2.3 Exercises. (a) F P(A). (Solution)

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1 2.3 Exercises 1. Analyze the logical forms of the following statements. You may use the symbols, /, =,,,,,,, and in your answers, but not,, P,,, {, }, or. (Thus, you must write out the definitions of some set theory notation, and you must use equivalences to get rid of any occurrences of.) (a) F P(A). By the definition of subset, the given statement is equivalent to x(x F x P(A)). Now, writing out the definition of power set, the whole statement would be x(x F y(y x y A)). (b) x F\ G. By the definition of \, the given statement is equivalent to (x F) (x / G). By the definition , x F is equivalent to A(A F x A), and similarly, x / G is equivalent to A(A G x A). By quantifier negation law, this is equivalent to A (A G x A), which is equivalent to A(A / G x / A) by the definition of and DeMorgan s law, which is equivalent to A(A G x / A) by conditional law. Therefore, the entire statement is equivalent to A(A F x A) A(A G x / A). (c) A {2n + 1 n N}. 1

2 By the definition of subset, this means x(x A x {2n + 1 n N}). Now writing out x {2n + 1 n N}, we get x(x A n N(x = 2n + 1). (d) {n 2 + n + 1 n N} {2n + 1 n N}. By the definition of subset, this means x(x {n 2 + n + 1 n N} x {2n + 1 n N}). Writing out x {n 2 + n + 1 n N} and x {2n + 1 n N}, we have x( n N((x = n 2 + n + 1) (x = 2n + 1))). (e) {x B x / C} P(A). {x B x / C} means {x B\C}. By the definition of power set, y(y x y A). Therefore, the given statement is equivalent to x(x B\C y(y x y A)). (f) x i I (A i B i ). By the alternative notation of intersection of an indexed family of sets, the statement is equivalent to i I x(x A i B i ). Writing out x A i B i, we get i I x(x A i x B i )). (g) x ( i I A i ) ( i I B i )). By the definition of union, the statement is equivalent to x i I A i x i I B i. 2

3 By the alternative notation of intersection, the above statement is equivalent to ( i I x(x A i )) ( i I x(x B i ). Putting these all together, we have i I x(x A i x B i )). (h) P( i I A i ) i I P(A i ). By the definition of, the statement is equivalent to x(x P( i I A i ) x i I P(A i )), which is equivalent to x (x P( i I A i ) x i I P(A i )) (quantifier negation law), which is equivalent to x (x / P( i I A i ) x i I P(A i )) (conditional law), which is equivalent to x(x P( i I A i ) x / i I P(A i )) (DeMorgan s law). By the definition of power set, x(x P( i I A i ) x / i I P(A i )) is equivalent to ( y(y x y i I A i ) (x / i I P(A i )). By the alternative notation of union, ( y(y x y i I A i ) (x / i I P(A i )) is equivalent to x( y(y x y i I A i ) i I(x / PA i )). By the alternative notation of union x( y(y x y i I A i ) i I(x / PA i )) is equivalent to x[ y(y x i I(y A i ) i I(x / P(A i )))]. By the definition of power set, x[ y(y x i I(y A i ) i I(x / P(A i )))] is equivalent to x( y(y x i I(y A i )) i I(y x y / A i )). 2. We ve seen that P( ) = { }, and { }. What is P({ })? 3

4 P( ) = {, { }}. 3. Suppose F ={{red, green, blue}, {orange, red, blue}, {purple, red, green, blue}}. Find F and F. F = {red, blue}. F = {red, green, blue, orange, purple}. 4. Let I = {2, 3, 4, 5} and for each i I let A i = {i, i + 1, i 1, 2i}. (a) List the elements of all the sets A i, for each i I. For i = 2, A i = A 2 = {2, 3, 1, 4} = {1, 2, 3, 4}. For i = 3, A i = A 3 = {3, 4, 2, 6} = {2, 3, 4, 6}. For i = 4, A i = A 4 = {4, 5, 3, 8} = {3, 4, 5, 8}. For i = 5, A i = A 5 = {5, 6, 3, 10} = {3, 5, 6, 10}. (b) Find i I A i and i I A i. i I A i = {1, 2, 3, 4, 5, 6, 8, 10}. i I A i = {3, 4}. 5. Let I = {2, 3}, and for each i I let A i = {i, 2i} and B i = {i, i + 1}. (a) List the elements of the sets A i and B i for i I. For i = 2, A i = A 2 = {2, 4} and B i = B 2 = {2, 3}. For i = 3, A i = A 3 = {3, 6} and B i = B 3 = {3, 4}. (b) Find i I (A i B i ) and ( i I A i ) ( i I B i ). Are they the same? 4

5 By (a), we know A 2 = {2, 4} and B 2 = {2, 3}. Therefore, A 2 B 2 = {2, 3, 4}. Similarly, A 3 B 3 = {3, 4, 6}. Thus, i I (A i B i ) = {3}. On the other hand, i I A i = A 2 A 3 = {2, 4} {3, 6} =, and i I B i = B 2 B 3 = {2, 3} {3, 4} = {3}. Therefore, ( i I A i ) ( i I B i ) = {3} = {3}. Thus, i I (A i B i ) and ( i I A i ) ( i I B i ) are the same. (c) In parts (f) and (g) of exercise 1 you analyzed the statements x i I (A i B i ) and x ( i I A i ) ( i I B i ). What can you conclude from your answer to part (b) about whether or not these statements are equivalent? Since the values of each statement match, they are equivalent to each other. 6. Show that for any sets A and B for which P(A B) = P(A) P(B), by showing that the statements x P(A B) and x P(A) P(B) are equivalent. (See Example ) Proof. From Example , x P(A B) means y(y x (y A y B)), which is equivalent to y(y y A) y(y y B), which x P(A) P(B) means. Therefore, P(A B) = P(A) P(B). 7. Give example of sets A and B for which P(A B) P(A) P(B). 5

6 As long as A B =, P(A B) P(A) P(B). For example, let A = {1} and B = {2, 3}. Then P(A B) = P{1, 2, 3} = {, {1}, {2}, {3}, {1, 2}, {1, 3 {2, 3}, {1, 2, 3}}. On the other hand, P(A) = P(1) = {, {1}}, and P(B) = P(2, 3) = {, {2}, {3}, {2, 3}}, so P(A) P(B) = {, {1}, {2}, {3}, {2, 3}}, which is clearly not equal to P(A B) ( {1, 2}, {1, 3}, and {1, 2, 3, } are missing!) 8. Verify the following identities by writing out (using logical symbols) what it means for an object x to be an element of each set and then using logical equivalences. (a) i I (A i B i ) = ( i I A i ) ( i I B i ). ( ) Let x i I (A i B i ). By the alternative notation of union, this is equivalent to i I(x A i Bi). By the definition of, this is equivalent to i I(x A i x B i ). Distributing the quantifiers, this is equivalent to i I(x A i ) i I(x B i ). By the definition of union, this is equivalent to x i I A i x i I B i. By the definition of, this is equivalent to x ( i I A i ) ( i I B i ). ( ) Let x ( i I A i ) ( i I B i ). By the definition of, this is equivalent to x i I A i x i I B i. By the alternative notation of union, this is equivalent to i I(x A i ) i I(x B i ). 6

7 Collectiong (?) the quantifiers, this is equivalent to i I(x A i x B i ). By the definition of, this is equivalent to i I(x A i B i ). By the definition of union, this is equivalent to x i I (A i Bi). (b) ( F) ( G) = (F G). ( ) Let x ( F) ( G). By the definition of, this is equivalent to x F x G. A(A F x A) ( A(A G x A)). By the conditional law, this is equivalent to A(A / F x A) A(A / G x A). Collecting the quantifiers (?), this is equivalent to A((A / F x A) (A / G x A)). By distributative law, this is equivalent to A((A / F A / G) x A). By De Morgan s law, this is equivalent to A( (A F) A G) x A. By the definition of, this is equivalent to A( (A F G) x A. By the conditional law, this is equivalent to A(A F G x A). By the definition of intersection, x (F G). ( ) Let x (F G). A(A F G x A). By the definition of, this is equivalent to A((A F A G) x A). By the conditional law, this is equivalent to A( (A F A G) x A). By the DeMorgan s law, this is equivalent to A((A / F A / G) x A). 7

8 By the distributive law, this is equivalent to A((A / F x A) (A / G x A)). By the conditional law, this is equivalent to A((A F x A) (A G x A). Distributing the quantifiers, this is equivalent to A(A F x A) A(A G x A). x ( F) ( G). (c) i I (A i \B i ) = ( i I A i )\( i I B i ). ( ) Let x i I (A i \B i ). i I(A i \B i ). By the definition of \, this is equivalent to i I(x A i x / B i ). Distributing of quantifiers, this is equivalent to i I(x A i ) i I(x / B i ). By the definition of /, this is equivalent to i I(x A i ) i I (x B i ). By the quantifier negation law, this is equivalent to i I(x A i ) i I(x B i )). By the definitions of intersection and union, this is equivalent to x i I A i x i I B i. By the definition of, this is equivalent to x ( i I A i ) ( i I B i ). ( ) Let x ( i I A i ) ( i I B i ). By the definition of, this is equivalent to x i I A i x i I B i. By the definitions of intersection and union, this is equivalent to i I(x A i ) i I(x B i )). By the quantifier negation law, this is equivalnet to i I(x A i ) i I (x B i ). 8

9 By the definition of /, this is equivalent to i I(x A i ) i I(x / B i ). Collecting (?) quantifiers, this is equivalent to i I(x A i x / B i ). By the definition of \, this is equivalent to i I(A i \B i ). x i I (A i \B i ). 9. Sometimes each set in an indexed family of sets has two indices. For this problem, use the following definitions: I = {1, 2}, J = {3, 4}. For each i I and j J, let A i,j = {i, j, i + j}. Thus, for example A 2,3 = {2, 3, 5}. (a) For each j J let B j = i I A i,j = A 1,j A 2,j. Find B 3 and B 4. A 1,3 = {1, 3, 4}. A 2,3 = {2, 3, 5}. Therefore, B 3 = {1, 3, 4} {2, 3, 5} = {1, 2, 3, 4, 5}. A 1,4 = {1, 4, 5}. A 2,4 = {2, 4, 6}. Therefore, B 4 = {1, 4, 5} {2, 4, 6} = {1, 2, 4, 5, 6}. (b) Find j J B j. (Note that, replacing B j with its definition, we could say that j J B j = j J ( i I A i,j ).) By (a), j J B j = {1, 2, 4, 5}. (c) Find i I ( j J A i,j ). (Hint: You may want to do this in two steps, corresponding to parts (a) and (b).) Are j J ( i I A i,j ) and i I ( j J A i,j ) equal? 9

10 A 1,3 = {1, 3, 4}. A 2,3 = {2, 3, 5}. Therefore, j J A i,3 = {3}. A 1,4 = {1, 4, 5}. A 2,4 = {2, 4, 6}. Therefore, j J A i,4 = {4}. Thus, i I ( j J A i,j ) = {3, 4}. Therefore, j J ( i I A i,j ) and i I ( j J A i,j ) are NOT equal. (d) Analyze the logical forms of the statements x j J ( i I A i,j ) and x i I ( j J A i,j ). Are they equivalent? Let x j J ( i I A i,j ). j J(x i I A i,j ). By the definition of union, this is equivalent to j J i I(x A i,j ). Now, let x i I ( j J A i,j ). By the definition of union, this is equivalent to i I(x j J A i,j ). i I j J(x A i,j ). Since the order of the quantifiers are different, these two expressions are not equivalent. 10

11 10. (a) Show that if F =, then the statement x F will be false no matter what x is. It follows that =. Proof. Suppose F =. Suppose x F. By the definition of x F, this is equivalent to A(A F x A). Since F =, it follows that A(A x A). But then A is false no matter what A is. Therefore, the whole statement will be always false no matter what x is. Then it follows that x F which is, in this case, x, and thus =. (b) Show that if F =, then the statement x F will be true no matter what x is. In a context in which it is clear what the universe of discourse U is, we might therefore want to say that = U. However, this has the unfortunate consequence that the notation will mean different things in different contexts. Furthermore, when working with sets whose elements are sets, mathematicians often do not use a universe of discourse at all. (For more on this, see the next exercise.) For these reasons, some mathematicians consider the notation to be meaningless. We will avoid this problem in this book by using the notation F only in contexts in which we can be sure that F. Proof. Suppose F =. Suppose x F. By the definition of x F, this is equivalent to A(A F x A). By the conditional law, this is equivalent to A(A / F x A). Since F =, it follows that A(A / x A). But then A / is true no matter what A is no matter what x is. Therefore, the entire statement will always be true. 11

12 11. In Section 2.3 we saw that a set can have other sets as elements. When discussing sets whose elements are sets, it might seem most natural to consider the universe of discourse to be the collection of all sets. However, as we will see in this problem, assuming that there is suc a universe leads to contradictions. Suppose U were the collection of all sets. Note that in particular U is a set, so we would have U U. This is not yet a contradiction; although most sets are not elements of themselves, perhaps some sets are elements of themselves. But it suggests that the sets in the universe U could be split into two categories: the unusual sets that, like U itself, are elements of themselves, and the more typical sets that are not. Let R be the set of sets in the second category. In other words, R = {A U A A}. This means that for any set A in the universe U, A will be an elements of R iff A / A. In other words, we have A U(A R A / A). (a) Show that applying this last fact to the set R itself (in other words, plugging in R for A) leads to a contradiction. This contradiction was discovered by Bertrand Russell in 1901, and is known as Russell s Paradox. Proof. Suppose A U(A R A / A). Applying this to the set R, we get R U(R R R / R). ( ) Suppose R U(R R R / R). This means that if R R, R / R, which is clearly a contradiction. ( ) Suppose R U(R R R / R). This means that if R / R, R R, which, again, is a contradiction. (b) Think some more about the paradox in part (a). What do you think it tells us about sets? 12

13 The statement the sets in the universe U could be split into two categories: the unusual sets that, like U itself, are elements of themselves, and the more typical sets that are not is not true. 13