Cantor Set Cantor set is a special subset of the closed interval [0; 1] invented by a German mathematician Georg Cantor in 1883. We have already dicussed the construction of this `ternary' set in the class but let me quickly recall it. Let I0 := [0; 1]. Remove the open middle third (1=3; 2=3) from I0. Let I1 := [0; 1=3] [ [2=3; 1], precisely I1 = I0n(1=3; 2=3). Remove middle third open intervals (1=9; 2=9) and (7=9; 8=9) from the respective closed intervals whose union is I1. 1 2 3 4 Let I2 := [0; 1=9] [ [2=9; 3=9] [ [6=9; 7=9] [ [8=9; 9=9] say, I2 [ I2 [ I2 [ I2 . k Continue this precedure of removal of open intervals from each closed interval Ij ; 1 ≤ j k ≤ 2 whose union is Ij, j = 3; 4; ··· and get Ij+1 by taking the union of whatever k is left from each Ij after the removal of open inetervals as before. 1 Define the Cantor set C := \n=1In. Observations/Facts about C (1) Cantor is non-empty: Clearly all end points of the closed intervals comprising In for every n = 1; 2; ··· are in C. Further it contains countably many points, for example points of the form 1=3n , n = 1; 2; 3; ··· . n (2) Cantor set is a closed subset in [0; 1] : Each In is a union of 2 closed inter- vals. As a finite union of closed intevals, it is a closed set in [0; 1] with the usual metric. Then C is a closed subset of [0; 1] being the intersection of closed sets In, n = 1; 2; ··· n (3) Cantor set has length 0 : Each In is a union of 2 closed intervals, each of n length 1=3 . Thus, length of I1 i.e. `(I1) = 2 × 1=3 = 2=3, `(I2) = 4 × (1=9) = 2 n n n (2=3) ··· `(In) = 2 × (1=3 ) = (2=3) . This implies that the length of the Cantor n set `(C) = limn!1(2=3) = 0. (4) Cantor set is a uncountable set: Suppose on contrary C is a countable set. Let k x 2 C. Thus x 2 In; n = 1; 2 ··· . Note that every closed interval Ij in Ij gives rise to two closed intervals L and R in the next step of forming Ij+1. As x 2 C ⊂ [0; 1], x 2 L or R in I1. Suppose it is in L in I1. Then x can either be in L or in R in I2. Suppose it is R in I2 and then in R in I3. Tracing down the intervals in which x lies in 0 0 each In, we get a unique infinite sequence of L s and R s, which in this case is LRR ··· . Therefore, given a x 2 C we get a unique sequence of L0s and R0s. 1 2 Now, suppose we consider this same sequence LRR ··· as above and go on choos- th ing random points yn at every stage in In from L or R as given in the n place of the sequence. Then it is easy to check that fyngn forms a Cauchy sequence and as n ! 1 it converges to the point x as above. Let us re-name L $ 0 and R $ 2. Then it will give a ternary expansion for ev- ery x 2 C. For example in the above case, x $ LRR · · · $ 022 ··· ( or to be precise 0:022 ··· ). As C is assumed to be a countable set we can list it's elements and write their respec- tive ternary exampnsion in 00s and 20s. a1 = a11a12a13 ··· a2 = a21a22a23 ··· a3 = a31a32a33 ··· . an = an1an2an3 ··· . We will now use Cantor's diagonal argument to construct an element b 2 C which is not in the list fa1; a2; a3; · · · g. Let b = b1b2b3 ··· where bi = 0 if aii = 2 and bi = 2 th if aii = 0. We can check that at every n stage, bn differs from an. As b also has a ternary expression in 00s and 20s it belongs to C (For this use the fact (8) ). Thus proving that C is an uncountable set. (5) Cantor set has no intervals: ( part of tutorial) (6) Cantor set is a no where dense subset of [0; 1] (part of tutorial) (7) Cantor set is a totally disconnected set . It is also compact in the usual topology on [0; 1]( Later) Ternary expansions and the Cantor Set (8) An element x in [0; 1] belongs to the Cantor set C if and only if x has only 00s and 20s in it's ternary expansion : Note that a real number x may not have a unique ternary expansion (for example 2 3 4 2 2 1=3 = (0:1)3. Also, (0:022)3 = 2=3 + 2=3 + 2=3 + ··· = 2=3 (1 + 1=3 + 1=3 + ··· ) = 1=3 ). But x can have at most one ternary expansion consisting of only 00s and 20s (Prove this fact as a exercise). Proof of (8). Now, let x 2 C. Then following the argument of (4), we know that x has a unique ternary expansion in 00s and 20s. Now let x 2 [0; 1] such that it has only 0 0 P1 −k 0 s and 2 s in it's ternary expansion i.e. x = k=1 ak3 , where ak = 0 or 2. 3 We will show by using induction on k that x 2 Ik for every k and thus is contained in C. Clearly x 2 I0 = [0; 1]. next if a1 = 0 then x ≤ (0:1)3 = 1=3; if a1 = 2 then x ≥ (0:2)3 = 2=3. So x 2 [0; 1=3] [ [2=3; 1] = I1. This gives base case for induction. k Assume that x 2 Ik. Then x is in one of the 2 disjoint closed intervals comprising Ik, say [a; b]. The construction process of removing middle third open intervals ensures that every left end point in Ik must either be a left end point of Ik−1 or is a sum of k a left end point of Ik−1 and 2=3 . Now removing open middle third set from [a; b] results in two disjoint closed intervals L = [a; a + 1=3k] and a right closed interval k+1 k+1 [b−1=3 ; b]. If ak+1 = 0 then x ≤ a+1=3 and is thus contained in L. If ak+1 = 2, k+1 then x ≥ a + 2=3 and is contained in the right interval R. Either way x 2 Ik+1. So the inductive step is complete. Thus x 2 C. .