MATH501_HW1 - Serge Ballif MATH 501 Homework 1 In all...

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Unformatted text preview: Serge Ballif MATH 501 Homework 1 September 26, 2007 In all problems below, measurable means Lebesgue measurable; m is the Lebesgue mea- sure and m * is the Lebesgue outer measure. 1. The Cantor Set. Denote by C = [0 , 1] . Let C 1 be the set obtained by removing the middle third open interval of C , that is, C 1 = [0 , 1 / 3] ∪ [2 / 3 , 1] . Next, let C 2 be the set obtained by removing the middle third open interval from each interval of C 1 . Continuing this way, we obtain a decreasing sequence of closed sets C ⊃ C 1 ⊃ C 2 ⊃ ... . The Cantor set C is the intersection of all C k ’s, C = ∩ ∞ k =0 C k . Show: (a) C is totally disconnected, that is, if x,y ∈ C and x < y , then there is z ∈ ( x,y ) such that z / ∈ C . Claim 1. Every element x ∈ C can be written in the form x = ∑ ∞ i =1 c i 3 i , where c i = 0 or c i = 2 for every whole number i . Proof. We first note that every number x ∈ [0 , 1] can be written in the form x = ∑ ∞ i =1 c i 3 i where c i ∈ { , 1 , 2 } . We adopt the convention, that we express x in the form listed in claim 1, if it is possible to do so. This policy removes the ambiguity of non-uniqueness that arises from the case where c i = 1 for all i ≥ n . The set C k consists precisely of those numbers x in for which c i ∈ { , 2 } for each i ≤ k (since c k = 1 places x in the open middle third of its interval). Therefore, every x of the prescribed form is in C n for all n and hence, in C . Furthermore, if x is not of the form in the claim, then c n = 1 for some n > 0, and thus, x is not in the set C n or in C . Now if x,y ∈ C and x < y , then x = ∑ ∞ i =1 c i 3 i and y = ∑ ∞ i =1 d i 3 i for c i ,d i ∈ { , 2 } . Since x < y , then we know that there is a smallest index j such that c j = 0 and d j = 2. Define z to be the number z = ∑ ∞ i 6 = j c i 3 i + 1 3 j . By claim 1, z / ∈ C . Then we have x < z < y such that z / ∈ C . Therefore, C is totally disconnected. (b) C is perfect, that is, it is closed and without isolated points. Each set C n is closed as they are defined to be complements of open sets. There- fore, C is closed as an intersection of closed sets. To see that C has no isolated points, consider an arbitrary element x = ∑ ∞ i =1 c i 3 i of C . Let > 0 and consider the open ball B ( x, ). There exists some number n such that 1 3 n- 1 < . Therefore, for any choice of numbers d i ∈ { , 1 , 2 } , we have ∞ X i = n d i 3 i ≤ ∞ X i = n 2 3 i = 2 3 n ∞ X i =0 1 3 i = 2 3 n 1 1- 1 / 3 = 1 3 n- 1 < ....
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This note was uploaded on 04/01/2008 for the course MATH 501 taught by Professor Wysocki,krzysztof during the Fall '08 term at Penn State.

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MATH501_HW1 - Serge Ballif MATH 501 Homework 1 In all...

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