HW2sol - Math 55 Homework 2 Solutions September 15, 2010...

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Math 55 Homework 2 Solutions September 15, 2010 § 2.1 6. For each of the following sets, determine whether { 2 } is an element of that set. a) { 2 } 6∈ { x R | x is an integer greater than 1 } . This set contains only numbers, and no sets. b) { 2 } 6∈ { x R | x is the square of an integer } . The reason is the same as above. c) { 2 } ∈ { 2 , { 2 }} . { 2 } is explicitly included in this set. d) { 2 } ∈ {{ 2 } , {{ 2 }}} . The reason is the same as above. e) { 2 } ∈ {{ 2 } , { 2 , { 2 }}} . The reason is the same as above. f) { 2 } 6∈ {{{ 2 }}} . { 2 } . This set contains only one element: {{ 2 }} . This element is not equal to { 2 } . 8. a) True. b) True. c) False. d) True. e) True. f) True. g) False; {{∅} , {∅}} = {{∅}} , so this is not a proper subset. 21. How many elements does each set have, given a 6 = b . The cardinality | P ( S ) | = 2 | S | , as noted in § 2.1 in the text. So the solutions are: a) 2 3 = 8, b) 2 4 = 16, and c) 2 2 0 = 2. In this case, P ( P ( )) = P ( {∅} ) = {∅ , {∅}} . 28. Let A = { a,b,c } , B = { x.y } and C = { 0 , 1 } . a) A × B × C = { ( a,x, 0) , ( a,x, 1) , ( a,y, 0) , ( a,y, 1) , ( b,x, 0) , ( b,x, 1) , ( b,y, 0) , ( b,y, 1) , ( c,x, 0) , ( c,x, 1) , ( c,y, 0) , ( c,y, 1) } . 1
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2 b) C × B × A = { (0 ,x,a ) , (0 ,x,b ) , (0 ,x,c ) , (0 ,y,a ) , (0 ,y,b ) , (0 ,y,c ) , (1 ,x,a ) , (1 ,x,b ) , (1 ,x,c ) , (1 ,y,a ) , (1 ,y,b ) , (1 ,y,c ) } . c) C × A × B = { (0 ,a,x ) , (0 ,a,y ) , (0 ,b,y ) , (0 ,b,y ) , (0 ,c,x ) , (0 ,c,y ) , (1 ,a,x ) , (1 ,a,y ) , (1 ,b,x ) , (1 ,b,y ) , (1 ,c,y ) , (1 ,c,y ) } . d) B × B × B = { ( x,x,x ) , ( x,x,y ) , ( x,y,x ) , ( x,y,y ) , ( y,x,x ) , ( y,x,y ) , ( y,y,x ) , ( y,y,y ) } . 38. Let S = { sets x | x 6∈ x } . a) Show that the assumption that S S leads to a contradiction. Proof. Suppose, by way of contradiction, that S S . Then S does not satisfy the criterion for membership in S , since S consists only of those sets which are not members of themselves. Hence we conclude S 6∈ S , contradicting our assumption. ± b) Show that the assumption that S 6∈ S leads to a contradiction. Proof. Suppose, by way of contradiction, that S 6∈ S . Then S satisfies the criterion for membership in S , since it is not a member of itself. Hence we conclude S S , contradicting our assumption. ± Note to the students: The definition of S is paradoxical. This problem should absolutely disturb you. I encourage you to look up Russell’s Paradox on Wikipedia. In short, the way around this is to say that S cannot be a set. So what follows is an alternate solution to parts a) and b). Proof. Suppose by way of contradiction that S is a set. S is defined by the proposition x ( x S x 6∈ x ), where we are quantifying over sets. We apply universal instantiation with x = S to obtain S S S 6∈ S . This proposition is false, a contradiction. We conclude that S is not a set. ±
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3 § 2.2
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HW2sol - Math 55 Homework 2 Solutions September 15, 2010...

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