SET THEORY - ASSIGNMENT 1 (1) Find the size of each of the following sets. (a) The even integers greater than 4 and smaller than 9. (b) P(f2; 3; 4g). (c) f1; f2; 3; 4g; 5; 6g. (d) f5; 6; 7g × f7; 8g. (e) P(P(f0g)). Solution. (a) 4. (b) 8. (c) 4 (note that f2; 3; 4g only counts for a single object. (d) 6. (e) jP(P(f0g))j = 221 = 4. (2) Let A = f1; 2; 3; 4; 5; 6g;B = f3; 4; 5; 6; 7; 8g;C = f1; 3; 5; 7; 9g. Find A [ B [ C; (A [ B) \ C; B n A; (A \ C) n (C n A); (A \ B) × (B \ C). Solution. A [ B [ C = f1; 2; 3; 4; 5; 6; 7; 8; 9g; (A [ B) \ C = f1; 3; 5; 7g;B n A = f7; 8g; (A \ C) n (C n A) = A \ C = f1; 3; 5g; (A \ B) × (B \ C) = f3; 4; 5; 6g × f3; 5; 7g = f(3; 3); (3; 5); (3; 7); (4; 3); (4; 5); (4; 7); (5; 3); (5; 5); (5; 7); (6; 3); (6; 5); (6; 7)g. (3) Determine whether each of the following statements is true or false. (a)0 2 f0; 1; 2g. (b)0 2 ;. (c)5 ⊆ f8; 5; 9g. (d) f2; 3g ⊆ Z n fZ n fThe even integersgg: Solution. (a)0 2 f0; 1; 2g. Yes. (b)0 2 ;. No, the empty set doesn't contain any object. (c)5 ⊆ f8; 5; 9g. No, 5 is a member of the set, but not a subset of it. (d) f2; 3g ⊆ Z n fZ n fThe even integersgg: The set on the right is just the odd integers. Since 2 isn't odd, the statement is false. (4) The following Venn diagram describes the TV preferences of students in a certain school. The set A describes those who like Atypical, the set R describes those who like Riverdale, and the set S describes those who like Stranger Things. What do the sets (S \ R) n A and (R [ A) n (R \ A) describe? What is the size of each set? 1 SET THEORY - ASSIGNMENT 1 2 10 5 4 A 6 7 R S 8 Solution. (S \ R) n A is the collection of students that watch Stranger things and Riverdale, but not Atypical. The size of this set is 7. (R [A)n(R \A) is the describes the students watching either Atypical or Riverdale but not both (in other words, it is the exclusive or between the two sets). There are 21 students in that set (5) Draw a Venn diagram that is consistent with the following data. • All cats are mammals. • Some mammals like carrots. • Some of the animals that like carrots are not mammals. Solution. There is more than one solution to this question. One possible solu- tion is Mammals Cats Carrot eaters (6) Show that for any two sets M and N, we have M ⊆ N if and only if M [N = N. Solution. If M ⊆ N then every element in M is already in N. Therefore, by taking the union of N with M, no new elements are added, so N [ M = N. On the other hand, suppose that M [ N = N. Let x 2 M. We need to show that x 2 N. If that was not true, then by the assumption N [M = N, it would follow that x is not in M [ N, and in particular not in M. That's a contradiction, so x must be in N. (7) Suppose that A; B ⊆ C. Can we conclude that C n A contains B? If it's true, then prove it. Otherwise, what other condition do we need to add to make it true? Solution. The statement is false. For example, let C = f1; 2; 3g;B = f1; 2g;A = f2; 3g. Then C n A = f1g does not contain B. However, if we add the condition that A \ B = ;, then the statement becomes correct. SET THEORY - ASSIGNMENT 1 3 (8) Show that for any three sets A; B; C, we have A \ (B [ C) = (A \ B) [ (A \ C). Solution. Let x 2 A \ (B [ C). We need to show that x 2 (A \ B) [ (A \ C). By assumption, x is both in A, and in either B or C. If x is a member of B, then it is in (A \ B). If x is a member of C, then it is in (A \ C). In any case, x is in (A \ B) [ (A \ C). Now, suppose that x 2 (A \ B) [ (A \ C). If x 2 (A \ B), then in particular it is in A and in B [ C, so x 2 A \ (B [ C). Similarly, if x 2 (A \ C) then x is in both A and B [ C, so it is in A \ (B [ C), and we are done. (9) Show that for any three sets A; B; C, we have A n (B \ C) = (A n B) [ (A n C). Solution. Let x 2 A n (B \ C). Then x 2 A, and x2 = B \ C. Therefore, x is either not in B or not int C. In particular, it is in the union of (A n B)[ and (A n C). Let x 2 (A n B) [ (A n C). Then x is either in (A n B) or in (A n C). If it is in A n B, then it is in A and not in B. In particular, it is not in B \ C, so it is in A n B \ C. A similar argument works if x is in A n C..