BACHELOR THESIS

Tom´aˇsYe

Rectangles inscribed in Jordan curves

Mathematical Institute of Charles University

Supervisor of the bachelor thesis: doc. RNDr. Zbynˇek S´ır,Ph.D.ˇ Study programme: Mathematics Study branch: General mathematics

Prague 2018 I declare that I carried out this bachelor thesis independently, and only with the cited sources, literature and other professional sources. I understand that my work relates to the rights and obligations under the Act No. 121/2000 Sb., the Copyright Act, as amended, in particular the fact that the Charles University has the right to conclude a license agreement on the use of this work as a school work pursuant to Section 60 subsection 1 of the Copyright Act.

In ...... date ...... signature of the author

i Title: inscribed in Jordan curves

Author: Tom´aˇsYe

Institute: Mathematical Institute of Charles University

Supervisor: doc. RNDr. Zbynˇek S´ır,Ph.D.,ˇ department

Abstract: We will introduce quotients, which are very special kinds of continuous maps. We are going to study their nice universal properties and use them to formalize the notion of topological gluing. This concept will allow us to define interesting topological structures and analyze them. Finally, the developed theory will be used for writing down a precise proof of the existence of an inscribed in any Jordan curve.

Keywords: Jordan curve, rectangle, topology

ii I would like to thank my supervisor doc. RNDr. Zbynˇek S´ır,Ph.D.ˇ for helping me willingly with writing of this bachelor thesis. It was a pleasure working under his guidance.

iii Contents

Introduction 2

1 Topological preliminaries 3 1.1 General topology ...... 3 1.2 Topological gluing ...... 6 1.3 Special quotient spaces ...... 14

2 Rectangles and curves 21 2.1 Jordan curves ...... 21 2.2 Rectangles inscribed in curves ...... 25 2.3 Remarks ...... 31

Conclusion 33

Bibliography 34

List of Figures 35

1 Introduction

The theory of curves has been a classical mathematical discipline for centuries. One of the reasons for that is that curves naturally show up in many areas of mathematics, for instance in differential , algebraic geometry, complex analysis, and so on. What makes the study of curves both fulfilling and depressing is the fact that most of the problems involving them are very hard. No exception to this is the inscribed rectangle problem, on which we will take a close look in this thesis. In 1977, mathematician Vaughan proved that every Jordan curve in the plane contains an inscribed rectangle. To do this, he used a gluing argument combined with the fact that projective plane cannot be embedded into three dimensional . The essential trick of the proof is that every simple closed curve has a hidden topological structure homeomorphic to the Mobius strip. To make this notion precise, one has to formalize the concept of topological gluing. This is best done via quotients, which are special kinds of surjective topological maps that naturally preserve continuity of other functions. However, even with the machinery of quotient maps, there are still some technical subtleties left that need to be resolved to make Vaughan’s argument bulletproof. The purposes of the thesis is to prepare all the theoretical arsenal needed for making Vaughan’s argument rigorous and then precisely proving the inscribed rectangle theorem. The entire content of the paper is divided into two chapters. The first chapter is purely topological whereas the second one is focused on curves and on the final proof. Starting off as elementary as possible, assuming onlythe basics of general topology, we are going to define quotients and study how they interact with other maps. We will see that any projection induced by an equiv- alence relation is a quotient which will be the key for understanding topological gluing. In that regard, we are going to discover some general truths about the re- lationship between quotients and equivalences. These results will then enable us to comfortably work with concrete topological spaces and prove useful identities. The second chapter will begin with some basic properties of curves followed by intuitive, yet extremely hard to prove, results like the theorems of Jordan and Schoenflies. In the last section, there can be found the desired proof ofthe inscribed rectangle theorem together with reasoning why the use of topology in the proof was necessary.

2 1. Topological preliminaries

The following section will serve as an intro to the topological notions we will use in our proofs. Although later we will be working exclusively with metric spaces and quotients of metric spaces, there is no reason to limit ourselves to these special cases right now since most of the theory we will use works in general and is actually easier to get a grasp on when done generally. The basic knowledge of general topology including metric spaces is assumed. Throughout the paper, we will use standard mathematical notation.

1.1 General topology

All the proofs of theorems in this section and much more can be found in a remarkable topological textbook Munkres [2000]. The entire section is contained within the chapters 2 and 3. We will only mention results that are going to directly impact the proofs done in latter sections. Definition 1.1. Let X be an abstract set. By topology on X we mean a collection τ ⊂ 2X that is closed under finite intersections and arbitrary unions and contains ∅ and X. Remark. Throughout the paper, we are going to follow this standard topological terminology: The pair (X, τ) is called a topological space. The sets in the col- lection τ are called open, their complements are called closed. If the topology τ is known from context, we will omit it and just talk about open subsets of X. Set U ⊂ X is called a neighborhood of a point x ∈ X if there exists an open set G ⊂ U such that x is in G. Subcollection B ⊂ τ is called a base of topology τ if any open set can be written as the union of some subcollection B˜ ⊂ B. If Y is a subset of X, then the collection τ˜ = {G ∩ Y : G ∈ τ} is a topology on Y and (Y, τ˜) is called a topological subspace of X. In the following pages, when we say that a subset Y of X has some topological property we mean that Y has such property as a space equipped with the subspace topology. Definition 1.2. Topological space is called Hausdorff if every two distinct points can be separated by disjoint open sets.

Hausdorffness is also known as the T2 separation axiom. There are other separation axioms with different separation conditions but for our purpose, the Hausdorff axiom will be enough. The reason for this is that continuous functions into Hausdorff spaces, in some sense, behave well. We will discuss precisely how ”well” in the following pages.

Figure 1.1: Visualization of the Hausdorff property

3 Lemma 1.3. Subspace of a Hausdorff space is Hausdorff. Theorem 1.4. Let (X, d) be a metric space and let B(x, δ) be the open ball centered at x with radius δ > 0, then the system τ defined by ∀G ⊂ XG ∈ τ ⇐⇒ ∀x ∈ G ∃δ > 0 : B(x, δ) ⊂ G is a Haussdorff topology on X and the collection {B(x, δ): x ∈ X, δ > 0} is a base of the induced topology. Topology τ will be referred to as the topology induced by metric d. Definition 1.5. Map f : X −→ Y between topological spaces X and Y is called continuous if one of the following equivalent conditions holds 1. f −1(G) is open for every open G ⊂ Y 2. f −1(F ) is closed for every closed F ⊂ Y Moreover, if f is a bijection and f, f −1 are both continuous then f is called a homeomorphism. Remark. Equivalence of 1. and 2. follows from the identity f −1(Y \ G) = f −1(Y ) \ f −1(G) = X \ f −1(G) Theorem 1.6. For continuous maps the following statements are true: 1. Composition of continuous maps is continuous 2. Restriction of a continuous map is again a continuous map with respect to the subspace topology Theorem 1.7. Let f : X −→ Y be a map between two metric spaces X and Y. Then f is continuous in the metric sense if and only if f is continuous in the topological sense with respect to topologies induced by the metrics. The motivation for creating topological spaces was to find a way to formalize the notions of closeness and continuity which is broader then the ϵ − δ approach with metrics, as these concepts also make sense in non-metrizable structures. According to the previous theorem, the given definition of topological spaces suc- ceeds in this regard.

Two topological spaces X and Y are said to be homeomorphic, if there exists a homeomorphism between them. This fact will be denoted X ≃ Y . The relation ≃ is an equivalence on the class of all topological spaces. If X ≃ Y and Y is a subspace of Z, then we say that X embeds into Z and the corresponding homeomorphism is called an embedding. Homeomorphism transforms open sets into open sets and since topological properties are defined through open sets, in a purely topological sense, homeomorphic spaces are the same. Example. The mapping

X idX :(X, 2 ) −→ (X, {∅,X}) is a continuous bijection which is never a homeomorphism (unless X is a one-point set).

4 We see, that a continuous bijection between two general topological spaces need not be a homeomorphism. However, if we include some additional properties we get a sufficient condition for homeomorphism.

Definition 1.8. Let (X, τ) be a topological space. Subcollection G ⊂ τ is called an open cover of X if ∪G = X

Definition 1.9. Hausdorff space X is called compact, if every open cover of X contains a finite subcover.

The definition of compactness might have sense without the Hausdorff prop- erty but most of the important properties of compact spaces require this assump- tion so we will add it to our definition from the beginning. Compact sets are in many cases similar to finite sets. The ability to transform infinite covers into finite ones has vast consequences even though they maynot be evident upon first sight. Thankfully, in certain topological spaces, there are other, more intuitive, conditions equivalent to this general concept.

Theorem 1.10. For a subset X of a Rn, the following conditions are equivalent 1. X is compact

2. Every infinite sequence of elements in X contains a convergent subsequence

3. (Heine-Borel theorem) X is complete and totally bounded

4. X is bounded and closed

Theorem 1.11. (Properties of compact spaces) Let f : X −→ Y be a continuous map between a compact space X and a Hausdorff space Y . Then it holds:

1. Any closed subset of X is compact

2. Any compact subset of Y is closed in Y

3. f(X) is compact.

4. If f is bijective, then f is a homeomorphism.

From now on, any subset of the Euclidean plane/space will be viewed as a topological space with topology inherited from the Euclidean metric.

5 1.2 Topological gluing

Generally, when given a set with some mathematical structure, it is often use- ful to divide the set into disjoint groups of elements sharing some property and operating only with such groups. In abstract algebra, for instance in the theory of rings, this is done by introducing an equivalence relation which preserves the ring structure and naturally defining addition and multiplication on the setof equivalence classes. This can also be done with topological spaces. However, unlike in algebraic structures, the choice for a topology on the factor set is not uniquely determined by the initial topology and we must choose. The purpose of this chapter is to make such choice and explain why is this choice natural and that it actually formalizes the concept of gluing points together. The content of this section is partially inspired by Chapter 22 of Munkres [2000] but the order of theorems, as well as the proofs, may vary. In addition to what can be found in the textbook, we will, from scratch, develop theory around equiv- alences on topological spaces and study how they interact with homeomorphisms. Though these results are not necessarily something new, they are also not some- thing commonly included in topological literature. Later, this might seem a bit surprising since, at least in our case, they will turn out to be very useful for proving important topological identities. Definition 1.12. A surjective map p: X −→ Y between topological spaces X and Y satisfying G ⊂ Y is open in Y ⇐⇒ p−1(G) ⊂ X is open in X is called a quotient. Quotient maps are obviously continuous but their importance stems from their ability to naturally preserve continuity of other maps. We will get back to what this precisely means later. For now, we will try to focus on recognizing when a surjective continuous map is a quotient. Generally, this is hardly the case. However, if one is lucky enough to work with Hausdorff and compact spaces (which is exactly our case), the situation becomes much more optimistic. Definition 1.13. A topological map is called open (resp. closed) if images of open (resp. closed) sets are open (resp. closed). Remark. Topological mapping is a homeomoprhism if and only if it is a continuous bijection and an open map (or a closed map). Lemma 1.14. For a map p: X −→ Y between topological spaces, the following conditions are equivalent: 1. p is a quotient 2. F ⊂ Y is closed in Y ⇐⇒ p−1(F ) ⊂ X is closed in X Proof. Take F ⊂ Y arbitrary. Implication ”1 ⇒ 2” follows immediately from the upcoming calculation, F is closed in Y ⇐⇒ Y \ F is open in Y ⇐⇒ p−1(Y \ F ) = X \ p−1(F ) is open in X ⇐⇒ p−1(F ) is closed in X The implication ”2 ⇒ 1” is done symmetrically.

6 Remark. This easy lemma allows us to immediately put together some sufficient conditions for a surjective map to be a quotient.

Corollary 1.15. Consider a continuous surjective map p: X −→ Y . If p is an open map (or closed map) then p is a quotient.

Proof. Because p is onto, it holds that

p(p−1(G)) = G (1.1) for every G ⊂ Y . If p is an open map, the condition for a quotient follows trivially from (1.1). If p is a closed map, then we use the previous lemma together with (1.1) It might seem that the condition is not really helpful, since being an open ( or a closed) map continuous map is a much more special property than just being a quotient. What is useful to realize is that a continuous maps between compact spaces and a Hausdorff space is always closed.

Lemma 1.16. If p: X −→ Y is a continuous map between a compact space X and a Hausdorff space Y, then p is closed.

Proof. Take F ⊂ X arbitrary closed set. Theorem 1.11 provides us with this sequence of arguments: Since X is compact, so is F . Then p(F ) is a compact subset of Y and since Y is Hausdorff, p(F ) is closed in Y. We started with an arbitrary closed set and showed that the image is again closed, hence p is a closed map.

Corollary 1.17. Continuous surjective map with compact domain and Hausdorff codomain is a quotient.

Proof. From previous lemma such maps must be closed and therefore quotients.

Obviously, homeomorphisms are quotients. On the other hand, it is impor- tant to realize that quotient maps are not that far from being homeomorphisms themselves. This is meant in the sense that if p−1 existed, it would have been continuous, which is the assertion of the following lemma.

Lemma 1.18. If a quotient map p: X −→ Y is injective, then it is a homeo- morphism.

Proof. The only thing to prove is that p is an open map. Take G an arbitrary open subset of X. Because p is a bijection, it holds that G = p−1(p(G)) and since p is quotient, p(G) is open in Y . Now we will clarify what was meant by the saying: ”Quotients naturally preserve continuity of other maps.”

Theorem 1.19. Let X, Y and Z be topological spaces and let p: X −→ Y and f : Y −→ Z be maps. Assume p is a quotient. Then f is continuous if and only if f ◦ p is continuous.

7 Remark. The assertion of Theorem 1.19 is best remembered by the following commutative diagram p X Y

f f◦p Z Proof. By definition, we need to show that preimages of open sets areopen sets. Take G arbitrary open subset of Z. First, assume f is continuous. Then (f ◦ p)−1(G) = p−1(f −1(G)) is open since both f and p preserve open sets through preimages. Conversely, if f ◦ p is continuous then (f ◦ p)−1(G) = p−1(f −1(G)) is open in X and since p is a quotient, then f −1(G) is open in Y which gives the continuity of f. Remark. The first argument works generally and proves that arbitrary composi- tion of continuous maps is continuous. The second implication relied heavily on the fact that p is a quotient and generally is not true. Imagine now that we have a topological space X and an abstract set A with no topological structure whatsoever. Let p: X −→ A be a surjection and consider the system τ = {G ⊂ A : p−1(G) is open in X} Since −1 ⋃ ⋃ −1 p ( Gi) = p (Gi) i i −1 ⋂ ⋂ −1 p ( Gi) = p (Gi), i i τ is a topology on A. More importantly, if we equip A with the topology τ then p is a quotient map. Definition 1.20. Let X, A, p and τ be as above. The topology τ is called strong topology induced by the map p. The term strong comes from the fact that τ is the largest (in topological terms finest), topology such that p is continuous. Continuity of p is easy to assure, for instance by taking the indiscrete topology {∅,A}. However, τ is special because by adding any new subset of A to the system τ, p becomes discontinuous. The above construction may be applied to any triplet (X, A, p) where X has a topology, A is an ordinary set and p is a surjection between them. Let us take a look at a very important special case. Consider an arbitrary topological space X and let ∼ be some equivalence on X.

Denote X∼ = {[x]∼ : x ∈ X}, where [x]∼ = {y ∈ X : x ∼ y}. Right now X∼ is just a factor set with no topology. We could literally choose any topology on X∼ and proclaim it to be the quotient topology. However, every equivalence comes with a surjective canonical projection map

π : X −→ X∼

x ↦−→ [x]∼ so it feels natural to apply the construction in definition 1.20 to the triplet (X,X∼, π).

8 Definition 1.21. Let X, ∼ and π be as above. The strong topology τ on X∼ induced by the projection π is called the quotient topology and (X∼, τ) is called the quotient space of X with respect to equivalence ∼.

Remark. An equivalence ∼ on a topological space X is basically a prescription on how to glue certain points of X together. The topology on X∼ is defined so that the following assertion holds. If f : X −→ Y is a continuous map and the derived map

fˆ: X∼ −→ Y

[x]∼ ↦−→ f(x) is well defined, then it is continuous. This is guaranteed by Theorem 1.19. This implies that continuous maps on X∼ are precisely those maps derived from con- tinuous maps on X. This fact, in a specific sense, means that the topology on X∼ is the closest it can be to the original topology on X. It is only natural to require the glued space to be topologically similar to the original space. By this reasoning, our definition of a quotient topology, as being the topology onthe factor set X∼ that is most similar to the initial topology on X with respect to continuous maps, is justified.

Lemma 1.22. Quotient of a compact space is compact.

Proof. Let X be compact and ∼ an equivalence on X. Then X∼ = π(X), where π is the canonical projection induced by ∼. Projection π is continuous and so by theorem 1.11, X∼ is compact as the continuous image of a compact space. In general, given two sets X, Y and a map f : X −→ Y , the relation

x1 ∼f x2 ⇐⇒ f(x1) = f(x2) (1.2) is an equivalence on X. This equivalence will be called the kernel of f. Let us denote Xf the set of kernel’s equivalence blocks and πf the corresponding canonical projection. Then the map

f˜: Xf −→ Y (1.3)

[x]f ↦−→ f(x) is well defined and injective. If f is onto, then f˜ is bijective. This is a general construction where we factor out the noninjectivity of f by identifying points which have the same image. When X and Y are topological spaces, it is tempting to ask, when is f˜ a homeomorphism.

Theorem 1.23. (Homeomorhpism theorem) Let X, Y be topological spaces and assume f : X −→ Y is a surjective map. Then f˜ is a homeomorphism if and only if f is a quotient.

πf X Xf

f f ˜ Y

9 Proof. We know that f˜ is a bijection. Let π = πf denote the canonical projection of X onto Xf . It holds that f = f˜ ◦ π. First, let us assume that f is a quotient. Since π is also a quotient, by theorem 1.19 f˜ is continuous. Take G an arbitrary open subset of X˜. It holds that

x ∈ f −1(f˜(G)) ⇐⇒ f(x) ∈ f˜(G)

⇐⇒ f˜([x]f ) ∈ f˜(G)

⇐⇒ [x]f ∈ G

⇐⇒ x ∈ π−1(G) which gives us f −1(f˜(G)) = π−1(G). The third equivalence is true because f˜ is injective. Since π−1(G) is open in X, f˜(G) must be open in Y because f is a quotient. This means that that f˜ is an injective open map which is the same as saying f˜−1 is continuous. Hence f˜ is a homeomorphism. Conversely, if f˜ is a homeomorphism then it is also a quotient and that means f is a composition of two quotients which is again a quotient. Corollary 1.24. Let X,Y be topological spaces and let f : X ↦−→ Y be a contin- uous map. If X is compact and Y is Hausdorff, then Xf ≃ Im(f). Proof. f is a surjective continuous map from X onto Im(f). X is compact and Im(f) is Hausdorff, as it is a subspace of Hausdorff space Y . By Corollary 1.14 f is a quotient and previous theorem therefore implies that f˜ is a homeomorphism.

Corollary 1.24 gives us a powerful tool that helps us determining the structure of certain quotient spaces. Example. Everybody knows that a cylinder can be obtained by gluing two sides of a rectangle together. Using the above developed theory we can make this concept topologically precise. Let us denote X = {(s, t): 0 ≤ s ≤ 2π , 0 ≤ t ≤ 1} Y = {(x, y, z): x2 + y2 = 1 , 0 ≤ z ≤ 1} Then X is a rectangle and Y is a cylinder. Consider the map φ: X −→ Y (s, t) ↦−→ (cos(s), sin(s), t) Obviously, φ is continuous and onto. Since both X and Y are Hausdorff and compact, by theorem above Xφ ≃ Y . Now we need to determine Xφ. By definition of kernel of φ we have

(s1, t1) ∼φ (s2, t2) ⇐⇒ φ(s1, t1) = φ(s2, t2) After few simple calculations we obtain: ⎧ ⎨{(s, t)} if 0 < s < 2π [(s, t)]φ = (1.4) ⎩{(0, t) , (2π, t)} if s ∈ {0, 2π} which means that Xφ is exactly the rectangle X with two opposite edges glued together.

10 Sometimes it might happen that on a single topological space X there are multiple meaningful equivalence relations. On the set of all equivalences on X, there is a simple ordering ≼ defined as follows:

( ∼1 ≼ ∼2 ) ⇐⇒ ( ∀x, y ∈ X : x ∼1 y =⇒ x ∼2 y ) (1.5) When we view equivalences as special subsets of X × X, then the order ≼ co- incides with the standard set-theoretic inclusion. Important observation is that when two equivalences are comparable, there is a natural connection between the corresponding quotient spaces.

Lemma 1.25. Let ∼1 and ∼2 be equivalences on a topological space X and assume ∼1 ≼ ∼2. For i = 1, 2 let us denote Xi the quotient space with respect to equivalence ∼i. Then the map

π : X1 −→ X2

[x]1 ↦−→ [x]2 is a well defined quotient. Proof. Condition (1.5) assures that the image of an equivalence class is indepen- dent of its representative and therefore π is well defined. Let us denote πi the corresponding canonical projection of X onto Xi, for i = 1, 2. By definition it holds that π2 = π ◦ π1. Since π1 is a quotient and π2 is continuous, by Theorem 1.19, π is continuous. To prove that π is a quotient, take G ⊂ X2 arbitrary and −1 assume π (G) is open in X1. Because π1 is continuous, −1 −1 −1 −1 π1 (π (G)) = (π ◦ π1) (G) = π2 (G) is open in X and since π2 is a quotient, G is open in X2.

π1 X X1

π π2

X2

The previous lemma gives us a tool to treat multiple equivalence relations on a single topological space. Now, consider the opposite case where we have two spaces X and Y and only a single equivalence ∼1 defined on X. Given a bijection f : X −→ Y , the relation ∼2 defined by −1 −1 y1 ∼2 y2 ⇐⇒ f (y1) ∼1 f (y2) is an equivalence on Y , as one may easily verify. This allows us to naturally transport the equivalence from X onto Y . Again, this is a completely general construction that can be done for any two abstract sets. However, if X and Y are topological spaces and f is homeomorphism, we shall investigate the relationship between the two corresponding quotient spaces X1 = X∼1 and Y2 = Y∼2 . First, let us notice, that

[x1]1 = [x2]1 ⇐⇒ x1 ∼1 x2 ⇐⇒ f(x1) ∼2 f(x1)

⇐⇒ [f(x1)]2 = [f(x2)]2

11 which makes the map

fˆ: X1 −→ Y2

[x]1 ↦−→ [f(x)]2 a well defined bijection. It comes as no surprise that continuity of fˆ strongly depends on continuity of f. The following theorem will describe this correlation precisely.

Theorem 1.26. Let X, Y , f and fˆ be as above. If f is a homeomorphism then fˆ is also a homeomorphism.

Proof. We already know that fˆ is one-to-one so it is sufficient to show that fˆ and −1 fˆ are continuous. Let us denote by πi the canonical projection induced by ∼i, for i = 1, 2. Then it holds that

π2 ◦ f = fˆ ◦ π1 which can be writen in a commutative diagram like this:

f X Y

π1 π2

X1 Y2 fˆ

Since f is continuous, π2 ◦ f is continuous, which gives us the continuity of fˆ ◦ π1 and because π1 is a quotient map, again, Theorem 1.19 implies that fˆ is continuous. To obtain continuity of fˆ−1, we can apply exactly the same arguments to the diagram

f −1 X Y

π1 π2

X1 Y2 fˆ−1 and use that π2 is a quotient.

The last few paragraphs of this general section will be devoted to gluing of topological maps. When searching for a continuous transformations, it is often useful to divide a topological space into parts and define continuous maps on each part separately. The following lemma gives a criterion on when such partition actually defines a continuous map on the whole space. Theorem 1.27. (Gluing lemma) Let X and Y be topological spaces and let A and B be closed subsets of X such that X = A ∪ B. If f : A → Y and g : B → Y are continuous and f(x) = g(x) ∀x ∈ A ∩ B, then the map h: X → Y given by formula ⎧ ⎨f(x) if x ∈ A h(x) = ⎩g(x) if x ∈ B is a well defined continuous map from X to Y .

12 Proof. Condition f = g on A ∩ B assures that h is well defined. To prove continuity of h, it suffices to show that preimages of closed sets are closed. Take F ⊂ Y arbitrary closed subset. Then

h−1(F ) = f −1(F ) ∪ g−1(F )

From continuity of f we get that f −1(F ) is a closed subset of A in the subspace topology. Therefore there exists a closed subset Fˆ of topological space X such that f −1(F ) = Fˆ ∩ A But because A is closed in X, f −1(F ) must also be closed in X since it is an intersection of two closed sets. Analogically we can prove that g−1(F ) is closed in X by using that B is closed. So h−1(F ) is a union of two closed sets and therefore a closed set. Because F was arbitrary h must be continuous. The map h will be referred to as the gluing map constructed from f and g and will be denoted h = f ∪ g. The gluing lemma is a nice tool and can be used almost always since the as- sumptions are very general. It is especially useful when dealing with quotients of compact spaces.

Corollary 1.28. Let X, Y , f, g and h be as in the gluing lemma. If X is compact, Y Hausdorff and f(A) ∪ g(B) = Y . Then Xh ≃ Y . Proof. Assumption f(A) ∪ g(B) = Y assures that h is onto and we may proceed analogically as in the proof of corollary 1.20. Using this last result, we can find a homeomorphism between a quotient space X∼ and a general topological space Y using the following procedure: 1. Assure X is compact and Y Hausdorff

2. Divide X into closed subsets A and B and construct continuous maps

f : A −→ Y and g : B −→ Y such that f(A)∪g(B) = Y and fA∩B = gA∩B

3. Define the gluing map h, which will be a quotient and prove ∼ = ∼h

4. h˜ is the wanted homeomorphism

In the following proofs we will use this technique several times.

13 1.3 Special quotient spaces

In this section we will use the machinery of quotients to define and examine topological structure of the Mobius strip and the Real projective plane. Our main concern will be to prove equivalence between different definitions of the Projective plane and then to show that that it can decomposed into a disk and a Mobius band. Even though these results are well known within the mathematical community, the proofs of these facts presented in standard text books, if any, are a lot of hand-wavy and rely heavily on geometric intuition. Since precision will be necessary in the second chapter, proofs in this section contain original formulas designed explicitly to describe the involved transformations. Notation. For the sake of avoiding confusion when it comes to using quotient spaces in practice, we will shorten the notation when talking about equivalences. When defining a concrete equivalence on a given space X, we will only describe nontrivial identification of elements and the left out elements will be thought of as single point equivalence classes.

Also, from now on we will use this simplified notation

3 S = S2 = {x ∈ R : ∥x∥ = 1}

2 B = B2 = {x ∈ R : ∥x∥ ≤ 1} Example. Using the convention above, we could describe the kernel of φ from (1.4) on page 10 more compactly like

∀t ∈ [0, 1]: (0, t) ∼φ (1, t)

In the following chapters, we will continue to use this convention. Most of the theory we have developed so far crucially depends on the Hausdorff property of participated spaces. Unfortunately, useful criteria describing when a quotient space is Hausdorff are quite complicated and would also require defining some additional concepts. Thankfully, all the equivalences that we are going to define have the beautiful property of preserving Hausdorffness onto the quotient space. This could be easily checked straight from definition in the case of every such space. For convenience, we will not do it but we shall keep at the back of our minds that it is true and that it is important.

Definition 1.29. Let ∼1 denote the equivalence on the unit defined as follows: (t, 0) ∼1 (1 − t, 1) ∀t ∈ [0, 1] The quotient space [0, 1]2 is called the Mobius strip and will be denoted . ∼1 M Remark. The Mobius strip, similarly like the cylinder, can be obtained from a rectangle by gluing opposite sides to each other. However, this time the gluing requires a continuous half-twist of one of the edges. This gluing can be visualized in R3. In fact, the Mobius strip is mostly known in this form.

14 Figure 1.2: Mobius strip embedded in R3

Lemma 1.30. The following topological spaces are homeomorphic:

1. ( 3 \{0}) where x ∼ y ⇐⇒ ∃λ ̸= 0 such that x = λy R ∼2 2

2. S∼3 where x ∼3 −x ∀x ∈ S

3. B∼4 where x ∼4 −x ∀x ∈ ∂B = S1

3 Proof. For the purpose of the proof, let us denote V = R \{0} and V2 = ( 3 \{0}) . R ∼2 1. ≃ 2. Consider the map

P : V −→ S x x ↦−→ ∥x∥

We see that P projects a vector x ∈ V onto the unit sphere S in the direction of x. Obviously, P is well defined, continuous and onto. Unlike in the case ofthe example from previous section, V is not compact and we need to show that P is a quotient mapping some other way. To do this, we will prove, that P is an open map. From basic geometric intuition about the projection P we can see that for any open ball B ⊂ V , the set P (B) is an open copula-like region on the sphere. ⋃ Additionally, any open G ⊂ V can be written as G = i Bi where {Bi}i is some collection of open balls. Therefore

⋃ ⋃ P (G) = P ( Bi) = P (Bi) i i which is open as the union of open sets. Hence, P is an open map. Corollary 1.15 grants us that P is a quotient and therefore, from the Homeomorphism Theorem (Theorem 1.23), we obtain S ≃ VP with the concrete homeomorphism

µ: x ↦−→ [x]P From the fact that ∥λx∥ = |λ| ∥x∥ we conclude that

x ∼P y ⇐⇒ ∃λ > 0 such that x = λy

15 This means, that ∼P ≼ ∼2, so by Lemma 1.25 we get a natural quotient

π :[x]P ↦−→ [x]2 from VP onto V2. If we compose π with the homeomorphism µ we get yet another quotient

ϕ: x ↦−→ [x]2 which maps S onto V2. This time the domain is compact so from Corollary 1.24, we readily deduce that Sϕ ≃ V2.

µ S VP

ϕ π

V2

It is now sufficient to prove that ∼ϕ = ∼3. Let us calculate:

x ∼ϕ y ⇐⇒ ϕ(x) = ϕ(y)

⇐⇒ [x]2 = [y]2 ⇐⇒ x ∼2 y ⇐⇒ ∃λ ̸= 0 such that x = λy

By taking the Euclidean norm of both sides in the last equality and using that

∥x∥ = 1 = ∥y∥ we obtain that |λ| = 1 ⇐⇒ λ = ±1 This gives us x ∼ϕ y ⇐⇒ x = ±y ⇐⇒ x ∼3 y which completes the proof of V2 ≃ S∼3

2. ≃ 3. 3 For a vector x ∈ R let us denote by x1, x2, x3 its Cartesian coordinates, meaning that x = [x1, x2, x3]. Define SN = {x ∈ S : x3 ≥ 0} and SS = {x ∈ S : x3 ≤ 0}. Both SN and SS are closed subsets of S and SN ∩ SS = S1 × {0} ⊂ S. On the set SN consider the equivalence ∼ defined by

x ∼ −x ∀x ∈ S1 × {0} ⊂ SN Now consider the maps

fN : SN −→ SN ∼

x ↦−→ [x]∼

fS : SS −→ SN ∼

x ↦−→ [−x]∼

16 Both of these maps are continuous because they can be written as composition of a continuous map with the canonical projection π∼. For a given point x = [x1, x2, 0] ∈ S1 × {0} we see that

fN (x) = [x]∼ = [−x]∼ = fS(x) from which we deduce that the gluing map f = fN ∪fS is well defined and contin- uous (Lemma 1.27). It is also a quotient because it is onto and the participating spaces are all Hausdorff and compact. From definition ofmaps fN and fS we quickly conclude that

x ∼f y ⇐⇒ x = ±y ⇐⇒ x ∼3 y

This, combined with the Theorem 1.24, gives us S∼3 ≃ SN ∼. Therefore, it suffices to show SN ∼ ≃ B∼4 . To see this, we just need to realize that the natural projection [x1, x2, x3] ↦−→ [x1, x2] is a continuous bijection between SN and B. Since both of these spaces are Hausdorff and compact, it is a homeomorphism. Furthermore, this bijection transports the equivalence ∼ onto ∼4 in the sense of Theorem 1.26 and therefore

SN ∼ ≃ B∼4 , which was desired. The topological spaces defined in the lemma above were proved to be home- omorphic. This basically means that even though they look different upon first sight, their topological properties are the same. The topological structure rep- resented by these spaces will be very important for us in the latter proofs so we will give it a name. Definition 1.31. The topological spaces from the previous Lemma 1.30 will be all referred to as the Real projective plane and denoted P2 Remark. Projective plane is an interesting entity that shows up in many areas of mathematics. As we can see from the previous lemma 1.30, P2 has many topologically equivalent definitions. According to the first one, P2 is the set of lines in R3 going through the origin. The notion of quotient spaces allows to give topological meaning to this purely geometric entity. The second form is a sphere with opposite points glued together. It is intuitive that these spaces are homeomorphic, because each line intersects the sphere in exactly two opposite points (called antipodal points). Therefore, a line can be represented by a pair of antipodal points on the sphere and vice versa. In the previous lemma, we had to show that this one-to-one correspondence is homeomorphic. Sometimes, using one definition of P2 may be more efficient than the other, de- pending on the specific problem that is being solved. In our case, the third form will do the trick. Also, to make the following proofs more elegant, instead of the unit ball, we shall work with a ball of radius 2. This can be done because the mapping x ↦−→ 2x is a self homeomorphism of R2. Definition 1.32. From now on, every time the symbol P2 shows up, we are referring to this specific quotient space:

{x ∈ 2 : ∥x∥ ≤ 2} where x ∼ −x ∀x ∈ 2 such that ∥x∥ = 2 (1.6) R ∼5 5 R

17 Corollary 1.33. P2 is Hausdorff and compact. Proof. Hausdorffness can be easily checked from definition. As for compactness, P2 is a quotient of a compact space. Then by lemma 1.22, it must be compact.

Theorem 1.34. (Decomposition theorem) There exist closed subsets X,Y ⊂ P2 satisfying the following conditions:

1. X ∪ Y = P2

2. X ∩ Y ≃ S1 3. X is homeomorphic to the unit ball B

4. Y is homeomorphic to the Mobius strip M Proof. For convenience, let ∼ denote the equivalence in definition 1.32. Let us define: 2 X = {x ∈ R : ∥x∥ ≤ 1}∼ 2 Y = {x ∈ R : ∥x∥ ≥ 1}∼ Then X and Y are compact as quotients of compact spaces. Since P2 is Hausdorff, X and Y must be closed. Obviously X ∪ Y = P2. Also

2 X ∩ Y = {x ∈ R : ∥x∥ = 1}∼ = S1∼ ≃ S1 The last equality follows from the fact that the corresponding quotient map is continuous and one-to-one when restricted to the compact set S1 which means that it is a homeomorphism (Theorem 1.11). Using this same argument we obtain

X = B∼ ≃ B where the desired homeomorphism is just the quotient map restricted to B. So now it is sufficient to find a homeomorphism between M and Y . Let us denote U = [0, 1] × [0, 2π]. Then

φ: U −→ Y

(s, t) ↦−→ [ ((1 + s) cos t, (1 + s) sin t)]∼ is a quotient map. By the homeomorphism theorem Uφ ≃ Y . It is now sufficient to show that Uφ ≃ M. We shall use the gluing procedure introduced in the previous chapter since all the spaces we are working with are Hausdorff and compact. First we will describe ∼φ. We know that:

(s1, t1) ∼φ (s2, t2) ⇐⇒ φ(s1, t1) = φ(s2, t2)

Using geometric interpretation of the map φ and the equivalence ∼ we conclude that only nontrivial identifications are

(s, 0) ∼φ (s, 2π) ∀s ∈ [0, 1] and (1.7)

(1, t) ∼φ (1, t + π) ∀t ∈ [0, π] (1.8)

18 Denote A = [0, 1] × [0, π] and B = [0, 1] × [π, 2π]. A, B are both closed subsets of U. Now we will define maps f and g. Recall ∼1 is the equivalence inducing M

f : A −→ M [( s t )] (s, t) ↦−→ , 2 π ∼1

g : B −→ M [( s t )] (s, t) ↦−→ 1 − , − 1 2 π ∼1 Both f, g are continuous since they are compositions of affine transformations and the cannonical projection. A ∩ B = [0, 1] × {π}. It holds that [( s )] [( s )] f(s, π) = , 1 = 1 − , 0 = g(s, π) ∀s ∈ [0, 1] 2 ∼1 2 ∼1 making sure that the gluing map h = f ∪ g is well defined. By checking where the vertices of rectangles A and B land, we obtain f(A) = ([0, 1 ] × [0, 1]) and 2 ∼1 g(B) = ([ 1 , 1] × [0, 1]) meaning that h is onto. According to Corollary 1.28, 2 ∼1 this means that Uh ≃ M. Hence, the only thing left to prove is that ∼h = ∼φ. Given arbitrary s ∈ [0, 1] we see that [( s )] [( s )] h(s, 0) = f(s, 0) = , 0 = 1 − , 1 = g(s, 2π) = h(s, 2π) 2 ∼1 2 ∼1 Analogically for any t ∈ [0, π] we get [(1 )] h(1, t) = f(1, t) = , 0 = g(1, t + π) = h(1, t + π) 2 ∼1

Combining these two identities with characterization of ∼φ we conclude ∼φ ≼ ∼h. To see that ∼φ ≽ ∼h, we just need to realize, that the affine transformations inducing f and g are regular, which means they are injective. Therefore two distinct points u,v in U have the same image through h if and only if one of the following situations happens

1. u ∈ A \ B, v ∈ B \ A and f(u) = g(u) or

2. h(u) ∼1 h(v) It can be easily verified that the first case corresponds to pairs identified in(1.8) and the second one to pairs identified in (1.7). This establishes ∼h = ∼φ. We have successfully proven that

Y ≃ Uφ = Uh ≃ M (1.9)

Remark. From the Theorem 1.23 we know exactly how the homeomorphisms in (1.9) look like. Using this, we can write the homeomorphism from Y onto M

19 explicitly. If we denote the map Φ, this is how it looks:

⎧[( )] s , t if t ∈ [0, π] ⎪ 2 π ⎨⎪ ∼1 Φ: [ ((1 + s) cos t, (1 + s) sin t)]∼ ↦−→ 5 ⎪[( )] ⎪ s t ⎩ 1 − 2 , π − 1 if t ∈ [π, 2π] ∼1 (1.10)

The homeomomorphism from X onto B is simpler and has the form:

Ψ:[x] ↦−→ x (1.11) ∼5 Similarly to the Mobius strip, the Projective plane is obtained by taking a simple subset of R2 and gluing some points together. In the case of M, the gluing can be easily visualized in three dimension as we have already seen in figure 1.2. However, the same cannot be said about P2. It hurts a little just trying to think about continuously gluing together opposite points on the edge of a disk. After thinking about this for a while, we may wonder if it can be done at all. A crucial result for us is that this gluing, actually, cannot performed in R3.

Theorem 1.35. The real projective plane cannot be embedded into R3 This is ,yet another, example of an intuitive statement that is true and belongs to general mathematical knowledge but at the same time is very hard to prove. To show that this fact is true, one has to go deep into algebraic topology which is a field of math far beyond the reach of this thesis. The key property of P2 is that, in addition to its topology, it possesses a structure of an unorientable smooth surface. These objects cannot be smoothly embedded into R3. The theory needed for proving this can be found in Milnor and Stasheff [2016]. Smooth version of Theorem 1.35 is an immediate consequence of corollary 10.14 on page 120. It can be shown that this implies our version of Theorem 1.35 but we shall omit this proof. Later we will understand that the previous lemmas form the pieces of a beau- tiful puzzle which we will solve in the final chapter.

20 2. Rectangles and curves

We are entering the more geometric part of the paper. Before we prove the desired theorem, we first need to clarify what curves are we going to be interested in and what special properties they possess. First, we will use quotients to show that the intuition: ”Simple closed curve is just a homeomorphicaly deformed ”, is correct. Unfortunately, other useful questions related to curves are very complicated, despite being very intuitive and easy to state. Thankfully, some powerful results about curves have already been proven and we shall mention them (without proof) and use them later.

2.1 Jordan curves

Definition 2.1. A continuous map γ :[a, b] → R2 is called a planar curve. The set γ([a, b[) is called the image of γ and denoted ⟨γ⟩. Additionally, planar curve is called closed, if γ(a) = γ(b) and it is called simple, if it is injective on [a, b). Simple closed curves are referred to as Jordan curves. By our definition, curve is a continuous mapping of an interval into the plane. However, what we really intend to study are the images of such maps since they are the geometric entities where the rectangles are to be found. The correspon- dence between a continuous function and a its image is far from being one-to-one. Given a curve γ :[a, b] −→ R2, consider, for example, the function ∀t ∈ [0, 1]: φ(t) = (1 − t)a + tb which is a homeomorphism between intervals [0, 1] and [a, b]. Then the map

γ˜ = γ ◦ φ (2.1) is also a curve with the same image as γ. When talking about a curve γ, we will typically be interested in the set ⟨γ⟩. If we find some other continuous mapping ν of some interval [c, d] into R2, satisfying ⟨γ⟩ = ⟨ν⟩, we will say that ν is a different parametrization of curve γ. The maps γ and ν are just tools for describing the same geometric set. The existence of a rectangle on a given curve is obviously independent of the choice of map we use to describe the curve. Therefore, we can choose whichever parametrization we find the best to solve the inscribed rectangle problem. For instance, in (2.1) we see that any curve has a parametrization on the interval [0, 1] so without any loss of generality we can limit ourselves to such parametrizations. Lemma 2.2. Let γ be a planar curve. Then γ is Jordan if and only if ⟨γ⟩ is homeomorphic to S1.

Proof. S1 is a Jordan curve with parametrization ∀t ∈ [0, 1]: α(t) = [cos(2πt), sin(2πt)]

Given a homeomorphism θ : S1 −→ ⟨γ⟩, we will show that

γ˜ = θ ◦ α: [0, 1] −→ ⟨γ⟩

21 is a parametrization of γ satisfying the conditions for a Jordan curve. The map γ˜ is obviously continuous. Since θ is one-to-one and α is injective on [0, 1), γ˜ must be injective on [0, 1) as well. Finally, we see that

γ˜(0) = θ([cos(0), sin(0)]) = θ([cos(2π), sin(2π)]) = γ˜(1) which means that γ˜ is a Jordan curve. Conversely, let γ be a Jordan curve with domain [0, 1]. On the unit interval, consider the equivalence 0 ∼ 1. Becaus γ is a continuous surjection from compact interval [0, 1] onto Hausdorff space ⟨γ⟩ we can use corollary 1.24 and get [0, 1]γ ≃ ⟨γ⟩. Since γ is closed and simple, it immediately follows that ∼φ = ∼ which means [0, 1]∼ ≃ ⟨γ⟩. Using the same argument for the map α described above, we get [0, 1]∼ ≃ S1 which gives ⟨γ⟩ ≃ S1 as seen from the diagram bellow.

[0, 1] γ α

⟨γ⟩ ≃ [0, 1]∼ ≃ S1

The definition of a simple closed curve in some sense formalizes thenotion of taking a pen and drawing a line on a piece of paper in such a way that we never lift the pen from the paper, never cross through a point more than once and eventually ending at the point we started from. When we actually take a pen and draw a curve like this, without any surprises we will notice that the curve actually splits the plane into two parts. One that is located ’within’ the curve and the other one outside of the curve. One would hope, that this property is possessed by all simple closed curves. The fact that this actually is true is called the and it is a fundamental result in the theory of planar curves.

Theorem 2.3. (Jordan curve theorem) Let γ be a Jordan curve in R2. Then the complement of ⟨γ⟩ has exactly two components one of which is bounded, the other is unbounded and ⟨γ⟩ is their mutual border. Proof. See Luisto [2011] where two proofs of the theorem can be found. One of the proofs is fairly elementary and can be followed by anyone with basic knowledge of complex analysis. Definition 2.4. Let γ be a Jordan curve. The unbounded component of the complement of ⟨γ⟩ is called the Exterior of γ (denoted Extγ) and the bounded component is called the Interior of γ (denoted Intγ). The closed set ⟨γ⟩ ∪ Intγ will be denoted Bγ

Since the image of γ is a closed subset of R2, the complement is an open set and therefore the components Intγ and Extγ are open. Actually, there exists even a stronger version of Theorem 2.3 confirming the intuition that the inside of a Jordan curve is toplogically the same as the inside of a circle

Theorem 2.5. (Schoenflies) If γ is a Jordan curve, then Bγ ≃ B.

22 Proof. Let γ be a Jordan curve. 2 2 Then there exists a homeomorphism Λ: R −→ R which maps S1 = ∂B2 onto ⟨γ⟩. The precise construction of Λ can be found in Cairns [1951]. The cited proof is very elementary and from the construction it is evident that Λ restricted to B has Bγ as its image and is ,therefore, the desired homeomorphism. A intuitive as assertions in theorems 2.3 and 2.5 might seem, proving them is very complicated and could serve as a theme for a whole new thesis. Remark. By the previous theorem, we get, for an arbitrary Jordan curve γ, a homeomorphism Λγ : B ↦−→ Bγ mapping S1 onto ⟨γ⟩. Analogously as in the proof of lemma 2.2, we obtain that

γ˜(t) = Λγ(cos(2πt), sin(2πt)) is a parametrization of γ on the interval [0, 1]. Since all the geometric properties of the curve are invariant under reparametrization, in the upcoming proofs, without any loss of generality, we can assume that γ = γ˜. In this case it especially holds that ( t ) γ = Λ (cos t, sin t) ∀t ∈ [0, 2π] (2.2) 2π γ So from now on, any time we talk about a Jordan curve, we will automatically assume it is parametrized according to (2.2). The major complication of proving the inscribed rectangle theorem for a gen- eral Jordan curve is that the assumptions on the curve are very weak. For in- stance, there exist continuous curves that are differentiable nowhere.

Figure 2.1: Koch snowflake: an example of a continuous but nowhere differen- tiable curve

In fact, in a certain sense, most Jordan curves are ugly and have the fractal- like structure as in the Figure 2.1. One might think, that he does not need to limit himself only to the general case. It seems that if one can prove the theorem for a dense enough family of well-behaved curves, the ugly curves can be expressed as limits of sequences of nicer curves with rectangles inscribed in them. Though it is true, that the limit of a sequence of rectangles is a rectangle, it, so far, seems impossible to make sure, that the limiting rectangles do not collapse into a single point or a , even in the case of smoother curves. Hence, we cannot use this limiting argument. This makes the problem unsolvable using

23 only the tools of differential geometry. Since basically the only property the curve possesses is continuity, it feels natural to look for help in the field of topology, because here continuous maps are the main objects of concern. This explains our preoccupation with topological theory in the first chapter.

24 2.2 Rectangles inscribed in curves

We are about to prove the inscribed rectangle theorem. The trick will be to transform the purely geometrical problem into a topological one which is going to be solvable using the machinery we have been building up in the previous chapters. Even though the crucial idea belongs to Vaughan, the proof given here shows how one can formalize the idea using quotient topology and how to finish the argument using decomposition of P2 with the explicit formulas from page 20. Ultimately, we are looking for rectangles. Yet, none of the theory so far had anything to do with rectangles. Therefore, it might be useful to actually have some geometrical criterion that would help us with our search.

Lemma 2.6. (Sufficient condition for a rectangle) Let A1,B1,A2,B2 be four, not necessarily distinct, points in the Euclidean plane. Suppose that the points satisfy the following conditions {A1 B1}= ̸ {A2 ,B2} (2.3)

A + B A + B 1 1 = 2 2 (2.4) 2 2

∥A1 − B1∥ = ∥A2 − B2∥ (2.5) Then the points are distinct from each other and they are the vertices of a rect- angle.

Proof. First we will prove that the four points are distinct. First of all, it holds that Ai ̸= Bi, i = 1, 2. To see this, for the sake of contradiction assume, for instance, that A1 = B1. Then from (2.5) it follows that A2 = B2. This means that both sets in (2.3) contain only a single element. Then, according to (2.4), the two elements must be equal which contradicts (2.3). Now to realize that actually all of the four points are distinct from each other, assume for example that A1 = A2. From basic geometric intuition we know that for i = 1, 2, Bi lies Ai+Bi on the ray stretching from point Ai through the point 2 . From the assumed equality and from (2.4) it follows that B1 and B2 lie on the same ray and by (2.5) they have the same distance from the initial point of the ray. This means they are equal, which contradicts (2.3) again. A1+B1 To see that the points form a rectangle, let C denote the mutual 2 ∥A1−B1∥ and denote r = 2 . By (2.4) and (2.5), all of the points lay on a circle around C with radius r. Finally, from (2.5) we deduce the equality of angles ∠(A1,C,A2) = ∠(B1,C,B2) and ∠(A1,C,B2) = ∠(A2,C,B1) from which the conclusion follows. Remark. Condition (2.3) is broader than just assuming that the four points are distinct. For instance if the quadruplet satisfies A1 = B1 ̸= B2 = A2, then it satisfies (2.3) even though the points are not distinct. The purpose of thelemma is to conclude that four points actually enclose a rectangle provided that only this weaker assumption holds. Given a Jordan curve γ : [0, 1] → R2, let us consider the set

Nγ = {{A, B} : A, B ∈ ⟨γ⟩}

25 Essentially Nγ is the set of all unordered pairs of points in ⟨γ⟩. Notice that we did not exclude single point subsets since these correspond to pairs where A = B. A two point subset of the Euclidean plane naturally represents a line segment connecting the two points (single point is a line segment!). In the lemma above, we have seen when two pairs from Nγ form a rectangle, depending on the length of the corresponding line segment and position of its midpoint. Now, consider the set M = {(x, y): 0 ≤ x ≤ 1, 0 ≤ y ≤ x}. The map

(x, y) ↦−→ {γ(x), γ(y)} (2.6) is a surjection from M onto Nγ However, γ(0) = γ(1) which makes this correspon- dence not exactly one-to-one. If we denote ∼ the kernel of (2.6), which is just the equivalence we obtain when we preform the abstract construction from page 9 to the concrete mapping defined above, we obtain a truly one-to-one mapping

[(x, y)]∼ ←→ {γ(x), γ(y)} (2.7) between the factor set M∼ and Nγ. Notice that M∼ is independent of the Jordan curve γ. To see how this set looks like, we need to precisely determine ∼. After some case checking, we can see that only nontrivial identifications done by ∼ are

(t, 0) ∼ (1, t) ∀t ∈ [0, 1]

The set M∼ will be very important for us, even though it is not yet evident how. In this regard, we shall once more simplify the notation to make the following pages more coherent. Definition 2.7. Let M and ∼ be as above. On M we consider the inherited Euclidean topology. From now on, by M˜ we will denote the quotient space M∼.

The bijection in (2.7) allows M˜ to topologically represent the set Nγ. This is important as we have given a precise topological context to Nγ, which is a purely geometrical entity. In the following lemma, we will find out that this representation is equivalent to a well known topological space.

Lemma 2.8. M˜ is homeomorphic to the Mobius strip.

Proof. Recall that ∼1 is the equivalence defining M from definition 1.29 and it is characterized by (t, 0) ∼1 (1 − t, 1) ∀t ∈ [0, 1] We will, again, proceed according to the gluing procedure from page 13 and construct a homeomorphism from M onto M˜. Divide the set M into two right angled triangles M1 = M ∩ {(x, y): y ≤ 1 − x} and M2 = M ∩ {(x, y): y ≥ 1 − x}. and also divide the unit square into other two right angled triangles 2 2 M1 = [0, 1] ∩ {(x, y): y ≥ x} and M2 = [0, 1] ∩ {(x, y): y ≤ x}. Now, consider the affine transformations of R2 given by the formulas (x + y y − x) T :(x, y) ↦−→ , 1 2 2

(y − x x + y ) T :(x, y) ↦−→ + 1, 2 2 2

26 Affine transformations are continuous. By looking where the vertices of trina- gles Mi land, we conclude that Ti(Mi) = Mi , i = 1, 2. Apparently, M1 ∩ M2 = {(t, t): t ∈ [0, 1]} and because

T1(t, t) = (t, 0) ∼ (1, t) = T2(t, t) ∀t ∈ [0, 1] the gluing map

⎧ ⎪ [T1(x, y)] if (x, y) ∈ M1 ⎨⎪ ∼ Ω:(x, y) ↦−→ ⎪ ⎩⎪ [T2(x, y)]∼ if (x, y) ∈ M2 is a well defined continuous surjection mapping the unit square onto M˜. This 2 and corollary 1.24 gives us [0, 1] ω ≃ M˜. Using very similar arguments as in the proof of Theorem 1.34, we conclude that ∼ω = ∼1. Therefore, M ≃ M˜ and the concrete homeomorphism is

⎧ ( x+y y−x ) ⎪ 2 , 2 if (x, y) ∈ M1 ⎨⎪ ∼ ω :[(x, y)] ↦−→ (2.8) ∼1 ⎪ ( y−x x+y ) ⎩⎪ + 1, if (x, y) ∈ 2 2 2 ∼ M

In the proofs of Theorem 1.34 and Lemma 2.8, the formulas for the important affine transformations, in some sense, fall from the sky. To understand thatthese formulas are not at all random, one can draw the concrete quotient spaces as geometric shapes situated in the Euclidean plane and coloring some edges to express identifications done by the equivalences. In both theorems, one quotient space can be cut and rearranged into the other one by using different affine transformations on the separated pieces. The formulas are then exactly what we get when we use standard linear algebra to describe such transformations. Example. In Figure 2.2, we have a geometric interpretation of the affine transfor- mation defined in the proof of Lemma 2.8. In the drawing, the red edges represent the equivalence ∼ and the blue edges represent the equivalence ∼1.

Figure 2.2: Transformation of M˜ onto the Mobius strip

This completes our arsenal and we are now ready to prove the final theorem.

27 Theorem 2.9. (Vaughan 1977) Let γ be a Jordan curve. Then there exist four distinct points in ⟨γ⟩, which are the vertices of a rectangle.

Proof. Let γ be a Jordan curve and let M, ∼ and M˜ be as above. Without loss of generality, we can assume that γ is parametrized on [0, 1] as in (2.2). Now, let us consider

2 + F : M −→ R × R 0 (γ(x) + γ(y) ) (x, y) ↦−→ , ∥γ(x) − γ(y)∥ 2 Continuity of γ makes F continuous as well. From the fact that γ is closed, we see that (γ(t) + γ(0) ) F (t, 0) = , ∥γ(t) − γ(0)∥ 2 (γ(1) + γ(t) ) = , ∥γ(1) − γ(t)∥ = F (1, t) 2 from which we deduce that the corresponding derived map

2 + F˜ : M˜ −→ R × R 0 (γ(x) + γ(y) ) [(x, y)] ↦−→ , ∥γ(x) − γ(y)∥ ∼ 2 is well defined. Then it must also be continuous, according to Theorem 1.19.We will prove that F˜ is not injective. For the sake of contradiction, assume it is. 2 Let ∼5 be the equivalence from definition 1.32 characterizing P with its canonical projection π and let X,Y be the subsets of P2 defined in Theorem 1.34 with the corresponding homeomorphisms Φ and Ψ described in (1.10) and (1.11) from page 20. Additionally, let us remind ourselves that ω : M ↦−→ M˜ is the homeomorphism from lemma 2.8. Now consider the map ϕ1 = F˜ ◦ ω ◦ Φ. This is a well defined continuous map taking Y onto some subset of R3 ω M M˜ Φ F˜ Y 3 ϕ1 R

Furthermore, we can define ϕ2 = ν ◦ Λγ ◦ Ψ where ν is the mapping [x, y] ↦−→ [x, y, 0]

2 3 which is an embedding of R into R . Let us also recall that Bγ = Intγ ∪ ⟨γ⟩ and Λγ : ↦−→ Bγ is the homeomorphishm from theorem 2.5. The mapping ϕ2 is evidently also continuous.

Λγ B Bγ

Ψ ν X 3 ϕ2 R

28 2 3 It is tempting to invoke the gluing lemma and construct ϕ = ϕ1 ∪ϕ2 : P ↦−→ R . In order to do so, we must prove that ϕ = ϕ . In other words, we need 1X∩Y 2X∩Y to show that the diagram

Φ ω Y M M˜ ⊆ F˜

3 (2.9) X ∩ Y R

⊆ ν X B Bγ Ψ Λγ commutes. We know, that

X ∩ Y = π(S ) = {[ (cos t, sin t)] : t ∈ [0, 2π]} 1 ∼5 so it is sufficient to show that

ϕ ([ (cos t, sin t)] ) = ϕ ([ (cos t, sin t)] ) ∀t ∈ [0, 2π] (2.10) 1 ∼5 2 ∼5 Take t ∈ [0, 2π] arbitrary. First, let us take the corresponding element through the lower branch of the diagram above, ergo

[ (cos t, sin t)] ↦−→Ψ (cos t, sin t) ∼5 Λγ ↦−→ Λγ(cos t, sin t) (2.2) ( t ) = γ 2π ( ( t ) ) ↦−→ν γ , 0 . 2π This gives us ( ( t ) ) ϕ2([ (cos t, sin t)] ) = γ , 0 ∀t ∈ [0, 2π] ∼5 2π To properly investigate the second branch, we shall distinguish two cases. Recall that M1 and M2 is the partition of M from Lemma 2.8. It is important to note that the output of the map ω : M −→ M˜ depends in which of the Mi the input lies. If t ∈ [0, π], then

[ (cos t, sin t)] = [ ((1 + 0) cos t, (1 + 0) sin t)] ∼5 ∼5 Φ [( t )] ↦−→ 0, ∈ M1 π ∼1 [( t t )] ↦−→ω , 2π 2π ∼ (γ( t ) + γ( t ) t t ) ↦−→F˜ 2π 2π , ∥γ( ) − γ( )∥ 2 2π 2π ( ( t ) ) = γ , 0 . 2π

29 Analogically, for t ∈ [π, 2π] we get [( t )] [ (cos t, sin t)] ↦−→Φ 1, − 1 ∈ ∼5 M2 π ∼1 [( t t )] ↦−→ω , 2π 2π ∼ ( ( t ) ) ↦−→F˜ γ , 0 . 2π

This proves the assertion 2.10 which means that the gluing map ϕ = ϕ1∪ϕ2 is well defined and continuous. Now we will derive that ϕ is also injective. It is evident that ϕ1 and ϕ2 are injective since they are composed of injective maps. This means that ϕ is injective on X and Y separately. By definition, ϕ1 = F˜ ◦ ω ◦ Φ. Also, the map α = ω ◦ Φ is a homeomorphism between Y and M˜, especially it is a bijection. Define

D = {[(t, t)]∼ : t ∈ [0, 1]} ⊂ M˜ (2.11) and let F˜3 denote the third component of the map F˜. The calculation of the upper branch of diagram (2.9), among other things, showed that α(X ∩ Y ) = D. It also holds that

F˜3([ (x, y)]∼) = 0 ⇐⇒ ∥γ(x) − γ(y)∥ = 0 ⇐⇒ γ(x) = γ(y)

⇐⇒ [x, y]∼ ∈ D

So if we take arbitrary p ∈ Y \ X, then α(p) ̸∈ D and therefore F˜3(α(p)) > 0. 2 + 2 Therefore ϕ1(Y \ X) ⊂ R × R . On the other hand, ϕ2(X) ⊂ R × {0}.A direct consequence of this is that ϕ is injective. From compactness of P2 and Hausdorffness of R3 it follows that ϕ is a topological embedding of P2 into R3. This is impossible due to theorem 1.35 so we have obtained a contradiction. The assumption leading to this contradiction was that F˜ is injective. Hence, F˜ cannot be injective. This means that there exist two equivalence classes

[(x1, y1)]∼ ̸= [ (x2, y2)]∼ (2.12) such that

F˜([ (x1, y1)]∼) = F˜([ (x2, y2)]∼) (2.13)

Rewriting (2.13) with respect to the two components of F˜ we get: γ(x ) + γ(y ) γ(x ) + γ(y ) 1 1 = 2 2 2 2 ∥γ(x1) − γ(y1)∥ = ∥γ(x2) − γ(y2)∥ Finally, if we plug the equivalence classes in (2.12) to the one-to-one corre- spondence in (2.7) on page 26, we get

{γ(x1), γ(y1)} ̸= {γ(x2), γ(y2)}

We can, therefore, happily apply Lemma 2.6 to the quadruplet γ(x1), γ(y1), γ(x2), γ(y2) to get the desired rectangle.

30 2.3 Remarks

In the proof of Theorem 2.9, we have, for a given curve γ, constructed a continuous transformation F˜ mapping M˜, which is one of the representations of the Mobius strip, into R3. It is quite easy to realize that the edge of the Mobius strip in the embedded form (Figure 1.2 on page 15) actually corresponds to the set D ⊂ M˜ from (2.11) which we have shown to be mapped by F˜ onto ⟨γ⟩ × {0}. So the image of F˜ is some surface S˜ in the upper Euclidean half-space that is glued to γ. The key step in the proof was that the mapping F˜ cannot be injective. This is equivalent to the statement that the induced surface S˜ contains a self intersection. For a concrete curve, we can draw the surface using computer software.

0.5

-1.0 -0.5 0.5 1.0

-0.5

-1.0

-1.5

Figure 2.3: Random curve γ in the plane

Figure 2.4: The surface above γ induced by the map F˜

In Figure 2.4, we can see some evident self-intersections. Moreover, we easily notice that there is an entire curve of these self-intersections. Recall that each such intersection corresponds to an inscribed rectangle due to Lemma 2.6. This means that there are infinitely many inscribed rectangles.

To wrap things up, the essence of the proof, even though it may have got overshadowed by the technicalities, was actually based on a fairly simple and in- tuitive argument, which is: ”It is impossible to glue the edge of a Mobius strip to a simple closed planar curve without forcing the strip to intersect itself”. If you do not feel that the assertion is intuitive, I encourage you to have watch Sanderson

31 [2016] which is a very entertaining YouTube video explaining the proof of the inscribed rectangle problem in a much more visual way. In fact, the video is so nicely done that it inspired the whole writing of this thesis.

There are related questions naturally succeeding the inscribed rectangle the- orem. For instance, there is the inscribed square problem, first proposed in 1911 by German mathematician Toeplitz. Interestingly enough, this problem remains an open question up to date, even though some special cases have been solved. Each of the solutions uses some additional structure of the curve, like smoothness, convexity or symmetry. For example, to see some nice integration arguments used to solve certain instances of the inscribed square problem, go have a look at Tao [2017].

But we do not have to stop here. It turns out, it is also interesting to study inscribing of other . One of my favorite theorems from this section is that all, but at most two, points on an arbitrary Jordan curve are the vertices of an inscribed . For the proof of this surprising result, see Meyerson [1980].

There is so much more that can be said about these inscribing problems. Some additional general overview of the topic with some more interesting results can be found in Matschke [2014] and Nielsen [2000].

32 Conclusion

In our quest of proving the inscribed rectangle theorem, we had to come a long way from abstract topology to a concrete simple closed curve in the plane. However, the purpose of the thesis was never just to prove the theorem but to illustrate the general procedure of solving a complicated mathematical problem. We took the intuitive idea and developed the tools required to turn this idea into a precise solution. In this regard, I believe the thesis succeeded. Another thing worth mentioning is that along the way to the solution, we rediscovered useful mathematical concepts like quotient spaces and the homeo- morphism theorem, and invented new mathematical techniques like the gluing procedure or transporting equivalences via continuous bijections. Since this the- ory is specifically designed to work with quotient spaces, such techniques canalso be used to tackle other problems related to them. This can be useful because questions related to quotient spaces appear frequently in topology, and other parts of math (like geometry) as well. Yet, I do not see similar techniques working in the case of the general inscribed square theorem. First of all, if it worked, somebody would probably have already solved it. Secondly, the key issue with is that analogous criteria to the Lemma 2.6 for squares require some information about angles. This is a problem, since angles between line segments with endpoints on a curve need not change continuously in the case of general curves. Therefore, no function similar to the map f˜ from the final proof can be defined. We can go further and askevena harder question, whether any Jordan curve contains an inscribed rectangle of a prescribed ratio for side lengths. It is evident that topology is the ”go to” theory when approaching these inscribing problems since continuity of the curve is the only assumption we have and the limiting argument cannot be used. However, there seems to be something unknown and complicated about such problems that no one, so far, has been able to effectively capture. That being said, I feel that the inscribed square problem has a positive answer and that it will be solved once. Just by taking a look at Figure 2.4 on page 31, we see that there are infinitely many rectangles (a whole continuum of them actually) inscribed in the curve. It would be a great surprise to me if none of these rectangles was a square. Therefore I see definite potential for future work involving these inscribing problems. If not solving them completely, I think even just finding out a little more about the correlation between curves and polygons laying inside them might be an interesting topic for future research.

33 Bibliography

Stewart S Cairns. An elementary proof of the jordan-schoenflies theorem. Pro- ceedings of the American Mathematical Society, 2(6):860–867, 1951.

Rami Luisto. Proof of the jordan curve theorem, 2011.

Benjamin Matschke. A survey on the square peg problem. Notices Am. Math. Soc., to appear, 2014.

Mark Meyerson. Equilateral triangles and continuous curves. Fundamenta Math- ematicae, 110(1):1–9, 1980.

John Milnor and James D Stasheff. Characteristic Classes.(AM-76), volume 76. Princeton university press, 2016.

James R Munkres. Topology. Prentice Hall, 2000.

Mark J Nielsen. Figures inscribed in curves, a short tour of an old problem. http://www.webpages.uidaho.edu/˜markn/squares/, June 2000. Grant Sanderson. Who cares about topology (inscribed rectangle problem). https://www.youtube.com/watch?v=AmgkSdhK4K8&t=0s, Nov 2016.

Terence Tao. An integration approach to the toeplitz square peg problem. In Forum of Mathematics, Sigma, volume 5. Cambridge University Press, 2017.

34 List of Figures

1.1 Visualization of the Hausdorff property ...... 3 1.2 Mobius strip embedded in R3 ...... 15 2.1 Koch snowflake: an example of a continuous but nowhere differen- tiable curve ...... 23 2.2 Transformation of M˜ onto the Mobius strip ...... 27 2.3 Random curve γ in the plane ...... 31 2.4 The surface above γ induced by the map F˜ ...... 31

35