m3113 - Math 311: Advanced Calculus Wolmer V. Vasconcelos...

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Math 311: Advanced Calculus Wolmer V. Vasconcelos Set 3 Spring 2008 Wolmer Vasconcelos (Set 3) Advanced Calculus Spring 2008 1 / 33
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Outline 1 Goals 2 Cantor Set 3 Open Sets 4 Compact Sets Wolmer Vasconcelos (Set 3) Advanced Calculus Spring 2008 2 / 33
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Real Functions Our main aim is to study interesting functions of the kind X f Y where X and Y are subsets of R . If f is a function and the sequence a 1 , a 2 , a 3 , . . . , a n , . . . lies in the domain of f , then the sequence f ( a 1 ) , f ( a 2 ) , f ( a 3 ) , . . . , f ( a n ) , . . . is contained in Y . Wolmer Vasconcelos (Set 3) Advanced Calculus Spring 2008 3 / 33
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Basic Topology of R We want f to have the following property: If ( a n ) is convergent then ( f ( a n )) convergent. This requires us to examine some sets of subsets of R : Open Sets Closed Sets Compact Sets Connected Sets Strange Sets These subsets have properties that will explain why continuous functions act as they do. Wolmer Vasconcelos (Set 3) Advanced Calculus Spring 2008 4 / 33
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Outline 1 Goals 2 Cantor Set 3 Open Sets 4 Compact Sets Wolmer Vasconcelos (Set 3) Advanced Calculus Spring 2008 5 / 33
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Cantor Set C 0 C 1 C 2 • • • • • • • • Rule: From each subinterval of C n remove the inner third, to obtain C n + 1 Cantor Set: C = T C n 0 Wolmer Vasconcelos (Set 3) Advanced Calculus Spring 2008 6 / 33
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Building the Cantor set in detail 1 C 0 = [ 0 , 1 ] , C 1 = C 0 \ ( 1 / 3 , 2 / 3 ) , that is C 1 is obtained by removing from the interval C 0 its mid third (leaving the endpoints): C 1 = [ 0 , 1 / 3 ] [ 2 / 3 , 1 ] 2 Iterate by removing from each closed subinterval above its mid third (and so on) C 2 = ([ 0 , 1 / 9 ] [ 2 / 9 , 1 / 3 ]) ([ 2 / 3 , 7 / 9 ] [ 8 / 9 , 1 ]) 3 This leads to a nested sequence of sets C 0 C 1 C 2 ⊃ ··· C n ⊃ ··· . 4 C = T C n 0 is called the Cantor set. Wolmer Vasconcelos (Set 3) Advanced Calculus Spring 2008 7 / 33
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Building the Cantor set–cont’d Note that C is obtained from [ 0 , 1 ] by repeatedly carving out the heart. At least, the endpoints of the various subintervals belong to C . What else? 1 We are going to argue C is very thin by adding the lengths of the intervals that were removed: 1 3 + 2 1 3 2 + 2 2 1 3 3 + ··· , a geometric series whose first term is 1 / 3 and whose ratio is 2 / 3, so it has for sum 1 / 3 1 - 2 / 3 = 1 ! So from [ 0 , 1 ] we took away a subset of measure 1! Wolmer Vasconcelos (Set 3) Advanced Calculus Spring 2008 8 / 33
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Q is very thin Exercise Given ± > 0, argue that any countable set A is contained in a countable union S n 1 [ a n , b n ] , such that X n 1 | b n - a n | < ±. Wolmer Vasconcelos (Set 3) Advanced Calculus Spring 2008 9 / 33
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Cardinality of C If C only contained the endpoints [all rational points] of the subintervals of its construction, it would be countable. Let us show otherwise: 1 We are going to code the elements of C by infinite strings of { 0 , 1 } as follows: If a C , we set a 1 = 0 if a belongs to the leftmost
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m3113 - Math 311: Advanced Calculus Wolmer V. Vasconcelos...

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