Notes on the Cantor set 1. Definition of the Cantor set 1 1 Given a set A ⊂ R, let 3 A = f 3 x : x 2 Ag; that is, the image of A under 1 2 2 the map x ! 3 x. Similarly, A + 3 = fx + 3 : x 2 Ag. Start with C0 = [0; 1], the closed unit interval in R. 1 1 2 1 2 Let C1 = 3 C0 [ 3 C0 + 3 = [0; 3 ] [ [ 3 ; 1]. 1 1 2 1 2 1 2 7 8 Let C2 = 3 C1 [ 3 C1 + 3 = [0; 9 ] [ [ 9 ; 3 ] [ [ 3 ; 9 ] [ [ 9 ; 1]. 1 1 2 Recursively, let Ck+1 = 3 Ck [ 3 Ck + 3 . 1 1 2 Note that if for any subset A ⊂ [0; 1], the sets 3 A and 3 A + 3 are disjoint 1 2 sets, respectively contained in [0; 3 ] and [ 3 ; 1]. k We now see by induction that Ck consists of 2 disjoint, closed intervals −k of length 3 . It is also easy to see that Ck+1 ⊂ Ck using induction: if Ck ⊂ Ck−1 then 1 1 2 1 1 2 Ck+1 = 3 Ck [ 3 Ck + 3 ⊂ 3 Ck−1 [ 3 Ck−1 + 3 = Ck: And we have that C1 ⊂ C0 to start the induction. T1 We now let C = k=1 Ck, which is a nonempty compact subset of [0; 1]. Also, 1 1 2 C = 3 C [ 3 C + 3 : The complement of C is an open subset of (0; 1) since 0 2 C and 1 2 C, so the complement is the countable union of disjoint open intervals. These are the \middle thirds" that you remove to construct C, and there are exactly 2k−1 open intervals of length 3−k for k ≥ 1. Note that 1 1 X 1 X2k 1 2 1 2k−13−k = = · · = 1 2 3 2 3 1 − 1 k=1 k=1 3 so the lengths of those open intervals add up to 1. 2. Ternary expansions We consider base 3 expansions of numbers x 2 [0; 1]. We will always take aj is equal to 0, 1, or 2 in what follows. Suppose we have a sequence (a1; a2; a3;:::) 2 f0; 1; 2gN. We associate to this sequence a number 1 X ak : 3k k=1 Pn −k The sequence bn = k=1 ak 3 is increasing in n, and bounded above by 1 X 1 2 1 2 = = 1 3k 3 1 − 1 k=1 3 so the infinite sum makes sense as the least upper bound of the bn. We next see that this map from f0; 1; 2gN ! [0; 1] is onto (but not 1{1). First we consider x 2 [0; 1). a • Suppose x 2 [0; 1), and let a be the largest integer such that 1 ≤ x. 1 3 Then since 0 ≤ x < 1 we have 0 ≤ a1 ≤ 2, and a 1 (1) 0 ≤ x − 1 < 3 3 a a Now let a be the largest integer so 2 ≤ x − 1 . Then by (1) we must have 2 32 3 0 ≤ a2 ≤ 2, and a a 1 0 ≤ x − 1 − 2 < 3 32 32 In general, we will take aj recursively so that n X ak 1 (2) 0 ≤ x − < 3k 3n k=1 n X ak Briefly, the a are the largest integers from 0, 1, or 2 so that ≤ x. j 3k k=1 The bound (2) implies that n 1 X ak X ak x = lim = n!1 3k 3k k=1 k=1 This shows the map is onto [0; 1). For the number 1, we take aj = 2 for all j, and note n n−1 1 X 2 2 X 1 2 1 − ( )n 1 = = 3 = 1 − 3k 3 3k 3 1 − 1 3n j=1 k=0 3 P1 2 Thus k=1 3k = 1. • We now determine how the map f0; 1; 2gN ! [0; 1] can fail to be 1{1. Suppose that we have two sequences (a1; a2;:::) and (b1; b2;:::) such that 1 1 X ak X bk = 3k 3k k=1 k=1 Suppose that n is the first position where an 6= bn, and assume an > bn. We then write 1 1 an − bn X ak X bk + = 3n 3k 3k k=n+1 k=n+1 Since 1 1 an − bn 1 X bk X 2 1 ≥ and ≤ = 3n 3n 3k 3k 3n k=n+1 k=n+1 The only way we can have equality is if an − bn = 1, and ak = 0 and bk = 2 for k ≥ n + 1 : That is, the sequence ak has terminal 0's and bk has terminal 2's, for example :1202000000 ··· = :1201222222 ··· Other than this kind of case, the ternary expansion of x 2 [0; 1] is unique. Note that the values of x for which the expansion has two posibilities are precisely those x of the form m x = for some m 2 f0;:::; 3m − 1g: 3n Observe: in the above argument, since an = bn + 1 we must have either an = 1 or bn = 1. So different sequences in f0; 2gN cannot give the same value of x. 3. Ternary expansions and the Cantor set We now claim that the Cantor set consists precisely of numbers of the form 1 X ak (3) x = 3k k=1 where each ak is either 0 or 2. The map f0; 2gN ! C is then a bijection by the above observation. Suppose x is given by (3). Then 1 X bk 1 x = where b = 0 ; b = a if k ≥ 2; 3 3k 1 k k−1 k=1 1 X bk 1 x + 2 = where b = 2 ; b = a if k ≥ 2: 3 3 3k 1 k k−1 k=1 Thus, x 2 C1 if and only if it equals a ternary expansion where either a1 = 0 1 1 2 or a1 = 2, since it is of the form 3 y or 3 y + 3 for some y 2 [0; 1]. Repeating the argument, we see that x 2 Cn iff it equals a ternary expansion where ak is either 0 or 2 for 1 ≤ k ≤ n. It follows that ternary expansions where every digit is either 0 or 2 belong to Cn for every n, hence give an element of C. This shows that f0; 2gN maps into C under the ternary expansion. Suppose now that x 2 C. If the ternary expansion of x is unique, the above argument shows that every ak is either 0 or 2. If x has two ternary expansions, let n be the first digit they differ. Then the above argument shows that ak is either 0 or 2 for 1 ≤ k < n, and we know ak is either identically 0 or identically 2 for k > n. And one of the two expansions has an = 1, the other must have an = 0 or 2. So x has one expansion where ak 6= 1 for every k. This shows that f0; 2gN maps onto C under the ternary expansion. 4. The Cantor map We now can construct a map f : C ! [0; 1] that is onto. To do this, expression x 2 C uniquely as 1 X ak x = where a 2 f0; 2g: 3k k k=1 Define 1 1 X ak f(x) = 2 2k k=1 That is, if x has ternary expansion (a1; a2;:::), then f(x) has binary expan- 1 1 1 sion ( 2 a1; 2 a2;:::). Since ak 2 f0; 2g we have 2 ak 2 f0; 1g, so the binary expansion gives an element of [0; 1], and the map is onto since every sequence in f0; 1gN is obtained this way. The map is not 1{1: for example 1 1 ! (0; 2; 2; 2; 2; 2 ··· ) ! (0; 1; 1; 1; 1; 1 ··· ) ! 3 2 and 2 1 ! (2; 0; 0; 0; 0; 0 ··· ) ! (1; 0; 0; 0; 0; 0 ··· ) ! 3 2.