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

Transcription

1 Geeralizatios ad Properties of the Terary Cator Set ad Exploratios i Similar Sets by Rebecca Stetti A capstoe project submitted i partial fulfillmet of graduatig from the Academic Hoors Program at Ashlad Uiversity May 207 Faculty Metor: Dr. Darre D. Wick, Professor of Mathematics Additioal Reader: Dr. Gordo Swai, Professor of Mathematics

2 Abstract Georg Cator was made famous by itroducig the Cator set i his works of mathematics. This project focuses o differet Cator sets ad their properties. The terary Cator set is the most well kow of the Cator sets, ad ca be best described by its costructio. This set starts with the closed iterval zero to oe, ad is costructed i iteratios. The first iteratio requires removig the middle third of this iterval. The secod iteratio will remove the middle third of each of these two remaiig itervals. These iteratios cotiue i this fashio ifiitely. Fially, the terary Cator set is described as the itersectio of all of these itervals. This set is particularly iterestig due to its uique properties beig ucoutable, closed, legth of zero, ad more. A more geeral Cator set is created by takig the itersectio of iteratios that remove ay middle portio durig each iteratio. This project explores the terary Cator set, as well as variatios i Cator sets such as lookig at differet middle portios removed to create the sets. The project focuses o attemptig to geeralize the properties of these Cator sets. i

3 Cotets Page The Terary Cator Set 2 The -ary Cator Set 9 3 The -ary Cator Set 24 4 Coclusio 35 Bibliography 40 Biography 4 ii

4 Chapter The Terary Cator Set Georg Cator, bor i 845, was best kow for his discovery of the Cator set. After writig a thesis i umber theory, Cator became iterested i topology ad wrote a series of papers o poit set topology. He coied the term everywhere dese, which is curretly used today. I this same series of papers, Cator later itroduced the idea of perfect sets. He made the terary Cator set famous i this same paper whe attemptig to prove that it was possible to have a set that is closed ad owhere dese. He oted that this was a perfect, ifiite set that is owhere dese i ay iterval, regardless of the size. (Flero [3]) The Cator set is created by startig with the closed iterval [0,]. The ext step is to remove a middle portio of the iterval. This iitial removal will the leave two itervals, amely [0, ] ad [ 2, ]. This process is cotiued a ifiite umber of times, with the fial set of 3 3 umbers, after the ifiite iteratio, beig the Cator set itself. This set is the itersectio of all iteratios, deoted as T = i= T i, where T i represets the i th iteratio. There are differet proportios that ca be removed with this set, the most well kow beig 3. This set, referred to as the terary Cator set, ca also be referred to as the classic set. This particular versio of the Cator set was the oe Georg Cator used to make his argumets about the ature of certai sets. (Aczel []) We will describe the costructio of the terary Cator set, T, i a similar maer as it is

5 described i Thomso [7]. Begi with the closed iterval [0, ] ad remove a dese ope set, G. The remaiig set, T = [0, ] \G will also be closed ad owhere dese i [0, ]. Based o our costructio of G, T will have o isolated poits. It is easiest to uderstad the set G if we costruct it i stages. Let G = ( 3, 2 3 ) ad let T be what is remaiig i [0, ] after removig G. Thus T = [0, ] [ 2, ] is what remais whe the middle third of the iterval 3 3 [0,] is removed. This is referred to as the first iteratio of the terary Cator set. We repeat this costructio for each of the two itervals of T. Let G 2 = (, 2) ( 7, 8 ). These itervals are the middle third of the previous 2 itervals. The T 2 = [0, ] [ 2, ] [ 2, 7] [ 8, ]. This completes the secod iteratio. We cotiue iductively, ad ultimately take the itersectio of each iteratio of T i to create the terary Cator set with the followig properties: For each i N. T i is a uio of 2 i pairwise disjoit closed itervals. 2. Each compoet of the set of ope itervals removed is the middle third of some compoet of T i. 3. The legth of each compoet of T i is. 3 i Some proofs of the properties of the terary Cator set will be explored i later chapters i regards to more geeralized Cator sets. The followig is a list of characteristics of the terary Cator set with some proofs to follow:. The Cator set is oempty. 2. The Cator set is closed. 3. The Cator set has legth The Cator set cotais all umbers that ca be writte i base-3 without s. 5. The Cator set is ucoutable. 6. The Cator set is owhere dese. 2

6 7. The Cator set has o isolated poits. 8. The Cator set is a perfect set. Theorem. The terary Cator set is oempty. Proof. Usig similar methods as Nelso [5], we must show that the terary Cator set cotais at least oe elemet. We ca see that durig each iteratio, the edpoits remai i the set. For example, durig T, everythig betwee (, 2) is removed. Note that it 3 3 is everythig betwee the two edpoits, which implies the two edpoits remai. Oce removed, T is [0, ] [ 2, ]. As see i here, the itervals are closed, thus icludig the 3 3 edpoits. This tred cotiues throughout each iteratio. For T 2, the middle thirds of those two previously metioed itervals are removed. To clarify, everythig betwee ( 9, 2 ) as well as ( 7, 8) is removed. This leaves [0, ] [ 2, ] [ 2, 7] [ 8, ], which is oce agai a uio of closed itervals, which iclude edpoits. Notice that 0,, 2, ad are 3 3 agai icluded i this iteratio. These iteratios cotiue ifiitely. Sice each iteratio of T cotais the edpoits of each iteratio before it, it becomes clear that the edpoits of each iteratio are icluded. Sice the previous edpoits are cosistetly icluded, there will clearly be elemets i the itersectio of each iteratio, provig the terary Cator set is oempty. Theorem 2. The terary Cator set is closed. Proof. As show earlier whe lookig at the legth of T, it is a itersectio of closed sets. For example, T is [0, ] [ 2, ]. This tred cotiues with each T 3 3 i also cosistig of a fiite uio of closed itervals. Each T i is closed sice it is a fiite uio of closed itervals. Sice T = i=t i it is a coutable itersectio of closed sets ad is therefore closed. Theorem 3. The terary Cator set has legth of 0. Proof. This proof is also based off of similar methods used i Nelso [5]. This proof ca be demostrated by otig everythig that is take out of the startig iterval has the same 3

7 legth as the iitial iterval. Lookig at what is take out durig each iteratio, a sequece emerges. Durig the first iteratio, a iterval of legth 3 is removed from the set. Durig the secod iteratio, there are 2 differet itervals, amely [0, ] ad [ 2,], to remove the 3 3 middle third from. I this case, however, the legth of what is removed is 3 of 3, or 9. This must occur twice because there are 2 itervals that ( 9 )th is beig removed from. Therefore, a legth of = 2 9 will be removed durig the secod iteratio of T. This tred cotiues throughout each iteratio resultig i the legth removed at each iteratio as 3, 2 9, 4 27, 8 8, The total legth removed ca also be represeted as i=0 2 i 3 i+ This sum ca be determied through the sum as a geometric series, or 3 ( 2 3 ) =. This meas that everythig removed has legth. As metioed previously, the startig iterval, T 0, has legth as well. A simple subtractio of the removed portio from the iitial legth will give the fial legth of T as 0. Therefore, T has legth 0. The followig theorem will be explored through examples, but will ot be prove. A similar proof ca be foud i Chapter 3. Theorem 4. The terary Cator set oly cotais umbers that have a base-3 represetatio excludig the digit. A brief review of base-3 represetatio is ecessary to proceed with this example. Base-3, or terary, represetatio cosists of 3 digits, 0,, ad 2. A umber of the form (.a a 2 a 3 ) 3 ca be thought of i terms of sums of fractios with umerators a ad deomiators 3. That is, (.a a 2 a 3 ) 3 = a 3 + a a where a must be either 0,, or 2. The claim is every umber that is i the terary Cator set ca be writte i base-3 represetatio with oly 2 s ad 0 s. That is, x T, x = (.a a 2 a 3 ) 3 such that a i {0, 2}. It is helpful to look at edpoits, startig with T = [0, ] [ 2, ]. This meas that it 3 3 4

8 cotais everythig betwee 0 ad ad the everythig betwee 2 ad. The edpoits 3 3 0,, 2, ad are icluded as well. Lookig at the edpoits i base-3 represetatio is 3 3 useful. 0 = (.0) 3, 3 = (.) 3 = (.02) 3, 2 3 = (.2) 3, ad = (.2) 3. The edpoits 3 ad are particularly iterestig. = (.2) 3 sice 2 is the largest digit i base-3. This is the same idea that is used i base-0 where = (.9) 0. = (.) 3 3 because it is simply *. This ca be 3 rewritte usig repeatig digits just like the umber. Ay umber i base-3 with a as a termiatig digit ca actually be rewritte by replacig the with 02. We ca see this whe covertig (.02) 3 ito base-0: (.02) 3 = which is a geometric series writte as j=2 2 3 = 2 j 3 j ( ) 9 = 2 3 j=2 ( ) 9 = = = 3 9 = 3 This coversio shows that 3 = (.) 3 = (.02) 3. The first iterval, [0, ] i base-3 ca be 3 writte as ay umber x where x = (.0a 2 ) 3, where a 2 represets ay of the base-3 digits. Similarly, the secod iterval, [ 2, ] i base-3 ca be writte as ay umber x where x = 3 (.2a 2 ) 3, agai where a 2 is ay base-3 digit. The same procedure ca be repeated for T 2. T 2 = [0, 9 ] [ 2 9, 3 ] [ 2 3, 7 9 ] [ 8 9, ]. I additio to the previous edpoits, 0,, 2, ad, there are 4 more edpoits. These 3 3 edpoits ca be writte i base-3 as follows: 9 = (.002) 3, 2 9 = (.02) 3, 7 9 = (.202) 3, ad 8 9 = (.22) 3. These edpoits ca agai create restrictios with what digits i base-3 represetatio are icluded. For example, the iterval [0, ] i base-3 ca be writte as ay umber x where 9 5

9 x = (.00a 3 ) 3, where a 3 represets ay of the base-3 digits. The ext ew iterval, [ 2 9, 3 ] i base-3 ca be writte as ay umber x where x = (.02a 3 ) 3, where a 3 represets ay of the base-3 digits. Usig this otatio, [ 2, 7] would have x = (.20a ) 3, ad [ 8, ] would 9 have x = (.22 ) 3. This shows that as each iteratio cotiues, the digits of the base-3 represetatios become more specific. Each iteratio adds a ew digit restrictio for this terary represetatio. A proof i oe of the followig chapters will give a similar proof of base represetatios i a geeralized Cator set. A set is cosidered ucoutable whe the umber of elemets caot be writte as some subset of the atural umbers. Theorem 5. The Cator set is ucoutable. Proof. This will be a proof by cotradictio usig similar methods as Shaver [6]. First assume that the terary set is coutable. This meas that there exists a oto fuctio that maps all elemets from the atural umbers to the set. Recall from Theorem 4 that the terary Cator set cosists of all umbers [0,] that ca be writte i base-3 represetatio usig oly 0 s ad 2 s. These umbers are of the form (.a a 2 a 3 ) 3 such that each a i is either 0 or 2. This set will be deoted as T = {x [0,] : x has base-3 represetatio of oly 0 s ad 2 s }. Assume that this set is coutable, f: N T, where f is oe to oe ad oto. Let f() = 0.a a 2 a 3 f(2) = 0.a 2 a 22 a 23 f(3) = 0.a 3 a 32 a 33. Each of a ij must be 0 s or 2 s, based o the base-3 represetatio theorem that was previously discussed. Defie a umber b as b =.b b 2 where 6

10 2 if a jj = 0 b j = 0 if a jj = 2 It is clear by this piecewise fuctio that b T sice its base-3 represetatio oly cotais 0 s ad 2 s. However b f() because they differ i the th terary positio. Therefore this fuctio is ot oto, which creates a cotradictio that there exists a oto fuctio that maps all elemets of N ito the set. A set is owhere dese whe every ope iterval cotais a subiterval such that the itersectio of the subiterval ad the origial set is equal to the empty set. Alteratively, a set is owhere dese if it cotais o ope itervals. Theorem 6. The terary Cator set is owhere dese. Proof. To prove that the terary Cator set is owhere dese, we must show that T has o ope itervals, similar to the methods used i Dimartio [3]. Let J be a ope iterval i [0,] with legth λ. Choose i such that < λ. T 3 i i is a uio of 2 i pairwise disjoit closed itervals, kow as the compoets of T i. Each compoet of T i has legth. Now let L 3 i be a compoet of T i. Defie l as the legth of a compoet. The l(l) = 3 i < λ = l(j). Therefore, J L T i. Sice the Cator set, T = i= T i, the J T, otherwise J would be a subset of all T i, but we have show that J T i. Therefore the Cator set cotais o ope itervals ad is owhere dese. Isolated poits are defied as ay poit that is some positive distace away from every other poit i the set. Theorem 7. The terary Cator set has o isolated poits. Proof. To prove this we must show that every poit i T is a limit poit. Suppose x T. We must show that ε > 0 y T with x - y < ε ad y does ot equal x. Let x T ad ε > 0. Choose i such that 3 i < ε. Sice x T ad T = T i, the x T i. This meas 7

11 that there is a compoet, L of T i such that x L ad l(l) = 3 i. Now, L T i+ has 2 compoets, L 0 ad L, ad x must be i oe of these. Say x L 0. L 0 also cotais a poit i T called y. I particular, y is a edpoit of L 0. The x - y 3 i < ε. A perfect set is a closed set with o isolated poits. Theorem 8. The terary Cator set is a perfect set. Theorem 2 ad Theorem 7 show that the Cator set is closed ad has o isolated poits, so this theorem has already bee proved. The terary Cator set is easily the most studied versio of the Cator set due to its uique properties. The ext chapters will explore more geeral Cator sets, where a differet proportio is removed durig each iteratio ad determie which properties hold i those particular sets. 8

12 Chapter 2 The -ary Cator Set The previous chapter described umerous characteristics ad properties of the terary Cator set. The word terary meat that the middle 3 was beig removed durig each iteratio. It is iterestig to look at variatios o this typical Cator set to see what ca be discovered. We will look at variatios of this costructio i the ext two chapters. These sets ca agai be described by what is beig removed durig i each iteratio. We first will look at removig the middle durig each iteratio. These sets ca be deoted as C i k, where i deotes the i th iteratio ad k describes the portio of the iterval beig removed at each iteratio. Usig this otatio, the terary Cator set would be writte as C 3 ad the 2 d iteratio of the terary Cator set would be writte as C 2 3. This chapter will begi by explorig the edpoits of the first iteratio of C ad evetually attemptig to create a base represetatio of all umbers that are i C, where C represets C. Lemma 9. The edpoits of C ca be represeted as a = 2 ad b = + 2 where C = [0, a] [b, ]. Proof. Let k = where is the middle portio of the iterval that is beig removed. The 9

13 after the first iteratio, [0, a] [b, ] is remaiig. Sice k is beig removed from the iitial iterval, the b a = k. The combied legths of [0, a] ad [b, ] must equal k. That is, (a 0) + ( b) = k. The legths of [0, a] ad [b, ] must be equal sice the iterval removed is i the middle of the iitial iterval. Sice the two itervals are equal, the they each must be half the legth of what is remaiig after removig the middle iterval. That is a 0 = a = 2 ( k) ad b = ( k) 2 These ca fially be simplified to a = k 2 ad b = + k 2 Substitutig for k yields a = 2 ad b = + 2 which simplifies to a = 2 ad b = + 2 The followig proofs relate back to the iitial list of properties of the terary Cator set listed i chapter that were ot proved. Sice is i the form of, where = 3, the 3 terary set falls uder these particular Cator sets. Therefore, the followig collectio of proofs also prove results for the terary Cator set. We start by lookig at the umber of itervals remaiig ad removed after each iteratio of C. Lemma 0. C i cosists of 2 i pairwise disjoit subitervals, [s, t]. Proof. We give a proof by iductio. Case i = : based o the previous theorem, C cosists of disjoit itervals of the form [0, 2 ] [ + 2, ]. It is clear that C cosists of 2 = 2 0

14 itervals, so this holds for i =. Assume true for C i. Cosider case C i+. By costructio, after removig the middle of ay iterval, 2 will remai. Therefore, each compoet [s, t] of C i creates 2 pairwise disjoit itervals to C i+. Therefore, the umber of subitervals i C i+ = 2 * the umber of subitervals i C i. With the iductive hypothesis, this is equal to 2(2 i ) = 2 i+. Lemma. 2 i itervals are removed from C i whe costructig C i+. Proof. Based o the previous theorem, we kow that C i cosists of 2 i pairwise disjoit itervals. Therefore, to costruct C i+, we will remove a middle portio of each iterval i C i. Sice there are 2 i itervals i C i, the that meas we will remove a middle portio from each of these itervals, so we will remove 2 i itervals from C i to costruct C i+. The ext property to explore i the -ary Cator set is the legth of these itervals that remai after each iteratio, as well as the legth of the itervals that are removed durig each iteratio of C. Lemma 2. The legth of each iterval remaiig i a compoet of C i is ( 2 ) i Proof. We give a proof by iductio. For the i = case, C is defied as [0, 2 ] [ + 2, ]. The legths of each compoet of C are equal sice the middle iterval is removed. This meas that each iterval i C has legth 2 case. 0 =. Therefore, this holds for the i = 2 Assume true for C i. Cosider C i+. Each iterval of this iteratio will have equal legth due to the costructio of removig the middle iterval. If the portio of the iterval beig removed is, the that meas that is remaiig. This must be split betwee the two remaiig itervals. Sice the middle portio is beig removed, this will be a eve split, so ay iterval remaiig i C i+ will be ( ) th of the previous iterval s legth, ( ) i, 2 2

15 which ca be writte as 2 ( 2 ) i ( = 2 ) i+ Lemma 3. The legth of each iterval that is removed durig iteratio C i to create C i+ is equal to ( 2 ) i Proof. Let [s, t] be a compoet of C i ad [s, t ] be the iterval that is removed from C i to costruct C i+. The legth of [s, t ] is equal to t s. This iterval ca be thought of as oe th of [s, t]. This expressio ca be writte as (t s). From the previous lemma, the legth of (t s) is equal to ( ) i (. Therefore, the legth of [s, t ] = ) i. 2 2 Usig the previous four lemmas i cojuctio, we ca ow explore the total legth removed ad the total legth remaiig i each iteratio of C. Lemma 4. The total legth removed i each iteratio of C i to create C i+ is ( )i i+. Proof. Based o Lemma, there are 2 i itervals removed durig each iteratio of C. Based o Lemma 3, each of these itervals have legth ( ) i. 2 The, the total legth removed would be 2 i itervals of this legth ( ) i, 2 which ca be writte as 2 i ( )i 2 i i = ( )i i+ Lemma 5. The total legth remaiig i each iteratio of C i is equal to ( ) i. Proof. We give a proof by iductio. Cosider i = case. C cosists of [0, 2 ] [ + 2, ]. The legth of each of these itervals are equal sice the middle iterval is removed. The first 2

16 iterval has legth 2 of C is - 0 = 2. Sice C cosists of two such itervals, the total legth 2 2 = Therefore, this holds for the i = case. ( = Assume true for C i. Cosider the i + case. To costruct C i+, we will remove of C i. Based o the iductive hypothesis, C i has legth of ( ) i. The C i+ will have legth ( ) i ( ) i ) = ( )i ( )i i i = ( )i ( )i i i+ = ( )i ( ) i i+ = ( )i ( ) i+ = ( )i+ i+ = ( ) i+ Therefore, each iteratio of C i has legth ( ) i. The previous six lemmas are useful i lookig at the total legth of the -ary Cator set, much like we did with the terary Cator set. Theorem 6. The -ary Cator set has legth 0. Proof. This ca be prove by lookig at everythig that is beig removed from C. Recall by Lemma 4 that ( )i i+ is removed durig each iteratio of C i. This meas that durig C 0, is removed, durig C, ( ) 2 is removed, ad durig C 2, ( )2 3 3 is removed. This patter

17 cotiues ad total legth removed ca be writte as j=0 ( ) j j+ = = = This meas that the total legth of is removed. The legth of C 0 is the legth of the iitial iterval, [0, ] which is. Therefore, the legth of C is = 0. = The ext few proofs utilize the otatio of the edpoits i C ad attempt to geeralize what umbers remai i C by givig a base represetatio, much like the terary Cator set has. These theorems must be give i cases based o parity of the edpoits as follows. Theorem 7. Whe the i beig removed is eve, a = 2 ad b = + 2 are both irreducible fractios. Proof. Whe is eve, the the edpoits a ad b are fractios with umerators ad + respectively. Sice is eve, + ad must both be odd. The 2 reduce if ad oly if GCD(, 2) > or GCD( +, 2) > respectively. Suppose r ad also r 2. There exists 2 itegers, x ad y, such that or + 2 will = rx ad 2 = ry The = rx + ad y = 2 r Usig substitutio, 2 = 2(rx + ) = 2rx + 2 4

18 Therefore, y = 2rx + 2 r = 2x + 2 r is a iteger. Sice y ad 2x are itegers, 2 r must also be a iteger. Therefore, r = or r = 2. Sice is eve, is odd. So 2. Therefore, r = ad GCD(, 2) = so a = 2 that is irreducible. For edpoit b, suppose r + ad also r 2. There exists 2 itegers, x ad y, such + = rx ad 2 = ry The = rx ad y = 2 r Usig substitutio, 2 = 2(rx ) = 2rx 2 Therefore, y = 2rx 2 r = 2x 2 r is a iteger. Sice y ad 2x are itegers, 2 r must also be a iteger. Therefore, r = or r = 2. Sice is eve, + is odd so 2 +. Therefore, r = ad GCD( +, 2) = so b = + 2 is irreducible. Theorem 8. Whe the i to the irreducible form a = k 2k+ beig removed is odd, a = 2 ad b = k+ 2k+ where = 2k +. ad b = + 2 ca be reduced Proof. Let = 2k + where k is a iteger. The edpoits a ad b ca the be rewritte as a = 2 = 2k + 2(2k + ) = 2k 2(2k + ) = k 2k + ad b = + 2 = 2k + + 2(2k + ) = 2k + 2 2(2k + ) = k + 2k + 5

19 Suppose r k ad also r 2k +. There exist 2 itegers, x ad y, such that k = rx ad 2k + = ry The y = 2k + r Therefore, y = 2rx + r = 2rx r + r = 2x + r is a iteger. Sice y is a iteger, the r must also be a iteger. Therefore, r = ad GCD(k, 2k + ) = ad edpoit a = k 2k+ is irreducible. Suppose r k + ad r 2k +. There exists 2 itegers, x ad y, such that k + = rx ad 2k + = ry The ad k = rx ad y = 2k + r y = 2(rx ) + r y = 2rx r y = 2x r is a iteger. Sice y is a iteger, the r must also be a iteger. Therefore, r = ad GCD(k +, 2k + ) = ad edpoit b = k+ 2k+ is irreducible. Theorem 9. Whe the i beig removed is eve, the edpoits after the first iteratio ca be represeted i base 2 as a = 2 = + 2 = (.( + )000 ) 2. = (.( - )0000 ) 2 = (.( - 2)2 ) 2 ad b 6

20 Proof. The proof follows by covertig base 2 represetatios to base 0 represetatios as follows: (.( 2)(2 )) 2 = (2) (2) 3 + = 2 2 = 2 2 = 2 2 = 2 2 = j=2 2 (2) j + (2 ) + (2 ) + (2 ) j=2 (2) j (2) 2 2 ( (2) ( + (2 ) (2) 2 = = 2 ) 2 2 ) Therefore, a = 2 = (( 2)(2 )) 2 Covertig b base 2 to b base 0 requires lookig at each digit ad dividig by 2 to the th power. The first digit i base 2 is simply equal to 2 This meas that (.( + )0000 ) 2 = + 2 = + 2 multiplied by the first digit. = b Example. We cosider the case of removig the middle where = 4 is eve. Thus, the edpoits are a = 2 ad b = + 2 : 7

21 The edpoits a ad b i the case would be 4 a = 4 2(4) = 3 8 ad b = 4 + 2(4) = 5 8 Edpoit a ca be represeted i base 2(4) = 8 as 3 8 = (.(4 2)(2(4) )) 8 = (.27) 8 which ca be see by covertig this to base 0 represetatio: (.27) 8 = Therefore, a = 3 8 = (.27) 8. = j=2 7 8 j = ( j=2 ) 8 j ( ) = ( ) = ( ) = = ( ) 2 = = 3 8 = a Edpoit b ca be represeted i base 2(4) = 8 as 5 8 = (.(4 + )0000 ) 8 = (.5) 8 8

22 which ca be see by covertig this to base 0 represetatio: (.5) 8 = 5 8 = 5 8 = b T herefore, b = 5 8 = (.5) 8. Theorem 20. Whe the i beig removed is odd, the edpoits after the first iteratio ca be represeted i base 2k + as a = k+ = (.(k + )000 ) 2k+ 2k+ where = 2k +. k = (.k0000 ) 2k+ 2k+ = (.(k - )2k) 2k+ ad b = Proof. Let = 2k +. The edpoits, a ad b, ca determie which digits will be icluded i the base represetatio. a = 2 = k 2k + ca be represeted i base 2k + as (.k0000 ) 2k+ = (.(k )2k) 2k+ The proof follows by covertig a i base 2k + to a i base 0: k 2k + + 2k (2k + ) 2 + 2k (2k + ) 3 + = k 2k + + 2k (2k + ) j j=2 = k 2k + + (2k) j=2 (2k + ) j = k ( ) 2k + + 2k (2k+) 2 2k+ ) ( = k 2k + + 2k (2k+) 2 2k 2k+ = k ( 2k + + 2k = k 2k + + 2k + = (2k + ) 2 2k + 2k 9 k 2k + = a )

23 Therefore, a = 2 = k 2k+ = (.(k - )2k) 2k+ b = + 2 = k + 2k + ca be represeted i base 2k + as (.(k + )0000 ) 2k+ We will covert b i base 2k + to b i base 0. The first digit i base 2k + is simply equal to 2k+ multiplied by the first digit. This meas that (.(k + )0000 ) 2k+ = k + 2k (2k + ) (2k + ) 3 + (.(k + )0000 ) 2k+ = k + 2k + = b Example. We cosider the case of removig the middle where = 5 is odd. Thus, the edpoits are a = 2 = k 2k+ ad b = + 2 = k+ 2k+ Solvig = 2k + for k = 2. I the 5 would be a = where = 2k + : case, k = 5 2 = 2 ad the edpoits a ad b 2 2(2) + = 2 5 ad b = 2 + 2(2) + = 3 5 Edpoit a ca be represeted i base 2(2) + = 5 as (.(2 )(2(2))) 5 = (.4) 5 which ca be see by covertig this to base 0 represetatio: (.4) 5 = = j=2 4 5 j

24 ( = ) j j=2 ( ) = ( ) = ( ) = = ( ) = = 2 5 = a Therefore, a = 2 5 = (.4) 5. Edpoit b ca be represeted i base 2(2) + = 5 as (.(2 + )0000 ) 5 = (.3) 5 which ca be see by covertig this to base 0 represetatio: (.3) 5 = 3 5 = 3 5 = b Therefore, b = 3 5 = (.3) 5. The previous two theorems will oly hold true for the first iteratio of C. These will ot help determie if a umber is i C. They are oly used to determie the base represetatios of the edpoits of the first iteratio of C. The purpose of these two theorems was a attempt to geeralize the base represetatios of umbers that are i the -ary Cator set. However, the base represetatios of umbers i the -ary Cator set are more complicated tha those of the terary Cator set. This will lead us ito the ext Cator set. To demostrate how a attempt to restrict the digits used i base represetatios i the -ary Cator set fails, we will look at 2 previous examples, oe where is eve ad oe 2

25 where is odd. Recall the -ary Cator set. Sice = 4 is eve, we will be workig i base 2(4) = 8 4 without usig the middle digits, which are 3 ad 4. We previously showed that C = [0, 3 8 ], [ 5 8, ] ad that 0 = (.0) 8, 3 8 = (.27) 8, 5 8 = (.5) 8, ad = (.7) 8. These edpoits all ca be writte i base 8 without usig the middle 2 digits. To elaborate, the base-8 digits are 0,, 2, 3, 4, 5, 6, ad 7. Therefore, the middle 2 digits are 3 ad 4, which are ot used i the base-8 represetatios of these edpoits. Although the Theorem 9 was oly referrig to the edpoits, we will see that we caot easily defie the base represetatios of what remais like we ca i the terary Cator set. Sice there are o issues regardig these middle digits i the first iteratio, we will look at the secod iteratio. C 2 = [0, 9 64 ] [5 64, 3 8 ] [5 8, ] [55 64, ] We must ow covert the ew edpoits 9 64, 5, , ad to base 8 to see if there are ay 3 s 64 or 4 s i their represetatios = (.) 8, 5 64 = (.7) 8, = (.6) 8, ad = (.67) 8. Noe of these edpoits use the digits 3 or 4, so i theory, a theorem restrictig the -ary Cator set whe is eve to base 2 without the middle 2 digits is still plausible. However, whe lookig at the third iteratio, umerous couterexamples surface. C 3 = [0, ] [ 45 52, 9 64 ] [5 64, ] [65 52, 3 8 ] [5 8, ] [365 52, ] [55 64, ] [485 52, ] We will ow covert the ew edpoits, 27 52, 45 52, 47 52, 65 52, , 365, , ad to base 8. The coversios are as follows: = (.033) 8, = (.055) 8, = (.223) 8, = (.245) 8, = (.533) 8, = (.555) 8, = (.723) 8, ad = (.745) 8. This shows that there are some edpoits i C 3 that use the digits 3 or 4 i their base 8 represetatio. We caot maipulate the base represetatio to elimiate all of the 3 s ad 4 s sice they are ot all termiatig digits. Sice some edpoits caot be writte without those two digits, this shows that we caot restrict the base represetatios of elemets i the -ary Cator set whe is eve. 22

26 We will ow look at a example of the -ary Cator set where is odd. Recall the 5 example. Sice = 5, we will be workig i base 5 without usig the middle digit, 2. We previously showed that C = [0, 2] [ 3, ] ad that 0 = (.0) 5 5 5, 2 = (.4) 5 5, 3 = (.3) 5 5, ad = (.4) 5. Noe of these edpoits use the digit 2 i their base 5 represetatio. Sice there are o issues i the first iteratio, we will look at the secod iteratio. C 2 = [0, 4 25 ] [ 6 25, 2 5 ] [3 5, 9 25 ] [2 25, ] We must ow covert the ew edpoits, 4 25, 6, , ad 2 25 to base 5 to see if there are ay 2 s i their represetatios = (.04) 5, 6 25 = (.) 5, 9 25 = (.34) 5, ad 2 25 = (.4) 5. Noe of these edpoits use the digit 2, so i theory, a theorem restrictig the -ary Cator set whe is odd to base without the middle digit is still plausible. However, whe lookig at the third iteratio, umerous couterexamples surface. C 3 = [0, 8 25 ] [ 2 25, 4 25 ] [ 6 25, ] [ 42 25, 2 5 ] [3 5, ] [ 87 25, 9 25 ] [2 25, 3 25 ] [7 25, ] We will ow covert the ew edpoits, 8 25, 2 25, 38 25, 42 25, 83 25, 87, , ad 7 25 to base 5. The coversios are as follows: 8 25 = (.03) 5, 2 25 = (.022) 5, = (.23) 5, = (.32) 5, = (.33) 5, = (.322) 5, 3 25 = (.423) 5, ad 7 25 = (.432) 5. This shows that there are some edpoits i C 3 that use the digit 2 i the base 5 represetatio. We caot maipulate the base represetatio to elimiate all of the 2 s sice they are ot all termiatig digits. Sice some edpoits caot be writte without this digit, this shows that we caot restrict the base represetatios of elemets i the -ary Cator set whe is odd. 23

27 Chapter 3 The -ary Cator Set The ext variatio of the Cator set ca be costructed by removig a proportio of durig each iteratio. Some examples of middle itervals that satisfy this requiremet iclude, 2, 3, ad 4. Like the terary Cator set, the actual set ca be described by the itersectio of the iteratios. That is, C = C i. This set will be explored i a attempt to determie a geeralized base represetatio of all elemets of C. Much like the previous chapter, we will start by lookig at the represetatio of the edpoits of the first iteratio of C, ad evetually determie the base represetatio of elemets of this set where C is C. ( ) th Lemma 2. The edpoits after removig the middle from the iterval [0, ] ca be represeted as where C = [0, a] [b, ]. Proof. Let k = where a = 2 ad b = 2 2 is the middle portio of the iterval that is beig removed. The after the first iteratio, [0, a] [b, ] is remaiig. Sice k is beig removed from the iitial iterval, the b a = k. The combied legths of [0, a] ad [b, ] must equal k. That is, (a 0) + ( b) = k. The legths of [0, a] ad [b, ] must be equal sice the iterval removed is the middle of the iitial iterval. Sice the two itervals are equal, the 24

28 they each must be half the legth of what is remaiig after removig the middle iterval. That is a 0 = a = 2 ( k) ad b = ( k) 2 These ca fially be simplified to a = k 2 ad b = + k 2 Substitutig for k yields a = 2 ad = + 2 which evetually simplifies to a = 2 ad b = 2 2 The followig proofs are similar to the iitial properties of the terary Cator set listed i Chapter ad proved i chapter 2 for C. I this case, however, they are strictly for the -ary Cator set. The first two lemmas will explore how may itervals remai ad are removed durig each iteratio of C. Lemma 22. C i cosists of 2 i pairwise disjoit subitervals. This proof follows the same structure as the proof for Lemma 0 i Chapter 2. Lemma i itervals are removed from C i whe costructig C i+. This proof follows the same structure as the proof for Lemma i Chapter 2. The ext two lemmas will look at the legths of the itervals remaiig i each iteratio of C as well as the legth of each iterval that is removed from C. 25

29 Lemma 24. The legth of each iterval remaiig i a compoet of C i is ( 2) i. Proof. Let [s, t] be a compoet of C i. For i =, C is defied as [0, ] [ 2, ]. The 2 2 legths of each compoet of C are equal sice the middle iterval is removed. This meas that each iterval i C has legth 2-0 = 2. This is also equal to ( 2). Therefore, this holds for the i = case. Assume true for C i. Cosider C i+. Each iterval of this iteratio will have equal legth due to the costructio of removig the middle iterval. If meas that is remaiig. This is beig removed, the that must be split betwee the two remaiig itervals. Sice the middle portio is beig removed, this will be a eve split, so ay iterval remaiig i C i+ will be ( ) th of the previous iterval s legth, ( i 2 2), which ca be writte as ( ) i ( ) i+ = Lemma 25. The legth of each iterval that is removed durig iteratio C i to C i+ is equal to 2 i i+ Proof. Let [s, t] be a compoet of C i with iterval [s, t ] removed whe costructig C i+. The the legth of [s, t ] is equal to t s. This iterval ca be thought of as oe th of [s, t]. This expressio ca be writte as (t s). The legth of (t s) is equal to ( 2) i. Therefore, the legth of [s, t ] is ( ) i = 2 2 i. i+ The ext two lemmas will use the previous four lemmas to determie the total legth removed ad remaiig of each iteratio of C. 26

30 Lemma 26. The total legth removed i each iteratio of C i to make C i+ is i+. Proof. Based o Lemma 23, there are 2 i itervals removed durig each iteratio of C. Based o Lemma 25, each of these itervals have legth would be 2 i itervals of this legth, which ca be writte as 2 i i+ 2 i 2 i = i+ i+. The, the total legth removed 2 i i+ Lemma 27. The total legth remaiig i each iteratio of C i is equal to ( ) i. Proof. We give a proof by iductio. Cosider i = case. C cosists of [0, 2 ] [ 2 2, ]. The legth of each of these itervals are equal sice the middle iterval is removed. The first iterval has legth 2 0 = 2. Sice C cosists of two such itervals, the total legth of C is 2 2 = 2 2 = Therefore, this holds for the i = case. Assume true for C i. Cosider the i + case. To costruct C i+, we will remove C i. Based o the iductive hypothesis, C i has legth of ( ) i. The Ci+ will have legth ( ) i 2 i 2 2i ( 2 ) i = 2 i ( 2 i 2 i ) i 2 i i = ( ) i i = i i+ 27 of

31 = i+ i+ = i+ = Therefore, each iteratio of C i has legth ( ) i. ( ) i+ The previous six lemmas will ow be used to determie the total legth of the -ary Cator set. Theorem 28. The -ary Cator set has legth 0. Proof. This ca be prove by lookig at everythig that is beig removed from C. Recall by Lemma 26 that i+ is removed, durig C 2 is removed durig each iteratio of C i. This meas that durig C 0 is removed, ad durig C 2 3 ad the total legth removed ca be writte as j=0 = ( ) j+ j=0 j+ ) = ( )( ) = ( )( = ( ) = is removed. This patter cotiues This meas that a legth of is removed. The startig legth of C is the legth of the iitial iterval, [0, ] =. Therefore, the legth of C = = 0. It is importat to ote that the terary, -ary, ad the -ary Cator sets all had total legth 0. I the coclusio, we will cosider if there are ay variatios i the Cator set that 28

32 allow ay legth. The ext few theorems are ecessary to fialize the base represetatio of all umbers i the -ary Cator set. Theorem 29. Whe is beig removed from the iterval [0, ], the edpoits a = 2 ad b = 2 2 are both irreducible fractios. 2 Proof. These edpoits, ad will reduce if ad oly if GCD(, 2) > or GCD(2 2 2, 2) >. For edpoit a, we attempt to show that the GCD(, 2) =. The GCD of ay umber ad will always be, therefore GCD(, 2) = so edpoit a = 2 is irreducible. For edpoit b, suppose r 2 ad r 2. There exists 2 itegers, x ad y, such that 2 = rx ad 2 = ry The = rx + 2 ad y = 2 r Usig substitutio, ( rx + ) 2 = 2 = rx + 2 Therefore, y = rx + r = x + r is a iteger. Sice y is a iteger, r GCD(2, 2) = so b = 2 2 is irreducible. must also be a iteger. Therefore r = ad Theorem 30. The left edpoits of the compoets i each C i ca be expressed i base 2 usig oly the digits 0 ad 2 -. If s is a left edpoit of a compoet of C i, the s = (.a a 2 a i 0) 2 with all a k {0, (2 - ) }. Proof. Let [s, t] be a compoet of C i. s ca be writte i the form of (.a a 2 a 3 a i ) 2. Cosider case i =. C = [0, ] [ 2, ]. The two left edpoits i this case are 0 ad

33 2. Lookig at 0, this ca be represeted i base 2 as (.00) 2 2. This ca easily be see because this would be (2) (2) 3 + = = 0 Therefore, this holds for the first edpoit i the i = case. The other left edpoit i this case is 2. This edpoit ca be writte as (.(2 - )0) 2 2. This ca also be see by lookig at the coversio from base 2 to base 0 as follows (2 ) (2) (2) 3 + = (2 ) = 2 2 Therefore, this holds for the secod edpoit i the i = case ad will hold for both edpoits i this case. Assume true for C i ad look at the C i+ case. If [s, t] is a compoet of C i the [s, s ] [t, t] [s, t] with s = (.a a 2 a i 0) 2, where all a k {0, (2 )}. The two left edpoits are s ad t. s will be writte i base 2 usig oly 0 ad (2 - ) by the iductive hypothesis sice s is a left edpoit i C i. Edpoit t = s + the legth of C i+ + the legth removed i C i. Based o the previous theorems, this will be writte as t = s + ( ) i+ + ( ) i 2 2 Let ( ) i+ + = x. We ca see that x is writte followig the base 2 rules as follows 2 2 i i+ x = ( ) ( ) i + ( ) i = ( ) i ( + )

34 = ( ) i = 2 ( ) 2 2 ( ) i ( ) (2 ) 2 2 = ( ) i+ (2 ) 2 This (2 ) is i the (i + ) th positio i the base represetatio sice it is multiplied by ( 2) i+. Recall t = s+x, where s = (.a a 2 a i 0) 2. Therefore, t = (.a a 2 a i (2 )0) 2 ad is of the form (.a a 2 a i+ 0) 2 where a k {0, (2 )}. Theorem 3. The right edpoit t of a compoet i C i ca be writte i base 2 usig oly the digits 0 ad (2 - ) i the form t = (.a a 2 a i (2 )) 2 with all a k {0, (2 - )}. Proof. Let [s, t] be a compoet of C i. The t is the right edpoit i this case. Edpoit t = s+ the legth remaiig. By Lemma 24, this ca be expressed as t = s + ( ) i 2 Which by Theorem 30 ca be writte as t = (.a a 2 a i 0) + ( ) i 2 with all a i {0, 2 } Clearly, the first part of t follows the requiremets for the theorem. We must show that ( ) i = (.0 0(2 )) 2 2 with (2 ) repeatig startig i the (i+) th positio. = j=i+ = (2 ) (2 ) (2) j 3 j=i+ (2) j

35 = (2 ) = (2 ) (( 2 ) i+ 2 ) (2) 2 i+ 2 = ( ) i (2) = i 2 Therefore, t = (.a a 2 a i 0) 2 + (.0 0(2 )) 2 where there are i 0 s i the secod compoet of t ad all a k {0, (2 )}. This ca fially be simplified to (.a a 2 a i (2 )) 2. Theorem 32. x C if ad oly if x = a j j= ( ) j 2 with a j {0, (2 - )}. Proof. We first show that if x C the it ca be writte i base 2 usig oly 0 ad (2 ). Sice x C the there exists compoets [s i, t i ] C i with x [s i, t i ]. This is true for all i because C = C i ad if a elemet is i C, the by defiitio of itersectios, it must be i all C i. Thus i x s i t i s i ad by Lemma 24, t i s i = ( 2) i. The, ( ) i lim x s i lim = 0 i i 2 Therefore the sequece {s i } coverges to x. We also kow from Theorem 30 that s i+ = s i + a i+ (2) i+ where a i+ is either 0 or 2. Applyig Theorem 30 to the left edpoits of s i, x = lim i s i = lim i i a j j= ( ) j 2 = a j j= ( ) j 2 32

36 where all a j {0, (2 )}. We ext show that if x ca be writte i base 2 usig oly 0 ad (2 ) the x C. Recall that every left edpoit s of ay compoet i C i is of the form (.a a 2 a i 0) 2 where a j {0, (2 )}. Sice there are 2 choices for each a j, amely 0 or (2 ), the there are exactly 2 i umbers of this form. By Lemma 22, there are also 2 i compoets of C i. Therefore, s is a left edpoit i C i if ad oly if s = (.a a 2 a i 0) 2 with all a j {0, (2 )}. Let x = j= ( ) j a j with all a j i {0, (2 )} 2 For all i s i = i j= ( ) j a j 2 j= ( ) j a j s i + 2 j=i+ ( ) j (2 ) = t i 2 where [s i, t i ] is a compoet of C i. Therefore, x [s i, t i ] C i i. Therefore, x C. This fial proof is useful sice it ca be used to determie if a give umber is i the -ary Cator set, or what umbers the -ary Cator set cotais. If a umber ca be writte i base 2 usig oly the digits 0 ad (2 - ), it is i C. Furthermore, if a umber is i C, its base 2 represetatio uses oly the digits 0 ad (2 - ). This is the sort of coclusio that was attempted i Chapter 2. However, the -ary Cator set does ot have this property, where the -ary Cator set does. Example. We cosider the case of removig the middle where = 4. Therefore, we will be removig 3 4 durig each iteratio of C. The claim is that every edpoit i C i ca be represeted as s = (.a a 2 a i 0) 2 with all a i i {0, (2 - )} if s is a left edpoit ad as t = (.a a 2 a i (2 )) 2 with all a i i {0, (2 - )} if t is a right edpoit. I this case, we will be workig i base 2(4) = 8, usig oly the digits 0 ad 2(4) = 7. The first iteratio of C is [0, ] [ 7, ]. We must ow covert each of these edpoits to 8 8 base-8. 0 = (.0) 8, 8 = (.) 8, 7 8 = (.7) 8, ad = (.7) 8. Although the base-8 represetatio of 33

37 cotais a, it ca be rewritte as (.07) 8 8. I fact, ay umber termiatig i ca be rewritte with a 0 followed by a repeatig 7. Sice all of these edpoits ca be writte i base-8 usig oly 0 ad 7, this holds for C. We will ow look at the ext iteratio. C 2 = [0, 64 ] [ 7 64, 8 ] [7 8, ] [63 64, ] We must covert each ew edpoit ito base-8 as follows: 64 = (.0) 8 = (.007) 8, 7 64 = (.07) 8, = (.7) 8 = (.707) 8, ad = (.77) 8. All of these edpoits ca be represeted i base-8 usig oly the digits 0 ad 7, so the theorem still holds. We will look at oe more iteratio, sice the -ary Cator set failed i the third iteratio, we will ed by lookig at that oe. C 3 = [0, 52 ] [ 7 52, 64 ] [ 7 64, ] [ 63 52, 8 ] [7 8, ] [455 52, ] [63 64, ] [5 52, ] Agai, we must covert the ew edpoits to base-8 as follows: 52 = (.00) 8 = (.0007) 8, 7 52 = (.007) 8, = (.07) 8 = (.0707) 8, = (.077) 8, = (.70) 8 = (.7007) 8, = (.707) 8, = (.77) 8 = (.7707) 8, ad 5 52 = (.777) 8. As the theorem states, each of these edpoits are represeted i the proper form usig the digits 0 ad 7, so this example helps illustrate the usefuless of this theorem. 34

38 Chapter 4 Coclusio After studyig a variety of Cator sets, some questios arise. The first questio would be regardig the -ary Cator set. Oe might woder why there is a base represetatio for elemets i the -ary Cator set but ot the -ary Cator set. I fact, the oly thig that has the desired base represetatio i the -ary Cator set are the edpoits durig the first iteratio. These edpoits could be writte i base 2 or base, depedig o the parity of, without usig the middle digits of that particular base. O the other had, the -ary Cator set has elemets that oly cotai the digits 0 or (2 - ) i base 2. It is easy to woder why the -ary Cator set does ot have a defied represetatio like the -ary Cator set, especially sice 3 is of the form with = 3. Lookig at > 3, all -ary Cator sets will ivolve either base, whe odd, or base 2, whe eve. The smallest eve umber greater tha 3 is 4, meaig base 2(4) = 8 will be used. This base has digits 0,, 2, 3, 4, 5, 6, ad 7. Although the edpoits after the first iteratio of C ca be writte without usig the digits 3 ad 4, which was show i a example i a previous chapter, it becomes clear that more edpoits i later iteratios will cotai some of these umbers sice ot every coflictig digit is the termiatig digit. The same is true for odd umbers. 5 is the smallest odd umber greater tha 3. This would require base 5, which uses digits 0,, 2, 3, ad 4. Based o Theorem 20, the edpoits ca 35

39 be writte without usig digit 2, which happes to be the middle digit. However, by the third iteratio, some edpoits will oly be able to be writte usig the digit 2, showig that we caot restrict the base represetatio of elemets i the -ary Cator set. It is the easy to woder why we ca, i fact, restrict digits i the terary Cator set. It is of the form. = 3 is a very special case of the -ary Cator set. Recall the costructio of the terary Cator set. T is costructed by removig 3 from the iterval [0, ]. This leaves the itervals [0, ] [ 2, ]. The legth of the first iterval is 0 =. The middle portio removed also has legth. Fially, the legth of the secod iterval is equal to 3 2 =. This shows that the legth of the itervals remaiig at the first iteratio are 3 3 also equal to the legth removed at the first iteratio. I ay other -ary Cator set, this is ot true. For example, i the -ary Cator set, the first iteratio cotais two itervals 4 with legth 3 ad the iterval removed has legth, which are ot all equal. This is also 8 4 true for whe is odd. For example, i the -ary Cator set, the first iteratio cotais 5 two itervals of legth 2 ad the legth of the iterval removed is legth. Therefore, the 5 5 terary Cator set is the oly -ary Cator set where all three itervals have equal legth. It is also importat to look at the digits i base 3. Sice the terary Cator set has = 3, ad is odd, the we will be workig i base. The digits i base 3 are 0,, ad 2. Whe maipulatig umbers i base 3, termiatig digits ca be writte with a repeatig digit istead. For example, (.) 3 = (.02) 3. This is comparable to how = (.9) 0. I base 3, the oly umber we are restrictig from our represetatio i the terary Cator set is. This digit also falls i the middle of the digits. Sice it is the oly oe beig restricted, we ca easily write ay termiatig umber with as the fial digit i the form of 02. Ay termiatig umber with 0 or 2 as the fial digit is allowed, so we ca maipulate problematic decimals to remove the digit. This idea relies primarily o the fact that is the middle digit ad there is oly digit o either side of whe ordered umerically. For example, the -ary Cator set caot use 5 this. The digits 0,, 2, 3, ad 4 are used i base 5. There will be istaces where we caot 36

40 write a elemet of this Cator set i base 5 without the digit 2, so we caot create a base represetatio requiremet of the geeral -ary Cator sets. I a attempt to defie the elemets of some geeral Cator set, the -ary Cator set was explored. This set is special because it is quite opposite of the -ary Cator set. Istead of removig of the previous iterval, we created a set that leaves of the previous iterval after the iteratio is complete. That is, after is removed, exactly is left remaiig. Each iterval is half of this legth as previously explored, which ca explai why two digits are used for the base represetatio of the -ary Cator set ad why it is i base 2. Although these sets are slight variatios of each other, both the -ary Cator set as well as the -ary Cator set have legth 0. This the causes oe to woder what has to be doe for a Cator set to have some sort of legth. A example of these ca be see i Lamb [4]. Smith-Volterra-Cator sets, or fat Cator sets, are otable because they actually have some legth. They start with the traditioal iterval [0, ]. Istead of removig a set portio durig each iteratio, this will chage. There is a particular way to chage the legth removed durig each iteratio to properly costruct a fat Cator set. For example, start with C 0 = [0, ]. Remove the middle 4 for the first iteratio of C. The, for the secod iteratio, remove the middle. This patter will 6 cotiue, removig the middle 4 i each time. I the previous example, the first few iteratios of this fat Cator set will look like C = [0, 3 8 ] [5 8, ] with 4 removed C 2 = [0, ] [ 5 256, 3 8 ] [5 8, ] [2 256, ] with 6 removed The total legth removed durig each iteratio is 2 i+. We will determie the legth of this fat Cator set like we did i previous examples by determiig the total legth removed, 37

41 which ca be writte as = ( j= ) 2 j+ ( ) 4 = = 2 2 Therefore, this particular fat Cator set has legth =, which is ot equal to These Cator sets are particularly iterestig due to the fact that they have a legth at the ed. Fially, below are some tables summarizig the results from this paper. -ary Cator Set Remaiig Removed Number of Itervals 2 i 2 i Legth of Itervals Total Legth at Iteratio ( ) i 2 ( ) i ( ) i i+ ( ) i 2 Edpoits of the First Iteratio i -ary Cator Set, Eve a b Edpoits 2 Irreducible Yes Yes Base Notatio (.( - 2)(2 )) 2 (.( + )000 ) Edpoits of the First Iteratio i -ary Cator Set, Odd a b Edpoits k 2k+ k+ 2k+ Irreducible Yes Yes Base Notatio (.(k - )(2k)) 2k+ (.(k + )000 ) 2k+ 38

42 Remaiig -ary Cator Set Removed Number of Itervals 2 i 2 i Legth of Itervals Total Legth at Iteratio ( ) i 2 2 i i+ ( ) i i+ a -ary Cator Set b Edpoits of First Iteratio 2 Irreducible Yes Yes 2 2 Base Notatio of Edpoits Right: (.a a 2 a i (2 )) 2 Left: (.a a 2 a i 0) 2 39

43 Bibliography [] Amir D. Aczel, A Strage Wilderess the Lives of the Great Mathematicias, Sterlig Publishig Co. 20. [2] Robert Dimartio ad Wilfredo O. Urbia, Excursios o Cator-Like Sets, [3] Julia F. Flero, A Note o the History of the Cator Set ad Cator Fuctio, Mathematics Magazie, Vol. 67, No. 2 (Apr., 994), pp [4] Evely Lamb, A Few of My Favorite Spaces: Fat Cator Sets, Scietific America, [5] Dyla R. Nelso, A Cator Set - A Brief Itroductio, delso/storage/delso.cator-set.pdf. [6] Christopher Shaver, A Exploratio of the Cator Set, [7] Bria S..Thomso, Judith B. Brucker, Adrew M. Brucker, Real Aalysis, Secod Editio, Pretice Hall

44 Biography Rebecca Stetti was bor i North Royalto, Ohio o August 4, 995. She graduated from North Royalto High School i 203. At Ashlad Uiversity, Rebecca is majorig i mathematics ad actuarial sciece. She served as a executive board member i Delta Zeta sorority ad as a member of Mathematics Associatio of America for the past four years. Rebecca is a member of the mathematics hoorary Pi Mu Epsilo, the Greek hoorary Order of Omega, the leadership hoorary Omicro Delta Kappa, the wome s Greek hoorary Rho Lambda, as well as the freshma hoorary Alpha Lambda Delta. She was o the Dea s list four semesters ad is a recipiet of Who s Who Amog College Studets. After graduatio i May 207, Rebecca plas to study for ad pass actuary exams so she ca begi her career as a actuary. 4