 # Definition the cantor set is the subset of 0 1

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Definition:The Cantor set is the subset of! =[0,1] defined as follows: LetF1 =[0,1/3] U [2/3, 1]be the subset of[O, 1] formed by removing the open middle third (1/3, 2/3). LetF2 =[0,1/9] U [2/9, 1/3] U [2/3, 7/9] U [8/9, 1]be the subset of F1 formed by removing the open middle thirds (1/9, 2/9) and(7/9, 8/9) of the two components ofF,. Continuing in this manner, letbe thesubset ofobtained by removing the open middle third of each of the componentsofThen the setis the Cantorset.Note that the Cantor set is a closed subset of P since it is the intersection ofclosed sets.For an alternative-definition of the Cantor set, recall that a ternary expansionof a real number x, 0x1,is an expression x =0.x1x2x3...representingx asa sum of powers of 3,x=
188SIX /COMPACTNESSF1I—PIo1213F21141o1212119399F31—II—II—IP—41—II—I1—401212_Lil92727332727992727FIGURE6.8The firstthreestages In the construction of the Cantor set.whereis restricted to the values 0, 1, and 2. Thus 1/3 has the two ternary ex-pansions1/3= 1/3+ 0/32 +0/3+and1/9 can be similarly expressed as1/9 =0/3+ 1/32 +0/3+0/32 ++ 2/3d +Observe that the numbers which absolutely require 1 in the first place of theirternary expansions are the numbers strictly between 1/3 and 2/3. As shown above,1/3 has a ternary expansion without 1 in the first place. Thus, F1 excludes allmembers of[0, 11 which require 1 in the first place of the ternary expansion. Sim-ilarly, F2 excludes those members of [0, 1J which require 1 in the second place. Ingeneral,excludes those members of [0, 1] which require 1 in the nth place oftheir ternary expansions. Thus the Cantor set is the set of real numbers xwhich have a ternary expansion x =0.x1x2x3...whereis restricted to the values0 and 2.
6.5 / The Cantor Set189Example6.5.1TheCantor set contains the end points 0, 1, 1/3, 2/3, 1/9, 2/9, 7/9, 8/9,of the open intervals deleted to define the setsHowever, these are notthe only members of the Cantor set. In particular, the number 1/4 belongs toC.Tosee this consider the convergent seriesx=2/32n=2/32+ 2/3d + 2/36 +Byfactoring out 1/32fromeach term, it follows that x satisfies the equationx= 1/9(2+x).Solving gives x =1/4.Thus 1/4 is a member of C since it has a ternary expansionwhich does not require 1 in any term.Actually, the Cantor set is uncountable. The proof of this fact, which followsfrom the ternary expansion representation, is left as an exercise.Example 6.5.2In the construction of the Cantor set from the intersection of the sets F1,F2,F3, ... ,theopen intervals removed were of lengthL=1/3+2/9+4/27+8/81+++Thesum of this geometric series, which represents the total length of the intervalsremoved, can be calculated by several different methods. First, there is the formula1+r+r2+...+rn+...=*, IrI<1.This givesL =1/3(1+ 2/3 + 4/9 +++ '•.) = 1/3(i_l2/3)=1.For a second method, note thatL=2/3(1/2+ 1/3 + 2/9 + '•'++.. .) = 2/3(1/2 + L)so thatL =1/3+ (2/3)L,whichagaingivesL= 1.

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