Homework 1 2/7/2018 SOLTIONS Eercise 1. (a) Graph the following sets (i) C = { R in Z} nswer: 0 R (ii) D = {(, y), y in R,, y 2}. nswer: = 2 y y = 2 (iii) C C nswer: y 1
2 (iv) (C C) D nswer: = 2 y y = 2 (v) (C C) D nswer: = 2 y y = 2
3 (vi) (C C) D, where = R 2. nswer: Since (C C) looks like y D looks like = 2 y y = 2 we have (C C) D looks like = 2 y y = 2 (vii) ((C C) D), where = R 2. nswer: From the picture of (C C) D in part (v), we can see that the picture of ((C C) D) is the same as the one in part (vi). (b) Pick a set two universal sets 1 2 that illustrate that Ā depends on the choice of universal set. nswer: Let = {1}, 1 = {1} 2 = {1, 2, 3}. Then in 1, =, in 2, = {2, 3}.
4 (c) Let be sets contained in a universal set. Decide whether the following identities are true or false. If false, give an eample where the identity doesn t hold. If true, eplain why (in complete sentences). (i) = nswer: TRE. The set is the set elements that are in both, which does not depend on the order of sets considered. (ii) = nswer: TRE. The set is the set elements that are in either or (or both), which does not depend on the order of sets considered. (iii) = nswer: FLSE. For eample, if = {1, 2} = {2, 3}, then = {1} = {3}. (iv) = nswer: FLSE. For eample, if = {1} = {2}, then = {(1, 2)} = {(2, 1)}. (v) = nswer: FLSE. See part (iii) for eample. (vi) If is finite, then so is P() nswer: TRE. If is finite, then P() = 2 < (see the bonus problem). (vii) = ( ) nswer: TRE. If is in in, then is neither in nor in, so is in the complement of, vice versa.
(ONS) se what we learned about Pascal s triangle to eplain why 2 n = n k=0 ( n k) by counting the size of the power set of a set of size n in two ways (the cardinality of a power set of only depends on the cardinality of ). nswer: Let be a set of size n. Step 1: Show P() = 2 n. For each element a, eactly half of the subsets include a: P() = P( {a}) {S {a} S P( {a})}. So P() = 2 P( {a}). Similarly, do the same thing for any element of {a} until you ve ehausted the elements of. For eample, the subsets of = {, y, z} are 5 {S {a} S P( {})} all subsets P( {}) subsets with subsets without with y without y with y without y with z: {, y, z} without z: {, y} with z: {, z} without z: {} with z: {y, z} without z: {y} with z: {z} without z: So P() = } 2 2 {{ 2} = 2 n. terms Step 2: Show P() = n k=0 ( n k). The set of subsets of can be broken into the disjoint union of { subsets of size 0 } (of which there are ( n 0) ), { subsets of size 1 } (of which there are ( n 1) ), { subsets of size 2 } (of which there are ( n 2) ),. { subsets of size n } (of which there are ( n n) ). So there are n k=0 ( n k) subsets of in total. Step 3: conclusion. So for a set of size n, 2 n = P() = n k=0 ( ) n. k.
6 Eercise 2. Fi two sets a universal set. Draw shade in diagrams to decide whether the following identities are true or false. For those that are false, give a concrete eample illustrating the failure of the identity. (a) ( ) = looks like so looks like which is the same as the union of : :
7 (b) ( ) = looks like so looks like which is the same as the intersection of as drawn in part (a). (c) ( ) = ( ( )) ( ) is as drawn in part (b), so ( ) = looks like ( ) On the other h, looks like so ( ) looks like Then subtracting ( ) (drawn in part (b)) gets us back to the picture in ( ).
8 Eercise 3. Let i = {1, 2,..., i} for i = 1, 2, 3,.... Calculate n i n i for n = 2, 4, 5. Make a hypothesis about what n i n i are for general n. Make a hypothesis about what i i are. Eplain in words why. We have 2 i = 1 2 = {1} {1, 2} = {1, 2} = 2, 3 i = 1 2 3 = {1} {1, 2} {1, 2, 3} = {1, 2, 3} = 3, 4 i = 1 4 = {1} {1, 2, 3, 4} = {1, 2, 3, 4} = 4. From this we can conjecture that n i = n. Since for every integer in Z >0, there is a big enough n such that n contains it, we hypothesize that i = Z >0. Net, we have 2 i = 1 2 = {1} {1, 2} = {1} = a 3 i = 1 2 3 = {1} {1, 2} {1, 2, 3} = {1} = 1 4 i = 1 4 = {1} {1, 2, 3, 4} = {1} = 1 From this we can conjecture that n i = 1. ut since 1 is a subset of every finite intersection, we conjecture that i = 1.
Eercise 4. Show that if i = {1, 2,..., i} for i = 1, 2, 3,..., then i = {1}. [Hint: Call i = for brevity. First argue that {1}. For the reverse, eplain why = 1 i=2 i, use that epression to show that {1}.] 9 Proof. Let i =. Since 1 i for all i, we have 1, so that {1}. Net, by the associative identity, we have = i = 1 i. i=2 ut = 1 i=2 i 1. In summary {1} {1}. So = {1}. Eercise 5. rgue formally that =. Proof. First, we ll show that : Let. This equivalent to /. This means that is not in (so that ) is not in (so that ). So. Thus. Net, show that : Let. This equivalent to / /. So is in neither nor, so. Thus. Therefore =.
10 2.1#4 For each of these pairs of sets, determine whether the first is a subset of the second, the second is a subset of the first, or neither is a subset of the other. (a) the set of people who speak English, the set of people who speak English with an ustralian accent: { people who speak English } { people who speak English with an ustralian accent } (b) the set of fruits, the set of citrus fruits: { fruits } { citrus fruits } (c) the set of students studying discrete mathematics, the set of students studying data structures: Normally, neither is a subset of the other. 2.1#10 Determine whether these statements are true or false. (a) { } TRE (b) {, { }} TRE (c) { } { } FLSE (the only element of the RHS is ) (d) { } {{ }} TRE (e) { } {, { }} TRE (f) {{ }} {, { }} TRE (g) {{ }} {{ }, { }} FLSE. e careful: {{ }} = {{ }, { }}, but the symbol means subset but not equal. This is a small bit of trickiness. 2.1#23 How many elements does each of these sets have where a b are distinct elements? (a) P ({a, b, {a, b}}) nswer: {a, b, {a, b}} = 3, so P({a, b, {a, b}}) = 2 3 = 8. (b) P ({, a, {a}, {{a}}}) nswer: {, a, {a}, {{a}}} = 4, so P({, a, {a}, {{a}}}) = 2 4 = 16. (c) P (P ( )) nswer: P ( ) = 1, so P( ) = 2. 2.1#30 Suppose that =, where are sets. What can you conclude? nswer: If both are nonempty, then this isn t possible. So at least one of or must be empty. nd in fact, if is empty, then so is ; similarly so when is empty. 2.2 #1 Let be the set of students who live within one mile of school let be the set of students who walk to classes. Describe the students in each of these sets. (a) ns: Students who both live within a mile of school walk to classes. (b) ns: Students who live within a mile of school /or who walk to classes. (c) ns: Students who live within a mile of school but don t walk to classes. (d) ns: Students who live farther that a mile away from school but who walk to classes anyway.
2.2 #14 Find the sets if = {1, 5, 7, 8}, = {2, 10}, { = {3, 6, 9}. ns: = ( ) ( ) = {1, 3, 5, 6, 7, 8, 9}, = ( ) ( ) = {2, 3, 6, 9, 10}. 11 2.2 #50 Find for the following. i (a) i = {i, i + 1, i + 2,... }: Claim: i = Z >0 i i = Proof. Let = i I = i. To show = Z >0, first we show Z >0 : For each i = 1, 2,..., i Z >0. So their union is also a subset of Z >0. Net, we show Z >0 : For each n Z >0, we have n n. So Z >0. So = Z >0. To show I =, we note that I 1 = Z 1. ut for each n Z 1, we have n is not an element of n+1. So n is not an element of I. Thus I =. (b) i = {0, i}: Claim: i = Z 0 i = {0} Proof. Let = i I = i. To show = Z 0, first we show Z 0 : For each i = 1, 2,..., we have i Z 0. So their union is also a subset of Z 0. Net, we show Z 0 : For each n Z 0, we have n n. Thus Z 0. So = Z >0. To show I = {0}, first we show I {0}: We have that I = 1 2 ( 3 4 ) 1 2. So since 1 2 = {0, 1} {0, 2} = {0}, we have I {0}. Net, we show I {0}: For each i = 1, 2,..., we have 0 i. Thus 0 I, so that I {0}. So I = {0}.
12 (c) i = (0, i): Claim: i = (0, ) i = (0, 1) Proof. Let = i I = i. To show = (0, ), first we show (0, ): For each i = 1, 2, 3,..., we have i (0, ). So the union is also a subset of (0, ). Net, we show (0, ): For each (0, ), let n be an integer greater than. Then n. Thus (0, ). So = (0, ). To show I = (0, 1), first we show I (0, 1): We have that I = 1 2 ( 3 4 ) 1 2. So since 1 2 = (0, 1) (0, 2) = (0, 1), we have I (0, 1). Net, we show I (0, 1): For each i = 1, 2, 3,..., we have (0, 1) i. So (0, 1) I. So I = (0, 1). (d) i = (i, ): Claim: i = (1, ) Proof. Let = i I = i. To show = Z >0, first we show (1, ): For each i = 1, 2, 3,..., we have i (1, ). subset of (1, ). Net, we show (1, ): This is because (1, ) = 1. So = (1, ). i = Therefore their union is also a To show I =, we note that I 1 = (1, ). ut for any (1, ), any integer n greater than, we have / n. So is not in the intersection I.