, 푦2(1) = 1.

If the two paths do not cross the function

푁(푠, 푡) = max (|푥1(푠) − 푥2(푡)| , |푦1(푠) − 푦2(푡)|) is strictly positive and the continuous map 푓 : 퐸 → Γ ⊂ 퐸 defined by (︂ )︂ 푥 (푡) − 푥 (푠) 푦 (푠) − 푦 (푡) 푓(푠, 푡) = 2 1 , 1 2 푁(푠, 푡) 푁(푠, 푡) can easily be checked to have no fixed point which contradicts the Brouwer fixed point theorem (see Exercise 1.1).

2 2 Definition 1.7. A closed set 퐹 separates the plane R if R − 퐹 has at least two components. Lemma 1.8. No arc 훼 separates the plane.

2 2 Proof. Suppose on the contrary that R − 훼 is not connected. Then R − 훼 has in addition to its unbounded component 푈∞ at least one bounded component 푊 . We have 퐹 푟(푈∞) ⊂ 훼 and 퐹 푟(푊 ) ⊂ 훼. Let 푥0 ∈ 푊 and 퐷 be a closed disc with centre 푥0 and radius 푅 which contains 푊 as well as a in its interior. By a straightforward application of the Tietze extension theorem (see Exercise 1.2) the identity map 퐼푑훼 on 훼 extends continuously to a retraction 푟 : 퐷 → 훼. Let us deﬁne ⌉︀ 푟(푥), if 푥 ∈ 푊¯ ; 푞(푥) = 푥, if 푥 ∈ 푊 푐 ∩ 퐷.

Then, 푞 : 퐷 → 퐷 is a continuous map whose values lie in 퐷 − {푥0}. Hence, the composed map 푝 ∘ 푞 : 퐷 → 퐹 푟(퐷) where 푞(푥) − 푥 푝(푥) = 푅 0 ‖푞(푥) − 푥0‖ is a retraction of the disc 퐷 onto its boundary, contradicting the no-retraction theorem (theorem 1.5).

Remark 1.9. For further use, note that using the same arguments we can show that no 2-cell (that is the homeomorphic image of the unit square 퐼2) separates the plane.

2 Proof of Jordan’s Theorem. We will divide the proof in three steps. First we will show that 퐽 separates the plane. Then we will prove that 퐽 is the boundary of each of its components and finally we will prove that the complement of 퐽 has exactly two components 푖푛푡(퐽) and 푒푥푡(퐽) such that 퐼푛푑(푥, 퐽) = 0 if 푥 ∈ 푒푥푡(퐽) and 퐼푛푑(푥, 퐽) = ±1 if 푥 ∈ 푖푛푡(퐽). Step 1 : Since 퐽 is compact, there exist two points 푎, 푏 in 퐽 such that 푎 − 푏 = diam(퐽). We may assume that 푎 = (−1, 0) and 푏 = (1, 0). Then the rectangular set 퐸(−1, 1; −2, 2) contains 퐽, and its boundary 퐹 meets 퐽 in exactly two points 푎 and 푏. Let 푛 be the middle point of the top side of 퐸, and 푠 the middle point of the bottom side. The segment 푛푠 meets 퐽 by lemma 1.6. Let 푙 be the 푦-maximal point in 퐽 ∩ 푛푠. Points 푎 and 푏 divide 퐽 into two arcs: we denote the one containing 푙 by 퐽푛 and the other by 퐽푠. Let 푚 be the 푦-minimal point in 퐽푛 ∩ 푛푠 then the segment 푚푠 meets 퐽푠; otherwise, the path 푛푙 + 푙푚 + 푚푠, (where 푙푚 denotes the subarc of 퐽푛 with end points 푙 and 푚) could not meet 퐽푠, contradicting lemma 1.6. Let 푝 and 푞 denote the 푦-maximal point and the 푦-minimal point in 퐽푠 ∩ 푚푠, respectively. Finally, let 푥0 be the middle point of the segment 푚푝 (see Figure 1.1). The choice of a homeomorphism ℎ : 푆1 → 퐽 permits us to define the index of a point with respect to 퐽 (it is well defined up to a sign). It is easy to see that 퐼푛푑(푥, 퐽) = 0 for any point in the unbounded complementary domain 푒푥푡(퐽) of 퐽. For fixing our ideas, we suppose now that we have oriented the curve 퐽 (resp. 퐹 ) in such a way that we cross successively 푎, 푝 and 푏 (resp. 푎, 푠 and 푏). Let Γ푛 (resp. Γ푠) be the subarc of 퐹 delimited by 푎 and 푏 and containing 푛 (resp. 푠). We set 휎푛 = Γ푛 − 퐽푛 (meaning we follow Γ푛, and then 퐽푛 with the reversed orientation) and 휎푠 = −퐽푠 + Γ푠. Since the half line 푥0푠 does not meet 휎푛, we have 퐼푛푑(푥0, 휎푛) = 0. By a similar argument 퐼푛푑(푥0, 휎푠) = 0 and since

Γ = 휎푛 + 퐽 + 휎푠, we have 퐼푛푑(푥0, 퐽) = 퐼푛푑(푥0, 휎푛) + 퐼푛푑(푥0, 퐽) + 퐼푛푑(푥0, 휎푠) = 퐼푛푑(푥0, Γ) = 1.

Therefore, 푥0 ̸∈ 푒푥푡(퐽) and 퐽 separates the plane. Step 2 : Let 푈 be any complementary domain of 퐽 and 푥 a point in 푈. Note that 퐹 푟(푈) ⊂ 퐽. Therefore, if 퐹 푟(푈) ⊂ 퐽, there is an arc 훼 which contains entirely 퐹 푟(푈). Because 퐽 separates 2 the plane, there is a point 푦 ∈ R − 퐽 such that 푥 and 푦 are separated by 퐹 푟(푈) hence by 훼 which contradicts lemma 1.8. Thus 퐹 푟(푈) = 퐽.

Step 3 : Let 푈 be the component of the point 푥0 deﬁned in (1). Suppose that there exists 2 another bounded component 푊 (̸= 푈) of R − 퐽. Clearly 푊 ⊂ 퐸. We denote by 훽 the path 푛푙 + 푙푚 + 푚푝 + 푝푞 + 푞푠, where 푝푞 is the subarc of 퐽푠, from 푝 to 푞. Hence, 훽 has no point of 푊 . Since 푎 and 푏 are not on 훽, there are circular neighborhoods 푉푎 and 푉푏, of 푎 and 푏, respectively, such that each of them contains no point of 훽. But 퐹 푟(푊 ) = 퐽, so there exist 푎1 ∈ 푊 ∩ 푉푎 and 푏1 ∈ 푊 ∩ 푉푏. Let 푎1푏1 be a path in 푊 from 푎푙 to 푏1. Then the path 푎푎1 + 푎1푏1 + 푏푏푙 fails to meet 훽. This contradicts lemma 1.6 and completes the proof.

Remark 1.10. Let 퐽 be a simple closed curve in the (oriented) plane. Then 퐼푛푑(푥, 퐽) = 1 for all points in 푖푛푡(퐽) or 퐼푛푑(푥, 퐽) = −1 for all points in 푖푛푡(퐽). In the ﬁrst case we will say that 퐽 is positively oriented otherwise it is negatively oriented. From now on, a subset 퐷 of the plane will be called a disc if it is the bounded complementary domain of a simple closed curve 퐽. 푈 being any connected open set of the plane, a cross-cut in 푈 is a simple arc 퐿 ⊂ 푈 which intersects 퐹 푟(푈) in exactly its two endpoints. If just one of the endpoints of 퐿 meets 퐹 푟(푈), then 퐿 is called an end-cut. A point 푎 of 퐹 푟(푈) which is the endpoint of an end-cut in 푈 is called accessible from 푈. We will prove later that every point of a simple closed curve is accessible from both of its complementary domains.

3 Figure 1.1: A Jordan curve

Lemma 1.11 (휃-curve Lemma). Let 퐽 be a simple closed curve in the plane. A cross-cut 퐿 in 푖푛푡(퐽) divides 푖푛푡(퐽) into exactly two domains. If 퐿1 and 퐿2 are the subarcs of 퐽 defined by the endpoints 푎 and 푏 of 퐿, these two domains are the discs 푖푛푡(퐿 ∪ 퐿1) and 푖푛푡(퐿 ∪ 퐿2).

Proof of 휃-curve Lemma. If 푋 denotes one of the sets 푒푥푡(퐽), 푖푛푡(퐿 ∪ 퐿1) or 푖푛푡(퐿 ∪ 퐿2) then 퐹 푟(푋) ⊂ 퐽 ∪ 퐿. Hence, (︀ 2 2 푋 ∩ R − (퐽 ∪ 퐿) = 푋 ∩ (R − (퐽 ∪ 퐿)).

2 2 Therefore, 푋 is open and closed in R − (퐽 ∪ 퐿) and is a component of R − (퐽 ∪ 퐿). We are going to show that

2 R − (퐽 ∪ 퐿) = 푒푥푡(퐽) ∪ 푖푛푡(퐿 ∪ 퐿1) ∪ 푖푛푡(퐿 ∪ 퐿2).

Suppose on the contrary that there exists another component 푈 distinct from the three above. ∘ ∘ ∘ ∘ Then 퐹 푟(푈) must meet each of the three (open) arcs 퐿 , 퐿1, 퐿2. Indeed, if 퐹 푟(푈 ∩ 퐿 ) = ∅, 2 2 2 then 퐹 푟(푈) ⊂ 퐽 and 푈 ∩ (R − 퐽) = 푈 ∩ (R − 퐽). Hence 푈 is open and closed in R − 퐽 and ∘ ∘ 푈 = 푖푛푡(퐽) ⊃ 퐿 , which gives a contradiction. Similarly, if 퐹 푟(푈) ∩ 퐿1 = ∅, then 푈 is open and 2 ∘ closed in R − (퐿 ∪ 퐿2) and thus 푈 = 푒푥푡(퐿 ∪ 퐿2) ⊃ 퐿1 which also gives a contradiction. ∘ ∘ Therefore, we can construct a cross-cut 휇 in 푈 with endpoints 푢1 ∈ 퐿1 and 푢2 ∈ 퐿2. We can also construct a cross-cut 휈 in 푒푥푡(퐽) with endpoints 푣1 and 푣2, where 푣1 and 푣2 belong to two non adjacent arcs among those of 퐽 − {푎, 푢1, 푏, 푢2}. Let 퐾 denote the simple closed curve

퐾 = 휇 ∪ 푢1푣1 ∪ 휈 ∪ 푢2푣2

∘ ∘ where 푢1푣1 ⊂ 퐿1 and 푢2푣2 ⊂ 퐿2. We are therefore in the following situation (see Figure 1.2). 2 1. 퐾 ∩ 퐿 = ∅ (since 휇 ⊂ 푈 ⊂ R − 퐿). 2. Each complementary domain of 퐾 meets 푖푛푡(퐽) and 푒푥푡(퐽) and hence 퐽 (since 휇∘ ⊂ 푖푛푡(퐽) and 휈∘ ⊂ 푒푥푡(퐽)).

4 3. The two arcs of 퐽 −퐾 are connected by 퐿 and are therefore in one complementary domain 푅 of 퐾.

4. 퐹 푟(푅) = 퐾 ⊃ 푢푖푣푖 (푖 = 1, 2). It follows from assertions (3) and (4) that 퐽 ⊂ 푅 which is a contradiction with assertion (2) (each complementary domain of 퐾 meets 퐽).

Figure 1.2: A 휃-curve

2 Theorem 1.12 (Invariance of Domain). Let 푈 be an open set in the plane and 푓 : 푈 → R be 2 a one-to-one and continuous map. Then 푓(푈) is an open set of R . 2 Remark 1.13. This theorem is not to be confused with the proposition that if R is mapped by a homeomorphism onto itself, open sets are mapped onto open sets, which follows immediately from the deﬁnition of a homeomorphism. In this theorem, it is not assumed that the map 푓 is deﬁned outside 푈, nor that there exists a homeomorphism of the whole space on to itself coinciding with the given one in 푈. The proof of this theorem is an immediate application of the following Lemma. As usual 퐷2 is the closed unit disc and 푆1 its boundary.

2 2 2 Lemma 1.14. Let 푓 : 퐷 → R be a continuous and one-to-one map. Then 푓(푖푛푡(퐷 )) = 푖푛푡(푓(푆1)).

Proof. Let 퐴 = 푓(퐷2) and 퐽 = 푓(푆1). Then 퐴 is a closed 2-cell and 퐽 is a simple closed curve. 푓(푖푛푡(퐷2)) is a connected set which does not meet 퐽 and is therefore contained in a 2 2 complementary domain 푅 of 퐽. If 푓(푖푛푡(퐷 )) ̸= 푅, there are some points of R − 퐴 in both 2 2 components of R − 퐽 and hence on 퐽 since R − 퐴 is connected (see remark 1.9). But this is impossible since 퐽 ⊂ 퐴. Therefore, 푓(푖푛푡(퐷2)) = 푅 = 푖푛푡(퐽) (since 푓(푖푛푡(퐷2)) is bounded).

The following corollaries of theorem 1.12 are straightforward and their proof will be omitted.

2 Corollary 1.15. Let 푈 an open subset of the plane and 푓 : 푈 → R be a continuous and one-to-one map. Then 푓 is a homeomorphism of 푈 onto the open set 푓(푈).

Corollary 1.16. Let 푓 be any homeomorphism between two plane sets 퐴 and 퐵. Then 푓(푖푛푡(퐴)) = 푖푛푡(퐵). Hence, if 퐴 and 퐵 are closed set 푓(퐹 푟(퐴)) = 퐹 푟(퐵).

The next application of the Jordan Curve Theorem is due to Kerékjártó[10, 16].

5 Figure 1.3: The intersection of two discs

Lemma 1.17 (Kerékjártó). Each component of the intersection of two (open) discs 퐷1 and 퐷2 is a disc.

Proof. If 퐷1 ∩ 퐷2 = ∅, there is nothing to prove. Otherwise, let 푅 be a component of 퐷1 ∩ 퐷2 and set 퐽푖 = 퐹 푟(퐷푖) for 푖 = 1, 2. It is clear that 퐹 푟(푅) ⊂ 퐽1 ∪ 퐽2. If 퐹 푟(푅) ⊂ 퐽푖 for some 푖, 2 then 푅 is open and closed in R − 퐽푖 hence 푅 = 퐷푖 and the Lemma is proved. Thus, we may assume that 퐹 푟(푅) ̸⊂ 퐽푖 for 푖 = 1, 2. Let 푥 ∈ 퐹 푟(푅), 푥 ̸∈ 퐽2. Then 푥 ∈ 퐽푙 ∩ 퐷2, and we can ﬁnd ∘ an arc 훾 in 퐽1 such that 푥 ∈ 훾 and

∘ 훾 ⊂ 퐷2, 훾 ⊂ 퐹 푟(푅), 휕훾 ⊂ 퐽1 ∩ 퐽2.

2 퐽2 ∪ 훾 is a 휃-curve and 푅 belongs to one of the complementary domains of R − 휃. Thus, the endpoints of 훾 deﬁne an arc 훿 on 퐽2 which doesn’t meet 푅 and such that 훿 ∩ 퐹 푟(푅) = 휕훿 (see Figure 1.3). There is at most a countable family of such arcs 훾푛 and diam(훾푛) → 0 as 푛 → ∞ (see ∘ Exercise 2.1). The corresponding arcs 훿푛 are pairwise disjoint (again an application of the 휃- curve lemma) and therefore diam(훿푛) → 0 as 푛 → ∞ as well. Let 퐽 be the simple closed curve obtained from 퐽2 by substituting to each arc 훿푛 the corresponding arc 훾푛 of 퐽1. It is then easy to verify that 퐹 푟(푅) ⊂ 퐽 and therefore 퐹 푟(푅) = 퐽 since 퐹 푟(푅) separates the plane but no proper closed subset of 퐽 does. Thus 퐹 푟(푅) is a simple closed curve and 푅 is a disc.

Exercises

Exercise 1.1. Let 퐾 ̸= ∅ be a compact and convex set in the plane and 푓 : 퐾 → 퐾 a continuous map. Deduce from Brouwer’s Fixed Point Theorem that 푓 has at least one ﬁxed point.

Exercise 1.2 (Tietze’s Extension Theorem). Let 퐹 be a non empty closed set of a metric space (푋, 푑) and 푓 : 퐹 → 퐼 = [0, 1] be a continuous map. Deﬁne 푓¯ : 푋 → 퐼 by 푓¯(푥) = 푓(푥), if 푥 ∈ 퐹 and 푑(푥, 퐹 ) 푓¯(푥) = sup 푓(푦) 푦∈퐹 푑(푥, 푦) if 푥 ∈ 푋 − 퐹 . Prove that 푓 is continuous. Extend the result when 푓 : 퐹 → 퐼푛,(푛 ≥ 1).

Exercise 1.3. Let 푈 be a connected open set in the plane. Prove that the accessible points of the boundary of 푈 are dense in 퐹 푟(푈).

6 Exercise 1.4. There is no homeomorphism from the 2-sphere onto one of its proper subset.

2 푛 Exercise 1.5. If 푛 ̸= 2, R and R are not homeomorphic. Exercise 1.6. Let 퐾 be a compact set and 퐹 a closed set in the plane with 퐾 ∩ 퐹 = ∅. Let 2 푎 ∈ 퐾 and 푏 ∈ 퐹 . Show that for any 휀 > 0, there exists a simple closed curve 퐽 in R − (퐾 ∪ 퐹 ) which separates 푎 and 푏 and such that 퐽 ⊂ 푉휀(퐾). Exercise 1.7. Deduce from Exercise 1.6 the following statements. The boundary of every bounded simply connected domain in the plane is connected. The boundary of each complementary domain of a compact subset of the plane is connected.

7 Chapter 2

The Schoenflies Theorem

The aim of this section is to prove the famous theorem of Schoenﬂies which asserts that any homeomorphism of the unit circle 푆1 onto a simple closed curve 퐽 can be extended to the whole plane. To do this, we now introduce an auxiliary notion which generalizes the concept of 휃-curve.

Definition 2.1. The sum 퐿0 ∪ 퐿1 ∪ · · · ∪ 퐿푛, of 푛 + 1 (푛 ≥ 1) arcs in the plane is called a network when:

1. 퐿0 ∪ 퐿1 is a simple closed curve,

2. 퐿푘 ∩ (퐿0 ∪ 퐿1 ∪ · · · ∪ 퐿푘−1) consists of just the endpoints of 퐿푘, for 푘 = 1, 2, . . . , 푛.

The network Γ푛 = 퐿0 ∪ 퐿1 ∪ · · · ∪ 퐿푛, extends Γ푚 = 퐿0 ∪ 퐿1 ∪ · · · ∪ 퐿푚 if 푚 < 푛 and we write then Γ푚 ≺ Γ푛.

2 Lemma 2.2. The set R − (퐿0 ∪ 퐿1 ∪ · · · ∪ 퐿푛) has exactly n distinct bounded components, each of them is a disc whose boundary lies in 퐿0 ∪ 퐿1 ∪ · · · ∪ 퐿푛. Proof. For 푛 = 1 it is just Jordan’s Curve Theorem and for 푛 = 2, the 휃-curve Lemma. So, let us suppose that the Lemma is true for some 푛 > 2 and let 퐷1, 퐷2, . . . , 퐷푛, be the bounded components of 퐷−(퐿0 ∪퐿1 ∪· · ·∪퐿푛). Let 퐿푛+1 be an arc such that 퐿푛+1 ∩(퐿0 ∪퐿1 ∪· · ·∪퐿푛) = ∘ 휕퐿푛+1. Without loss of generality we can suppose that 퐿푛+1 lies in 퐷푛. By the 휃-curve Lemma, + − ± we know that 퐿푛+1 divides 퐷푛 into exactly two discs 퐷푛 and 퐷푛 and 퐹 푟(퐷푛 ) ⊂ 퐹 푟(퐷푛)∪퐿푛+1, which proves the theorem.

Let 퐽 = 퐿0 ∪퐿1 be a simple closed curve which bounds a disc 퐷 and Γ = 퐿0 ∪퐿1 ∪· · ·∪퐿푛, an ¯ extension of 퐽 such that 퐿푘 ⊂ 퐷 for 푘 = 0, . . . , 푛. Then⋃︀ each bounded complementary domain ¯ 푛 ¯ ⌋︀퐷푘 (푘 = 1, .{︀ . . , 푛) of Γ is a subset of 퐷 and 퐷 = 푘=1 퐷푘. We will call the ﬁnite collection 퐷¯1,..., 퐷¯푛 a subdivision of 퐷¯ and we will say that Γ subdivides 퐷¯. The number

푚(Γ) = max {diam(퐷푘); 푘 = 1, . . . , 푛} will be called the mesh of Γ.

Lemma 2.3. Let 퐽 be a simple closed curve bounding a disc 퐷 and 휀 > 0. Then there exists a subdivision of 퐷¯ by a network Γ of mesh less than 휀.

Proof. Consider the family 풱 of vertical lines

푉푘 = {(푥, 푦); 푥 = 푘휀/2} , (푘 ∈ Z).

Only a ﬁnite number of these lines meet 퐷 (since 퐷 is bounded). If 푉푘 meets 퐷, then 푉푘 ∩ 퐷 is 푘 푘 the union of an at most countable family of pairwise disjoint segments (퐼푛) with diam(퐼푛) → 0 as 푛 → +∞.

8 Let 훿 > 0 be such that any two points of 퐽 with distance less than 훿 determine an arc on 퐽 푘 of diameter less than 휀/4 (see Exercise 2.1). Denote 퐿2, . . . , 퐿푝 those of the segments (퐼푛) such 푘 that diam(퐼푛) ≥ 훿 and let Γ1 = 퐽 ∪ 퐿2 ∪ · · · ∪ 퐿푝. Then each of the bounded complementary domains of Γ1 is contained in the union of at most 3 adjacent vertical strips. Indeed, suppose on the contrary that there exists a bounded complementary domain of Γ1 say 퐷1 which meets three consecutive vertical lines 푉푘−1, 푉푘, 푉푘+1. Let 푎 (resp. 푏) a point of 푉푘−1 ∩퐷1 (resp. 푉푘+1 ∩퐷1). Choose a polygonal arc 훾 (with no vertical segments !) which joins 푎 and 푏 in 퐷1. This arc meets 푉푘 in a ﬁnite number of points which belong to some segments 푘 푘 퐼푛1 , . . . , 퐼푛푟 deﬁned above. It is then easy to show by induction on 푟 that at least one of these 푘 segments say 퐼푛1 separates the two points 푎 and 푏 in 퐷. But 푑(푎, 푉푘) = 푑(푏, 푉푘) = 휀/2 so that

푘 length(퐼푛1 ) ≥ 훿, which leads to a contradiction. Next, we do the same construction with the family ℋ of horizontal lines 퐻푘 : 푦 = 푘.휀/2 (푘 ∈ Z) and we extends Γ1 into a network Γ by adding horizontal segments 퐿푝+1, . . . , 퐿푞 such that each bounded complementary domain of Γ lies in the union of 3 horizontal strips. By this way we have constructed a network Γ which subdivides 퐷¯, of mesh less than 3휀.

Definition 2.4. A metric space 푋 is uniformly locally connected provided that for any 휀 > 0 there exists 훿 > 0 so that any two points of 푋 with distance 푑(푥, 푦) < 훿 are contained in a connected subset of diameter less than 휀.

Corollary 2.5. The interior of any simple closed is uniformly locally connected.

Proof. Let 퐽 be a simple closed curve, 퐷 its bounded complementary domain and 휀 > 0 (we can suppose 2휀 < diam(퐽)). By the above Lemma, we can ﬁnd a network Γ of mesh < 휀 with subdivides 퐷¯. Since 2휀 < diam(퐽) there exists 푖, 푗 such that 퐷¯푖 ∩ 퐷¯푗 = ∅ (where 퐷1, . . . , 퐷푛 are the bounded complementary domains of Γ). Therefore, ⌋︀ {︀ 훿 = min 푑(퐷¯푖, 퐷¯푗); 퐷¯푖 ∩ 퐷¯푗 = ∅ > 0.

It is then easy to verify that two points of 퐷 with distance less than 훿 can be joined in 퐷 by an arc of diameter less than 2휀. Indeed, if 퐷¯푖 and 퐷¯푗 meet 퐽 then 퐷¯푖 and 퐷¯푗 have a common arc or there intersection is empty.

Corollary 2.6. A simple closed curve 퐽 is accessible from both of its complementary domains.

Proof. We are going to show first that every point of 퐽 is accessible from 퐷 = 푖푛푡(퐽). Let 푎 be any point of 퐽 and for each 푛 = 1, 2,..., let 훿푛 > 0 so that any two points in 퐵(푎, 훿푛) ∩ 퐷 can be joined by a polygonal arc in 퐵(푎, 1/푛) ∩ 퐷. For each 푛 choose a point 푥푛 ∈ 퐵(푎, 훿푛) ∩ 푖푛푡(퐽) and then a polygonal arc 훾푛, joining 푥푛, and 푥푛+1 in 퐵(푎, 1/푛) ∩ 퐷. We define this way, a sequence of linear segments 푠푛 = 푎푛푎푛+1 such that 푠푛 ∩ 푠푛+1 ̸= ∅ for all 푛 ≥ 1 and lim(푎푛) = 푎. By subdividing some of the segments, we can suppose that if 푠푖 ∩ 푠푗 ̸= ∅ and 푠푖 ̸= 푠푗, then 푠푖 ∩ 푠푗 is reduced to a common endpoint of 푠푖 and 푠푗. Hence, we define an end-cut 훾 connecting 푥1 and 푎 in 푖푛푡(퐽) by the following. First, we define 푘0 = max {푖; 푥1 ∈ 푠푖} and inductively, 푘푛+1 = max {푖; 푠푖 ∩ 푠푘푛 ̸= ∅}. Then we set ⎧

⎨ 푠푘0 , if 푡 ∈ [0, 1/2]; 푛 푛+1 훾(푡) = ⎩ 푠푘푛 , if 푡 ∈ [1/2 , 3/2 ]; 푎, t=1.

The accessibility from 푒푥푡(퐽) follows from the following observation. A fractional linear trans- 2 formation ℎ of the two sphere R ∪ {∞} which permutes a point 푥0 ∈ 푖푛푡(퐽) and ∞, sends 푒푥푡(퐽) onto 푖푛푡(ℎ(퐽)) − {푥0}.

9 Using similar arguments, we can show that the complement of an arc in the plane is also uniformly locally connected. In particular the endpoints of an arc are accessible from its complementary domain and we get the following result.

Lemma 2.7. Each arc of the plane lies in a simple closed curve.

A homeomorphism ℎ :Γ → Γ* between two networks is regular if it is possible to index the * * * bounded complementary domains of Γ (resp. Γ ) by 퐷1, . . . , 퐷푛, (resp. 퐷1, . . . , 퐷푛) in such a way that: * ℎ(퐹 푟(퐷푖)) = 퐹 푟(퐷푖 ), 푖 = 1, . . . , 푛. * * In that case we get also that ℎ(퐹 푟(퐷0)) = 퐹 푟(퐷0) where 퐷0 (resp. 퐷0) is the unbounded complementary domain of Γ (resp. Γ*) (see Exercise 2.2).

Lemma 2.8. Let Γ0 and Γ1 be two networks where Γ1 extends Γ0. Then any regular homeo- * morphism ℎ0 between Γ0 and a network Γ0 can be extended into a regular homeomorphism ℎ1 * * between Γ1 and an extension Γ1 of Γ0.

Proof. It is enough to prove the Lemma when Γ1 = Γ0 ∪ 퐿 where 퐿 is a simple arc. Let 푎, 푏 be the endpoints of 퐿, hence 퐿 ∩ Γ0 = {푎, 푏}. Without loss of generality, we can suppose that ¯ * * 퐿 ⊂ 퐷푛 so that {ℎ(푎), ℎ(푏)} ⊂ 퐹 푟(퐷푛). By corollary 2.6 we can construct a cross-cut 퐿 in * * 퐷푛 connecting the points ℎ(푎) and ℎ(푏). Hence we can extend ℎ on Γ1 by deﬁning ℎ1(퐿) = 퐿 . * * Next, we let 퐸푘 = 퐷푘, 퐸푘 = 퐷푘 for 푘 < 푛. Then we denote by 퐸푛 and 퐸푛+1 the two components * * * of 퐷푛 − 퐿 (resp. 퐷푛 − 퐿 ) indexed in such a way that ℎ1(퐹 푟(퐸푖)) = 퐹 푟(퐸푖 ) for 푖 = 푛, 푛 + 1.

Remark 2.9. In the case where Γ1 = Γ0 ∪ 퐿, we have

¯ ¯ * ℎ1(Γ1 ∩ 퐷푖) ⊂ 퐷푖 , for 푖 = 0, . . . , 푛.

This is also true by induction in the general case.

Theorem 2.10 (Schoenﬂies Theorem). Let 퐽 and 퐾 be simple closed curves. Then any homeomorphism ℎ : 퐽 → 퐾 can be extended to give a homeomorphism of 푖푛푡(퐽) ∪ 퐽 onto 푖푛푡(퐾) ∪ 퐾.

Proof. We are going to construct inductively two sequences of networks

* * Γ0 ≺ Γ1 ≺ · · · Γ0 ≺ Γ1 ≺ · · ·

* * * and two sequences of regular homeomorphisms ℎ푛 :Γ푛 → Γ푛 and ℎ푛 :Γ푛 → Γ푛 which are the * * inverse of each other and such that 푚(Γ푛) < 1/푛, 푚(Γ푛) < 1/푛 and Γ푛 (resp. Γ푛) subdivides 퐷¯퐽 = 푖푛푡(퐽) ∪ 퐽 (resp. 퐷¯퐾 = 푖푛푡(퐾) ∪ 퐾). * * We set Γ0 = 퐽,Γ0 = 퐾 and ℎ0 = ℎ. Assume inductively that Γ푛,Γ푛 and ℎ푛 have already * * been defined so that ℎ푛 is a regular homeomorphism of Γ푛 onto Γ푛 with inverse ℎ푛 and that Γ푛 * ¯ ¯ and Γ푛 subdivide 퐷퐽 and 퐷퐾 respectively. ¯ * * * If 푛 is even, we construct a subdivision Γ푛+1 of 퐷퐾 with Γ푛 ≺ Γ푛+1 and 푚(Γ푛+1) < 1/(푛+1). * * * * * We then extend ℎ푛 into a regular homeomorphism ℎ푛+1 :Γ푛+1 → Γ푛+1 = ℎ푛+1(Γ푛+1) by virtue of lemma 2.8. If 푛 is odd we do the same construction but we extend first this time Γ푛 into Γ푛+1 with 푚(Γ푛+1) < 1/(푛 + 1).ᨀ ᨀ * * * * * Then we set Γ = 푛 Γ푛, and Γ = 푛 Γ푛, and define ℎ :Γ → Γ (resp. ℎ :Γ → Γ) by * * * * ℎ(푥) = ℎ푛(푥) if 푥 ∈ Γ푛 (resp. ℎ (푥) = ℎ푛(푥) if 푥 ∈ Γ ). Then ℎ (resp. ℎ ) is uniformly continuous. Indeed, let 휀 > 0 and 푛 > 0 be such that 1/푛 < 휀. Let 퐷1, . . . , 퐷푝푛 be the bounded complementary domains of Γ푛 and set ⌋︀ {︀ 훿푛 = min 푑(퐷¯푖, 퐷¯푗); 퐷¯푖 ∩ 퐷¯푗 = ∅

10 ¯ * ¯ * 훿푛 is deﬁned for 푛 large enough). Then recalling that ℎ(Γ ∩ 퐷푖) ⊂ Γ ∩ 퐷푖 , it is straightforward to show that for any two points of Γ with distance less than 훿푛, we have 푑(ℎ(푥), ℎ(푦)) < 2휀. * Since Γ is dense in 퐷¯퐽 , ℎ extends uniquely to a continuous map ℎ : 퐷¯퐽 → 퐷¯퐾 . Similarly ℎ * * * extends to a map ℎ : 퐷¯퐾 → 퐷¯퐽 and we get ℎ ∘ ℎ = 퐼푑 and ℎ ∘ ℎ = 퐼푑 which completes the proof.

Corollary 2.11. Let 퐽 and 퐾 be simple closed curves. Then any homeomorphism ℎ : 퐽 → 퐾 2 2 can be extended into a homeomorphism ℎ : R → R which is the identity outside a compact set. Proof. We can suppose that 퐾 = 푆1 and that the origin 푂 belongs to 푖푛푡(퐽). Then, we choose a circle 퐶 such that 퐽 and 푆1 are contained in 푖푛푡(퐶). The half line 푂푥− meets 퐽 a last time at a point a before meeting 퐶 at a point 푎′ and the + ′ half line 푂푥 meets 퐽 a last time at a point 푏 before meeting 퐶 at a point 푏 . Let 퐿0 and 퐿1 be * −1 the two arcs of 퐽 defined by 푎 and 푏 and 퐿푖 = ℎ (퐿푖)(푖 = 0, 1). * * 푙 * We choose two disjoint arcs 퐿푎 and 퐿푏 in the annulus defined by 푆 and 퐶 so that 퐿푎 joins −1 ′ * −1 ′ * ′ ℎ (푎) and 푎 and 퐿푏 joins ℎ (푏) and 푏 (see Figure 2.1). Then, we extend ℎ by ℎ(퐿푎) = 푎푎 , * ′ ℎ(퐿푏 ) = 푏푏 and by the identity on 퐶. This extension of ℎ is a regular homeomorphism between the two networks * * * * 퐶 ∪ 퐿푎 ∪ 퐿0 ∪ 퐿푏 ∪ 퐿1, and ′ ′ 퐶 ∪ 푎푎 ∪ 퐿0 ∪ 푏푏 ∪ 퐿1. Now by Schoenflies theorem, it can be extended into a homeomorphism of the closed disc bounded by 퐶. Since ℎ = 퐼푑 on 퐶, we can then extend ℎ to the whole plane by letting ℎ = 퐼푑 on 푒푥푡(퐶).

Figure 2.1:

Remark 2.12. In fact we have also proved that if 퐽 and 퐾 are two simple closed curve such that 퐽 ⊂ 푖푛푡(퐾), there exists a homeomorphism of the plane (with compact support) sending 푖푛푡(퐾) ∩ 푒푥푡(퐽) onto the annulus ⌋︀ {︀ 2 2 2 퐴 = (푥, 푦) ∈ R ; 1 ≤ 푥 + 푦 ≤ 2 .

2 Let ℎ : R → R be a proper embedding and 퐿 = ℎ(R). The closure of 퐿 in the extended 2 plane R ∪ {∞} is a simple closed curve and therefore:

11 2 Corollary 2.13. Any proper embedding ℎ : R → R can be extended into a homeomorphism of the whole plane.

Since any arc of the plane belongs to a simple closed curve (lemma 2.7), we also get:

2 Corollary 2.14. Any homeomorphism ℎ : [0, 1] → 훼 ⊂ R can be extended to a homeomorphism of the whole plane (with compact support).

We suppose now that we have oriented the plane. Let 푓 : 푈 → 푉 be a homeomorphism between two open sets of the plane. Choose a point 푎 ∈ 푈 and 휀 > 0 with 퐵(푎, 휀) ∈ 푈 and let 퐽 be any simple closed curve with 푎 ∈ 푖푛푡(퐽) and 퐽 ⊂ 퐵(푎, 휀). From Schoenﬂies theorem we know that 퐷¯퐽 = 푖푛푡(퐽) ∪ 퐽 is a 2-cell and from theorem 1.12 we know that the inner domain of 퐽 is mapped by 푓 onto the inner domain of 푓(퐽). Hence there exists 푒 ∈ {−1, +1} such that

Ind(푓(푎), 푓(퐽)) = 푒 Ind(푎, 퐽).

The number 푒 is independent of the choice of 퐽 since two simple closed curves, 퐽 and both containing 푎 in their inner domain and both contained in 퐵(푎, 휀) are isotopic in 푈 − {푎}. The number 푒 = 푑(푎, 푓) is the degree of 푓 at 푎. Since 푎 ↦→ 푑(푎, 푓) is locally constant it is constant on any domain.

Definition 2.15. We say that a homeomorphism 푓 : 푈 → 푉 between two domains is orientation- preserving or orientation-reversing according to 푑(푓) is +1 or −1.

This deﬁnition extends easily when 푈 and 푉 are two oriented connected 2-dimensional topological manifolds or when 푓 is an embedding.

Exercises

Exercise 2.1. Let 퐽 be a simple closed curve. Prove that for all 휀 > 0, there exists 훿 > 0 such that if 푥 and 푦 are two distinct points such that 푑(푥, 푦) < 훿 then at least one of the two arcs of 퐽 delimited by 푥 and 푦 has diameter < 휀. Give and prove a similar statement for a simple arc.

* * Exercise 2.2. Let ℎ :Γ → Γ be a regular homeomorphism and 퐷0 (resp. 퐷0) be the unbounded * * component of Γ (resp. Γ ). Show that 푓(퐹 푟(퐷0)) = 퐹 푟(퐷0). Exercise 2.3. Let 푓 : 푈 → 푉 be a 퐶푙 diﬀeomorphism between two open set in the plane. Show that 푑(푎, 푓) = +1 if det[푓 ′(푎)] > 0 and 푑(푎, 푓) = −1 if det[푓 ′(푎)] < 0.

12 Chapter 3

Locally Connected Continua

Definition 3.1. A continuum is a nonempty, compact, connected metric space. It is non degenerate provided it contains more than one point. A point 푥 of a continuum 푋 is a cut point of 푋 if 푋 − {푥} is not connected. If 푌 is any subset of 푋, the notation 푌 = 푈|푉 means that 푌 = 푈 ∪푉 is a partition of 푌 into two nonempty sets both open and closed in 푌 . The proof of the following statement is straightforward and will be omitted. Lemma 3.2. If 푥 is a cut point of a continuum 푋 and 푋 − {푥} = 푈|푉 , then 푈¯ = 푈 ∪ 푥, 푉¯ = 푉 ∪ 푥 and 푈¯, 푉¯ are both sub-continua of 푋. Lemma 3.3. Let 푋 be a nondegenerate continuum. If 푋 has a cut point 푥 and 푋 −{푥} = 푈|푉 , then both of 푈 and 푉 contain a non-cut point of 푋. In particular every nondegenerate continuum has at least two non-cut points. Proof. Suppose that 푈 contains only cut points of 푋 and choose a countable dense set in 푋,

퐸 = {푥푛; 푛 ∈ N} .

Let 푛1 be the ﬁrst index 푛 for which 푥푛 ∈ 푈. Since 푥푛1 is a cut point, we can ﬁnd two open sets 푈1 and 푉1 such that

푋 − {푥푛1 } = 푈1 ∪ 푉1, 푥 ∈ 푉1 ¯ so that 푈1 = 푈1 ∪ {푥푛1 } ⊂ 푈 = 푈0. ¯ Assume inductively that 푥푛푘 , 푈푘 and 푉푘 have already been deﬁned with 푈푘 ⊂ 푈푘−1. Then we let 푛푘+1 be the smallest⌋︀ {︀ integer 푛 so that 푥푛 ∈ 푈푘 and we choose two open sets 푈푘+1 and ¯ 푉푘+1 such that 푋 − 푥푛푘+1 = 푈푘+1|푉푘+1 with 푥푘 ∈ 푉푘+1 and hence 푈푘+1 ⊂ 푈푘. Then we set ⋂︁ ⋂︁ 퐾 = 푈¯푛 = 푈푛 푛≥0 푛≥0 and choose a point 푥∞ ∈ 퐾. We can ﬁnd two open sets 푈∞ and 푉∞ such that 푋 − {푥∞} =

푈∞|푉∞. Since 푥∞ ̸= 푥푛푘 for all 푘 ≥ 1, there is an inﬁnity of values of 푘 for which, say, 푥푛푘 ∈ 푉∞ and so 푈¯∞ = 푈∞ ∪ {푥∞} ⊂ 푈푘. Hence 푈¯∞ ⊂ 퐾. But then, since 푈∞ is a nonempty open set, there must be a point 푥푚 ∈ 푈∞ which leads to a contradiction since 퐾 has been constructed so that it contains no point of 퐸.

Theorem 3.4. A continuum 푋 is an arc if it has only two non-cut points Proof. Denote by 푎 and 푏 the two non-cut points of 푋. It follows from lemma 3.3 that if 푥 ∈ 푋 − {푎, 푏}, then we can ﬁnd two open sets 퐴푥 ∋ 푎 and 퐵푥 ∋ 푏 with 푋 − {푥} = 퐴푥|퐵푥. It is then easy to check that for any two such decompositions 푋 −{푥} = 퐴푥|퐵푥 and 푋 −{푦} = 퐴푦|퐵푦 where 푥 and 푦 are distinct points, we have:

13 ∙ 푦 ∈ 퐵푥 ⇒ 퐴¯푥 ⊂ 퐴푦 and 퐵¯푦 ⊂ 퐵푥,

∙ 푦 ∈ 퐴푥 ⇒ 퐴¯푦 ⊂ 퐴푥 and 퐵¯푥 ⊂ 퐵푦.

From this it follows first that the sets 퐴¯푥 and 퐵¯푥 are uniquely defined for all 푥 ∈ 푋 − {푎, 푏} and that we define a simple ordering on 푋 if we set: 1. 푥 ≺ 푏, for all 푥 ∈ 푋 − {푏}, 2. 푥 ≺ 푎, for all 푥 ∈ 푋 − {푎},

3. 푥 ≺ 푦, if and only if 푦 ∈ 퐵푥, for all 푥, 푦 ∈ 푋 − {푎, 푏}. Moreover, if 푥 and 푦 are any two points of 푋 such that 푥 ≺ 푦, then there exists 푧 ∈ 푋 so that 푥 ≺ 푧 ≺ 푦. Choose a countable dense set {푥푙, 푥2,... } ⊂ 푋. An order preserving homeomorphism ℎ : [0, 1] → 푋 can now be constructed inductively as follows. We set ℎ(0) = 푎 and ℎ(1) = 푏. For each dyadic fraction 0 < 푚/2푘 < 1 with 푚 odd, let us assume inductively that ℎ((푚 − 1)/2푘) andℎ((푚 + 1)/2푘) have already been deﬁned. Then choose the smallest index 푗 so that 푚 − 1 푚 + 1 ℎ( ) ≺ 푥 ≺ ℎ( ) 2푘 푗 2푘 and set ℎ(푚/2푘) = 푥푗. It is not diﬃcult to show that the ℎ constructed in this way on dyadic fractions extends uniquely to an order preserving map from [0, 1] to 푋, which is necessarily a homeomorphism.

Corollary 3.5. A nondegenerate continuum whose connection is destroyed by the removal of two arbitrary points is a simple closed curve. Proof. Let 푥 and 푦 be two non-cut points of 푋 and choose two open sets 푈 and 푉 so that 푋 − {푥, 푦} = 푈|푉 . First, it is straightforward to check that 푈¯ = 푈 ∪ {푥, 푦}, 푉¯ = 푉 ∪ {푥, 푦} that there are both connected. We are going to show that 푈¯ and 푉¯ are both arcs with endpoints 푥, 푦 and hence 푋 = 푈¯ ∪ 푉¯ is a simple closed curve. At least one of the two continuum 푈 and 푉 is an arc with endpoints 푥, 푦. If not, we can find 푧 ∈ 푈¯ − {푥, 푦} and 푡 ∈ 푉¯ − {푥, 푦} which are non-cut points of 푈¯ and 푉¯ respectively. Hence 푋 − {푧, 푡} = (푈¯ − {푧}) ∪ (푉¯ − {푡}) is connected which is a contradiction. In fact both of them are arcs with endpoints 푥, 푦. If not, suppose for example that 푈¯ is not an arc. Then we can find 푧 ∈ 푈 so that 푈¯ − {푧} is connected. Hence, since 푉¯ is an arc, we can find 푡 ∈ 푉 such that 푉¯ − {푡} = 푅 ∪ 푆 where 푅 and 푆 are two connected sets with 푥 ∈ 푅 and 푦 ∈ 푆. Therefore 푋 − {푧, 푡} = (푈¯ − {푧}) ∪ 푅 ∪ 푆 is connected, contrary to the assumption.

Theorem 3.6 (Converse of Jordan’s Curve Theorem). Let 푋 be a plane continuum which has two distinct complementary domains 푅0, 푅1 from which it is at every point accessible. Then 푋 is a simple closed curve. Proof. In view of what we have just shown, it is enough to prove that the connection of 푋 is destroyed by the removal of two arbitrary points. Let 푥 and 푦 be two distinct points of 푋. Since both 푥 and 푦 are accessible from 푅0 and 푅1, it is possible to ﬁnd a cross-cut 훼푖 with endpoints 푥, 푦 in 푅푖 for 푖 = 0, 1. Hence 퐽 = 훼0 ∪ 훼1 is a simple closed curve. But 푖푛푡(퐽), like 푒푥푡(퐽), meets 푅0 and 푅1 and thus also 푋 = 퐹 푟(푅0) = 퐹 푟(푅1). Therefore 푋 − {푥, 푦} = (푖푛푡(퐽) ∩ 푋) ∪ (푒푥푡(퐽) ∩ 푋) is not connected.

14 Let 푋 be a metric space. We recall that 푋 is locally connected at a point 푥 ∈ 푋 if there exist arbitrary small connected neighborhoods of 푥 in 푋. We leave as an exercise the proof of the following.

Lemma 3.7. The following conditions are equivalent.

1. 푋 is locally connected at every point 푥 ∈ 푋,

2. Every point of 푋 possesses arbitrary small open connected neighborhoods,

3. Every connected component of an open subset of 푋 is itself open in 푋.

A space is arcwise connected if any two given points can be joined by an arc.

Theorem 3.8. In a compact and locally connected space 푋, each connected open set is arcwise connected.

In other words, two points of a connected open set can be joined by an arc lying entirely in 푈. The desired arc will be obtained as an intersection of a sequence 퐾1 ⊃ 퐾2 ⊃ · · · of “thinner and thinner” sub-continua of 푈. To do this carefully, let us first define an auxiliary notion. A simple chain is a finite sequence 풞 = (퐶푙, . . . , 퐶푛) of nonempty sets such that every two adjacent terms of the sequence intersect but no two non-adjacent terms do so. The sets 퐶푙, . . . , 퐶푛 are called the links of 풞, and if 푎 ∈ 퐶1 and 푏 ∈ 퐶푛, then 풞 is said to be a chain from 푎 to 푏. If 풟푙 and 풟2 are simple chains such that the concatenation of the sequence 풟푙 followed by the sequence 풟2 is a simple chain 풟, we will write 풟 = 풟푙 + 풟2. Note that if 풟 = 풟푙 + ··· + 풟푘, then for 푖 < 푗 each link of 풟푖 precedes each link of 풟푗 in 풟. Proof of theorem 3.8. Let 푈 be a nonempty connected open subset of 푋 and 푎, 푏 ∈ 푈, 푎 ̸= 푏. 푙 푘푛 It is easy to obtain a sequence 풞1, 풞2,... of simple chains from 푎 to 푏, 풞푛 = (퐶푛, . . . , 퐶푛 ), such that for each 푛:

1. every link of 풞푛 is a connected open subset of 푈 of diameter less than 1/푛,

2. the closure of each link of 풞푛+1 is a subset of some link of 풞푛, 3. there exists chains 풟푛,..., 풟푛 such that 풞푛+1 = 풟푛 + ··· + 풟푛 and such that each link 푙 푘푛 푙 푘푛 푛 푛 of 풟푖 is a subset of 풞푖 for 푖 = 1, . . . , 푘푛. 푛 For each 푛, let 퐾푛 denote the union of the closures of all the links of 풞 and let ⋂︁∞ 퐾 = 퐾푛. 푛=1 It is clear that 퐾 is a continuum lying in 푈 and containing 푎 and 푏. To prove that 퐾 is an arc it is suﬃcient to show that each point of 퐾 − {푎, 푏} is a cut point of 퐾. To this end, suppose 푝 ∈ 퐾 − {푎, 푏} and for each 푛 let 퐴푛 (respectively 퐵푛) be the set of all 푥 ∈ 퐾 such that every link of 풞푛 containing 푥 precedes (follows) every link of 풞푛 containing 푝. Then

⋃︁∞ ⋃︁∞ 퐴 = 퐴푛 and 퐵 = 퐵푛 푛=1 푛=1 are two open sets of 퐾 such that 푎 ∈ 퐴, 푏 ∈ 퐵, 퐴 ∩ 퐵 = ∅ (since 퐴1 ⊂ 퐴2 ⊂ ... and 퐵1 ⊂ 퐵2 ⊂ ... ) and 퐾 − {푝} = 퐴 ∪ 퐵. Lemma 3.9. Any continuous image of a compact locally connected space is compact and locally connected.

15 Proof. let 푈 be an open set of 푌 = 푓(푋) and 퐶 be a connected component of 푈. We have to −푙 −1 show that 퐶 is itself⋃︀ open in 푌 . Let (푁훼) be those components of 푓 (푈) which meets 푓 (퐶). −1 −1 Then 푓 (퐶) = 훼 푁훼, is an open set of 푋 and hence 푓(푋) − 퐶 = 푓(푋 − 푓 (퐶)) is closed in 푌 = 푓(푋).

An immediate corollary of these results is the following: Corollary 3.10. In a metric space, each path joining two distinct points 푎, 푏 contains an arc with endpoints 푎, 푏. Theorem 3.11. Every nonempty compact metric space is a continuous image of the Cantor set. Proof. Let 푋 be a nonempty compact metric space. Choose a covering of 푋by compact subsets 푛 1 푛 1 (not necessarily distinct) 퐴1 , . . . , 퐴2푝1 of diameter less than 1. Let 퐼1 , . . . , 퐼2푝1 be all the subin- tervals of the subdivision of [0, 1] of length 3−푝1 which meet the middle-third Cantor set 퐾 and 1 1 푝1 let 퐷푖 = 퐾 ∩ 퐼푖 for 푖 = 1,..., 2 . We deﬁne then a map 푋 퐹1 : 퐾 → 2

푋 1 1 where 2 is the set of all nonempty closed subset of 푋, by 퐹1(푡) = 퐴푖 if 푡 ∈ 퐷푖 . Since each set 1 푋 퐷푖 is open (in 퐾), the map 퐹1 is continuous (for the Hausdorﬀ metric on 2 ). 1 푝2 1 Next, we cover each compact set 퐴푖 by 2 (not necessarily distinct) compact subsets of 퐴푖 2 2 2 1 2 of diameter less than 1/2 and let 퐹2(푡) = 퐴푗 if 푡 ∈ 퐷푗 = 퐾 ∩ 퐼푗 where 퐼2 , . . . , 퐼2푝1+푝2 are the subinterval of the subdivision of [0, 1] of length 3−푝1+푝2 which meets 퐾. Inductively, we construct, this way, a sequence (퐹푛) of maps such that for each 푛, 푋 1. 퐹푛 : 퐾 → 2 is continuous,

2. 퐹푛(푡) ⊃ 퐹푛+1(푡) for all 푡 ∈ 퐾, ⋃︀ 3. 푋 = 푡∈퐾 퐹푛(푡),

4. the diameter of 퐹푛(푡) is less than 1/푛. ⋂︀ ∞ For all 푡 ∈ 퐾 let 푓(푡) be the unique point of 푛=1 퐹푛(푡). It is then straightforward to check that 푓 is continuous.

Theorem 3.12 (Hahn-Mazurkiewicz). A continuum is locally connected if and only if it is the continuous image of the unit interval. Proof. That a continuous image of the unit interval is a locally continuum follows from lemma 3.9. To prove the converse, we chose ﬁrst a continuous map 푓 : 퐾 → 푋 from the Cantor set onto our locally connected continuum 푋. It is easy to check that a locally connected continuum is in fact uniformly locally connected. Using the uniform continuity of 푓 and theorem 3.8, it is then straightforward to show that 푓 can be extended continuously on each of the subinterval ]푎푖, 푏푖[ of [0, 1] − 퐾.

Exercises

Exercise 3.1. Let (푋푛) a nested sequence of continuum, show that +⋂︁∞ 푋푛 푛=1 is a continuum.

16 Exercise 3.2. Show that the boundary of a simply connected and uniformly locally connected plane bounded domain is a simple closed curve (Second Form of the Converse of Jordan’s Curve Theorem).

Exercise 3.3. Let 푋 be a compact metric space and 2푋 be the set of all nonempty closed subset of 푋. For 퐴, 퐵 ∈ 2푋 , we let ⌉︀

푑퐻 (퐴, 퐵) = max sup 푑(푎, 퐵), sup 푑(푏, 퐴) . 푎∈퐴 푏∈퐵

푋 Show that (2 , 푑퐻 ) is a compact metric space and that the subset 퐶(푋) of all sub-continua of 푋 is closed in this topology. [Hint: Show that 2푋 is complete and totally bounded (for any 휀 > 0, 2푋 can be written as a ﬁnite union of sets of diameter less than 휀).]

Exercise 3.4. Show that theorem 3.8 is still true if 푋 is only supposed to be locally compact or complete.

Exercise 3.5. Using the proof of theorem 3.11, show that a compact, metric, perfect and totally disconnected set is homeomorphic with the Cantor set.

17 Chapter 4

Indecomposable Continua

Definition 4.1. A continuum 푋 is decomposable provided that 푋 can be written as the union of two proper sub-continua. A continuum which is not decomposable is said to be indecomposable. We have the following: Lemma 4.2. A continuum 푋 is indecomposable if and only if each of its sub-continua has empty interior. Proof. The proof is straightforward, and will be left to the reader.

Let 푎, 푏 ∈ 푋 two distinct points of a continuum. We say that 푋 is irreducible between 푎 and 푏 if there exists no proper sub-continuum of 푋 which contains both 푎 and 푏 There is a surprisingly simple argument to check that a continuum is indecomposable. Lemma 4.3. If there exist three points in a continuum 푋 such that 푋 is irreducible between each two of these three points then 푋 is indecomposable. Again the proof is straightforward and will be omitted. We will see later that this condition is also necessary if 푋 is nondegenerate (lemma 4.11). Remark 4.4. An indecomposable continuum is clearly nowhere locally connected. Moreover, if 퐶 is any proper sub-continuum of 푋, then 푋 −퐶 is connected. In particular, an indecomposable continuum has no cut-point. Example 4.5. Let 푎, 푏 and 푐 be distinct points in the plane. It is easy to construct simple chains 푛 −푛 풞 = (퐶1, . . . , 퐶푘푛 ), 푛 = 1, 2,... whose links are discs of diameter < 2 satisfying conditions below:

1. For each 푛 = 1, 2,... the closure of each link of 풞푛+1 is a subset of a link of 풞n,

2. For each 푛 = 1, 2,... ,C3푛+1 goes from 푎 to 푐 through 푏,C3푛+2 goes from 푏 to 푐 through 푎,C3푛+3 goes from 푎 to 푏 through 푐.

We set ⋃︁푘푛 ⋂︁∞ ¯푛 퐾푛 = 퐶푖 and 퐾 = 퐾푛 푖=1 푛=1 It is then easy to check that 퐾 is irreducible between each two of the three points 푎, 푏 and 푐 (See Figure 4.1). Surprisingly, “almost” every continuum is indecomposable. Lemma 4.6. In the space 퐶(퐷2) of sub-continua of a the unit disc 퐷2 (with the Hausdorff metric), the set of nondegenerate indecomposable sub-continua is a dense countable intersection of open sets (dense 퐺훿).

18 Figure 4.1: An indecomposable continuum

2 Proof. Let ℐ be the set of non degenerate indecomposable continua of 퐷 and 퐹푛 be the set of all the sub-continua which are the union of two sub-continua 퐴 and 퐵 such that 퐴 ̸⊂ 푉 (퐵, 1/푛) 2 2 and 퐵 ̸⊂ 푉 (퐴, 1/푛) 퐹푛 is a closed subset of 퐶(퐷 ),⋃︀ the set of all sub-continua of 퐷 . Hence, 2 ∞ the complement of ℐ (in 퐶(퐷 )) can be written as 푛=1 퐹푛. Furthermore, ℐ is a dense set in 퐶(퐷2) Indeed, let 퐶 ∈ 퐶(퐷2) and 휀 > 0. It is easy to 2 construct a polygonal arc 훼 in 퐷 with 푑퐻 (훼, 퐶) < 휖/2 (where 푑퐻 is the Hausdorﬀ metric). Using the construction of Example 1, it is then possible to construct an indecomposable continuum 퐾 such that 푑퐻 (퐾, 퐶) < 휀. Example 4.7 (The Pseudo-arc). Let 풞 and 풟 be two simple chains where 풞 strongly refines 풟 (meaning the closure of each link of 풞 = (퐶1, . . . , 퐶푛) is contained in a link of 풟 = (퐷1, . . . , 퐷푚)). We say that 풞 is crooked in 풟 provided that if 푘 − ℎ > 2 and 퐶푖 and 퐶푗 are links in 퐷ℎ and 퐷푘 of 퐷, respectively, then there are links 퐶푟 and 퐶푠 of 풞 in links 퐷푘−1 and 퐷ℎ+1 respectively, such that either 푖 > 푟 > 푠 > 푗 or 푖 < 푟 < 푠 < 푗. In the plane, let 풞1, 풞2,... be a sequence of chains between two distinct points 푎 and 푏 such that for each 푛 = 1, 2,... we have

1. 풞푛+1 is crooked in 풞푛,

2. the diameter of each link of C푛 is less than 1/푛.

Set ⋃︁푘푛 푛 퐾푛 = 풞푖 . 푖=1

19 ⋂︀ ∞ Then 퐾 = 푛=1 퐾푛 is called a pseudo-arc. Is is an indecomposable continuum. It has the amaz- ing following property. If 퐻 is a nondegenerate sub-continuum of 퐾, then 퐻 is indecomposable. That is the pseudo-arc is hereditarily indecomposable. In particular 퐾 contained no arc. Moise [14] proved that every two pseudo-arcs are homeomorphic and Bing [5] gave a topological characterization of the Pseudo-arc. Let 푋 a continuum and 푥 ∈ 푋. The composant of 푥 is the union of all proper sub-continua of 푋 which contain 푥. It is also the set of all points 푦 such that 푋 is not irreducible between 푥 and 푦. Two composants are not necessarily disjoint. For example an arc ab has exactly three composants, namely, 푎푏, 푎푏 − {푎} and 푎푏 − {푏} corresponding to 푥 ̸= 푎, 푏, or 푥 = 푎, or 푥 = 푏. In fact it is easy to show that every decomposable continuum is a composant for some point. In a space which is not connected the composants are the same as the components and this notion has no interest, but this notion is very useful in the study of indecomposable continua as we will see later.

Lemma 4.8. In a nondegenerate continuum 푋, each composant is a dense union of a countable family of proper subcontinua.

Proof. Let 퐶 be a composant deﬁned by a point 푝 and 푈, 푉 be nonempty open sets in 푋 with 푉¯ ⊂ 푈. If 푝 ̸∈ 푉 , the closure of the component 푁 of 푋 − 푉 which contains 푝 is contained in 퐶. But 푁 ∩ 퐹 푟(푉 ) ̸= ∅ (see Exercise 4.4), so that 퐶 ∩ 푈 ̸= ∅. Hence 퐶¯ = 푋. To prove that 퐶 is a countable union of proper sub-continua of 푋, choose a⋃︀ countable basis 푈1, 푈2,... for 푋 − {푝} and let 푁푖 be the component of 푝 in 푋 − 푈 푖 then C = 푖 푁 푖. Corollary 4.9. Every composant of an indecomposable continuum has empty interior.

The proof is a straightforward application of Baire’s theorem. In fact it is easy to verify, the following.

Lemma 4.10. If a continuum has a composant with empty interior then it is in decomposable.

Lemma 4.11. If 푋 is an indecomposable continuum, the composants of 푋 partition 푋 and there are uncountably many of them.

Proof. Let 퐶1 and 퐶2 be two composants deﬁned respectively by the points 푝1 and 푝2 and suppose that there exists 푥 ∈ 퐶1 ∩ 퐶2. Then we can ﬁnd

∙ a proper sub-continuum 퐾1 ⊂ 퐶1 such that 푥, 푝1 ∈ 퐾1,

∙ a proper sub-continuum 퐾2 ⊂ 퐶2 such that 푥, 푝2 ∈ 퐾2 .

Thus 퐾 = 퐾1 ∪ 퐾2 is a proper sub-continuum of 푋. Let 푦 be any point of 퐶1. There ′ exists a proper sub-continuum 퐾3 ⊂ 퐶1 such that 푦, 푝1 ∈ 퐾3, thus 퐾 = 퐾 ∪ 퐾3 is a proper sub-continuum of 푋 and therefore 푦 ∈ 퐶2. Hence, 퐶1 ⊂ 퐶2 and similarly 퐶2 ⊂ 퐶1.⋃︀ Suppose there are at most a countable many distinct composants. Then 푋 = 푛 퐶푛 is also an at most countable union of proper sub-continua, each of which having empty interior. But this is not possible by Baire’s theorem.

Example 4.12 (Knaster’s Continuum). We begin with the Cantor middle-third set 퐶. Then we add to 퐶 all the semicircles lying in the upper half plane with center at (1/2, 0) that connect each pair of points in the Cantor set that are equidistant from 1/2. Next we add all semicircles in the lower half plane which have for each 푛 ≥ 1 centers at (5/(2 · 3푛), 0) and pass through each point in the Cantor set lying in the interval

2/3푛 ≤ 푥 ≤ 1/3푛−1

20 The resulting set is partially depicted in Figure 4.2. The fact that this set is indecomposable results from the following observation [11]. The composant of the origin is the curve obtained by starting from the origin and then following successively all the successive semicircles. This curve has empty interior.

Figure 4.2: Knaster’s Continuum

We are now going to describe another way to look at continua, namely via inverse limits. This gives in some case, a parametrization of a continuum by a simpler space. Let (푋, 푑) be a metric space and 푓 : 푋 → 푋 a continuous map, the inverse limit space, (푋, 푓) is the space

(퐾, 푓) = {x = (푥0, 푥1,... ); 푥푛 ∈ 푋 and 푓(푥푛+1) = 푥푛 for all 푛 ≥ 0} with metric d given by ∑︁∞ 푑(푥푛, 푦푛) d(x, y) = 푛 . 푛=0 2 If 푋 is a continuum, then (푋, 푓) is also a continuum. The map 푓 induces a homeomorphism

푓ˆ :(푋, 푓) → (푋, 푓) given by 푓ˆ((푥0, 푥1,... )) = (푓(푥0), 푥0,... ) from which the inverse is just the usual shift map. Lemma 4.13. Let 푋 be a continuum and 푓 : 푋 → 푋 be a surjective continuous map such that whenever 푋 = 퐴 ∪ 퐵 where 퐴 and 퐵 are sub-continua of 푋, we have 푓(퐴) = 푋 or 푓(퐵) = 푋. Then, (푋, 푓) is an indecomposable continuum.

Proof. Let 푋∞ = (푋, 푓) and suppose there exist two sub-continua 퐴∞ and 퐵∞ of 푋∞ with 푋∞ = 퐴∞ ∪ 퐵∞. Let 휋푛 : 푋∞ → 푋 be the natural projection onto the nth coordinate. From the relation 휋푛(푋∞) = 휋푛(퐴∞) ∪ 휋푛(퐵∞) = 푋 for all 푛 ≥ 0, and from the property of 푓, it follows that there are inﬁnite many 푛 for which we have, say 휋푛(퐴∞) = 푋. It follows then from the relation 푓푛 ∘ 휋푛+1 = 휋푛 which is true on 푋∞, that 휋푛(퐴∞) = 푋 for all 푛 ≥ 0. Hence 퐴∞ is dense in 푋∞ and thus 퐴∞ = 푋∞. Remark 4.14. The preceding lemma is still true if we replace the hypothesis of the lemma by the weaker one : there exists 푁 > 0 such that for all sub-continua 퐴, 퐵 of 푋 such that 푋 = 퐴 ∪ 퐵, then 푓 푁 (퐴) = 푋 or 푓 푁 (퐵) = 푋.

21 Example 4.15 (The Solenoid). Let 푋 = 푆1, the unit circle and 푓 : 푆1 → 푆1 the map given by 푓(휃) = 푝휃 (푝 ≥ 2). Then (푋, 푓) is an indecomposable continua which is called the p-adic 1 solenoid Σ푝. There is a natural group structure on Σ푝 inherited from the one on 푆 . We can deduce from this fact that Σ푝 is homogenous (meaning that from any two points there is a homeomorphism of Σ푝 onto Σ푝 taking one of these point to the other). We can show also that each proper sub-continuum of Σ푝 is an arc. It is interesting to note that the last two properties actually characterize solenoid [8]. A continuum Λ which separates the plane into exactly two components and which is irreducible with respect to this property will be called a cofrontier. Let 푈푖 be the bounded complementary domain of Λ and 푈푒 be the unbounded one. Then 퐹 푟(푈푖) and 퐹 푟(푈푒) are sub- continua of Λ which separate the plane thus, 퐹 푟(푈푖) = 퐹 푟(푈푒) = Λ. Conversely, if a plane continuum Λ separates the plane into exactly two complementary domains and is there common frontier, then Λ irreducibly separates the plane and is therefore a cofrontier. Remark 4.16. Indecomposable continuum have been discovered by Brouwer to show that Schoen- ﬂies’ conjecture that every cofrontier could be decomposed into the union of two proper subcontinua was wrong.

Figure 4.3: Wada Lakes

Note that a continuum which is the common boundary of two simply connected domains is not necessary a cofrontier as we will see below. The ﬁrst explicit example of a such continuum was reported in a paper by Yoneyama in 1917 [23]. This beautiful example is known as the Lakes of Wada continuum. In his paper Yoneyama states that the example was reported to him

22 by a Mr. Wada, but no one seems to know anything else about Mr. Wada. In 1924, Kuratowski [11] proved that if 퐾 is a plane continuum is the common boundary of more than 3 disjoint open sets then 퐾 is either indecomposable or the union of two indecomposable continua. To preserve the poetic ﬂavour of the original example [9], we take a double annulus, as shown in Figure 4.3, to be a island in the ocean with two lakes, one having blue water and the other red water. At time 푡 = 0, we dig a canal from the ocean, which brings salt water to within a distance of 1 unit of every point of land. At time 푡 = 1/2, we dig a canal from the blue lake, which bring blue water to within a distance 1/2 of every point of land.At time 3/4, we dig a canal to bring red water to within a distance of 1/3 at every point of land. A time 푡 = 7/8, we dig a canal from the end of the ﬁrst canal to bring salt water to within a distance 1/4 of every point of land and so forth. If we think of these canals as open sets, at time 푡 = 1, the “land” remaining is a plane continuum which bounds three open domains in the plane.

Exercises

Exercise 4.1. In Theorem 4.5, give a meaning to the statement “the simple chains are as straight as possible in each other” (see the construction of the arc in the proof of theorem 3.8). Obtain, this way, an indecomposable continuum such that every of its proper nondegenerate sub-continua is an arc.

Exercise 4.2. Show that Theorem 4.7 (the Pseudo-arc) is hereditarily indecomposable.

Exercise 4.3. Show that most of the sub-continua of a closed 2-cell are hereditarily indecomposable.

Exercise 4.4. Let 푋 be a continuum, 푈 an open set of 푋 and 푁 a component of 푈. Then, 푁¯ ∩ 퐹 푟(푈) ̸= ∅.

Exercise 4.5. Suppose Λ is a locally connected cofrontier, then Λ is a simple closed curve.

23 Chapter 5

Attractors and Indecomposable Continua

Let 푋 be a compact metric space and 푓 : 푋 → 푋 a continuous map. A nonempty compact subset Λ ⊂ 푋 is an attractor for 푓 if there is an open neighbourhood 푈 of Λ such that 푓(푈) ⊂ 푈 and ⋃︁ Λ = 푓 푛(푈¯). 푛≥0 An attractor is clearly an invariant set for 푓 (푓(Λ) = Λ). There are other definitions of attractors in common use, but this one is perhaps the simplest. However this definition suffers the defect that it does not produce a single, indecomposable attractor To remedy this, we introduce the following terminology Λ is a transitive attractor for 푓 if 푓 is topologically transitive on Λ, that is for any pair of open sets 푈, 푉 ⊂ 푋 there exist 푘 > 0 such that 푓 푘(푈) ∩ 푉 ̸= ∅. Given a point 푝, the stable set of 푝 is given by

푊 푠(푝) = {푧 ∈ 푋; 푑(푓 푛(푝), 푓 푛(푧)) → 0 as 푛 → +∞} .

Equivalently, a point 푧 lies in 푊 푠(푝) if 푝 and 푧 are forward asymptotic. Similarly, we deﬁne the unstable set of 푝 to be ⌋︀ {︀ 푊 푢(푝) = 푧 ∈ 푋; 푑(푓 −푛(푝), 푓 −푛(푧)) → 0 as 푛 → +∞ .

We say that a periodic point of 푓 (ie a point 푝 for which there exists 푛 > 0 with 푓 푛(푝) = 푝) is stable (resp unstable) if 푊 푠(푝) (resp 푊 푢(푝)) is a neighbourhood of 푝. See [7], for a beautiful introduction to Dynamical Systems Example 5.1 (The Horseshoe Map). Let 푆 be the square 퐼 × 퐼 where 퐼 = [0, 1], and let 퐷 be the two dimensional disc 퐷 = 푆 ∪ 퐴 ∪ 퐵 where 퐴 and 퐵 are half disc attached to opposite sides {0} × 퐼 and {1} × 퐼 of the square 푆 as depicted in Figure 5.1. The standard horseshoe map 퐹 takes 퐷 inside itself according to the following prescription. First, linearly contract 푆 in the vertical direction by a factor 훿 < 1/2 and expand it in the horizontal direction by a factor 1/훿 so that 푆 is long and thin. Then put 푆 back inside 퐷 in a horseshoe shaped ﬁgure as in Figure 5.2. The semicircular regions 퐴 and 퐵 are contracted and mapped inside 퐴 as depicted. We remark that 퐹 (퐷) ⊂ 퐷 and that 퐹 is one-to-one. However, since 퐹 is not onto, 퐹 −1 is not globally deﬁned. Now let 휋 : 퐷 → 퐼 given by 휋(퐴) = {0}, 휋(퐵) = {1} and 휋((푥, 푦)) = 푥 for all (푥, 푦) ∈ 푆 = 퐼 × 퐼. The map 퐹 has the following properties:

1. 퐹 (휋−1(휋(푧))) ⊂ 휋−1(휋(퐹 (푧))) for all 푧 ∈ 퐷,

2. 퐹 (퐴) ⊂ 푖푛푡(퐴) and 퐹 (퐵) ⊂ 푖푛푡(퐴),

24 Figure 5.1: Building the Horseshoe

Figure 5.2: The Horseshoe

3. For all 푥 ∈ 퐼, 휋−1(푥) ∩ 퐹 (퐷) has exactly two components,

4. diam(퐹 푛(휋−1(휋(푧))) → 0 uniformly in 푧 ∈ 퐷 as 푛 → +∞.

The attractor for the horseshoe map 퐹 is the set ⋂︁ 푛 Λ퐻 = 퐹 (퐷). 푛≥0

It is possible to show that Λ퐻 is homeomorphic to Knaster’s continuum (Theorem 4.12). How- ever, we are going to show the indecomposability of Λ퐻 by other means. 퐹 induces a continuous map of the interval 푓 : 퐼 → 퐼 deﬁned by 푓(푥) = 휋(퐹 (휋−1(푧))). The map 푓 has the following properties 푓(0) = 푓(1) = 0 and there exist 0 < 푎1 < 푎2 < 푎3 < 푎4 < 1 such that 푓 is strictly increasing on [푎1, 푎2], 푓 is strictly decreasing on [푎3, 푎4], 푓([0, 푎1]) = 푓([푎4, 1]) = {0} and 푓([푎2, 푎3]) = {1} (see Figure 5.3).

Theorem 5.2. Let 휋ˆ :Λ퐻 → (퐼, 푓) defined by

휋ˆ(푧) = (휋(푧), 휋(퐹 −1(푧)),... ). ˆ ˆ Then 휋ˆ is a homeomorphism such that 퐹 ∘ 휋ˆ = 푓 ∘ 휋ˆ. That is, 퐹|Λ퐻 and 푓 are topologically conjugate. Proof. The following proof is due to Barge [3]. Since

푓(휋(퐹 −(푛+1)(푧))) = 휋(퐹 (퐹 −(푛+1)(푧))) = 휋(퐹 −푛(푧)),

25 Figure 5.3: we see that휋 ˆ(푧) is indeed an element of (퐼, 푓). 휋ˆ is clearly continuous. To see that휋 ˆ is one-to-one and onto, let x = (푥0, 푥1,... ) ∈ (푋, 푓) and let 푛 −1 퐶푛 = 퐹 (휋 (푥푛)) for 푛 = 0, 1,...

Then 퐶푛 is a closed, nonempty subset of 퐷 for each 푛 ≥ 0, and since

−1 −1 −1 퐹 (휋 (푥푛+1)) ⊂ 휋 (푓(푥푛+1)) = 휋 (푥푛), ⋂︀ we have 퐶푛+1 ⊂ 퐶푛 for all 푛 ≥ 0. Thus 푛≥0 퐶푛 is⋂︀ a nonempty compact set. But since diam(⋂︀퐶푛) → 0 as 푛 → +∞ (condition (4)), we have 푛≥0 퐶푛 = {푧} is a singleton. Now if 푧 ∈ 푛≥0 퐶푛, then −1 휋(푧) = 푥0, 휋(퐹 (푧)) = 푥1,... ⋂︀ That is,휋 ˆ(푧) = x, but then 푧 must be in 푛≥0 퐶푛.휋 ˆ is one to one as well as onto.

It follows then from Lemma 4.13 that Λ퐻 is an indecomposable continuum. We note also, that 푓 has two unstable fixed point 푝0 and 푝1 and one stable fixed point 0. These points correspond to fixed points for 푓,ˆ p0 = (푝0, 푝0,... ) and p1 = (푝1, 푝1,... ) and thus to fixed points for 퐹 in 푆. Given an arbitrary point 푥 ∈ 퐼, there exists a unique 푥1 ∈ [푎1, 푎2] such that 푓(푥1) = 푥, then a unique 푥2 ∈ [푎1, 푎2] such that 푓(푥2) = 푥1 and so on. In this way, −푛 we construct a sequence x = (푥, 푥1, 푥2,... ) ∈ (푋, 푓) with the property that 푓ˆ (x) → 0 as 푛 ˆ 푢 푢 푢 → +∞. Moreover, since 푓(푊 (p0)) ⊂ 푊 (p0), this shows that 푊 (p0) is dense in Λ퐻 . 1 2 3 Example 5.3 (The Solenoid from dynamical viewpoint). Let 푇 = 푆 × 퐷 be a solid torus in R and consider the map 1 1 퐹 (휃, 푥) = (2휃, 푥 + 푒2휋푖휃) 10 2 where 푥 ∈ 퐷2 and 휃 ∈ 푆1. Globally, 퐹 may be interpreted as follows. In the 휃 coordinate, 퐹 is simply the doubling map 휃 ↦→ 2휃. In the 퐷2-direction, 퐹 is a strong contraction, with image a disc whose center depends on 휃. Thus the image of 푇 is another solid torus inside 푇 which wraps twice around 푇 (see Figure 5.4).

26 Figure 5.4: the Solenoid

Figure 5.5: the Plykin attractor

⋃︀ 푛 1 2 1 We set Λ푆 = 푛≥0 퐹 (푇 ). Let 휋 : 푆 × 퐷 → 푆 be the projection onto the ﬁrst factor. We 1 can show as in the case of the horseshoe map that the induced map휋 ˆ :Λ푆 → (푆 , 푔) given by

휋ˆ(푧) = (휋(푧), 휋(퐹 −1(푧)),... ).

1 where 푔 : 휃 ↦→ 2휃 gives a conjugacy between 퐹 on Λ푆 and푔 ˆ on (푆 , 푔). This construction gives 3 us also a way to embed the solenoid Σ2 defined in Theorem 4.12 into the 3-space 푅 . In fact, the inverse limit construction works well for a class of attractors known as “expanding” attractors. These attractors are characterized by uniform expansion within the attractor itself. As in the case of the solenoid, such attractors can be suitably modeled by an inverse limit of a lower dimensional expanding map like 휃 ↦→ 2휃 on 푆1. The main difference is that the model space is more complicated than 푆1 ; usually it is a “branched one-manifold”, a continuum which is the union of a finite number of arcs each of which intersecting by pairs only in there endpoints. This concept was introduced by Williams [22]. He showed that every hyperbolic, one-dimensional, expanding attractor for a discrete dynamical system is topologically conjugate to the induced map on an inverse limit space based on a branched one-manifold. We will illustrate it via an example of an attractor due to Plykin. Rather than give a formula for this map, we will define it geometrically, as we did it for the horseshoe.

27 Figure 5.6: the Plykin attractor mapped to itself

Example 5.4 (Plykin attractor). Consider the region 푅 in the plane depicted in Figure 5.5. 푅 is a region with three open half disc removed. We equip 푅 with a foliation whose leaves are intervals as shown in Figure 5.5. Deﬁne a map 푃 : 푅 → 푅 as shown in Figure 5.6. We require that 푃 preserves and contracts the⋂︀ leaves of the foliation. Note that 푃 (푅) is contained in the interior of 푅, so that Λ푃 = 푛 훾1≥0 푃 (푅) is an attractor; the Plykin attractor. Since the leaves are contracted, any two points on the same leaf tend to the attractor in the same asymptotic manner. Thus to understand the action of 푃 globally, it suﬃces to understand the action of 푃 on the leaves. We thus collapse each leaf to a point as in Figure 5.7, and examine the induced map on this space. Observe that the collapsed space Γ has ”branched” points along the singular leaves 푙1 and 푙2. As in the example of the horseshoe map, we introduce a projection 휋 : 푅 → Γ where Γ = 훼 ∪ 훽 ∪ 훾 ∪ 훿. Since 푃 preserves the foliation, 푃 induces a continuous map 푔 :Γ → Γ which preserves the two vertices and maps the other intervals this way:

훼 → 훽, 훽 → 훽 + 훾 − 훿 − 훽, 훾 → 훼, 훿 → 훿 − 훾 − 훿, where the signs indicate orientations in which the image crosses the given arc. We may construct such a map so that 푔 expands all distances in the branched one-manifold Γ. Using lemma 4.13, we see then that (Γ, 푔) is an indecomposable continuum and as before it can be shown that the restriction of 푃 to Λ푃 is topologically conjugate to the induced map푔 ˆ on (Γ, 푔). However there exist more complicated situations as will be shown below. A ﬁxed point 푝 for a 퐶1-diﬀeomorphism 퐹 :F푛 → R푛 is called hyperbolic if 푓 ′(푝) has no eigenvalues on the unit circle. We note 퐸푠(푝) (resp. 퐸푢(푝)) the sum of the characteristic spaces corresponding to eigenvalues of modulus less (resp. greater) than 1. We have then the two following fundamental results [17].

Theorem 5.5 (Grobman-Hartman Theorem). Let 퐹 : 푅푛 → 푅푛 be a 퐶1-diffeomorphism with a hyperbolic fixed point 푝. Then there exists a homeomorphism ℎ defined on some neighbourhood 푈 of 푝 such that ℎ ∘ 푓 = 푓 푙(푝) ∘ ℎ.

28 Figure 5.7:

Theorem 5.6 (Stable Manifold Theorem). Let 퐹 : 푅푛 → 푅푛 be a 퐶1-diffeomorphism with a 푠 푢 hyperbolic fixed point 푝. Then there are local stable and unstable manifolds 푊푙표푐(푝) and 푊푙표푐(푝) 푠 푢 ′ tangent to the characteristic spaces 퐸푝 and 퐸푝 of 푓 (푝) at 푝 and of corresponding dimensions. Global stable and unstable manifolds are deﬁned by taking the union of backward and forward iterates of the local manifolds. We have

푠 푛 푛 푊푙표푐(푝) = {푥 ∈ 푈|퐹 (푥) → 푝 as 푛 → +∞, and 퐹 (푥) ∈ 푈, ∀푛 ≥ 0}

푢 −푛 푛 푊푙표푐(푝) = {푥 ∈ 푈|퐹 (푥) → 푝 as 푛 → +∞, and 퐹 (푥) ∈ 푈, ∀푛 ≥ 0} and ⋃︁ 푠 −n 푠 푊 (푝) = 퐹 (푊푙표푐(푝)) 푛≥0 ⋃︁ 푢 푛 푢 푊 (푝) = 퐹 (푊푙표푐(푝)) 푛≥0 푠 푢 푛 푛 푊 (푝) and 푊 (푝) are images of immersions from 푅 푠 and R 푢 but may not be submanifolds of 푅푛. They may be rather complicated set as was shown by Barge [3] in the two following results. 2 2 1 2 Let 퐹 : R → R be a 퐶 -diffeomorphism and 푝 ∈ R a hyperbolic fixed point for 퐹 with one dimensional stable and unstable manifolds 푊 푠(푝) and 푊 푢(푝). Let 푊 푢+(푝) be one of the branches of 푊 푢(푝) and assume that 퐹 (푊 푢+(푝)) = 푊 푢+(푝) (otherwise, replace 퐹 by 퐹 2). We are going to show that under certain conditions the closure of 푊 푢+(푝) is an indecomposable continuum, as it was the case of the unstable set of the fixed point p0 in the horseshoe map. Let 퐵1 = [−1, 1] × [0, 1]. By virtue of Grobman-Hartman theorem 5.5 there exists an embedding

2 Ψ: 퐵1 → R 푠 푢+ −1 such that Ψ([−1, 1] × {0}) = 푊푙표푐(푝) and Ψ({0} × [0, 1]) = 푊푙표푐 (푝) and 퐴 = Ψ ∘ 퐹 ∘ Ψ is linear where deﬁned. We consider the following conditions on 퐹 :

푢+ (퐻1) 푊 (푝) is compact,

29 푢+ 푠 푠 (퐻2) There is an arc 훼 in Ψ(퐵1) ∩ 푊 (푝) such that 훼 ∩ 푊푙표푐(푝) ̸= ∅ and 훼 ̸⊂ 푊푙표푐(푝),

푢+ (퐻3) There exist −1 < 푎 < 0 < 푏 < 1 such that Ψ(푎, 0)), Ψ(푏, 0)) ̸∈ 푊 (푝).

푠 푢+ Remark 5.7. 1. Conditions (퐻1) and (퐻2) forbid the trivial situation where 푊 (푝) ⊂ 푊 (푝) and 푊 푢+(푝) is a simple closed curve.

2. The three conditions are satisﬁed when 푊 푢+(푝) is compact and 푊 푢+(푝) contains a trans- verse homoclinic point.

2 2 1 Theorem 5.8 (Barge). Suppose that 퐹 : R → R be a 퐶 -diffeomorphism with hyperbolic fixed 푢+ point 푝 as above that satisfies conditions퐻 ( 1), (퐻2) and (퐻3). Then 푊 (푝) is an indecomposable continuum.

2 푢+ Proof. Let 휑 : [0, +∞[→ R be a continuous one-to-one, onto parametrization of 푊 (푝). (1) ∀푡 ∈ F, we have 휑([푡, +∞)) = 푊 푢+(푝) 푠 푠 Indeed, let 훼 = 휑([푢, 푣]) where 휑(푢) ∈ 푊 (푝) and 훼 ̸⊂ 푊푙표푐(푝) be as in (퐻2). For 푛 ≥ 0, let 훼푛 푛 푛 be the component of 퐹 (훼) ∩ Ψ(퐵1) which contains 퐹 (휑(푡)). Hyperbolicity of 퐹 at 푝 implies 푢+ that lim sup(훼푛) = 푊푙표푐 (푝). But for all suﬃciently large 푛, 훼푛 ⊂ 휑([푡, +∞[) so that

푢+ 푊푙표푐 (푝) ⊂ 휑([푡, +∞)) and hence (1). (2) Suppose now that 푊 푢+(푝) is decomposable. Then there exists a proper sub-continuum 퐻 of 푊 푢+(푝) with nonempty interior (relatively to 푊 푢+(푝)). That is

∙ For all 푡 ≥ 0, 푖푛푡(퐻) ∩ 휑([푡, +∞[) ̸= ∅,

∙ For all 푡 ≥ 0, 휑([푡, +∞[) ̸⊂ 퐻 (since 퐻 is proper).

Therefore we can ﬁnd 0 ≤ 푡1 < 푡2 < 푡3 < 푡4 such that 휑(푡1), 휑(푡3) ̸∈ 퐻 but 휑(푡2), 휑(푡4) ∈ 퐻. Let 푁 be large enough so that

−푁 푢+ 퐹 (휑(푡푖)) ∈ 푊푙표푐 (푝)

−푁 for 푖 = 1, 2, 3, 4 and let 0 ≤ 푦1 < 푦2 < 푦3 < 푦4 be such that Ψ(0, 푦푖) = 퐹 (휑(푡푖)). Since 퐹 −푁 (퐻) is compact, it is possible to ﬁnd 0 < 푟 ≤ 1 such that

−푁 Ψ([−푟, 푟] × {푦푖}) ∩ 퐹 (퐻) = ∅ for 푖 = 1 and 푖 = 3. Choose 휖 > 0 such that if 퐿푎 = {푎} × [0, 휖] and 퐿푏 = {푏} × [0, 휖] we have

푢+ 푢+ Ψ(퐿푎) ∩ 푊 (푝) = Ψ(퐿푏) ∩ 푊 (푝) = ∅.

푛 푛 푛 푛 Then if we let 퐿푎 = 퐴 (퐿푎) ∩ 퐵1 and 퐿푏 = 퐴 (퐿푏) ∩ 퐵1,we have

푛 푢+ 푛 푢+ Ψ(퐿푎 ) ∩ 푊 (푝) = Ψ(퐿푏 ) ∩ 푊 (푝) = ∅, since 푊 푢+(푝) is invariant under 퐹 . 푛 Again hyperbolicity of 퐹 at 푝 guarantees that there exists 푛(푟) such that 퐿푎 separates 푛 {0} × [0, 1] from {푟} × [0, 1] in 퐵1 and 퐿푏 separates {0} × [0, 1] from {−푟} × [0, 1] in 퐵1. 푛 Let Γ be the rectangle made of the union of the four segments [−푟, 푟] × 푦푖 (푖 = 1, 3), 퐿푎 ∩ 푛 [−1, 1] × [푦1, 푦3] and 퐿푏 ∩ [−1, 1] × [푦1, 푦3] (see Figure 5.8). Then Γ separates the points (0, 푦2) −푁 and (0, 푦4), which contradicts the connectedness of 퐹 (퐻) and therefore of H.

30 Figure 5.8:

Theorem 5.9 (Barge). Let 퐹 : 푅2 → 푅2 be a 퐶1-diffeomorphism with hyperbolic fixedpoint 푝 such that 푊 푢+(푝) is an indecomposable continuum and 푊 푢+(푝) ∩ 푊 푠(푝) ⊂ 푊 푢+(푝) Then for no continuous map 푓 of a branched one-manifold 퐾 is the restriction of 퐹 to 푊 푢+(푝) topologically conjugate to the induced map 푓ˆ. Proof. Suppose there is a branched one-manifold 퐾, a continuous map 푓 : 퐾 → 퐾 and a 푢+ −1 homeomorphism ℎ from 푊 (푝) onto (퐾, 푓) such that ℎ ∘ 푓 ∘ ℎ ⋂︀= 퐹 . Without loss of generality, we may assume that 푓 is onto (otherwise replace 퐾 by 퐿 = 푓 푛(퐾)). ⋃︀ 푛≥0 푢+ 푢+ 푢+ (1) Since 푊 (푝) = 푛≥0 휑([0, 푛]), the composant 퐶푝 of 푝 in 푊 (푝) contains 푊 (푝). 푠 푢+ 푢+ Hence 푊 (푝)∩푊 (푝) which is the stable set of 푝 in 푊 (푝) is contained in 퐶푝 and thus,푆(ℎ(푝)), the stable set of ℎ(푝) in (퐾, 푓) is contained in 퐶ℎ(푝). (2) It is clear from the deﬁnition of a branched one-manifold, that there exist an 푙 ∈ N such any collection of 푙 distinct points in 퐾 destroys the connectivity of 퐾. Let ℎ(푝) = (푝0, 푝0,... ) 푙−1 (푝0 ∈ 퐾) and x = (푥0, 푥1,... ) ∈ 푆(ℎ(푝)) − {ℎ(푝)}. Then x, 푓ˆ(x),..., 푓ˆ (x) are 푙 distinct points in 푆(ℎ(푝)). Thus, there exists 푁 ≥ 0 such that ˆ푖 ˆ푗 휋푁 (푓 (x)) ̸= 휋푁 (푓 (x)) for 0 ≤ 푖 < 푗 ≤ 푙 − 1 (where 휋푁 :(푋, 푓) → 퐾 is the projection onto the 푁-th coordinate). Therefore, ˆ ˆ푙−1 푙−1 퐾 − {휋푁 (x), 휋푁 (푓(x)), . . . , 휋푁 (푓 (x))} = 퐾 − {푥푁 , 푓(푥푁 ), . . . , 푓 (푥푁 )} is not connected. (3) Since (퐾, 푓) is indecomposable, there exists at least one composant 퐶 in (퐾, 푓)such that 퐶 ∩ 퐶ℎ(푝) = ∅. But 퐶 is dense in (퐾, 푓) and connected so 휋푛(퐶) is dense and connected 푙−1 in 퐾. Hence 휋푁 (퐶) ∩ {푥푁 , 푓(푥푁 ), . . . , 푓 (푥푁 )}= ̸ ∅. Therefore, it is possible to ﬁnd y = 푖 (푦0, 푦1,... ) ∈ 퐶 with 휋푁 (y) = 푦푁 = 푓 (푥푁 ) for some 0 ≤ 푖 ≤ 푙 − 1. But then 푛 푛+푁 푖 푛+푖 lim 푓 (푦0) = lim 푓 (푓 (푥푁 )) = lim 푓 (푥0) = 푝0 푛→+∞ 푛→+∞ 푛→+∞

31 so that y ∈ 푆(ℎ(푝)) ⊂ 퐶푝 which gives a contradiction.

Exercises

Exercise 5.1. Show that the unstable set of 0 in Theorem 5.3 is a continuous, one-to-one image of the real line.

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Thanks

I express my gratitude to Lucien Guillou and Marie-Christine P´erou`emefor their help during the preparation of these notes and William Bonnaz for the drawings.