Generalizations and Properties of the Ternary Cantor Set and Explorations in Similar Sets

Home , Cantor function, Cantor set, Perfect set

Rebecca Stettin

A capstone project submitted in partial fulﬁllment of graduating from the Academic Honors Program at Ashland University May 2017

Faculty Mentor: Dr. Darren D. Wick, Professor of Mathematics Additional Reader: Dr. Gordon Swain, Professor of Mathematics Abstract

Georg Cantor was made famous by introducing the Cantor set in his works of mathematics. This project focuses on different Cantor sets and their properties. The ternary Cantor set is the most well known of the Cantor sets, and can be best described by its construction. This set starts with the closed interval zero to one, and is constructed in iterations. The first iteration requires removing the middle third of this interval. The second iteration will remove the middle third of each of these two remaining intervals. These iterations continue in this fashion infinitely. Finally, the ternary Cantor set is described as the intersection of all of these intervals. This set is particularly interesting due to its unique properties being uncountable, closed, length of zero, and more. A more general Cantor set is created by tak- ing the intersection of iterations that remove any middle portion during each iteration. This project explores the ternary Cantor set, as well as variations in Cantor sets such as looking at different middle portions removed to create the sets. The project focuses on attempting to generalize the properties of these Cantor sets.

i Contents

Page

1 The Ternary Cantor Set 1

1 2 The n -ary Cantor Set 9

n−1 3 The n -ary Cantor Set 24

4 Conclusion 35

Bibliography 40

Biography 41

ii Chapter 1

The Ternary Cantor Set

Georg Cantor, born in 1845, was best known for his discovery of the Cantor set. After writing a thesis in number theory, Cantor became interested in topology and wrote a series of papers on point set topology. He coined the term everywhere dense, which is currently used today. In this same series of papers, Cantor later introduced the idea of perfect sets. He made the ternary Cantor set famous in this same paper when attempting to prove that it was possible to have a set that is closed and nowhere dense. He noted that this was a perfect, inﬁnite set that is nowhere dense in any interval, regardless of the size. (Fleron [3]) The Cantor set is created by starting with the closed interval [0,1]. The next step is to remove a middle portion of the interval. This initial removal will then leave two intervals, namely

1 2 [0, 3 ] and [ 3 , 1]. This process is continued an infinite number of times, with the final set of numbers, after the infinite iteration, being the Cantor set itself. This set is the intersection

∞ th of all iterations, denoted as T = ∩i=1 Ti, where Ti represents the i iteration. There are

1 diﬀerent proportions that can be removed with this set, the most well known being 3 . This set, referred to as the ternary Cantor set, can also be referred to as the classic set. This particular version of the Cantor set was the one Georg Cantor used to make his arguments about the nature of certain sets. (Aczel [1])

We will describe the construction of the ternary Cantor set, T , in a similar manner as it is

1 described in Thomson [7]. Begin with the closed interval [0, 1] and remove a dense open set, G. The remaining set, T = [0, 1] \G will also be closed and nowhere dense in [0, 1]. Based on our construction of G, T will have no isolated points. It is easiest to understand the set

1 2 G if we construct it in stages. Let G1 = ( 3 , 3 ) and let T1 be what is remaining in [0, 1] after 1 2 removing G1. Thus T1 = [0, 3 ] ∪ [ 3 , 1] is what remains when the middle third of the interval [0,1] is removed. This is referred to as the ﬁrst iteration of the ternary Cantor set. We repeat

1 2 7 8 this construction for each of the two intervals of T1. Let G2 = ( 9 , 9 ) ∪ ( 9 , 9 ). These intervals 1 2 1 2 7 8 are the middle third of the previous 2 intervals. Then T2 = [0, 9 ] ∪ [ 9 , 3 ] ∪ [ 3 , 9 ] ∪ [ 9 , 1]. This completes the second iteration. We continue inductively, and ultimately take the intersection of each iteration of Ti to create the ternary Cantor set with the following properties: For each i ∈ N

i 1. Ti is a union of 2 pairwise disjoint closed intervals.

2. Each component of the set of open intervals removed is the ”middle third” of some

component of Ti.

1 3. The length of each component of Ti is 3i .

Some proofs of the properties of the ternary Cantor set will be explored in later chapters in regards to more generalized Cantor sets. The following is a list of characteristics of the ternary Cantor set with some proofs to follow:

1. The Cantor set is nonempty.

2. The Cantor set is closed.

3. The Cantor set has length 0.

4. The Cantor set contains all numbers that can be written in base-3 without 1’s.

5. The Cantor set is uncountable.

6. The Cantor set is nowhere dense.

2 7. The Cantor set has no isolated points.

8. The Cantor set is a perfect set.

Theorem 1. The ternary Cantor set is nonempty.

Proof. Using similar methods as Nelson [5], we must show that the ternary Cantor set contains at least one element. We can see that during each iteration, the endpoints remain

1 2 in the set. For example, during T1, everything between ( 3 , 3 ) is removed. Note that it is everything between the two endpoints, which implies the two endpoints remain. Once

1 2 removed, T1 is [0, 3 ] ∪ [ 3 , 1]. As seen in here, the intervals are closed, thus including the

endpoints. This trend continues throughout each iteration. For T2, the middle thirds of

1 those two previously mentioned intervals are removed. To clarify, everything between ( 9 , 2 7 8 1 2 1 2 7 8 9 ) as well as ( 9 , 9 ) is removed. This leaves [0, 9 ] ∪ [ 9 , 3 ] ∪ [ 3 , 9 ] ∪ [ 9 , 1], which is once 1 2 again a union of closed intervals, which include endpoints. Notice that 0, 3 , 3 , and 1 are again included in this iteration. These iterations continue inﬁnitely. Since each iteration of T contains the endpoints of each iteration before it, it becomes clear that the endpoints of each iteration are included. Since the previous endpoints are consistently included, there will clearly be elements in the intersection of each iteration, proving the ternary Cantor set is nonempty.

Theorem 2. The ternary Cantor set is closed.

Proof. As shown earlier when looking at the length of T, it is an intersection of closed sets.

1 2 For example, T1 is [0, 3 ] ∪ [ 3 , 1]. This trend continues with each Ti also consisting of a

ﬁnite union of closed intervals. Each Ti is closed since it is a ﬁnite union of closed intervals.

∞ Since T = ∩i=1Ti it is a countable intersection of closed sets and is therefore closed.

Theorem 3. The ternary Cantor set has length of 0.

Proof. This proof is also based oﬀ of similar methods used in Nelson [5]. This proof can be demonstrated by noting everything that is taken out of the starting interval has the same

3 length as the initial interval. Looking at what is taken out during each iteration, a sequence

1 emerges. During the ﬁrst iteration, an interval of length 3 is removed from the set. During 1 2 the second iteration, there are 2 diﬀerent intervals, namely [0, 3 ] and [ 3 ,1], to remove the 1 1 1 middle third from. In this case, however, the length of what is removed is 3 of 3 , or 9 . This 1 th must occur twice because there are 2 intervals that ( 9 ) is being removed from. Therefore, 2 2 a length of 32 = 9 will be removed during the second iteration of T. This trend continues 1 2 4 8 throughout each iteration resulting in the length removed at each iteration as 3 , 9 , 27 , 81 ,··· The total length removed can also be represented as

∞ X 2i 3i+1 i=0

1 1 This sum can be determined through the sum as a geometric series, or 3 2 = 1. This 1− 3 means that everything removed has length 1. As mentioned previously, the starting interval,

T0, has length 1 as well. A simple subtraction of the removed portion from the initial length will give the ﬁnal length of T as 0. Therefore, T has length 0.

The following theorem will be explored through examples, but will not be proven. A similar proof can be found in Chapter 3.

Theorem 4. The ternary Cantor set only contains numbers that have a base-3 representation excluding the digit 1.

A brief review of base-3 representation is necessary to proceed with this example. Base-3, or ternary, representation consists of 3 digits, 0, 1, and 2. A number of the form (.a1a2a3··· )3

n can be thought of in terms of sums of fractions with numerators an and denominators 3 .

a1 a2 a3 That is, (.a1a2a3··· )3 = 3 + 32 + 33 + ··· where an must be either 0, 1, or 2. The claim is every number that is in the ternary Cantor set can be written in base-3 representation with only 2’s and 0’s. That is, ∀ x ∈ T, x = (.a1a2a3··· )3 such that ai ∈ {0,

1 2 2}. It is helpful to look at endpoints, starting with T1 = [0, 3 ] ∪ [ 3 , 1]. This means that it

4 1 2 contains everything between 0 and 3 and then everything between 3 and 1. The endpoints 1 2 0, 3 , 3 , and 1 are included as well. Looking at the endpoints in base-3 representation is 1 2 1 useful. 0 = (.0)3, 3 = (.1)3 = (.02)3, 3 = (.2)3, and 1 = (.2)3. The endpoints 3 and 1 are

particularly interesting. 1 = (.2)3 since 2 is the largest digit in base-3. This is the same idea

1 1 that is used in base-10 where 1 = (.9)10. 3 = (.1)3 because it is simply 1 * 3 . This can be rewritten using repeating digits just like the number 1. Any number in base-3 with a 1 as a terminating digit can actually be rewritten by replacing the 1 with 02. We can see this

when converting (.02)3 into base-10:

1 1 1 (.02) = 0 ∗ + 2 ∗ + 2 ∗ + ··· 3 31 32 33

which is a geometric series written as

∞ ∞ X 2 X 1 = 2 ∗ 3j 3j j=2 j=2

1 1 9 9 = 2 ∗ 1 = 2 ∗ 2 1 − 3 3 1 3 3 1 = 2 ∗ ∗ = = 9 2 9 3

1 1 This conversion shows that 3 = (.1)3 = (.02)3. The ﬁrst interval, [0, 3 ] in base-3 can be

written as any number x where x = (.0a2··· )3, where a2 represents any of the base-3 digits.

2 Similarly, the second interval, [ 3 , 1] in base-3 can be written as any number x where x =

(.2a2··· )3, again where a2 is any base-3 digit.

1 2 1 2 7 8 The same procedure can be repeated for T2. T2 = [0, 9 ] ∪ [ 9 , 3 ] ∪ [ 3 , 9 ] ∪ [ 9 , 1]. 1 2 In addition to the previous endpoints, 0, 3 , 3 , and 1, there are 4 more endpoints. These 1 2 7 8 endpoints can be written in base-3 as follows: 9 = (.002)3, 9 = (.02)3, 9 = (.202)3, and 9 =

(.22)3. These endpoints can again create restrictions with what digits in base-3 representation

1 are included. For example, the interval [0, 9 ] in base-3 can be written as any number x where

5 2 1 x = (.00a3··· )3, where a3 represents any of the base-3 digits. The next new interval, [ 9 , 3 ] in base-3 can be written as any number x where x = (.02a3··· )3, where a3 represents any of

2 7 8 the base-3 digits. Using this notation, [ 3 , 9 ] would have x = (.20a3··· )3, and [ 9 , 1] would

have x = (.22··· )3. This shows that as each iteration continues, the digits of the base-3 representations become more speciﬁc. Each iteration adds a new digit restriction for this ternary representation. A proof in one of the following chapters will give a similar proof of base representations in a generalized Cantor set. A set is considered uncountable when the number of elements cannot be written as some subset of the natural numbers.

Theorem 5. The Cantor set is uncountable.

Proof. This will be a proof by contradiction using similar methods as Shaver [6]. First assume that the ternary set is countable. This means that there exists an onto function that maps all elements from the natural numbers to the set. Recall from Theorem 4 that the ternary Cantor set consists of all numbers ∈ [0,1] that can be written in base-3 representation using only 0’s and 2’s. These numbers are of the form (.a1a2a3··· )3 such that each ai is either 0 or 2. This set will be denoted as T = {x ∈ [0,1] : x has base-3 representation of only 0’s and

2’s }. Assume that this set is countable, f: N → T, where f is one to one and onto. Let

f(1) = 0.a11a12a13 ···

f(2) = 0.a21a22a23 ···

f(3) = 0.a31a32a33 ···

. .

Each of aij must be 0’s or 2’s, based on the base-3 representation theorem that was previously discussed. Deﬁne a number b as b = .b1b2··· where

6   2 if ajj = 0 bj =  0 if ajj = 2

It is clear by this piecewise function that b ∈ T since its base-3 representation only contains 0’s and 2’s. However b 6= f(n) ∀ n because they diﬀer in the nth ternary position. Therefore this function is not onto, which creates a contradiction that there exists an onto function that maps all elements of N into the set.

A set is nowhere dense when every open interval contains a subinterval such that the intersection of the subinterval and the original set is equal to the empty set. Alternatively, a set is nowhere dense if it contains no open intervals.

Theorem 6. The ternary Cantor set is nowhere dense.

Proof. To prove that the ternary Cantor set is nowhere dense, we must show that T has no open intervals, similar to the methods used in Dimartino [3]. Let J be an open interval in

1 i [0,1] with length λ. Choose i such that 3i < λ. Ti is a union of 2 pairwise disjoint closed 1 intervals, known as the components of Ti. Each component of Ti has length 3i . Now let L 1 be a component of Ti. Deﬁne ` as the length of a component. Then `(L) = 3i < λ = `(J). T∞ Therefore, J * L ⊆ Ti. Since the Cantor set, T = i=1 Ti, then J * T, otherwise J would

be a subset of all Ti, but we have shown that J * Ti. Therefore the Cantor set contains no open intervals and is nowhere dense.

Isolated points are deﬁned as any point that is some positive distance away from every other point in the set.

Theorem 7. The ternary Cantor set has no isolated points.

Proof. To prove this we must show that every point in T is a limit point. Suppose x ∈ T. We must show that ∀ ε > 0 ∃ y ∈ T with |x - y| < ε and y does not equal x. Let x ∈ T

1 and ε > 0. Choose i such that 3i < ε. Since x ∈ T and T = ∩Ti, then x ∈ Ti. This means

7 1 that there is a component, L of Ti such that x ∈ L and `(L) = 3i . Now, L ∩Ti+1 has 2 components, L0 and L1, and x must be in one of these. Say x ∈ L0. L0 also contains a point

1 in T called y. In particular, y is an endpoint of L0. Then | x - y | ≤ 3i < ε.

A perfect set is a closed set with no isolated points.

Theorem 8. The ternary Cantor set is a perfect set.

Theorem 2 and Theorem 7 show that the Cantor set is closed and has no isolated points, so this theorem has already been proved. The ternary Cantor set is easily the most studied version of the Cantor set due to its unique properties. The next chapters will explore more general Cantor sets, where a diﬀerent proportion is removed during each iteration and determine which properties hold in those particular sets.

8 Chapter 2

1 The n-ary Cantor Set

The previous chapter described numerous characteristics and properties of the ternary

1 Cantor set. The word ternary meant that the middle 3 was being removed during each iteration. It is interesting to look at variations on this typical Cantor set to see what can be discovered. We will look at variations of this construction in the next two chapters. These sets can again be described by what is being removed during in each iteration. We ﬁrst will

1 k look at removing the middle n during each iteration. These sets can be denoted as Ci , where i denotes the ith iteration and k describes the portion of the interval being removed

1 at each iteration. Using this notation, the ternary Cantor set would be written as C 3 and

nd 1 the 2 iteration of the ternary Cantor set would be written as C2 3 . This chapter will begin by exploring the endpoints of the ﬁrst iteration of C and eventually attempting to create a

1 base representation of all numbers that are in C, where C represents C n .

Lemma 9. The endpoints of C can be represented as

n − 1 n + 1 a = and b = 2n 2n

where C1 = [0, a] ∪ [b, 1].

1 1 Proof. Let k = n where n is the middle portion of the interval that is being removed. Then

9 after the ﬁrst iteration, [0, a] ∪ [b, 1] is remaining. Since k is being removed from the initial interval, then b − a = k. The combined lengths of [0, a] and [b, 1] must equal 1 − k. That is, (a − 0) + (1 − b) = 1 − k.

The lengths of [0, a] and [b, 1] must be equal since the interval removed is in the middle of the initial interval. Since the two intervals are equal, then they each must be half the length of what is remaining after removing the middle interval. That is

1 1 a − 0 = a = (1 − k) and 1 − b = (1 − k) 2 2

These can ﬁnally be simpliﬁed to

1 − k 1 + k a = and b = 2 2

1 Substituting n for k yields 1 − 1 1 + 1 a = n and b = n 2 2

which simpliﬁes to n − 1 n + 1 a = and b = 2n 2n

The following proofs relate back to the initial list of properties of the ternary Cantor

1 1 set listed in chapter 1 that were not proved. Since 3 is in the form of n , where n = 3, the ternary set falls under these particular Cantor sets. Therefore, the following collection of proofs also prove results for the ternary Cantor set. We start by looking at the number of intervals remaining and removed after each iteration of C.

i Lemma 10. Ci consists of 2 pairwise disjoint subintervals, [s, t].

Proof. We give a proof by induction. Case i = 1: based on the previous theorem, C1 consists

n−1 n+1 1 of disjoint intervals of the form [0, 2n ] ∪ [ 2n , 1]. It is clear that C1 consists of 2 = 2

10 intervals, so this holds for i = 1.

Assume true for Ci. Consider case Ci+1. By construction, after removing the middle

1 n of any interval, 2 will remain. Therefore, each component [s, t] of Ci creates 2 pairwise

disjoint intervals to Ci+1. Therefore, the number of subintervals in Ci+1 = 2 * the number

i i+1 of subintervals in Ci. With the inductive hypothesis, this is equal to 2(2 ) = 2 .

i Lemma 11. 2 intervals are removed from Ci when constructing Ci+1.

i Proof. Based on the previous theorem, we know that Ci consists of 2 pairwise disjoint

intervals. Therefore, to construct Ci+1, we will remove a middle portion of each interval in

i Ci. Since there are 2 intervals in Ci, then that means we will remove a middle portion from

i each of these intervals, so we will remove 2 intervals from Ci to construct Ci+1.

1 The next property to explore in the n -ary Cantor set is the length of these intervals that remain after each iteration, as well as the length of the intervals that are removed during each iteration of C.

Lemma 12. The length of each interval remaining in a component of Ci is

n − 1i 2n

n−1 n+1 Proof. We give a proof by induction. For the i = 1 case, C1 is deﬁned as [0, 2n ] ∪ [ 2n , 1].

The lengths of each component of C1 are equal since the middle interval is removed. This

n−1 n−1 means that each interval in C1 has length 2n − 0 = 2n . Therefore, this holds for the i = 1 case.

Assume true for Ci. Consider Ci+1. Each interval of this iteration will have equal length due to the construction of removing the middle interval. If the portion of the interval being

1 n−1 n−1 removed is n , then that means that n is remaining. This n must be split between the two remaining intervals. Since the middle portion is being removed, this will be an even split,

n−1 th n−1 i so any interval remaining in Ci+1 will be 2n of the previous interval’s length, 2n ,

11 which can be written as n − 1n − 1i 2n 2n

n − 1i+1 = 2n

Lemma 13. The length of each interval that is removed during iteration Ci to create Ci+1

1 n−1 i is equal to n 2n

0 0 Proof. Let [s, t] be a component of Ci and [s , t ] be the interval that is removed from Ci to

0 0 0 0 construct Ci+1. The length of [s , t ] is equal to t − s . This interval can be thought of as

1 th 1 one n of [s, t]. This expression can be written as n ∗ (t − s). From the previous lemma, the n−1 i 0 0 1 n−1 i length of (t − s) is equal to 2n . Therefore, the length of [s , t ] = n 2n .

Using the previous four lemmas in conjunction, we can now explore the total length removed and the total length remaining in each iteration of C.

(n−1)i Lemma 14. The total length removed in each iteration of Ci to create Ci+1 is ni+1 .

Proof. Based on Lemma 11, there are 2i intervals removed during each iteration of C. Based

1 n−1 i on Lemma 13, each of these intervals have length n ∗ 2n . Then, the total length removed i 1 n−1 i would be 2 intervals of this length n ∗ 2n , which can be written as

1 (n − 1)i (n − 1)i 2i ∗ ∗ = n 2ini ni+1

n−1 i Lemma 15. The total length remaining in each iteration of Ci is equal to n .

n−1 n+1 Proof. We give a proof by induction. Consider i = 1 case. C1 consists of [0, 2n ] ∪ [ 2n , 1]. The length of each of these intervals are equal since the middle interval is removed. The ﬁrst

12 n−1 n−1 interval has length 2n - 0 = 2n . Since C1 consists of two such intervals, the total length

of C1 is n − 1 n − 1 n − 11 2 ∗ = = 2n n n

Therefore, this holds for the i = 1 case.

1 Assume true for Ci. Consider the i + 1 case. To construct Ci+1, we will remove n of Ci. n−1 i Based on the inductive hypothesis, Ci has length of n . Then Ci+1 will have length

n − 1i 1 n − 1i − ∗ n n n

(n − 1)i 1 (n − 1)i = − ∗ ni n ni (n − 1)i (n − 1)i = − ni ni+1 n(n − 1)i − (n − 1)i = ni+1 (n − 1)i(n − 1) = ni+1 (n − 1)i+1 n − 1i+1 = = ni+1 n

n−1 i Therefore, each iteration of Ci has length n .

1 The previous six lemmas are useful in looking at the total length of the n -ary Cantor set, much like we did with the ternary Cantor set.

1 Theorem 16. The n -ary Cantor set has length 0.

Proof. This can be proven by looking at everything that is being removed from C. Recall by

(n−1)i Lemma 14 that ni+1 is removed during each iteration of Ci. This means that during C0, 1 (n−1) (n−1)2 n1 is removed, during C1, n2 is removed, and during C2, n3 is removed. This pattern

13 continues and total length removed can be written as

∞ X (n − 1)j nj+1 j=0

1 n = n−1 1 − n

1 1 n n = n n−1 = 1 = 1 n − n n

This means that the total length of 1 is removed. The length of C0 is the length of the initial interval, [0, 1] which is 1. Therefore, the length of C is 1 − 1 = 0.

The next few proofs utilize the notation of the endpoints in C1 and attempt to generalize what numbers remain in C by giving a base representation, much like the ternary Cantor set has. These theorems must be given in cases based on parity of the endpoints as follows.

1 n−1 n+1 Theorem 17. When the n in n being removed is even, a = 2n and b = 2n are both irreducible fractions.

Proof. When n is even, then the endpoints a and b are fractions with numerators n − 1 and

n−1 n+1 n + 1 respectively. Since n is even, n + 1 and n − 1 must both be odd. Then 2n or 2n will reduce if and only if GCD(n − 1, 2n) > 1 or GCD(n + 1, 2n) > 1 respectively.

Suppose r | n − 1 and also r | 2n. There exists 2 integers, x and y, such that

n − 1 = rx and 2n = ry

Then 2n n = rx + 1 and y = r

Using substitution, 2n = 2(rx + 1) = 2rx + 2

14 Therefore, 2rx + 2 2 y = = 2x + r r

2 is an integer. Since y and 2x are integers, r must also be an integer. Therefore, r = 1 or r = 2. Since n is even, n − 1 is odd. So 2 - n − 1. Therefore, r = 1 and GCD(n − 1, 2n) = 1 n−1 so a = 2n is irreducible. For endpoint b, suppose r | n + 1 and also r | 2n. There exists 2 integers, x and y, such that n + 1 = rx and 2n = ry

Then 2n n = rx − 1 and y = r

Using substitution, 2n = 2(rx − 1) = 2rx − 2

Therefore, 2rx − 2 2 y = = 2x − r r

2 is an integer. Since y and 2x are integers, r must also be an integer. Therefore, r = 1 or r = 2. Since n is even, n + 1 is odd so 2 - n + 1. Therefore, r = 1 and GCD(n + 1, 2n) = 1 n+1 so b = 2n is irreducible.

1 n−1 n+1 Theorem 18. When the n in n being removed is odd, a = 2n and b = 2n can be reduced k k+1 to the irreducible form a = 2k+1 and b = 2k+1 where n = 2k + 1.

Proof. Let n = 2k + 1 where k is an integer. The endpoints a and b can then be rewritten as

n − 1 2k + 1 − 1 2k k a = = = = 2n 2(2k + 1) 2(2k + 1) 2k + 1

and n + 1 2k + 1 + 1 2k + 2 k + 1 b = = = = 2n 2(2k + 1) 2(2k + 1) 2k + 1

15 Suppose r | k and also r | 2k + 1. There exist 2 integers, x and y, such that

k = rx and 2k + 1 = ry

Then 2k + 1 y = r

Therefore, 2rx + 1 2rx 1 1 y = = + = 2x + r r r r

1 is an integer. Since y is an integer, then r must also be an integer. Therefore, r = 1 and k GCD(k, 2k + 1) = 1 and endpoint a = 2k+1 is irreducible. Suppose r | k + 1 and r | 2k + 1. There exists 2 integers, x and y, such that

k + 1 = rx and 2k + 1 = ry

Then 2k + 1 k = rx − 1 and y = r

and 2(rx − 1) + 1 y = r 2rx − 1 y = r 1 y = 2x − r

1 is an integer. Since y is an integer, then r must also be an integer. Therefore, r = 1 and k+1 GCD(k + 1, 2k + 1) = 1 and endpoint b = 2k+1 is irreducible.

1 Theorem 19. When the n in n being removed is even, the endpoints after the ﬁrst iteration n−1 can be represented in base 2n as a = 2n = (.(n - 1)0000··· )2n = (.(n - 2)2n − 1)2n and b n+1 = 2n = (.(n + 1)000··· )2n.

16 Proof. The proof follows by converting base 2n representations to base 10 representations as follows: n − 2 2n − 1 2n − 1 (.(n − 2)(2n − 1))2n = + + + ··· 2n (2n)2 (2n)3

∞ n − 2 X 2n − 1 = + 2n j j=2 (2n)

∞ n − 2 X 1 = + (2n − 1) ∗ 2n j j=2 (2n)

1 n − 2 (2n)2 = + (2n − 1) ∗ 1 2n 1 − 2n 1 ! n − 2 (2n)2 = + (2n − 1) ∗ 2n−1 2n 2n

n − 2 1 2n = + (2n − 1) ∗ ∗ 2n (2n)2 2n − 1 n − 2 1 n − 1 = + = 2n 2n 2n

Therefore, n − 1 a = = ((n − 2)(2n − 1)) 2n 2n

Converting b base 2n to b base 10 requires looking at each digit and dividing by 2n to

th 1 the n power. The ﬁrst digit in base 2n is simply equal to 2n multiplied by the ﬁrst digit. This means that n + 1 n + 1 (.(n + 1)0000 ··· ) = = = b 2n 2n 2n

1 Example. We consider the case of removing the middle n where n = 4 is even. Thus, the n−1 n+1 endpoints are a = 2n and b = 2n :

17 1 The endpoints a and b in the 4 case would be

4 − 1 3 4 + 1 5 a = = and b = = 2(4) 8 2(4) 8

Endpoint a can be represented in base 2(4) = 8 as

3 = (.(4 − 2)(2(4) − 1)) = (.27) 8 8 8 which can be seen by converting this to base 10 representation:

2 7 7 (.27) = + + + ··· 8 8 82 83

∞ 2 X 7 = + 8 8j j=2

∞ ! 2 X 1 = + 7 ∗ 8 8j j=2

1 ! 2 64 = + 7 ∗ 1 8 1 − 8

1 ! 2 64 = + 7 ∗ 7 8 8 ! 2 1 8 = + 7 ∗ ∗ 8 64 7

2 1 2 7 3 = + 7 ∗ = + = = a 8 56 8 56 8

3 Therefore, a = 8 = (.27)8.

Endpoint b can be represented in base 2(4) = 8 as

5 = (.(4 + 1)0000 ··· ) = (.5) 8 8 8 18 which can be seen by converting this to base 10 representation:

5 5 (.5) = = = b 8 81 8

5 T herefore, b = 8 = (.5)8.

1 Theorem 20. When the n in n being removed is odd, the endpoints after the ﬁrst iteration k can be represented in base 2k + 1 as a = 2k+1 = (.k0000··· )2k+1 = (.(k - 1)2k)2k+1 and b = k+1 2k+1 = (.(k + 1)000··· )2k+1 where n = 2k + 1.

Proof. Let n = 2k + 1. The endpoints, a and b, can determine which digits will be included in the base representation.

n − 1 k a = = can be represented in base 2k +1 as (.k0000 ··· ) = (.(k −1)2k) 2n 2k + 1 2k+1 2k+1

The proof follows by converting a in base 2k + 1 to a in base 10:

k − 1 2k 2k + + + ··· 2k + 1 (2k + 1)2 (2k + 1)3

∞ k − 1 X 2k = + 2k + 1 j j=2 (2k + 1)

∞ k − 1 X 1 = + (2k) ∗ 2k + 1 j j=2 (2k + 1)

1 k − 1 (2k+1)2 = + 2k ∗ 1 2k + 1 1 − 2k+1

1 ! k − 1 2 = + 2k ∗ (2k+1) 2k + 1 2k 2k+1 k − 1 1 2k + 1 = + 2k ∗ ∗ 2k + 1 (2k + 1)2 2k k − 1 1 k = + = = a 2k + 1 2k + 1 2k + 1

19 n−1 k Therefore, a = 2n = 2k+1 = (.(k - 1)2k)2k+1

n + 1 k + 1 b = = can be represented in base 2k + 1 as (.(k + 1)0000 ··· ) 2n 2k + 1 2k+1

We will convert b in base 2k + 1 to b in base 10. The ﬁrst digit in base 2k + 1 is simply equal

1 to 2k+1 multiplied by the ﬁrst digit. This means that

k + 1 0 0 (.(k + 1)0000 ··· )2k+1 = + + + ··· 2k + 1 (2k + 1)2 (2k + 1)3

k + 1 (.(k + 1)0000 ··· ) = = b 2k+1 2k + 1

1 Example. We consider the case of removing the middle n where n = 5 is odd. Thus, the n−1 k n+1 k+1 endpoints are a = 2n = 2k+1 and b = 2n = 2k+1 where n = 2k + 1:

n−1 1 5−1 Solving n = 2k + 1 for k = 2 . In the 5 case, k = 2 = 2 and the endpoints a and b would be 2 2 2 + 1 3 a = = and b = = 2(2) + 1 5 2(2) + 1 5

Endpoint a can be represented in base 2(2) + 1 = 5 as

(.(2 − 1)(2(2)))5 = (.14)5

which can be seen by converting this to base 10 representation:

1 4 4 (.14) = + + + ··· 5 5 52 53

∞ 1 X 4 = + 5 5j j=2

20 ∞ ! 1 X 1 = + 4 ∗ 5 5j j=2

1 ! 1 25 = + 4 ∗ 1 5 1 − 5

1 ! 1 25 = + 4 ∗ 4 5 5 ! 1 1 5 = + 4 ∗ ∗ 5 25 4

1 1 1 4 2 = + 4 ∗ = + = = a 5 20 5 20 5

2 Therefore, a = 5 = (.14)5.

Endpoint b can be represented in base 2(2) + 1 = 5 as

(.(2 + 1)0000 ··· )5 = (.3)5

which can be seen by converting this to base 10 representation:

3 3 (.3) = = = b 5 51 5

3 Therefore, b = 5 = (.3)5.

The previous two theorems will only hold true for the ﬁrst iteration of C. These will not help determine if a number is in C. They are only used to determine the base representations of the endpoints of the ﬁrst iteration of C. The purpose of these two theorems was an attempt

1 to generalize the base representations of numbers that are in the n -ary Cantor set. However, 1 the base representations of numbers in the n -ary Cantor set are more complicated than those of the ternary Cantor set. This will lead us into the next Cantor set.

To demonstrate how an attempt to restrict the digits used in base representations in the

1 n -ary Cantor set fails, we will look at 2 previous examples, one where n is even and one

21 where n is odd.

1 Recall the 4 -ary Cantor set. Since n = 4 is even, we will be working in base 2(4) = 8 without using the middle digits, which are 3 and 4. We previously showed that C1 =

3 5 3 5 [0, 8 ], ∪[ 8 , 1] and that 0 = (.0)8, 8 = (.27)8, 8 = (.5)8, and1 = (.7)8. These endpoints all can be written in base 8 without using the middle 2 digits. To elaborate, the base-8 digits are 0, 1, 2, 3, 4, 5, 6, and 7. Therefore, the middle 2 digits are 3 and 4, which are not used in the base-8 representations of these endpoints. Although the Theorem 19 was only referring to the endpoints, we will see that we cannot easily deﬁne the base representations of what remains like we can in the ternary Cantor set. Since there are no issues regarding these middle digits in the ﬁrst iteration, we will look at the second iteration.

9 15 3 5 49 55 C = [0, ] ∪ [ , ] ∪ [ , ] ∪ [ , 1] 2 64 64 8 8 64 64

9 15 49 55 We must now convert the new endpoints 64 , 64 , 64 , and 64 to base 8 to see if there are any 3’s 9 15 49 55 or 4’s in their representations. 64 = (.11)8, 64 = (.17)8, 64 = (.61)8, and 64 = (.67)8. None 1 of these endpoints use the digits 3 or 4, so in theory, a theorem restricting the n -ary Cantor set when n is even to base 2n without the middle 2 digits is still plausible.

However, when looking at the third iteration, numerous counterexamples surface.

27 45 9 15 147 165 3 5 347 365 49 55 467 485 C = [0, ] ∪ [ , ] ∪ [ , ] ∪ [ , ] ∪ [ , ] ∪ [ , ] ∪ [ , ] ∪ [ , 1] 3 512 512 64 64 512 512 8 8 512 512 64 64 512 512

27 45 147 165 347 365 467 485 We will now convert the new endpoints, 512 , 512 , 512 , 512 , 512 , 512 , 512 , and 512 to base 8. The 27 45 147 165 347 conversions are as follows: 512 = (.033)8, 512 = (.055)8, 512 = (.223)8, 512 = (.245)8, 512 = 365 467 485 (.533)8, 512 = (.555)8, 512 = (.723)8, and 512 = (.745)8. This shows that there are some endpoints in C3 that use the digits 3 or 4 in their base 8 representation. We cannot manipulate the base representation to eliminate all of the 3’s and 4’s since they are not all terminating digits. Since some endpoints cannot be written without those two digits, this shows that we

1 cannot restrict the base representations of elements in the n -ary Cantor set when n is even.

22 1 1 We will now look at an example of the n -ary Cantor set where n is odd. Recall the 5 example. Since n = 5, we will be working in base 5 without using the middle digit, 2. We

2 3 2 3 previously showed that C1 = [0, 5 ] ∪ [ 5 , 1] and that 0 = (.0)5, 5 = (.14)5, 5 = (.3)5, and

1 = (.4)5. None of these endpoints use the digit 2 in their base 5 representation. Since there are no issues in the ﬁrst iteration, we will look at the second iteration.

4 6 2 3 19 21 C = [0, ] ∪ [ , ] ∪ [ , ] ∪ [ , 1] 2 25 25 5 5 25 25

4 6 19 21 We must now convert the new endpoints, 25 , 25 , 25 , and 25 to base 5 to see if there are any 4 6 19 21 2’s in their representations. 25 = (.04)5, 25 = (.11)5, 25 = (.34)5, and 25 = (.41)5. None of 1 these endpoints use the digit 2, so in theory, a theorem restricting the n -ary Cantor set when n is odd to base n without the middle digit is still plausible. However, when looking at the third iteration, numerous counterexamples surface.

8 12 4 6 38 42 2 3 83 87 19 21 113 117 C = [0, ] ∪ [ , ] ∪ [ , ] ∪ [ , ] ∪ [ , ] ∪ [ , ] ∪ [ , ] ∪ [ , 1] 3 125 125 25 25 125 125 5 5 125 125 25 25 125 125

8 12 38 42 83 87 113 117 We will now convert the new endpoints, 125 , 125 , 125 , 125 , 125 , 125 , 125 , and 125 to base 5. The 8 12 38 42 83 conversions are as follows: 125 = (.013)5, 125 = (.022)5, 125 = (.123)5, 125 = (.132)5, 125 = 87 113 117 (.313)5, 125 = (.322)5, 125 = (.423)5, and 125 = (.432)5. This shows that there are some

endpoints in C3 that use the digit 2 in the base 5 representation. We cannot manipulate the base representation to eliminate all of the 2’s since they are not all terminating digits. Since some endpoints cannot be written without this digit, this shows that we cannot restrict the

1 base representations of elements in the n -ary Cantor set when n is odd.

23 Chapter 3

n−1 The n -ary Cantor Set

The next variation of the Cantor set can be constructed by removing a proportion of

n−1 n during each iteration. Some examples of middle intervals that satisfy this requirement 1 2 3 4 include 2 , 3 , 4 , and 5 . Like the ternary Cantor set, the actual set can be described by the intersection of the iterations. That is, C = ∩Ci. This set will be explored in an attempt to determine a generalized base representation of all elements of C. Much like the previous chapter, we will start by looking at the representation of the endpoints of the ﬁrst iteration of C, and eventually determine the base representation of elements of this set where C is

n−1 C n .

th n−1 Lemma 21. The endpoints after removing the middle n from the interval [0, 1] can be represented as 1 2n − 1 a = and b = 2n 2n

where C1 = [0, a] ∪ [b, 1].

n−1 n−1 Proof. Let k = n where n is the middle portion of the interval that is being removed. Then after the ﬁrst iteration, [0, a] ∪ [b, 1] is remaining. Since k is being removed from the initial interval, then b − a = k. The combined lengths of [0, a] and [b, 1] must equal 1 − k. That is, (a − 0) + (1 − b) = 1 − k. The lengths of [0, a] and [b, 1] must be equal since the interval removed is the middle of the initial interval. Since the two intervals are equal, then

24 they each must be half the length of what is remaining after removing the middle interval. That is 1 1 a − 0 = a = (1 − k) and 1 − b = (1 − k) 2 2

These can ﬁnally be simpliﬁed to

1 − k 1 + k a = and b = 2 2

n−1 Substituting n for k yields

1 − n−1 1 + n−1 a = n and = n 2 2

which eventually simpliﬁes to

1 2n − 1 a = and b = 2n 2n

The following proofs are similar to the initial properties of the ternary Cantor set listed

1 in Chapter 1 and proved in chapter 2 for C n . In this case, however, they are strictly for the

n−1 n -ary Cantor set. The ﬁrst two lemmas will explore how many intervals remain and are removed during each iteration of C.

i Lemma 22. Ci consists of 2 pairwise disjoint subintervals.

This proof follows the same structure as the proof for Lemma 10 in Chapter 2.

i Lemma 23. 2 intervals are removed from Ci when constructing Ci+1.

This proof follows the same structure as the proof for Lemma 11 in Chapter 2.

The next two lemmas will look at the lengths of the intervals remaining in each iteration of C as well as the length of each interval that is removed from C.

25 1 i Lemma 24. The length of each interval remaining in a component of Ci is 2n .

1 2n−1 Proof. Let [s, t] be a component of Ci. For i = 1, C1 is deﬁned as [0, 2n ] ∪ [ 2n , 1]. The

lengths of each component of C1 are equal since the middle interval is removed. This means

1 1 1 1 that each interval in C1 has length 2n - 0 = 2n . This is also equal to 2n . Therefore, this holds for the i = 1 case.

Assume true for Ci. Consider Ci+1. Each interval of this iteration will have equal length

n−1 due to the construction of removing the middle interval. If n is being removed, then that 1 1 means that n is remaining. This n must be split between the two remaining intervals. Since the middle portion is being removed, this will be an even split, so any interval remaining in

1 th 1 i Ci+1 will be 2n of the previous interval’s length, 2n , which can be written as

1 1 i 1 i+1 = 2n 2n 2n

Lemma 25. The length of each interval that is removed during iteration Ci to Ci+1 is equal

n−1 to 2ini+1

0 0 Proof. Let [s, t] be a component of Ci with interval [s , t ] removed when constructing Ci+1.

0 0 0 0 n−1 th Then the length of [s , t ] is equal to t − s . This interval can be thought of as one n of n−1 1 i [s, t]. This expression can be written as n ∗ (t − s). The length of (t − s) is equal to 2n . Therefore, the length of [s0, t0] is

n − 1 1 i n − 1 = . n 2n 2ini+1

The next two lemmas will use the previous four lemmas to determine the total length removed and remaining of each iteration of C.

26 n−1 Lemma 26. The total length removed in each iteration of Ci to make Ci+1 is ni+1 .

Proof. Based on Lemma 23, there are 2i intervals removed during each iteration of C. Based

n−1 on Lemma 25, each of these intervals have length 2ini+1 . Then, the total length removed i n−1 would be 2 intervals of this length 2ini+1 , which can be written as

n − 1 n − 1 2i ∗ = 2ini+1 ni+1

1 i Lemma 27. The total length remaining in each iteration of Ci is equal to n .

1 2n−1 Proof. We give a proof by induction. Consider i = 1 case. C1 consists of [0, 2n ] ∪ [ 2n , 1]. The length of each of these intervals are equal since the middle interval is removed. The ﬁrst

1 1 interval has length 2n − 0 = 2n . Since C1 consists of two such intervals, the total length of

C1 is 1 2 2 ∗ = 2n 2n 1 = n

Therefore, this holds for the i = 1 case.

n−1 Assume true for Ci. Consider the i + 1 case. To construct Ci+1, we will remove n of 1 i Ci. Based on the inductive hypothesis, Ci has length of n . Then Ci+1 will have length

1 i n − 1 1 i 2i − 2i 2n n 2n

1 n − 1 1 = 2i ∗ − ∗ 2i ∗ 2ini n 2ini

1 n − 1 1 = − ∗ ni n ni 1 n − 1 = − ni ni+1 27 n n − 1 = − ni+1 ni+1 1 1 i+1 = = ni+1 n

1 i Therefore, each iteration of Ci has length n .

n−1 The previous six lemmas will now be used to determine the total length of the n -ary Cantor set.

n−1 Theorem 28. The n -ary Cantor set has length 0.

Proof. This can be proven by looking at everything that is being removed from C. Recall by

n−1 n−1 Lemma 26 that ni+1 is removed during each iteration of Ci. This means that during C0 n1 n−1 n−1 is removed, during C1 n2 is removed, and during C2 n3 is removed. This pattern continues and the total length removed can be written as

∞ X n − 1 nj+1 j=0

∞ X 1 = (n − 1) nj+1 j=0

1 ! n = (n − 1) 1 1 − n

1 n = (n − 1) n−1 n 1 n = (n − 1) ∗ ∗ = 1 n n − 1

This means that a length of 1 is removed. The starting length of C is the length of the initial interval, [0, 1] = 1. Therefore, the length of C = 1 − 1 = 0.

1 n−1 It is important to note that the ternary, n -ary, and the n -ary Cantor sets all had total length 0. In the conclusion, we will consider if there are any variations in the Cantor set that

28 allow any length. The next few theorems are necessary to ﬁnalize the base representation of

n−1 all numbers in the n -ary Cantor set.

n−1 1 Theorem 29. When n is being removed from the interval [0, 1], the endpoints a = 2n 2n−1 and b = 2n are both irreducible fractions.

1 2n−1 Proof. These endpoints, 2n and 2n will reduce if and only if GCD(1, 2n) > 1 or GCD(2n− 1, 2n) > 1. For endpoint a, we attempt to show that the GCD(1, 2n) = 1. The GCD of any number

1 and 1 will always be 1, therefore GCD(1, 2n) = 1 so endpoint a = 2n is irreducible. For endpoint b, suppose r | 2n − 1 and r | 2n. There exists 2 integers, x and y, such that

2n − 1 = rx and 2n = ry

Then rx + 1 2n n = and y = 2 r

Using substitution, rx + 1 2n = 2 = rx + 1 2

Therefore, rx + 1 1 y = = x + r r

1 is an integer. Since y is an integer, r must also be an integer. Therefore r = 1 and 2n−1 GCD(2n − 1, 2n) = 1 so b = 2n is irreducible.

Theorem 30. The left endpoints of the components in each Ci can be expressed in base 2n using only the digits 0 and 2n - 1. If s is a left endpoint of a component of Ci, then s =

(.a1a2··· ai0)2n with all ak ∈ {0, (2n - 1) }.

Proof. Let [s, t] be a component of Ci. s can be written in the form of (.a1a2a3 ··· ai ··· )2n.

1 2n−1 Consider case i = 1. C1 = [0, 2n ] ∪ [ 2n , 1]. The two left endpoints in this case are 0 and

29 2n−1 2n . Looking at 0, this can be represented in base 2n as (.00)2n. This can easily be seen because this would be 1 1 1 0 ∗ + 0 ∗ + 0 ∗ + ··· 2n (2n)2 (2n)3

= 0 + 0 + 0 + ··· = 0

Therefore, this holds for the ﬁrst endpoint in the i = 1 case. The other left endpoint in this

2n−1 case is 2n . This endpoint can be written as (.(2n - 1)0)2n. This can also be seen by looking at the conversion from base 2n to base 10 as follows

1 1 1 (2n − 1) ∗ + 0 ∗ + 0 ∗ + ··· 2n (2n)2 (2n)3

1 2n − 1 = (2n − 1) ∗ + 0 = 2n 2n

Therefore, this holds for the second endpoint in the i = 1 case and will hold for both

endpoints in this case. Assume true for Ci and look at the Ci+1 case. If [s, t] is a component

0 0 of Ci then [s, s ] ∪ [t , t] ⊂ [s, t] with s = (.a1a2 ··· ai0)2n, where all ak ∈ {0, (2n − 1)}. The two left endpoints are s and t0. s will be written in base 2n using only 0 and (2n - 1) by the

0 inductive hypothesis since s is a left endpoint in Ci. Endpoint t = s + the length of Ci+1 +

the length removed in Ci. Based on the previous theorems, this will be written as

1 i+1 n − 1 1 i t0 = s + + ∗ 2n n 2n

1 i+1 n−1 Let 2n + 2ini+1 = x. We can see that x is written following the base 2n rules as follows

1 1 i n − 1 1 i x = ∗ + ∗ 2n 2n n 2n

1 i n − 1 1 = ∗ + 2n n 2n

30 1 i 2n − 1 = ∗ 2n 2n

1 i 1 = ∗ ∗ (2n − 1) 2n 2n

1 i+1 = ∗ (2n − 1) 2n

This (2n − 1) is in the (i + 1)th position in the base representation since it is multiplied by

1 i+1 0 0 2n . Recall t = s+x, where s = (.a1a2 ··· ai0)2n. Therefore, t = (.a1a2 ··· ai(2n−1)0)2n

and is of the form (.a1a2 ··· ai+10)2n where ak ∈ {0, (2n − 1)}.

Theorem 31. The right endpoint t of a component in Ci can be written in base 2n using

only the digits 0 and (2n - 1) in the form t = (.a1a2··· ai(2n − 1))2n with all ak ∈ {0, (2n - 1)}.

Proof. Let [s, t] be a component of Ci. Then t is the right endpoint in this case. Endpoint t = s+ the length remaining. By Lemma 24, this can be expressed as

1 i t = s + 2n

Which by Theorem 30 can be written as

1 i t = (.a a ··· a 0) + 1 2 i 2n

with all ai ∈ {0, 2n − 1} Clearly, the ﬁrst part of t follows the requirements for the theorem.

1 i th We must show that 2n = (.0 ··· 0(2n − 1))2n with (2n−1) repeating starting in the (i+1) position. ∞ X (2n − 1) = (2n)j j=i+1

∞ X 1 = (2n − 1) ∗ (2n)j j=i+1

31 1 i+1 ! 2n = (2n − 1) ∗ 1 1 − 2n 1 2n = (2n − 1) ∗ ∗ (2n)i+1 2n − 1

1 1 i = = (2n)i 2n

0 Therefore, t = (.a1a2 ··· ai0)2n + (.0 ··· 0(2n − 1))2n where there are i 0 s in the second com-

ponent of t and all ak ∈ {0, (2n−1)}. This can ﬁnally be simpliﬁed to (.a1a2 ··· ai(2n − 1))2n.

Theorem 32. x ∈ C if and only if

∞ j X 1 x = a ∗ j 2n j=1

with aj ∈ {0, (2n - 1)}.

Proof. We ﬁrst show that if x ∈ C then it can be written in base 2n using only 0 and (2n−1).

Since x ∈ C then there exists components [si, ti] ⊆ Ci with x ∈ [si, ti]. This is true for all i because C = ∩Ci and if an element is in C, then by deﬁnition of intersections, it must be in

1 i all Ci. Thus ∀ i |x − si| ≤ ti − si and by Lemma 24, |ti − si| = 2n . Then,

1 i lim |x − si| ≤ lim = 0 i→∞ i→∞ 2n

Therefore the sequence {si} converges to x. We also know from Theorem 30 that si+1 = s + ai+1 where a is either 0 or 2n − 1. Applying Theorem 30 to the left endpoints of s , i (2n)i+1 i+1 i

i j X 1 x = lim si = lim aj ∗ i→∞ i→∞ 2n j=1

∞ j X 1 = a ∗ j 2n j=1

32 where all aj ∈ {0, (2n − 1)}.

We next show that if x can be written in base 2n using only 0 and (2n − 1) then x ∈ C.

Recall that every left endpoint s of any component in Ci is of the form (.a1a2 ··· ai0)2n where

aj ∈ {0, (2n − 1)}. Since there are 2 choices for each aj, namely 0 or (2n − 1), then there are

i i exactly 2 numbers of this form. By Lemma 22, there are also 2 components of Ci. Therefore, s is a left endpoint in Ci if and only if s = (.a1a2 ··· ai0)2n with all aj ∈ {0, (2n − 1)}. Let

∞ j X 1 x = a with all a in {0, (2n − 1)} j 2n j j=1

For all i ≥ 1

i j ∞ j ∞ j X 1 X 1 X 1 s = a ≤ a ≤ s + (2n − 1) = t i j 2n j 2n i 2n i j=1 j=1 j=i+1

where [si, ti] is a component of Ci. Therefore, x ∈ [si, ti] ⊆ Ci ∀ i ≥ 1. Therefore, x ∈ C.

This ﬁnal proof is useful since it can be used to determine if a given number is in the

n−1 n−1 n -ary Cantor set, or what numbers the n -ary Cantor set contains. If a number can be written in base 2n using only the digits 0 and (2n - 1), it is in C. Furthermore, if a number is in C, its base 2n representation uses only the digits 0 and (2n - 1). This is the sort of

1 conclusion that was attempted in Chapter 2. However, the n -ary Cantor set does not have n−1 this property, where the n -ary Cantor set does.

n−1 Example. We consider the case of removing the middle n where n = 4 . Therefore, we 3 will be removing 4 during each iteration of C. The claim is that every endpoint in Ci can

be represented as s = (.a1a2 ··· ai0)2n with all ai in {0, (2n - 1)} if s is a left endpoint and as t = (.a1a2 ··· ai(2n − 1))2n with all ai in {0, (2n - 1)} if t is a right endpoint. In this case, we will be working in base 2(4) = 8, using only the digits 0 and 2(4) − 1 = 7.

1 7 The ﬁrst iteration of C is [0, 8 ] ∪ [ 8 , 1]. We must now convert each of these endpoints to 1 7 base-8. 0 = (.0)8, 8 = (.1)8, 8 = (.7)8, and 1 = (.7)8. Although the base-8 representation of

33 1 8 contains a 1, it can be rewritten as (.07)8. In fact, any number terminating in 1 can be rewritten with a 0 followed by a repeating 7. Since all of these endpoints can be written in base-8 using only 0 and 7, this holds for C1. We will now look at the next iteration.

1 7 1 7 57 63 C = [0, ] ∪ [ , ] ∪ [ , ] ∪ [ , 1] 2 64 64 8 8 64 64

1 7 We must convert each new endpoint into base-8 as follows: 64 = (.01)8 = (.007)8, 64 = 57 63 (.07)8, 64 = (.71)8 = (.707)8, and 64 = (.77)8. All of these endpoints can be represented in base-8 using only the digits 0 and 7, so the theorem still holds. We will look at one more

1 iteration, since the n -ary Cantor set failed in the third iteration, we will end by looking at that one.

1 7 1 7 57 63 1 7 449 455 57 63 505 511 C = [0, ] ∪ [ , ] ∪ [ , ] ∪ [ , ] ∪ [ , ] ∪ [ , ] ∪ [ , ] ∪ [ , 1] 3 512 512 64 64 512 512 8 8 512 512 64 64 512 512

1 7 Again, we must convert the new endpoints to base-8 as follows: 512 = (.001)8 = (.0007)8, 512 = 57 63 449 455 505 (.007)8, 512 = (.071)8 = (.0707)8, 512 = (.077)8, 512 = (.701)8 = (.7007)8, 512 = (.707)8, 512 = 511 (.771)8 = (.7707)8, and 512 = (.777)8. As the theorem states, each of these endpoints are represented in the proper form using the digits 0 and 7, so this example helps illustrate the usefulness of this theorem.

34 Chapter 4

Conclusion

After studying a variety of Cantor sets, some questions arise. The ﬁrst question would

1 be regarding the n -ary Cantor set. One might wonder why there is a base representation n−1 1 for elements in the n -ary Cantor set but not the n -ary Cantor set. In fact, the only thing 1 that has the desired base representation in the n -ary Cantor set are the endpoints during the ﬁrst iteration. These endpoints could be written in base 2n or base n, depending on the parity of n, without using the middle digits of that particular base. On the other hand, the

n−1 n -ary Cantor set has elements that only contain the digits 0 or (2n - 1) in base 2n. It 1 is easy to wonder why the n -ary Cantor set does not have a deﬁned representation like the n−1 1 1 n -ary Cantor set, especially since 3 is of the form n with n = 3.

1 Looking at n > 3, all n -ary Cantor sets will involve either base n, when odd, or base 2n, when even. The smallest even number greater than 3 is 4, meaning base 2(4) = 8 will be used. This base has digits 0, 1, 2, 3, 4, 5, 6, and 7. Although the endpoints after the ﬁrst iteration of C can be written without using the digits 3 and 4, which was shown in an example in a previous chapter, it becomes clear that more endpoints in later iterations will contain some of these numbers since not every conﬂicting digit is the terminating digit. The same is true for odd numbers. 5 is the smallest odd number greater than 3. This would require base 5, which uses digits 0, 1, 2, 3, and 4. Based on Theorem 20, the endpoints can

35 be written without using digit 2, which happens to be the middle digit. However, by the third iteration, some endpoints will only be able to be written using the digit 2, showing

1 that we cannot restrict the base representation of elements in the n -ary Cantor set.

It is then easy to wonder why we can, in fact, restrict digits in the ternary Cantor set. It

1 1 is of the form n . n = 3 is a very special case of the n -ary Cantor set. Recall the construction 1 of the ternary Cantor set. T1 is constructed by removing 3 from the interval [0, 1]. This 1 2 1 1 leaves the intervals [0, 3 ] ∪ [ 3 , 1]. The length of the first interval is 3 − 0 = 3 . The middle 1 portion removed also has length 3 . Finally, the length of the second interval is equal to 2 1 1 − 3 = 3 . This shows that the length of the intervals remaining at the first iteration are 1 also equal to the length removed at the first iteration. In any other n -ary Cantor set, this 1 is not true. For example, in the 4 -ary Cantor set, the first iteration contains two intervals 3 1 with length 8 and the interval removed has length 4 , which are not all equal. This is also 1 true for when n is odd. For example, in the 5 -ary Cantor set, the first iteration contains 2 1 two intervals of length 5 and the length of the interval removed is length 5 . Therefore, the 1 ternary Cantor set is the only n -ary Cantor set where all three intervals have equal length.

It is also important to look at the digits in base 3. Since the ternary Cantor set has n = 3, and n is odd, then we will be working in base n. The digits in base 3 are 0, 1, and 2. When manipulating numbers in base 3, terminating digits can be written with a

repeating digit instead. For example, (.1)3 = (.02)3. This is comparable to how 1 = (.9)10. In base 3, the only number we are restricting from our representation in the ternary Cantor set is 1. This digit also falls in the middle of the digits. Since it is the only one being restricted, we can easily write any terminating number with 1 as the ﬁnal digit in the form of 02. Any terminating number with 0 or 2 as the ﬁnal digit is allowed, so we can manipulate problematic decimals to remove the digit 1.

This idea relies primarily on the fact that 1 is the middle digit and there is only 1 digit

1 on either side of 1 when ordered numerically. For example, the 5 -ary Cantor set cannot use this. The digits 0, 1, 2, 3, and 4 are used in base 5. There will be instances where we cannot

36 write an element of this Cantor set in base 5 without the digit 2, so we cannot create a base

1 representation requirement of the general n -ary Cantor sets.

n−1 In an attempt to deﬁne the elements of some general Cantor set, the n -ary Cantor set 1 was explored. This set is special because it is quite opposite of the n -ary Cantor set. Instead 1 1 of removing n of the previous interval, we created a set that leaves n of the previous interval n−1 1 after the iteration is complete. That is, after n is removed, exactly n is left remaining. Each interval is half of this length as previously explored, which can explain why two digits

n−1 are used for the base representation of the n -ary Cantor set and why it is in base 2n. 1 Although these sets are slight variations of each other, both the n -ary Cantor set as well as n−1 the n -ary Cantor set have length 0.

This then causes one to wonder what has to be done for a Cantor set to have some sort of length. An example of these can be seen in Lamb [4]. Smith-Volterra-Cantor sets, or fat Cantor sets, are notable because they actually have some length. They start with the traditional interval [0, 1]. Instead of removing a set portion during each iteration, this will change.

There is a particular way to change the length removed during each iteration to properly

1 construct a fat Cantor set. For example, start with C0 = [0, 1]. Remove the middle 4 for the 1 ﬁrst iteration of C. Then, for the second iteration, remove the middle 16 . This pattern will 1 continue, removing the middle 4i each time.

In the previous example, the ﬁrst few iterations of this fat Cantor set will look like

3 5 1 C = [0, ] ∪ [ , 1] with removed 1 8 8 4

45 51 3 5 205 211 1 C = [0, ] ∪ [ , ] ∪ [ , ] ∪ [ , 1] with removed 2 256 256 8 8 256 256 16

1 The total length removed during each iteration is 2i+1 . We will determine the length of this fat Cantor set like we did in previous examples by determining the total length removed,

37 which can be written as ∞ X 1 2j+1 j=1

1 ! 1 22 4 1 = 1 = 1 = 1 − 2 2 2

1 1 Therefore, this particular fat Cantor set has length 1 − 2 = 2 , which is not equal to 0. These Cantor sets are particularly interesting due to the fact that they have a length at the end.

Finally, below are some tables summarizing the results from this paper.

1 n -ary Cantor Set Remaining Removed

Number of Intervals 2i 2i

n−1 i 1 n−1 i Length of Intervals 2n n 2n n−1 i (n−1)i Total Length at Iteration n ni+1

1 Endpoints of the First Iteration in n -ary Cantor Set, Even a b

n−1 n+1 Endpoints 2n 2n Irreducible Yes Yes

Base Notation (.(n - 2)(2n − 1))2n (.(n + 1)000··· )2n

1 Endpoints of the First Iteration in n -ary Cantor Set, Odd a b

k k+1 Endpoints 2k+1 2k+1 Irreducible Yes Yes

Base Notation (.(k - 1)(2k))2k+1 (.(k + 1)000··· )2k+1

38 n−1 n -ary Cantor Set Remaining Removed

Number of Intervals 2i 2i

1 i n−1 Length of Intervals 2n 2ini+1 1 i n−1 Total Length at Iteration n ni+1

n−1 n -ary Cantor Set a b

1 2n−1 Endpoints of First Iteration 2n 2n Irreducible Yes Yes

Base Notation of Endpoints Right: Left: (.a1a2 ··· ai0)2n

(.a1a2 ··· ai(2n − 1))2n

39 Bibliography

[1] Amir D. Aczel, A Strange Wilderness the Lives of the Great Mathematicians, Sterling Publishing Co. 2011.

[2] Robert Dimartino and Wilfredo O. Urbina, Excursions on Cantor-Like Sets, https://arxiv.org/pdf/1411.7110.pdf

[3] Julian F. Fleron, A Note on the History of the Cantor Set and Cantor Function, Math- ematics Magazine, Vol. 67, No. 2 (Apr., 1994), pp. 136 - 140.

[4] Evelyn Lamb, A Few of My Favorite Spaces: Fat Cantor Sets, Scientiﬁc Amer- ican, https://blogs.scientiﬁcamerican.com/roots-of-unity/a-few-of-my-favorite-spaces- fat-cantor-sets/.

[5] Dylan R. Nelson, A Cantor Set - A Brief Introduction, http://wwwmpa.mpa- garching.mpg.de/ dnelson/storage/dnelson.cantor-set.pdf.

[6] Christopher Shaver, An Exploration of the Cantor Set, https://www.missouriwestern.edu/orgs/momaa/ChrisShaver-CantorSetPaper4.pdf.

[7] Brian S. .Thomson, Judith B. Bruckner, Andrew M. Bruckner, Real Analysis, Second Edition, Prentice Hall. 1997.

40 Biography

Rebecca Stettin was born in North Royalton, Ohio on August 4, 1995. She graduated from North Royalton High School in 2013. At Ashland University, Rebecca is majoring in mathematics and actuarial science. She served as an executive board member in Delta Zeta sorority and as a member of Mathematics Association of America for the past four years. Rebecca is a member of the mathematics honorary Pi Mu Epsilon, the Greek honorary Order of Omega, the leadership honorary Omicron Delta Kappa, the women’s Greek honorary Rho Lambda, as well as the freshman honorary Alpha Lambda Delta. She was on the Dean’s list four semesters and is a recipient of Who’s Who Among College Students. After graduation in May 2017, Rebecca plans to study for and pass actuary exams so she can begin her career as an actuary.