DIMENSION THEORY by DANIELLE WALSH a Thesis Submitted to The

Home , Dimension function, Hausdorff dimension, Inductive dimension

DIMENSION THEORY

DANIELLE WALSH

A Thesis Submitted to the Graduate Faculty of

WAKE FOREST UNIVERSITY GRADUATE SCHOOL OF ARTS AND SCIENCES

in Partial Fulﬁllment of the Requirements

for the Degree of

MASTER OF ARTS

Mathematics

May 2014

Winston-Salem, North Carolina

Approved By:

Sarah Raynor, Ph.D., Advisor

Jason Parsley, Ph.D., Chair Brian Pigott, Ph.D. Acknowledgments

There are many people who helped to make this thesis possible. First I would like to thank my family and friends, speciﬁcally my parents, for always being there for me and supporting me throughout my academic career. Next I would like to thank Dr. Raynor for all the time and direction she has given me throughout this process. I greatly appreciate how she has put up with me during her year on sabbatical. Moreover, I would also like to thank Dr. Parsley for tolerating me through all the classes we have had together, and even adding an extra class to his already extremely busy schedule. Additionally, I would like to thank him for serving as one of my thesis committee members. Furthermore, I would like to express my gratitude for Dr. Pigott for letting me bully him into teaching measure theory. I also really appreciate him serving as one of my thesis committee members. I deeply appreciate all of the professors at Wake Forest for being an important part of my mathematical journey.

ii Table of Contents

Acknowledgments ...... ii

Abstract ...... iv

Chapter 1 Introduction ...... 1 1.1 Basic Deﬁnitions and Notation ...... 3 1.1.1 Topology ...... 3 1.1.2 Measure Theory ...... 5

Chapter 2 Dimension ...... 8 2.1 Topological Dimension ...... 8 2.1.1 Small Inductive Dimension ...... 8 2.1.2 Large Induction Dimension ...... 12 2.2 Hausdorﬀ Dimension ...... 20

Chapter 3 Embedding Theorems ...... 33 3.1 The Whitney Embedding Theorem ...... 33 3.2 The Van Kampen-Flores Theorem ...... 37

Chapter 4 Conclusion ...... 51 4.1 Further Study ...... 51

Bibliography ...... 52

Curriculum Vitae ...... 54

iii Abstract

Danielle Walsh

We explore the concept of dimension using mathematical tools. We start by providing deﬁnitions, examples, and basic properties for two types of topological dimension: small inductive dimension and large inductive dimension. We then repeat this process to investigate a type of fractal dimension called Hausdorﬀ dimension.

We then focus on embedding n-dimensional topological objects into Euclidean space. This discussion brings up three main theorems. The ﬁrst theorem states that any n-dimensional topological object can be embedded into R2n+1. This is a very well known result and will be used to spark the conversation for the next two theorems. The next theorem was made famous by Hassler Whitney in 1944 and states that any smooth n-dimensional manifold may be embedded into R2n. The last theorem is called the Van Kampen-Flores theorem, and it states that in general, the best bound for an embedding of an n-dimensional object into Euclidean space is 2n + 1. We compare these theorems and why particular hypotheses are necessary to ensure these embeddings exist.

iv Chapter 1: Introduction

Dimension is a very broad topic that most mathematicians use without thinking about the details. Dimension can lead to the study of fractals, embeddings, and many other theories in math. One simple question can lead you down a path of combinatorial topology that you think will never end. Other ideas lead to the basics of measure theory or metric space topology.

A. R. Pears states that dimension theory “reveals the properties of dimension functions”, with a dimension function deﬁned as a function that takes the class of topological spaces to the nonnegative extended integers and function values are topological invariants [10].

Henri Poincaréfirst proposed defining dimension inductively in 1912 [10]. The next year Luitzen Brouwer acted on Poincaré’sproposition and rigorously defined inductive dimension [10]. He defined what is now known as small inductive dimension. This is the definition of topological dimension that is the most intuitive, and we encounter basic properties and examples of this topological dimension. We then define large inductive dimension and come across similar properties and examples. The biggest result of this section is that for separable metric spaces, the small inductive dimension and large inductive dimension coincide. This section is guided by [2].

Although useful, dimension functions with a codomain of the nonnegative extended integers are limited. Theorem 2.1.1 states that the Cantor set has small inductive dimension 0, while Lemma 2.1.1 states that ﬁnite point sets also have small inductive dimension 0. The Cantor set has much more complex structure than any ﬁnite point set, but we are unable to deduce this fact from a topological dimension function.

1 Broadening the range of a dimension function to the positive extended real num- bers allows for further distinction between sets. In section 2.2 we define a new dimension function called Hausdorff dimension that is based on Hausdorff measure. Finite point sets still have dimension zero, while the Cantor set has Hausdorff dimension log 2/ log 3 [2]. This section is also guided by [2].

The ideas behind topological dimension follow us to the next chapter on embedding theorems. Although this is a slight detour from deﬁning types of dimensions, the transition into embedding theorems is a smooth one. We can compute the dimension of simple geometric objects, but an interesting question to note is how can these objects be realized in Euclidean space. This question leads us to embeddings.

The ﬁrst embedding theorem encountered comes from Hurewicz and Wallman. This theorem states that any arbitrary n-dimensional object (in the topological sense) can be embedded into R2n+1 [5]. This is a very well known result, and is proved many diﬀerent ways (see [7], [5], [8], [4], [1]). We try to make this bound on embedding objects into Euclidean spaces more exact by confronting the Whitney embedding theorem. This theorem states that n-dimensional smooth manifolds can be embedded into R2n. We use Hassler Whitney’s original paper along with [1] to direct us through this result.

Even though the Whitney embedding theorem gives us a better bound, it excludes certain topological objects. The Van Kampen-Flores theorem states that n-skeletons of (2n + 2)-dimensional simplices, which are n-dimensional, cannot be embedded into

R2n. These n-skeletons are not manifolds and need the extra dimension described by the ﬁrst embedding theorem to be realized in Euclidean space. This section is guided by [7] through the means of topological combinatorics and is supplemented by [6].

2 1.1 Basic Deﬁnitions and Notation

Dimension theory is a subject that is dense with ideas from point-set topology and measure theory. The main background information from these areas are listed here for reference. All of these basic deﬁnitions and notation come from [2], [8], and [12].

1.1.1 Topology

Deﬁnition 1. A topology on a set X is a collection τ of subsets of X such that

(i) both X and Ø are in τ,

(ii) the ﬁnite intersection of elements in τ is in τ, and

(iii) the arbitrary union of elements in τ is in τ.

Deﬁnition 2. If X is a set, then a basis for a topology on X is a collection, B, of subsets of X such that

(i) for all x ∈ X, there exists B ∈ B such that x ∈ B, and

(ii) if x ∈ B1 ∩ B2 for B1,B2 ∈ B, then there exists B3 ∈ B such that x ∈ B3 and

B3 ⊂ B1 ∩ B2.

Deﬁnition 3. A metric space is a pair (X, d) where X is a set and d : X ×X → R+ is a function with the following properties:

(ia) d(x, y) ≥ 0 for all x, y ∈ X

(ib) d(x, y) = 0 if and only if x = y

(ii) d(x, y) = d(y, x) for all x, y ∈ X

(iii) d(x, z) ≤ d(x, y) + d(y, z) for all x, y, z ∈ X

The function d is called a metric.

3 Let (X, d) be a metric space. Then the topology generated by the metric d is the topology whose basis is the collection of the open balls

Br(x) := {y ∈ X | d(x, y) < r}.

Deﬁnition 4. A topological space X is called a Hausdorﬀ space if for each pair x, y of distinct points of X, there exist neighborhoods U and Y of x and y respectively such that U ∩ V = Ø.

Lemma 1.1.1. All metric spaces are Hausdorﬀ spaces.

1 Proof. Let (X, d) be a metric space. Let x, y ∈ X with x 6= y. Let r = 3 d(x, y). Then x ∈ Br(x), y ∈ Br(y) and Br(x) ∩ Br(y) = Ø.

Deﬁnition 5. Let X and Y be topological spaces. A function f : X → Y is continuous if for each open set V ⊂ Y , the set f −1(V ) is open in X.

Deﬁnition 6. A function h : X → Y is a homeomorphism if h is bijective, continuous, and has a continuous inverse. A property preserved by every homeomorphism is called a topological property.

Deﬁnition 7. A function f : X → Y is regular if every non-zero element is mapped to a non-zero element.

Deﬁnition 8. An embedding is a continuous function f : X → Y such that fˆ : X → f(X) is a homeomorphism. We say that X embeds into Y .

Deﬁnition 9. A subset A of a space X is dense in X if the closure of A equals the set X.

Deﬁnition 10. A metric space X is separable if there is a countable dense subset in X.

4 Deﬁnition 11. A space X is connected if, except for Ø and X, there are no sets with empty boundary.

Lemma 1.1.2. The real line R is connected and so are intervals and rays in R.

The proof of this is in [8] p.153-154.

Deﬁnition 12. A collection A of subsets of a space X is a cover of X if the union of elements of A is equal to X.

Deﬁnition 13. A space X is compact if every open covering A of X has a ﬁnite subcover of X.

Deﬁnition 14. A function f is proper if inverse images of compact sets are compact.

Deﬁnition 15. An n-dimensional manifold X is a Hausdorﬀ space with a countable basis such that for all x ∈ X there exists a neighborhood U of x that is homeomorphic to an open set in Rn.

1.1.2 Measure Theory

Deﬁnition 16. A collection F of subsets of a set X is called a σ-algebra on X if

1. Ø,X ∈ F;

2. if A ∈ F, then X − A ∈ F; and

3. if A ,A , · · · ∈ F, then S A ∈ F. 1 2 i∈N i

Theorem 1.1.1. Let X be a set, and let D be any set of subsets of X. Then there is a set F of subsets of X such that

1. F is a σ-algebra on X;

2. F ⊃ D; and

5 3. if G is any σ-algebra on X with G ⊃ D, then G ⊃ F.

This proof is in [2] on p. 132-133.

Deﬁnition 17. We say that F from Theorem 1.1.1 is the σ-algebra generated by D.

Deﬁnition 18. Let S be a metric space. A subset of X is a Borel set if it belongs to the σ-algebra on S generated by the open sets.

Deﬁnition 19. A set function is a function whose domain is a family of sets.

Deﬁnition 20. A measure on a collection F of subsets of a space X is a set function M : F → [0, ∞] such that

1. M(Ø) = 0 and

2. if An ∈ F is a disjoint sequence of sets, then

! ∞ [ X M An = M(An), n∈N n=1

i.e., countable additivity.

The elements of the set F are called M-measurable.

Deﬁnition 21. Let X be a set. An outer measure on X is a set function M deﬁned on all subsets of X with values in [0, ∞] satisfying:

1. M(Ø) = 0;

2. if A ⊂ B, then M(A) ≤ M(B); and

3. ! ∞ [ X M An ≤ M(An), n∈N n=1 i.e., countable subaddivity.

6 Deﬁnition 22. We say that M is a metric outer measure iﬀ M(A ∪ B) = M(A) + M(B) for any pair of sets with positive separation (dist(A, B) > 0).

Deﬁnition 23. Let M be a metric outer measure on a metric space S. A set E ⊂ S is said to be M-measurable if E satisﬁes the condition

M(A) = M(A ∩ E) + M(A − E) for every set A ⊂ X.

This condition is called the Carath´eodory condition.

Theorem 1.1.2. Let M be a metric outer measure on a metric space S. Then every Borel subset of S is M-measurable.

This proof is in [2] on p. 139.

7 Chapter 2: Dimension

Dimension is a word that is used in many different contexts. When thinking of dimension we typically think that discrete sets have dimension zero, curves have dimension one, and surfaces have dimension two. This convention is useful when working with these normal geometric objects, but what is the rigorous definition behind the term dimension? In this chapter we discover two distinct branches of dimension and their formal definitions.

2.1 Topological Dimension

Topological dimension is the ﬁrst category of dimension we encounter. When thinking of dimension intuitively, topological dimension is what we use to describe the dimension of typical objects, i.e., the dimension of a rectangle is 2, the dimension of a line is 1, etc. Topological dimension is named such since it is a topological invariant. Here we will deﬁne two types of topological dimension, which happen to be the same when dealing with separable metric spaces [2].

2.1.1 Small Inductive Dimension

The two types of topological dimension we will talk about in this section are deﬁned inductively. Small inductive dimension uses induction and the topological boundaries of sets to measure dimension. We will start with two basic deﬁnitions from point-set topology.

Deﬁnition 24. A set X is clopen if X is both open and closed.

Deﬁnition 25. The boundary of a set X is deﬁned as ∂X = X¯ − X˚.

8 Thus if a set X is clopen, X˚ = X = X¯, so ∂X = X¯ − X˚ = X − X = Ø, and vice versa.

Deﬁnition 26. The small inductive dimension of a metric space X is denoted ind X, where ind X ∈ {−1, 0, 1, 2, ··· , ∞}. We set ind Ø = −1. If X is non-empty and k ∈ {0, 1, 2, ··· , ∞}, then ind X ≤ k if there is a basis for the open sets of X consisting of sets U with ind ∂U ≤ k − 1. If ind X ≤ k, but ind X k − 1, then ind X = k. And if ind X k for all k ∈ Z, then ind X = ∞.

From the deﬁnition above, a metric space X is zero-dimensional if and only if X has a basis for its open sets consisting of sets with empty boundary, i.e. X has a basis consisting of clopen sets.

Lemma 2.1.1. If X is a non-empty set of ﬁnitely many points, then ind X = 0.

Proof. The standard topology on a ﬁnite set is the discrete topology, so every subset is clopen. In particular, the singleton sets form a clopen basis, so ind X ≤ 0. X is non-empty, so ind X 6= −1. Thus ind X = 0.

The Cantor set, denoted C, is a subset of the interval [0, 1] that is a nowhere

1 2 dense, uncountable set with no isolated points [2]. Let C0 = [0, 1], C1 = [0, 3 ] ∪ [ 3 , 1], 1 2 1 2 7 8 C2 = [0, 9 ] ∪ [ 9 , 3 ] ∪ [ 3 , 9 ] ∪ [ 9 , 1], and continue this pattern by removing the middle thirds from each interval. This gives us

2n−1 [ ak bk C = I such that I = , n n,k n,k 3n 3n k=0 where ak, bk follow the pattern indicated above. Now deﬁne the Cantor set as

∞ \ C := Cn. n=0

9 Theorem 2.1.1. The Cantor set C has topological dimension 0.

Proof. 0 ∈ C. Hence C 6= Ø, so ind C= 6 −1. It needs to be shown that C has a basis consisting of clopen sets. First, it will be shown that for all open sets U ⊂ C, for all x ∈ U, there exists n, k ∈ N such that x ∈ In,k ⊂ U.

Let > 0. Let x ∈ C. Choose n ∈ N such that 3−n < and choose k such that x ∈ In,k. Such an In,k exists since C is a nested intersection of the Cn so C ⊂ Cn for all n ∈ N. The open ball B(x) centered around x of radius is a basis element of

R that contains x. We know that In,k ⊂ B(x) since the length of In,k is less than the radius of B(x) and B(x) is centered at x. So x ∈ C ∩ In,k ⊂ C ∩ B(x) because x ∈ C ∩ In,k and In,k ⊂ B(x).

Now we will show that the {C ∩ In,k} are clopen. We know that In,k is closed in

R. Thus C ∩ In,k is closed in C in the subspace topology.

−n −n Finally, each In,k has length 3 and dist(In,k,In,j) ≥ 3 for all k 6= j ∈

n−1 ak−1 bk+1 {1,..., 2 } by construction. Therefore C ∩ In,k = C ∩ ( 3n , 3n ) which is open in C, so each C ∩ In,k is clopen in C. Therefore ind C = 0 since ind C 6= −1 and C has a basis consisting of clopen sets.

Theorem 2.1.2. Let X be a connected metric space with at least two elements. Then X does not have topological dimension 0.

Proof. Let x, y ∈ X and x 6= y. Then there exist disjoint open sets U, V ⊂ X such that x ∈ U and y ∈ V since metric spaces are Hausdorﬀ. Both U and V are non- empty open subsets of X that are not all of X. So there must exist basis elements

Bu 3 x and Bv 3 y that are not Ø or X such that Bu ⊂ U and Bv ⊂ V . So any basis

B for X contains the elements Bu and Bv that are nonempty and are not the entire set X since U and V are disjoint subsets of X. Since X is connected, Bu and Bv are

10 not clopen. Thus a basis for the open sets in X consisting of only clopen sets does not exist, so ind X 6= 0.

Corollary 2.1.1. R does not have topological dimension 0.

Proof. R is connected by Lemma 1.1.2, thus ind R 6= 0.

Theorem 2.1.3. R has small inductive dimension 1.

Proof. The standard basis for R is the collection of open intervals (x − r, x + r) for x ∈ R with r > 0. The boundary of each of these intervals is a two point set {x − r, x + r}. Finite sets are zero dimensional by Lemma 1, so ind R ≤ 1. By Corollary 2.1.1, ind R 6= 0. Thus ind R = 1.

Corollary 2.1.2. Rn has small inductive dimension n.

Proof. This can be shown by induction. Let n = 1. Then the statement is true by the theorem above. Now let us assume ind Rk = k for all k ∈ {1, 2, . . . , n − 1}. Consider n n R . One basis for R is the collection of all n-dimensional rectangles Rn. We know n−1 ∂Rn is the union of 2n rectangles Rn−1, each embedded in a copy of R . These rectangles are basis elements in Rn−1, so by the induction hypothesis, each rectangle n Rn−1 has dimension of exactly n − 1. So there exists a basis for R consisting of sets with boundaries of dimension n − 1. Thus ind Rn = n.

11 2.1.2 Large Induction Dimension

Small inductive dimension is just one way to deﬁne topological dimension. The name suggests the possibility of there being a “larger” way to characterize dimension. For small inductive dimension, boundaries of basis elements are used to calculate dimension. We can think of a boundary as just being a fence or a county line. These objects show the distinction between two regions. To actually separate two objects takes something that might need to be larger.

Deﬁnition 27. Let A and B be disjoint subsets of a metric space X. A set L ⊂ X separates A and B if there exists disjoint open sets U and V in X such that A ⊂ U, B ⊂ V , and L = X − (U ∪ V ).

Separation is just as natural as boundaries of sets, especially for metric spaces. Using this term to deﬁne a new type of topological dimension will work just as well as using boundaries for small inductive dimension.

Deﬁnition 28. The large inductive dimension of a metric space X is denoted Ind X, where Ind X ∈ {−1, 0, 1, 2, ··· , ∞}. Again, we set Ind Ø = −1. If X is non-empty and k ∈ {0, 1, 2, ··· , ∞}, then Ind X ≤ k if any two disjoint closed sets in X can be separated by a set L with Ind L ≤ k−1. If Ind X ≤ k, but Ind X k−1 then Ind X = k. If Ind X k for all k ∈ Z, then Ind X = ∞.

Theorem 2.1.4. Finite point sets have large inductive dimension 0.

Proof. Consider X = {x1, . . . , xk}. We know that Ind X 6= −1 since X is non-empty.

Let A = {xn1 , . . . , xni } and B = {xm1 , . . . , xmj } such that A ∩ B = Ø. A and B are open in X since X has the discrete topology. So let U = A and V = X − A. We see that U and V are disjoint and open in X and U ⊃ A, V ⊃ B. Let L = X − (U ∪ V ). L separates U and V . But U ∪ V = A ∪ (X − A) = X, so L = Ø. Thus Ind X = 0.

12 Finite point sets having large inductive dimension 0 provides the ﬁrst sense of similarity between large and small inductive dimension. To prove more of a relationship between them requires a few more lemmas.

Lemma 2.1.2. Let X be a metric space, let A and B be disjoint closed sets, and let T ⊂ X.

1. Let U and V be open sets with U ⊃ A, V ⊃ B, and U ∩ V = Ø. If L0 ⊂ T separates the sets T ∩ U and T ∩ V , then there is a set L ⊂ X that separates A and B in X with L ∩ T ⊂ L0.

2. Suppose, in addition, that T is closed. Then, for any set L0 ⊂ T separating T ∩A and T ∩ B in T , there is a set L ⊂ X separating A and B with L ∩ T ⊂ L0.

Proof.

1. Since L0 separates U ∩ T and V ∩ T we have T − L0 = U 0 ∪ V 0 with U 0 and V 0 open and disjoint in T , U ∩ T ⊂ U 0 and V ∩ T ⊂ V 0. We claim that A ∩ V 0 = Ø. We know that V 0 = V 0 ∩ T is open in T . This gives us U ∩ V 0 = U ∩ T ∩ V 0 ⊂ U 0 ∩ V 0 = Ø since U ∩ T ⊂ U 0. U is open, so U ∩ V 0 = Ø. By hypothesis, A ⊂ U, thus A ∩ V 0 = Ø. Similarly, B ∩ U 0 = Ø.

U 0 and V 0 are disjoint and open in U 0 ∪ V 0, so U 0 ∩ V 0 = Ø = U 0 ∩ V 0. Now consider (A ∪ U 0) ∩ (B ∪ V 0).

(A ∪ U 0) ∩ (B ∪ V 0) = (A ∪ U 0) ∩ (B ∪ V 0)

= (A ∩ B) ∪ (A ∩ V 0) ∪ (U 0 ∩ B) ∪ (U 0 ∩ V 0)

= Ø.

Similarly, (A ∪ U 0) ∩ (B ∪ V 0) = Ø. So, there exist disjoint open sets U 00 and V 00 in X with A ∪ U 0 ⊂ U 00 and B ∪ V 0 ⊂ V 00. Let L = T − (U 00 ∪ V 00). By deﬁnition

13 L separates A and B, and

T ∩ L = T − (U 00 ∪ V 00) ⊂ T − (U 0 ∪ V 0) = L0.

2. Since L0 separates T ∩ A and T ∩ B in T , there are disjoint sets U 0 and V 0 open in T with T ∩ A ⊂ U 0, T ∩ B ⊂ V 0, and T − (U 0 ∪ V 0) = L0. Because of this the sets A and (T − U 0) ∪ B are disjoint and closed, so there exists an open set U 00 with A ⊂ U 00 ⊂ U 00 ⊂ X − ((T − U 0) ∪ B).

Similarly, B and (T − V 0) ∪ U 00 are disjoint and closed, so there exists an open set V 00 with B ⊂ V 00 ⊂ V 00 ⊂ X − ((T − V 0) ∪ U 00).

This gives us A ⊂ U 00, B ⊂ V 00, and U 00 ∩ V 00 = Ø. Now let L = X − (U 00 ∪ V 00). Thus by part (1) of this lemma we get the desired result.

A L0 X B L

Figure 2.1: Visualization of Lemma 2.1.2

Now we can revisit a familiar idea for metric spaces. Recall the deﬁnition of separable. A metric space X is separable if there is a countable dense subset in X.

An equivalent condition to separability for a metric space X is that every open cover of X has a countable subcover. This is called the Lindel¨ofproperty [2]. We

14 can use this property to establish a relationship between large and small inductive dimension.

Lemma 2.1.3. Let X be a separable metric space with ind X = 0. Then Ind X = 0.

Proof. Let A and B be closed and disjoint sets in X. Let ind X = 0, so there exists a basis B for X consisting of clopen sets. For all x ∈ X choose Ux ∈ B such that x ∈ Ux and either Ux ∩ A = Ø or Ux ∩ B = Ø. We can do this since if x∈ / A, then X − A is an open set containing x since A is closed. Similarly if x∈ / B, then X − B is an open set containing x. Every x ∈ X satisﬁes one of these since A and B are disjoint. Thus

{Ux| x ∈ X} is an open cover of X. X is separable, so by the Lindel¨ofproperty there

exists a countable subcover of {Ux| x ∈ X}, {Ux1 ,Ux2 , ···} of X. Deﬁne V1 = Ux1 and

Vn = Uxn −(V1 ∪V2 ∪· · ·∪Vn−1). Each Vn is clopen since they are constructed from Uxn which are clopen basis elements. Also S V = S U = X. Each V satisﬁes n∈N n n∈N xn n S either Vn ∩A = Ø or Vn ∩B = Ø. Let U = {Vn| Vn ∩B = Ø}. U is open since each Vn

c S is clopen and U ∩B = Ø by construction. So U = V = X −U = {Vn| Vn ∩B 6= Ø}.

c V is open since each Vn is clopen. So U is closed since U = V . V ∩ A = Ø because for all Vn ⊂ V , Vn ∩ B 6= Ø and A, B are disjoint. Thus A ⊂ U and U ∩ B = Ø, which implies Ø separates the disjoint sets A and B. Hence Ind X ≤ 0. Also since ind X = 0, X 6= Ø. Therefore Ind X = 0.

Now we can use the previous two lemmas and the idea of separability to get a result that relates small inductive dimension and separation.

Corollary 2.1.3. Let X be a separable metric space with disjoint closed subsets A and B. Let T ⊂ X such that ind T = 0. Then there exists a set L that separates A and B such that L ∩ T = Ø.

15 Proof. Choose U and V open in X with A ⊂ U, B ⊂ V , and U ∩ V = Ø. We can choose U and V this way since X is separable. So U ∩ T and V ∩ T are disjoint closed sets in T , which is a zero dimensional set. By Lemma 2.1.3, U ∩ T and V ∩ T can be separated by a set L0 such that ind L0 = −1. So by Lemma 2.1.2, there exists a set L separating A and B with L ∩ T ⊂ L0 = Ø.

x − x y 0 x − 2

Figure 2.2: Example 2.1.1

Example 2.1.1. Let > 0 and X = [0, y]. Let A = [0, x − 2 ] and B = [x + 2 , y]. Then U = [0, x − ) and V = (x − , y] are open in X with the subspace topology inherited from R. Also A ⊂ U, B ⊂ V , and U ∩ V = Ø. Let T = {x}. Hence ind T = 0. Then L = X − (U ∪ V ) = {x − } separates A and B, and L ∩ T = Ø.

Lemma 2.1.4. Let X be a separable metric space, and let A, B ⊂ X. Then

ind(A ∪ B) ≤ 1 + indA + ind B.

Proof by Induction. If either ind A = ∞ or ind B = ∞, then ind (A ∪ B) ≤ ∞. So suppose ind A = m and ind B = n for some ﬁnite n, m ∈ N. We will induct on the quantity m + n. First, if m + n = −2, then ind A = −1 and ind B = −1. So both A and B are empty and A ∪ B is empty. Thus ind (A ∪ B) = −1 = 1 + −1 + −1.

Now assume the result holds for all m + n ∈ {−2, −1, 0, ··· , k − 1}. Consider ind (A)+ ind (B) = k. Let x ∈ A ∪ B. Then either x ∈ A or x ∈ B. Suppose x ∈ A. Let V be open in A ∪ B with x ∈ V . Then the sets {x} and A − V are separated in A by a set L0 with ind L0 ≤ m − 1. So by Lemma 2.1.2 there exists a set L that

16 separates {x} and (A ∪ B) − V , and L ∩ A ⊂ L0. Now L = (L ∩ A) ∪ (L ∩ B), so by the induction hypothesis ind L ≤ (m−1)+n = m+n. Thus ind (A∪B) ≤ m+n+1 since L separates (A ∪ B).

The previous lemma lends to the idea that topological dimension has a subaddi- tivity property. The next result shows that this is not necessarily the case for large inductive dimension.

Lemma 2.1.5. Let X be a separable metric space and let the closed sets Tn ⊂ X satisfy ind T ≤ k for all n ∈ . Then ind T = ind S T ≤ k and T = V ∪ Z n N n∈N n such that ind V ≤ k − 1 and ind Z ≤ 0.

Proof. Let k = 0. Let E and F be disjoint closed sets in T . We want to ﬁnd a clopen set U with E ⊂ U and F ∩ U = Ø.

Start with T1. Ind T1 = 0, so there exists a clopen set A1 ⊂ T1 such that

T1 ∩ E ⊂ A1 and T1 ∩ F ⊂ T1 − A1. The sets E ∪ A1 and F ∪ (T1 − A1) are disjoint and closed. So there exist open disjoint sets U1 and V1 in X with E ∪ A1 ⊂ U1 and

F ∪ (T1 − A1) ⊂ V1.

Next, consider T2. Ind T2 = 0, so there is a clopen set A2 ⊂ T2 such that

T2 ∩ U1 ⊂ A2 and T2 ∩ V1 ⊂ T2 − A2. Then the sets U1 ∪ A2 and V1 ∪ (T2 − A2) are disjoint and closed. So there exists disjoint open sets U2 and V2 in X such that

U1 ∪ A2 ⊂ U2 and V1 ∪ (T2 − A2) ⊂ V2.

Continue this pattern to get the sequences (U ) and (V ). Let U = S U and n n n∈N n V = S V . Then by construction U and V are disjoint open sets such that U ⊃ E n∈N n and V ⊃ F . Also, U ∪ V ⊃ S T = T , so U and V are clopen in T . n∈N n By Corollary 2.1.3 , we get the same result for small inductive dimension. Thus the base case holds. Assume that the result holds for all k ∈ {−1, 0, ··· , d − 1}.

17 Now let ind Tn ≤ d for all n ∈ N and for some large d.

For each n, let Bn be a basis for the open sets of Tn consisting of sets with boundary of dimension less than d. We may assume that each Bn is countable by the Lindel¨of

property. For all n and for all U ∈ Bn we have ind ∂Tn U ≤ d − 1. By the inductive hypothesis, [ [ V = ∂Tn U n∈N U∈Bn has small inductive dimension less than or equal to d − 1. But Zn = Tn − V has

Z = {U − V | U ∈ Bn} as a basis for its open sets, and the sets U − V are clopen in Z . Hence ind Z ≤ 0. Now let Z = S Z . Each Z = T − V = T ∩ Z is closed n n n∈N n n n n in Z, so by the base case, ind Z ≤ 0. Thus by the previous lemma,

ind T = ind (V ∪ Z) ≤ 1 + (d − 1) + 0 = d.

D C

A B

Figure 2.3:

Let the square in Figure 2.3 be our space T . Let Z be the set of vertices A, B, C, D. Let V be the intervals in between the vertices. We see T = Z ∪ V and ind Z = 0 and ind V = k − 1 = 1. Let each Tn be subintervals which can include the vertices. Since each Tn can be thought of as a subset of the real line R, ind Tn ≤ 2 (in fact we know ind T ≤ 1). Then ind T = ind S T ≤ 2. n n∈N n

18 If our space T was the solid square in Figure 2.3, then Z would still be the set of vertices. V would be all of the intervals contained by the square T including intervals between each line segment, but not including the vertices. Again T = V ∪ Z, ind

Z = 0, and ind V = k − 1 = 1. This time we let each Tn be any subset of the solid

2 square T . Each Tn can be thought of as a subset of R , hence ind Tn ≤ 2. Then ind T = ind S T ≤ 2. n∈N n This leads to our ﬁrst major result.

Theorem 2.1.5. If X is a separable metric space, then ind X = Ind X.

Proof.

Let Ind X = n. We want to show that ind X ≤ Ind X. The case when n = 0 is

Lemma 2.1.3. Let n ≥ 1. A basis of X is B = {B(x)| > 0, x ∈ X}. Suppose V is a collection of open sets satisfying that for all x ∈ X, for all > 0, there exists a

V ∈ V such that x ∈ V and V ⊂ B(x), then V is also a basis for the same topology

c on X. Let x ∈ X, > 0, and A = B¯ (x) and B = B (x) . Both A and B are closed 2 and disjoint. Thus there exists a set L with Ind L = n − 1 such that there exists sets V ⊃ A and U ⊃ B with V and U open and V ∪ U ∪ L = X with V, U, L pairwise disjoint. Let V be the collection of all such V . For all B ∈ B and for all x ∈ X there exists a V ∈ V such that x ∈ V ⊂ B. We claim that ∂V = L and ind ∂V ≤ n − 1.

Let y ∈ ∂V . Then there exists a δ > 0 such that y ∈ Bδ(y) ∩ U, hence y∈ / V because U and V are disjoint and open. Hence y ∈ L.

Now we will show by induction that Ind X ≤ ind X.

If ind X = ∞, then Ind X ≤ ind X. Suppose ind X = k is ﬁnite. Let A, B be disjoint closed sets in X.

The case when k = 0 is Lemma 2.1.3. Now let k ≥ 1. Then by Lemma 5 we can represent X as the union of two sets Y,Z such that ind Y = k − 1 and ind Z = 0.

19 By Corollary 3 there exists a set L that separates A and B with L ∩ Z = Ø. So L ∩ T ⊂ Y , and hence ind L ≤ ind Y ≤ k − 1. By the induction hypothesis, Ind L ≤ k − 1. Thus Ind X ≤ k = ind X.

This theorem does not hold if the metric space X is not separable. We will always get that for any metric space X, ind X ≤ Ind X since this half of the proof uses nothing about the separability of a space. See chapter 7 section 4 in [10] for an example of a space X such that ind X 6= Ind X.

2.2 Hausdorﬀ Dimension

The next type of dimension we encounter is a fractal dimension. Unlike topological dimension, fractal dimensions do not have integer values. Hausdorﬀ dimension is a fractal dimension and depends on the Hausdorﬀ measure of a set, so basic knowledge of measure theory and Lesbegue measure is useful.

Deﬁnition 29. Let A be a countable cover of a metric space F . Let > 0. The cover A is an -cover of F if diam(A) ≤ for all A ∈ A.

The Hausdorﬀ outer measure of a set is deﬁned from the set function C(A) = diam(A)s with s ∈ R+ and

diam(A) = sup{dist(x, y): x, y ∈ A}.

We need to build up to Hausdorﬀ outer measure by deﬁning an intermediate measure

¯s H . Let F be a set and s be a positive real number, then for > 0

( ) ¯s X s H (F ) = inf diam(A) : A -cover of F . A∈A

20 Deﬁnition 30. Let S be a metric space. Let F ⊂ S. Consider a positive real number s. The s-dimensional Hausdorﬀ outer measure is written H¯s and

¯s ¯s H (F ) = lim H (F ). →0

The restriction to Borel sets is called s-dimensional Hausdorﬀ measure and written Hs.

¯s The limit in this deﬁnition exists since H is a monotonic function of . Now we ¯s ¯s need to prove that H and H are outer measures.

¯s Lemma 2.2.1. Let S be a metric space. For all s ≥ 0 and for all > 0, H is an outer measure.

Proof. We want to show

¯s (i) H (Ø) = 0,

¯s ¯s (ii) if A ⊂ B ⊂ S, then H (A) ≤ H (B), and

(iii) H¯s(S A ) ≤ P∞ H¯s(A ) for all A ,A ,A , · · · ⊂ S. n∈N n n=1 n 1 2 3

¯s P s (i) By deﬁnition H (Ø) = inf A∈A diam(A) such that A is an -cover of Ø. But every A is an -cover of Ø since Ø is contained in all sets A ∈ A. So there exists

A0 -cover such that diam(A) = 0 for all A ∈ A0. Thus

¯s X s H (Ø) = inf diam(A) = 0. A∈A

∞ (ii) Let A ⊂ B ⊂ S. Let B = {Ci}i=1 be an -cover of B. Then B is an -cover of A. So A ⊂ B ⊂ S C . Let i ∈ I if A ∩ C 6= Ø and let A = {C } . We see i∈N i A i i i∈IA that A is an -cover of A, and

X s X s ¯s diam(Ci) ≤ diam(Ci) ≤ H (B). Ci∈A Ci∈B

21 This implies that

( ) X s ¯s inf diam(Ci) | A -cover of A ≤ H (B). Ci∈A

¯s ¯s So H (A) ≤ H (B).

S∞ ¯s (iii) Let A1,A2,A3, · · · ⊂ S. Let A = n=1 An. We may assume that H (An) < ∞ ¯s for all n ∈ N because if H (An) = ∞ for some n ∈ N, then clearly

! ∞ ¯s [ X ¯s H An ≤ H (An) = ∞. n∈N n=1

By deﬁnition,

( ) ¯s X s H (An) = inf diam(C) : An an -cover of An . C∈An

Therefore for all δ > 0, for all n ∈ N there exists an -cover An of An such that

X δ diam(C)s < H¯s(A ) + . n 2n C∈An

Let A = S A . Denote A = {C : m ∈ }, so A = S S C . n∈N n n m,n N n∈N m∈N m,n Clearly A is an -cover of A, since a countable union of countable sets is countable. So we get that

X s X X s diam(Cm,n) = diam(Cm,n) Cm,n∈A n∈N m∈N

X δ < H¯s(A ) + n 2n n∈N

X ¯s = H (An) + δ. n∈N

22 So for all δ > 0

( ) X s [ X ¯s inf diam(C) | A -cover of An < H (An) + δ. C∈A n∈N n∈N

Let δ → 0, then

( ) ¯s X s [ X ¯s H (A) = inf diam(C) | A -cover of An ≤ H (An). C∈A n∈N n∈N

¯s Therefore for all s ≥ 0 and > 0, H is an outer measure.

Lemma 2.2.2. Let S be a metric space. For all s ≥ 0, H¯s is an outer measure.

Proof. Again, we want to show

(i) H¯s(Ø) = 0,

(ii) if A ⊂ B ⊂ S, then H¯s(A) ≤ H¯s(B), and

(iii) H¯s(S A ) ≤ P∞ H¯s(A ) for all A ,A ,A , · · · ⊂ S. n∈N n n=1 n 1 2 3

Let S be a metric space. Let s ≥ 0.

¯s (i) By Lemma 2.2.1 we know that for all > 0, H (Ø) = 0. Therefore,

¯s ¯s H (Ø) = lim H (Ø) = lim 0 = 0. →0 →0

¯s ¯s (ii) Let A ⊂ B ⊂ S. By Lemma 2.2.1 for all > 0, H (A) ≤ H (B). Let → 0 and we get H¯s(A) ≤ H¯s(B).

(iii) Let A ,A , ... ⊂ S. By Lemma 2.2.1 for all > 0, H¯s(S A ) ≤ P H¯s(A ). 1 2 n∈N n n∈N n Let → 0 and we get H¯s(S A ) ≤ P∞ H¯s(A ). n∈N n n=1 n Therefore for all s ≥ 0, H¯s is an outer measure.

23 Example 2.2.1. Consider the finite set F with |F | = k. Let > 0, s > 0, and A be an -cover of F . Since F is finite, we can cover F with finitely many sets A of diam(A) ≤ , each containing one element of F . Thus

X diam(A)s ≤ sk. A∈A

So ( ) ¯s X s s H (F ) = inf diam(A) : A -cover of F ≤ k. A∈A

We take the limit as goes to 0 to get

H¯s(F ) ≤ lim sk = 0. →0

Therefore for all ﬁnite sets F and for all s > 0, Hs(F ) = 0.

Theorem 2.2.1. Any Borel set is Hausdorﬀ measurable.

Proof. Theorem 1.1.2 states that every Borel set is measurable with respect to a metric outer measure. So we will prove that s-dimensional Hausdorﬀ measure is a metric outer measure. First, we will recall what it means to be a metric outer measure.

Deﬁnition 31. We say that M¯ is a metric outer measure iﬀ M¯ (A ∪ B) = M¯ (A) + M¯ (B) for any pair of sets with positive separation (dist(A, B) > 0).

¯s ¯s ¯s ¯s Since H is an outer measure, we know H (A) + H (B) ≥ H (A ∪ B) for all ¯s ¯s ¯s A, B ⊂ S. So we want to show that H (A) + H (B) ≤ H (A ∪ B) for all A, B ⊂ S such that dist(A, B) > 0.

Let S be a metric space. Let s ≥ 0. Let A, B ⊂ S have positive separation. Let

¯s P s 0 < < dist(A, B). For some δ < , consider Hδ(A ∪ B) = inf C∈C diam(C) with

24 ∞ C a δ-cover of A ∪ B. For all η > 0 there exists a C = {Ci}n=1 δ-cover of A ∪ B such ¯s P s that Hδ(A ∪ B) + η ≥ C∈C diam(C) . Let

IA := {i ∈ N | Ci ∩ A 6= Ø} and

IB := {i ∈ N | Ci ∩ B 6= Ø}.

We claim that IA ∩ IB = Ø. If not, there exists an i ∈ N such that Ci ∩ A 6= Ø and Ci ∩ B 6= Ø, but diam(Ci) < δ < ≤ dist(A, B). Hence there cannot exist such a Ci that has a nonempty intersection with both A and B since the diameter of Ci is strictly less than the distance between A and B. So i is in either IA or IB or not in either index set, and we get that IA ∩ IB = Ø.

Let CA = {Ci}i∈IA and CB = {Ci}i∈IB . These are δ-covers of A and B respectively

∞ since {Ci}n=1 ≥ {Ci}i∈IA ∪ {Ci}i∈IB is a δ- cover of A ∪ B. So

X s X s X s ¯s diam(C) + diam(C) ≤ diam(C) ≤ Hδ(A ∪ B) + η.

C∈CA C∈CB C∈C

We know that

( ) X X inf diam(C)s| C -cover of A ≤ diam(C)s

C∈C C∈CA and ( ) X X inf diam(C)s| C -cover of B ≤ diam(C)s.

C∈C C∈CB

Then

( ) ( ) X X inf diam(C)s| C -cover of A + inf diam(C)s| C -cover of B C∈C C∈C

¯s ≤ Hδ(A ∪ B) + η

25 and hence ¯s ¯s ¯s Hδ(A) + Hδ(B) ≤ Hδ(A ∪ B) + η.

Let η → 0, so ¯s ¯s ¯s Hδ(A) + Hδ(B) ≤ Hδ(A ∪ B).

Now, let δ → 0 and we get

H¯s(A) + H¯s(B) ≤ H¯s(A ∪ B).

¯s ¯s Thus Hδ and H are metric outer measures.

Lemma 2.2.3. Let 0 < s < t.

¯t ¯s (a) For any set F and for all 0 < ≤ 1, H(F ) ≤ H (F ).

(b) Let F be a Borel set. For t > s, if Hs(F ) < ∞, then Ht(F ) = 0.

Proof.

(a) Fix 0 < ≤ 1. Let A be an -cover of F . For all A ∈ A,

diam(A)t = diam(A)sdiam(A)t−s ≤ diam(A)st−s ≤ diam(A)s

P s P t since diam(A) ≤ ≤ 1. This implies A∈A diam(A) ≥ A∈A diam(A) , and thus

( ) ( ) X X inf diam(A)s| A -cover of F ≥ inf diam(A)t| A -cover of F . A∈A A∈A

¯s ¯t Therefore H (F ) ≥ H(F ).

26 s ∞ (b) Let 0 < ≤ 1. Suppose H (F ) < ∞. Let A = {An}n=1 be an -cover of F . We have

diam(An) ≤ for all n ∈ N and

t s t−s s t−s diam(An) = diam(An) diam(An) ≤ diam(An) .

So ∞ ∞ ∞ X t X s t−s t−s X s diam(An) ≤ diam(An) = diam(An) n=1 n=1 n=1

and

( ∞ ) ¯t X t H(F ) = inf diam(An) | A -cover of F n=1

( ∞ ) t−s X s ≤ inf diam(An) | A -cover of F n=1

t−s ¯s = H (F ).

Thus

¯t t−s ¯s t−s ¯s ¯s lim H(F ) ≤ lim H (F ) = lim lim H (F ) = 0 · lim H (F ) = 0 →0 →0 →0 →0 →0

¯s ¯t since H (F ) < ∞. Therefore H(F ) = 0.

The above lemma says that for a given set F there is a unique value s ∈ [0, ∞] such that Ht(F ) = ∞ for all t < s and Ht(F ) = 0 for all t > s.

27 This unique value s is called the Hausdorﬀ dimension of F and is written dimF = s. Note that it is possible that Hs(F ) = 0 or Hs(F ) = ∞ for all s. In these cases dimF = 0 or dimF = ∞ respectively.

In example 2.2.1, we showed that for any ﬁnite set F and any for all s > 0, Hs(F ) = 0. Hence dimF = 0. Consider H¯0 of any metric space X.

¯0 ¯0 H (X) = lim H (X) →0 ( )! X = lim inf diam(A)0 : A -cover of X →0 A∈A ( )! X = lim inf 1 : A -cover of X . →0 A∈A

As → 0, H¯0 is indicating how many elements the set X contains. Thus H¯0 is named the counting measure.

Now we will compute the Hausdorﬀ dimension of R.

Example 2.2.2. First, consider H¯1([0, 1]).

¯1 ¯1 H ([0, 1]) = lim H ([0, 1]) →0 ( )! X = lim inf diam(A)1 : A -cover of [0, 1] →0 A∈A

( n X = lim inf diam([xi−1, xi]) : diam([xi−1, xi]) < and →0 i=1

n )! [ [xi−1, xi] ⊃ [0, 1] since closed sets in R have the form [a, b]. i=1

28 ( n n )! X [ = lim inf (xi − xi−1): xi − xi−1 < and [xi−1, xi] ⊃ [0, 1] →0 i=1 i=1

since diam([xi−1, xi]) = xi − xi−1

n ! [ = lim 1 because diam [xi−1, xi] ≥ diam([0, 1]) = 1 →0 i=1 = 1.

So H¯1([0, 1]) = 1 = H1([0, 1]) since [0, 1] is a Borel set. By the monotonicity of

Hausdorﬀ measure, [0, 1] ⊂ R, so H1([0, 1]) ≤ H1(R). Thus 1 = dim[0, 1] ≤ dimR. If s > 1, Hs([0, 1]) = 0. Notice [0, 1] is isomorphic to [n, n + 1] for all n. Then H1([n, n + 1]) = 1 and Hs([n, n + 1]) = 0 for all s > 1. So

∞ s X s H (R) ≤ H ([n, n + 1]) = 0. n=−∞

This implies dimR ≤ s for all s > 1. Therefore dimR ≤ 1, and thus dimR = 1.

Computing the Hausdorff dimension of something more complicated, such as the Cantor set C, can be quite difficult. Hausdorff dimension is hard to compute directly. By a great deal of work, one can get the Hausdorff dimension through the similarity dimension, another fractal dimension. The similarity dimension relies on iterated function systems to describe sets, and the value comes from the ratios of the iterated function system. Because the Cantor set satisfies a special condition called the Open Set Condition, the Hausdorff dimension and similarity dimensions are the same for C, and we get dim C = log 2/ log 3 [3].

Lemma 2.2.4. Let S and T be metric spaces and suppose f : S → T is a function. Let A ⊂ S be a Borel set. If f has a bounded increase, then dimf[A] ≤ dimA.

Proof. Let s ≥ 0. Let ρS and ρT be the metrics for S and T respectively and suppose

29 there exists an element b ∈ R+ such that for all x, y ∈ S,

ρT (f(x), f(y)) ≤ bρS(x, y).

Let A be an -cover of A ⊂ S. Then diam(f[B]) ≤ b for all B ∈ A. Now we get

X X diam(f[B])s ≤ bs diam(B)s B∈A B∈A since each f[B] with B ∈ A has a diameter of b · diam(B). So

¯s s ¯s Hb(f[A]) ≤ b H (A).

Thus dimf[A] ≤ dimA.

Lemma 2.2.5. Let S and T be metric spaces and suppose f : S → T is a function. If f has a bounded decrease, then dimf[A] ≥ dimA.

Proof. Let s ≥ 0. Let ρS and ρT be the metrics for S and T respectively and suppose there exists an element b ∈ R+ such that for all x, y ∈ S, s bρT (f(x), f(y)) ≥ ρS(x, y). We know dim(A) = inf{s |H (A) = 0}. What needs to be shown is {s |Hs(f[A]) = 0} ⊂ {s |Hs(A) = 0}, i.e., for any s ≥ 0 such that

s s ¯s H (f[A]) = 0 implies H (A) = 0. We know that H is a monotone function of . So ¯s ¯s it suﬃces to show that if for all > 0, H (f[A]) = 0, then for all δ > 0, Hδ(A) = 0. ¯s Let > 0, s ≥ 0 such that H (f[A]) = 0. Now we know that

( ) X s inf diam(Ai) | AT -cover of f[A] = 0.

Ai∈AT

For all η > 0, there exists an -cover AT of f[A] such that

X s diam(Ai) ≤ η.

Ai∈AT

30 −1 −1 For all Ai ∈ AT , let Bi = f [Ai]. Let AS = {Bi| f [Ai] = Bi}. Clearly AS is a cover of A, what needs to be shown is that AS is a δ-cover of A. Let’s compute the diameter for some Bi ∈ AS. For all i, diam(Ai) < and diam(Bi) = sup{ρS(x, y)| x, y ∈ Bi}.

Let x, y ∈ Bi. 1 ρ (x, y) ≤ ρ (f(x), f(y)) ≤ . S b T b

1 So diam(Bi) ≤ b diam(Ai) ≤ b .

Let δ > 0. Let = bδ. Choose an -cover AT of f[A] such that

X s diam(Ai) ≤ η.

Ai∈AT

As above, the corresponding AS is a δ-cover of A, and we get that

s X X 1 X 1 1 diam(B )s ≤ diam(A ) = diam(A )s ≤ η. i b i bs i bs Bi∈AS Ai∈AT Ai∈AT

P s 1 As η → 0, diam(Bi) → 0 since bs is a ﬁxed constant. Thus for all δ > 0, Bi∈AS ¯s Hδ(A) = 0.

Theorem 2.2.2. Suppose S is a metric space and dimS < ∞. Then S is separable.

Proof. Let S be a metric space with dimS < ∞. Then there exists an s ∈ R+ such that H¯s(S) = 0. By deﬁnition,

( )! X lim inf diam(A)s| A -cover of A = 0. →0 A∈A

¯s We know that H is a monotone function of , so this implies that for all > 0, P s inf{ A∈A diam(A) = 0| A -cover of A} = 0. So for all > 0, for all δ > 0 there

31 P s 1 exists an -cover A such that A∈A diam(A) ≤ δ. Let = n and choose An = ∞ {Anm}m=1 an -cover of S such that

X s diam(Anm) < δ.

Anm∈An

Using the Axiom of Choice we can pick a single element from each Anm. Let

Sn = {anm : anm ∈ Anm}

S∞ be the set containing those elements. Let S0 = n=1 Sn. We claim that S0 ⊂ S is countable and dense.

Clearly S0 is countable by construction since it is taking a single element from each set in the countable collection.

Let > 0. Let x ∈ S. Let B = B(x). It needs to be shown that there exists

1 an x0 ∈ S0 such that x0 ∈ B. Let N be large enough such that N < . Since AN is a cover of S, there exists a set ANm ∈ AN such that x ∈ ANm. We know that

1 diam(ANm) < N < , so aNm ∈ ANm ⊂ B since B has radius and is centered at x.

Thus S0 is dense in S.

32 Chapter 3: Embedding Theorems

The motivation behind this chapter is to compare two classic theorems and to see why the Whitney embedding theorem, which states that smooth n-manifolds can be embedded into R2n, cannot be generalized from manifolds to all topological spaces1. We will see that if an object S with topological dimension n is not a manifold, it may not be possible for it to be embedded into R2n. To see that this provision is necessary, we will dive into combinatorial topology to understand the Van Kampen- Flores theorem which states that for certain n-dimensional topological objects to be embedded into Euclidean space, 2n + 1 dimensions are required. In this section when referring to the dimension of an object, we are using topological dimension.

3.1 The Whitney Embedding Theorem

An embedding can be thought of as a way to take a mathematical object and put it somewhere else. When moving an object, the deﬁning properties of the object should stay unchanged and the object should physically be the same. A signiﬁcant problem when embedding an object into an ambient space is self-intersections. If f : X → RN is not injective, then we say the image of f intersects itself when f(x) = f(y) and x 6= y. If the image of f has self-intersections, then f is not an embedding. Whitney tries to eliminate self-intersections of functions f that take manifolds into Euclidean space.

We start out small by getting a bound for embedding any topological metric space into Euclidean space.

1This chapter will be talking about topological dimension when referring to dimension. The objects in mind will have small inductive dimension and large inductive dimension that coincide.

33 Theorem 3.1.1. Let X be a space with topological dimension n. Then X is homeomorphic to a subset of R2n+1.

This theorem states that any n-dimensional object can be embedded into R2n+1. The proof for this can be found in [7], [8], [5], [4], and [1], so we will leave this proof to the reader. Instead, we focus our attention to try to make this bound one better by changing the restrictions on our space X.

Theorem 3.1.2 (The Whitney Embedding Theorem).

Any smooth n-dimensional manifold may be embedded into R2n.

The proof of the Whitney embedding theorem relies on concepts from diﬀerential topology. The smooth manifold is the main focus in this area of topology, but to properly deﬁne this term, we will need a new type of map.

Deﬁnition 32. A diﬀeomorphism is a homeomorphism that is smooth2 and has a smooth inverse.

Thus a smooth manifold is a manifold such that each neighborhood is diﬀeomorphic to an open set in Euclidean space. The proof of the Whitney embedding theorem is very involved, and most books just ignore the proof altogether. We will attempt to outline the major steps of the process here. For full proof see [14] or [1].

The proof relies heavily on the existence of a special map from a manifold into Euclidean space.

Deﬁnition 33. A map f : X → Y is completely regular if

(i) f has no more than two distinct points in M that have the same image in N and

2Smooth in this document will refer to functions with continuous ﬁrst partial derivatives.

34 (ii) the self-intersections of the image of f are transverse, i.e. ‘bumping’ the intersection still leaves an intersection.

Outline for the Proof of the Whitney Embedding Theorem.

Let M be an n-dimensional smooth manifold. First, we look at Whitney’s earlier work to say that there always exists a proper completely regular map f : M → R2n [13]. Although this theorem is the launching point for the rest of the proof for the Whitney embedding theorem, the proof of this theorem is extremely technical and provides no intuition for the rest of the Whitney proof. It just guarantees that this map f exists, so it is left to the reader to further investigate the details in [13].

Next, we want to show that if M is a compact manifold, then there exists a proper completely regular map f : M → R2n with the image of f having any desired number of self-intersections, denoted If . An intersection can be positive or negative, depending on the orientation of the manifold.

2n To show this, we let f0 : M → R be a proper completely regular map. We know this exists as above. Take a small piece of M, take it out and replace it by a piece with a single intersection, like in Figures 3.1 and 3.2. Depending on which direction the intersection is going in denotes whether the intersection is positive or negative,

thus changing If0 by +1 or −1. This means that If can reach any integer by repeating this process. This leads us to the following lemma.

Figure 3.1: Let this curve represent the manifold in the proof of the Whitney embedding theorem.

n 2n Lemma 3.1.1. Let n ≥ 3, M be closed, and f0 : M → R be a proper regular map

35 with |If0 | < ∞. Then there is a regular deformation ft of f into a proper completely regular map f1 such that If1 = If0 + 2.

This lemma states that If can be considered mod 2. We can show this result by smoothly ‘pushing’ sections of f(M) until we get the desired number of self- intersections. Lemma 3.1.1 is known as the Whitney Trick.

Finally, we can show that M n embeds into R2n. For n = 1, the result is trivial since by the 1-manifold classiﬁcation theorem, all 1-manifolds are either the line or S1 [8], [14], [4]. For n = 2, M 2 must be a combination of the sphere, n-genus torus, projective plane, or Klein bottle, all of which can be embedded into R4 [8], [8], [4]. Now let n ≥ 3. We know there exists a proper completely regular map f : M →

2n R . If M is closed, then by the second step we can choose an f1 such that If1 = 0. We notice that If1 = 0 implies that f(M) has the same number of positive and negative self intersections. Thus by the Whitney trick, we can get rid of all intersections in a smooth way, leaving an embedding of M into R2n. Let us suppose now that M is open. Choose two opposite self-intersections (if there is only one intersection, we can add another by the process similar to that of step two). Two opposite self-intersections allows for the construction of a closed path connected by the two points of self-intersection. R2n is connected, so this closed path bounds a disk. This disk is embedded in R2n and only intersects the image of f on the boundary of the disk. We then ‘push’ the manifold along this disk, this is actually called a regular deformation. Doing this removes the two opposite self-intersections that created the disk initially. We repeat this process until all self-intersections are removed. This leaves us with f being injective, and thus f is an embedding of M into R2n.

36 Figure 3.2: We can add an opposite self-intersection to the curve.

Figure 3.3: We then slide the curve along the path that is connected by the opposite self-intersections to remove those two self-intersections.

3.2 The Van Kampen-Flores Theorem

The Whitney embedding theorem is a useful tool, but the hypotheses of the theorem limit the topological spaces to which it applies. We cannot always assume the objects involved are smooth, or even that they are manifolds. In 1933 Egbert Van Kampen showed that there are some n-dimensional topological objects that cannot be embedded into 2n-dimensional Euclidean space [5]. The following section builds up to what is now know as the Van Kampen-Flores theorem and it shows explicitly that Theorem 3.1.1 is the best bound for embeddings into Euclidean space.

d Definition 34. Let v0, v1, ··· , vk be points in R . We call these points affinely Pk dependent if there are α0, ··· , αk ∈ R with αj 6= 0 for all j such that i=0 αivi = 0 Pk and i=0 αi = 0. Otherwise v0, v1, ··· , vk are called affinely independent.

Example 3.2.1. For k = 1, v0 and v1 are aﬃnely independent if v0 6= v1. For k = 2, v0, v1, v2 are aﬃnely independent if v0, v1, v2 are not all collinear. Figures 3.2.1 and

3.2.1 are aﬃne dependent and aﬃne independent sets in R2 respectively. For Figure P2 3.2.1, let α0 ∈ R, α1 = −2, and α2 = 1. Then i=0 αivi = 0. For Figure 3.2.1, no P2 matter the choice of αi, i=0 αivi 6= 0.

37 (2, 2) (1, 1) (0, 0)

Figure 3.4: An aﬃnely dependent set in R2.

(1, 2)

(0, 0) (2, 1)

Figure 3.5: An aﬃnely independent set in R2.

Deﬁnition 35. A set X is said to be convex if for all vi, vj ∈ X, X contains the line segment connecting vi and vj. The convex hull of X is the intersection of all convex sets containing X.

Figure 3.6: An illustration of a convex hull of a set of points in R2

Definition 36. A simplex σ is the convex hull of a finite affinely independent set A in Rd and we define dim σ = |A| − 1. We say that σ is an (|A| − 1)-simplex.

Deﬁnition 37. The convex hull of an arbitrary subset of vertices of a simplex σ is a face of σ. Thus every face is itself a simplex.

Example 3.2.2. A vertex is a 0-simplex. A line segment is a 1-simplex. A triangle is a 2-simplex. A tetrahedron is a 3-simplex. The faces of a tetrahedron are the tetrahedron, triangles, lines, and vertices.

38 a a a

c b b b b

d d

Figure 3.7: Some faces of a tetrahedron

Lemma 3.2.1. The small inductive dimension of a d-simplex K is d.

Proof. The d-simplex K is the convex hull of d + 1 aﬃnely independent points in Rd. An appropriate topology for K is the subspace topology inherited from Rd generated by the basis of d-dimensional rectangles, such as in Corollary 2.1.2.

Basis elements of K from the subspace topology inherited from Rd are d-rectangles intersected with the simplex K. The intersection of K with a d-rectangle is homeomorphic to closed d-rectangle of Rd. We know these sets have small inductive dimension of exactly d from Corollary 2.1.2. Thus ind K = d.

Deﬁnition 38. Simplicial Complex

1. A nonempty family ∆ of simplices is a geometric simplicial complex if the following two conditions hold:

(a) Each face of any simplex σ ∈ ∆ is also a simplex of ∆.

(b) Let σ1, σ2 ∈ ∆. Then σ1 ∩ σ2 is a face of both σ1 and σ2.

2. An abstract simplicial complex is a pair (V,K), where V is a set and K ⊂ 2V is a set such that F ∈ K and G ⊂ F implies G ∈ K. The sets K are called simplices and dim K = max {|F | − 1 : F ∈ K}.

39 The union of all simplices in a geometric simplicial complex ∆ is the polygon of ∆ and is denoted ||∆||. Thus the tetrahedron in Figure 3.7 is the polygon of the simplicial complex

∆ = {Ø, {a}, {b}, {c}, {d},

{a, b}, {a, c}, {a, d}, {b, c}, {b, d}, {c, d},

{a, b, c}, {a, c, d}, {b, c, d}, {a, b, c, d}}

For an abstract simplicial complex (V,K) we may assume V = S K since V ⊃ S K, and if v ∈ V \ S K then v is not in any simplex of K and v is irrelevant to the structure of (V,K). Thus for convenience we write (V,K) as K with the understanding that V = S K.

The vertex set of a geometric simplicial complex ∆, denoted by V (∆), is the union of the vertex sets A of all simplices of ∆. The tetrahedron from the example above along with its triangles, line segments, and vertices is a simplicial complex, and is denoted by ∆. The vertex set of ∆ is V (∆) = {a, b, c, d}.

Each geometric simplicial complex ∆ determines an abstract simplicial complex. Let V = V (∆) and let K be the set containing all simplices of ∆. Thus (V,K) obtained from ∆ in this way is an abstract simplicial complex.

Now start with an abstract simplicial complex (V,K) with V finite (for future reference we will always assume V is finite). Then we force V to be an affinely independent set using a homeomorphism, and let this set be our V (∆). The sets in K become the vertex sets of the simplices for ∆. We know that the geometric realization of our abstract simplicial complex K satisfies the first condition of being a geometric simplicial complex immediately from the definition of an abstract simplicial complex. The second condition follows from forcing V to be affinely independent.

For example, let ∆ be represented below.

40 2

4 3 1

Then V = {1, 2, 3, 4} and

K = {Ø, {1}, {2}, {3}, {4}, {1, 2}, {1, 3}, {2, 3}, {3, 4}, {1, 2, 3}}.

Thus we will use the geometric simplicial complex and abstract simplicial complex terms interchangeably and will just call this structure a simplicial complex.

Corollary 3.2.1. The small inductive dimension of a simplicial complex ∆ is

ind ∆ := max{K | K ∈ ∆}.

This follows directly from Lemma 3.2.1.

Deﬁnition 39. A subcomplex of a simplicial complex ∆ is a subset of ∆ that is also a simplicial complex.

Example 3.2.3. The simplicial complex

δ = {Ø, {a}, {b}, {d}, {a, b}, {a, d}, {b, d}, {a, b, d}} is a subcomplex of ∆ from Figure 3.7, which represents a triangle. This is also referred to as the subcomplex of ∆ by the vertex set {a, b, d}.

Deﬁnition 40. A k-skeleton of a simplicial complex ∆, denoted (∆, ≤ k + 1), consists of all simplices of ∆ of dimension at most k.

Example 3.2.4. The 3-skeleton of T is T . The 2-skeleton of T is the set of all the triangles, line segments, and vertices of T , i.e. the surface of T or the hollow tetrahedron.

41 Now we are ready to explicitly state the Van Kampen-Flores theorem.

Theorem 3.2.1 (The Van Kampen-Flores Theorem). For all d ≥ 1, the d-skeleton of the (2d + 2)-dimensional simplex cannot be embedded into R2d.

These simplicial complices cannot be embedded into R2d like a smooth manifold could be, according to the Whitney embedding theorem. The simplicial complices fail to meet the main hypothesis of the theorem: being a manifold. Before we can prove the Van Kampen-Flores theorem we will need a few more tools from combinatorial topology.

Let F and G be sets. Then deﬁne F ] G as the disjoint union of the sets F and G and can be described combinatorially as F ] G := (F × {1}) ∪ (G × {2}).

Deﬁnition 41. Let K and L be simplicial complices. The join K ∗L is the simplicial complex with vertex set V (K) ] V (L) and set of simplices {F ] G : F ∈ K,G ∈ L}. dim (K ∗ L) = dim (K)+ dim (L) + 1. For K ∗ K we write K∗2. The deleted join of K with itself has the vertex set V (K) × {1, 2} and is given by

∗2 ∗2 K∆ := {F ] G : F,G ∈ K,F ∩ G = Ø} ⊂ K .

∗2 Let D = {x0, x1}. The join D is pictured in the center below and the deleted

∗2 join D∆ is pictured to the right below.

x0 x1 x0 x1 x0 x1

0 0 0 0 0 0 x0 x1 x0 x1 x0 x1

Figure 3.8: The join and deleted join of the two point set D with itself. The join (center) attaches to all points in the set. The deleted join (right) attaches to those points in the disjoint union of each simplex.

42 A point p ∈ X ∗ Y is written as tx ⊕ (1 − t)y for x ∈ X, y ∈ Y , and t ∈ [0, 1]. We write points of X ∗ Y in this way since an element of X ∗ Y does not have to be entirely in either X or Y . As is depicted in Figure 3.8 an element in X ∗ Y can be on the line segment connected parts of X to Y , so an element in X ∗ Y is a ’weighted average’ of an element in X and an element in Y .

The join of maps can also be deﬁned. Consider f : X1 → Y1 and g : X2 → Y2.

Then f ∗ g : X1 ∗ X2 → Y1 ∗ Y2 is given by tx1 ⊕ (1 − t)x2 7→ tf(x1) ⊕ (1 − t)g(x2) for x1 ∈ X1, x2 ∈ X2, and t ∈ [0, 1].

Spheres are an important part of the proof of the Van Kampen-Flores theorem, but when we refer to an n-sphere we mean up to homeomorphism. Thus a hollow triangle will often be referred to as S1. It should be mentioned that the join of two spheres is also a sphere. As an example, we will show that S0 ∗ S1 = S2.

0 1 ∼ Example 3.2.5. Let S be represented by the two point set {x0, x1}, and S = {1, 2, 3}.

x0 2 x1

1 3

Then S0 ∗ S1 is represented below.

x0 2 x1

1 3

Remember that the triangle {1, 2, 3} is still hollow. This object can be adjusted to an object equivalent to S2, where the triangle {1, 2, 3} is, again, still hollow.

Deﬁnition 42. Let K ⊂ 2{1,2,...,n} be a simplicial complex. The Alexander dual of K is the simplicial complex B(K) ⊂ 2{1,2,...,n} that consists

43 x0 2

1 3

Figure 3.9: S0 ∗ S1 ∼= S2. The triangle {1, 2, 3} remains hollow. of the complements of the nonsimplices of K:

B(K) := {G ⊂ {1, 2, . . . , n} : Gc ∈/ K} = {Hc : H ∈ 2{1,2,...,n}\K}.

{1,2,3} Example 3.2.6. Let K0 = {Ø, {1}, {2}, {3}} ⊂ 2 . K0 is a simplicial com-

{1,2,3} plex. The nonsimplices of K0 in 2 are {{1, 2, 3}, {2, 3}, {1, 3}, {1, 2}}. Thus

B(K0) = {Ø, {1}, {2}, {3}}. We see B(K0) is a simplicial complex since B(K0) = K0.

{1,2,3} Example 3.2.7. Let K1 = {Ø, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}} ⊂ 2 . Again

{1,2,3} K1 is a simplicial complex. The only nonsimplex of K1 in 2 is {1, 2, 3}, so

B(K1) = {Ø} which is the trivial simplicial complex.

Deﬁnition 43. The Bier sphere associated with K is a simplicial complex with vertex set {1, 2, . . . , n} × {1, 2} deﬁned as the deleted join

BiernK := (K ∗ B(K))∆ = {F ] G : F ∈ K,G ∈ B(K),F ∩ G = Ø}

= {F ] G : F ∈ K,Gc ∈/ K,F ∩ G = Ø}

= {F ] Hc : F ∈ K,H/∈ K,F ⊂ H}.

Example 3.2.8. The Bier sphere associated with K0 from example 3.2.6 is

∗2 Bier3(K0) = (K0 ∗ B(K0))∆ = (K0 )∆

= {Ø, {1}, {2}, {3}, {10}, {20}, {30}, {1, 20},

{1, 30}, {2, 30}, {10, 2}, {10, 3}, {20, 3}}.

44 Bier3(K0) is a simplicial complex and is shown below.

1 3

10 30 20

Example 3.2.9. Consider the simplicial complex {Ø} ⊂ 2{1,··· ,n}. The nonsimplices of {Ø} in 2{1,··· ,n} are everything except {Ø}, so B({Ø}) = 2{1,··· ,n}\{1, ··· , n}. Thus

∼ n−2 Biern({Ø}) = ({Ø} ∗ B({Ø}))∆ = B({Ø}) and ||Biern({Ø})|| = S .

Theorem 3.2.2 (The Bier spheres are spheres). For every simplicial complex K ⊂

{1,2,...,n} 2 , the simplicial complex Biern(K) is an (n − 2)-sphere with at most 2n vertices.

The proof of the Van Kampen-Flores theorem relies heavily on the Bier spheres theorem, so we will summarize the proof here. For the full proof, see [7].

The proof uses induction on the number of simplices of K ⊂ 2{1,··· ,n} with n ﬁxed. The base case is K = {Ø}, which we have already computed in example 3.2.9. We want to know how adding a simplex to K eﬀects Biern(K). So we want to show that ∼ n−2 ∼ n−2 if ||Biern(K)|| = S , then ||Biern(K ∪ {F })|| = S with K ∪ {F } a simplicial complex, F ⊂ 2{1,··· ,n}\K, and every proper subset of F is in K3. If K did not contain every proper subset of F , then unioning F to K will not be adding only one simplex to K.

If L is a simplicial complex, then Lk denotes the set of all k-dimensional faces of

L. Now consider Biern(K ∪ {F }).

n−2 n−2 c c (Biern(K∪{F })) = ((Biern(K) \{(F \{i})]F : i ∈ F }∪{F ](F ∪{j}) : j∈ / F }.

3The last two conditions of this statement make F a minimal non-face.

45 The vertices aﬀected by the change in the Bier sphere are contained in

c VF = {{i}] Ø: i ∈ F } ∪ {Ø ]{j} : j ∈ F }.

So the subcomplex of Biern(K) by the vector set VF is

F F c L1 = (2 \{F }) ∗ 2 and ∼ k−2 n−k−1 ∼ (k−2)+(n−k−1)+1 n−2 ||L1|| = S ∗ S = S = S by the induction hypothesis. The subcomplex of Biern(K ∪ {F }) by the vector set

VF is

F F c c L2 = 2 ∗ (2 \{F }) and ∼ k−1 n−k−2 ∼ (k−1)+(n−k−2)+1 n−2 ||L2|| = S ∗ S = S = S

F F c F by the induction hypothesis. Then L0 = L1 ∩ L2 = ((2 \{F }) ∗ 2 ) ∩ (2 ∗ F c c F F c c ∼ k−2 n−k−2 ∼ (2 \{F })) = (2 \{F }) ∗ (2 \{F }). Let f = |F |, then ||L0|| = S ∗ S = Sn−3.

Note that a simplex that has a vertex that is not contained in VF never contains a simplex in L1\L0 or L2\L0. So ||L1|| and ||L2|| are glued to Biern(K) by ||L0||. ∼ ∼ n−2 Thus ||Biern(K ∪ {K})|| = ||Biern(K)|| = S .

Deﬁnition 44. The relative interior of a simplex σ arises from removing all faces of dimension smaller than dim σ.

Example 3.2.10. Let σ = {1, 2}. Then the relative interior of σ is the edge {1, 2} without its endpoints. Let σ = {1, 2, 3} Then the relative interior of σ is the open triangle.

46 For each point x ∈ ||∆|| with ∆ a simplicial complex, there exists exactly one simplex σ ∈ ∆ containing x in its relative interior. This simplex σ is denoted supp(x) and is called the support of the point x.

Deﬁnition 45. A Z2-space is a pair (X, ν) where X is a topological space and 2 ν : X → X is a homeomorphism called the Z2-action on X such that ν = ν◦ν = idx, where idx is the identity map.

n n If ν represents the antipodal map, then (S , ν) and (R , ν) are Z2-spaces. These

Z2-actions will be referred to as the standard Z2-actions and we will refer to these n n Z2-spaces as just S and R respectively.

Deﬁnition 46. If (X, ν) and (Y, ω) are Z2-spaces, a Z2-map f :(X, ν) → (Y, ω) is a continuous map from X to Y that commutes with the Z2-actions. So for all x ∈ X, f(ν(x)) = ω(f(x)), or f ◦ ν = ω ◦ f.

Deﬁnition 47. Let (X, ν) be a Z2-space. The Z2-index of (X, ν) is

Z2 n indZ2 := min{n ∈ N : X → S }

Z2 with → denoting that there exists a Z2-map.

The Z2-index is understood as the smallest value of n such that there exists a n Z2-map from X to S . This leads us to introducing another famous theorem.

Theorem 3.2.3 (The Borsuk-Ulam Theorem). If f : Sn → Sm is a continuous antipodal map, then n ≤ m.

n The Borsuk-Ulam theorem tells us that indZ2 (S ) = n [6]. This will be a very useful fact for proving the Van Kampen-Flores theorem.

d Let K be a simplicial complex. We will call f : ||K|| → R a bad map if f(x1) 6= f(x2) whenever supp(x1)∩ supp(x2) = Ø for x1, x2 ∈ ||K||. So a bad map takes every

47 simplex to something diﬀerent. A very useful example of a bad map is an embedding. An embedding is a bad map since it is a homeomorphism onto its image, and hence is always bijective onto its image.

d Lemma 3.2.2. Let f : ||K|| → R be a bad map. Then there exists a Z2-map ∗2 d ∗2 g : ||K∆ || → (R )∆ .

Proof. Start by letting f : ||K|| → Rd be a bad map. Then f ∗2 : ||K||∗2 → (Rd)∗2 is given by f ∗2(tx ⊕ (1 − t)y) = tf(x) ⊕ (1 − t)f(y) for x, y ∈ ||K|| and t ∈ [0, 1]. Now

∗2 ∗2 ∗2 restrict the domain of f to ||K∆ ||, and call this map g. Elements in ||K∆ ||, the domain of g, have the form tx ⊕ (1 − t)y) for t ∈ [0, 1], x, y ∈ ||K||, and supp(x) ∩ supp(y) = Ø. So elements in the image of g have the form f ∗2(tx ⊕ (1 − t)y) = tf(x) ⊕ (1 − t)f(y) for t ∈ [0, 1], x, y ∈ ||K||, and supp(x) ∩ supp(y) = Ø. Because we assumed that f is a bad map, f(x) 6= f(y) when supp(x) ∩ supp(y) = Ø. This implies the image of g is contained in

d {ty1 ⊕ (1 − t)y2 : t ∈ [0, 1], y1, y2 ∈ R , y1 6= y2} which is contained in the set

1 1 ( d)∗2\ y ⊕ : y ∈ d =: ( d)∗2. R 2 2 R R ∆

∗2 d ∗2 Thus g : ||K∆ || → (R )∆ is a Z2-map.

d ∗2 Lemma 3.2.3. indZ2 ((R )∆ ) ≤ d.

d ∗2 Proof. To show that indZ2 ((R )∆ ) ≤ d, we want to show that there exists a Z2 map d ∗2 d d d+1 from (R )∆ into S . For x = (x1, x2, . . . , xd) ∈ R letx ¯ = (1, x1, x2, . . . , xd) ∈ R . d ∗2 d Deﬁne h :(R )∆ → S by

tx¯ − (1 − t)¯y h(tx ⊕ (1 − t)y) = ||tx¯ − (1 − t)¯y||

48 d ∗2 for t ∈ [0, 1]. Then h is continuous by deﬁnition of (R )∆ and h commutes with the d d d ∗2 standard Z2-actions for R and S . Thus h is a Z2-map and indZ2 ((R )∆ ) ≤ d.

d ∗2 Theorem 3.2.4. If f : ||K|| → R is a bad map, then indZ2 (K∆ ) ≤ d.

Proof. Let f : ||K|| → Rd be a bad map. Then by Lemma 3.2.2 and Lemma 3.2.3, ∗2 d h ◦ g : ||K∆ || → S is a Z2-map with the functions g and h being those as described ∗2 in lemmas 3.2.2 and 3.2.3 respectively. Hence indZ2 (K∆ ) ≤ d.

∗2 The contrapositive of Theorem 3.2.4 tells us that if indZ2 (K∆ ) > d, then there does not exist a bad map f : ||K|| → Rd. Thus K does not embed in Rd. Now we are ready to prove the Van Kampen-Flores Theorem.

Proof of the Van Kampen-Flores Theorem. Let n = 2d + 3 and K = ({1, 2, . . . , n}, ≤ d + 1), the d-skeleton of the (2d + 2)- dimensional simplex. The nonsimplices of K are those with dimension strictly greater than d, so the complements of the nonsimplices of K are those with dimension less

∗2 than or equal to d. Thus B(K) = K. By the Bier spheres theorem, Biern(K) = K∆

∗2 2d+1 is a (2d + 1)-sphere. So indZ2 (K∆ ) = indZ2 (S ) = 2d + 1 > 2d. Thus K does not embed into R2d.

The most well known case of the Van Kampen-Flores theorem is when d = 1. When we set d = 1, that forces n = 5. This means we are trying to show that we cannot embed the 1-skeleton of the 4-simplex into R2. A 4-simplex is the convex hull of five affinely independent vertices, and so the line segments between all five vertices are contained in the 4-simplex. The 1-skeleton of the 4-simplex contains all of the

49 line segments and vertices of the 4-simplex. This implies that the 1-skeleton of the 4-simplex contains all ﬁve vertices and all of the line segments connecting the vertices.

Hence the case when d = 1 becomes the theorem that K5, the complete graph on

ﬁve vertices, is nonplanar. We can see in Figure 3.2 that K5 is the 1-skeleton of the

4-simplex since graphs are just vertices and line segments. The nonsimplices of K5 are those with dimension strictly greater than 1, so the complements of the nonsimplices of K5 are those with dimension less than or equal to 1, which are just vertices and

∗2 line segments. Thus B(K5) = K5. By the Bier spheres theorem, Biern(K5) = (K5)∆

∗2 3 3 2 is a 3-sphere. So indZ2 ((K5)∆ ) = indZ2 (S ). We know S does not embed into R , thus K5 does not embed into the plane.

1 3

4 5

Figure 3.10: K5

The simplicial complex K in the proof of the Van Kampen-Flores theorem is not a smooth manifold. The vertices of this simplicial complex cannot be made smooth through a homeomorphism onto Euclidean space. K5 is not a manifold because at each vertex K5 is not locally homeomorphic to R. This is why the Whitney embedding theorem does not hold for such a simplicial complex K. Thus the best possible bound for general n-dimensional topological spaces beings embedded into Euclidean space is 2n + 1.

50 Chapter 4: Conclusion

Dimension is a vital part of mathematics. Dimension arises in every area of math. It is a useful topological invariant which can help analyze mathematical objects such as fractals, a growing area in math. Topological dimension also leads to knowing when certain objects can be realized in Euclidean space.

4.1 Further Study

There are many different ways to categorize and define dimension. One way to continue study in this area would be to find and define all the various types of dimension such as the covering dimension, packing dimension, and similarity dimension.

We could also investigate further into other embedding theorems. We could try to make the bounds to embed into Euclidean space better by focusing on certain types of topological objects. One theorem that comes to mind is the Nash embedding theorem which states that any n-dimensional Riemannian manifold can be embedded into Rm with m ≥ n + 1 [9].

Also, dimension theory has recently become very useful in the ﬁeld of dynamical systems [11]. Going into this branch of dimension theory would be completely diﬀerent than anything presented in this paper.

51 Bibliography

[1] Masaisa Adachi, Embeddings and Immersions. American Mathematical Society, 1993.

[2] Gerald A. Edgar, Measure, Topology, and Fractal Geometry. Springer-Verlag, New York, 1990.

[3] Melinda Glass, Dimensions of Self-Similar Fractals. MA Thesis. Wake Forest University, 2011. Print.

[4] Victor Guillemin and Alan Pollack, Diﬀerential Topology. Prentice-Hall, Engle- wood Cliﬀs, 1974.

[5] Witold Hurewicz and Henry Wallman, Dimension Theory. Princeton University Press, Princeton, 1974.

[6] Mark de Longueville, A Course in Topological Combinatorics. Springer, New York, 2013.

[7] Jiri Matousek, Using the Borsuk-Ulam Theorem. Springer, Berlin, 2003.

[8] James R. Munkres, Topology. Prentice Hall, New Jersey, 2000.

[9] John Nash, The Imbedding Problem for Riemannian Manifolds. The Annals of Mathematics, 63 No. 1: 20-63, (1956).

[10] A. R. Pears, Dimension Theory of General Spaces. Cambridge University Press, Cambridge, 1975.

[11] Ya B. Pesin, Dimension Theory in Dynamical Systems. The University of Chicago Press, Chicago, 1997.

52 [12] Terrence Tao, An Introduction to Measure Theory. American Mathematical So- ciety, 2011.

[13] Hassler Whitney, Diﬀerentiable Manifolds. The Annals of Mathematics, 37 No. 3: 645-680, (1936).

[14] Hassler Whitney, The Self-Intersections of a Smooth n-Manifold in 2n-Space. The Annals of Mathematics, 45 No. 2: 220-246, (1944).

53 Curriculum Vitae

• Education

– PhD, Mathematics, Temple University, In Progress

– M.A., Mathematics, Wake Forest University, May 2014

– B.S., Mathematics, Honors Program, Summa Cum Laude, Florida South- ern College, May 2012

• Academic Awards and Organizations

– Member of AMS

– Member of Kappa Mu Epsilon, Mathematics Honor Society (FSC)

– Member of Phi Eta Sigma, Honor Society (FSC)

– Sunshine State Conference Academic Award, 2009-2012

– Division II Athletics Directors Association Academic Achievement Award, 2009-2012

• Research History

– Master’s thesis at Wake Forest University

∗ Topic: Dimension theory, embedding theorems

∗ Advisor: Dr. Sarah Raynor

– Honors thesis at Florida Southern College

∗ Topic: Knot theory, p-colorings

∗ Advisor: Dr. Daniel Jelsovsky

54 – Senior seminar project at Florida Southern College

∗ Topic: Point-set topology, Alexandroﬀ spaces

∗ Advisor: Dr. David Rose

∗ Paper Alexandroﬀ Spaces published in The Journal of Advanced Stud- ies in Topology, 2013

• Work History

– Teaching Assistant - Wake Forest University, Department of Mathematics (August 2012 - May 2014)

– Discrete Event Simulation Intern - NASA, Cape Canaveral, FL (May 2013 - August 2013)

• Campus Involvement

– Member of Florida Southern College Women’s Basketball Team (2008- 2012)

– Member of Wake Forest University Club Ultimate Frisbee Team (2012- 2014)