Exercise practice: Sets

5.1 How the exercises are laid out

Exercise 1.1 covers representation, Exercise 1.2 empty, finite, infinite and equal sets, Exercise 1.3 subsets, intervals and universal sets, Exercise 1.4 union, intersection and difference, and Exercise 1.5 complement and De Morgan laws. The miscellaneous exercise mixes them and adds set algebra proofs.

This page works one representative question from each section. Solve the rest from the book with the same steps; the pattern repeats and the marks come from clear working rather than a clever shortcut.

5.2 Exercise 1.1: representation

Question 1 asks which collections are sets. The months beginning with J pass the test, because each month can be checked. The ten most talented writers fail it, because talent has no fixed test. Your class roll gives a set, provided the roll itself is fixed. The collection of most dangerous animals fails for the same reason as the writers.

Question 3(i) hides a boundary trap. Set A holds the integers x that satisfy -3 <= x < 7, so the roster runs from -3 up to 6 and stops before 7. Note that -3 is taken in because of the weak inequality and 7 is left out because of the strict one.

Question 4(ii) asks for the set-builder form of the list 2, 4, 8, 16, 32. Each entry is a power of 2 from the first to the fifth, so one correct answer is the collection of 2 to the power n with n in N and 1 <= n <= 5. Another correct answer stops at n < 6. Multiple valid rules are accepted as long as they produce exactly this list.

5.3 Exercise 1.2: null, finite and equal sets

Question 1 asks which collections are null. The odd natural numbers divisible by 2 form the empty set, because a number divisible by 2 is even. The set of even prime numbers is {2}, so it is not empty. The set {x : x is natural, x < 5 and x > 7} is empty, because no natural number can satisfy both inequalities. The points common to two parallel lines form the empty set as well.

Question 6 gives eight sets named A to H and asks which pairs are equal. Sorting each roster settles the question. Two pairs coincide: B matches D, and E matches G. The remaining sets differ in at least one member, so they stay unequal.

Equal-set questions are decided by membership, not by the way the set is described. Once both sides are in roster form, sorting settles the question in one line.

5.4 Exercise 1.3: subsets, intervals and universal sets

Question 3 uses A = {1, 2, {3, 4}, 5}. The statement {3, 4} in A is correct, because the set {3, 4} is listed as a single element. The statement {3, 4} subset of A is wrong, because 3 alone is not an element of A. Likewise {1, 2, 5} subset of A is correct, while {1, 2, 5} in A is wrong because that set is not an element. The empty set is a subset of A but not an element of A.

Question 4 asks for all subsets of {a}, {a, b}, {1, 2, 3} and phi. The counts are 2, 4, 8 and 1. For {1, 2, 3} the eight subsets are phi, the three singletons, the three pairs and {1, 2, 3}. Listing them by size keeps the count honest.

Question 5 converts set-builder descriptions to intervals. The set {x : x is real and -4 < x <= 6} becomes (-4, 6], and {x : x is real and -12 < x < -10} becomes (-12, -10). Question 8 asks for a universal set for A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8}. The set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} works, and so does any larger set containing 0 to 8, but {0, 1, 2, 3, 4, 5, 6} fails because it misses 7 and 8.

5.5 Exercise 1.4: union, intersection and difference

Question 6 uses A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C = {11, 13, 15} and D = {15, 17}. Then A intersection (B union C) is computed inside out. B union C is {7, 9, 11, 13, 15}, and overlapping it with A leaves {7, 9, 11}. The expression A intersection C intersection D is empty, because 11 is in A and C but not in D.

Question 9 asks for differences among A = {3, 6, 9, 12, 15, 18, 21}, B = {4, 8, 12, 16, 20}, C = {2, 4, 6, 8, 10, 12, 14, 16} and D = {5, 10, 15, 20}. A - B keeps multiples of 3 that are not multiples of 4, so it is {3, 6, 9, 15, 18, 21}. B - C is {20}, because 4, 8, 12 and 16 also lie in C. C - A leaves {2, 4, 8, 10, 14, 16}.

Question 11 asks for R - Q. The real numbers that are not rational are exactly the irrational numbers, so R - Q = T.

5.6 Exercise 1.5: complement and De Morgan

Question 1 uses U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4}, B = {2, 4, 6, 8} and C = {3, 4, 5, 6}. The complement of A is {5, 6, 7, 8, 9}. A union C is {1, 2, 3, 4, 5, 6}, so its complement is {7, 8, 9}. The complement of B - C first needs B - C, which is {2, 8}, and then the complement is {1, 3, 4, 5, 6, 7, 9}.

Question 4 asks for a verification of both De Morgan laws with U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8} and B = {2, 3, 5, 7}. Compute A prime = {1, 3, 5, 7, 9} and B prime = {1, 4, 6, 8, 9}. Then A prime intersection B prime is {1, 9}, and A union B is {2, 3, 4, 5, 6, 7, 8} whose complement is also {1, 9}. The second law is checked the same way with A intersection B = {2}. Its complement is {1, 3, 4, 5, 6, 7, 8, 9}, which equals A prime union B prime.

Question 6 sets the universal frame as every triangle in a plane. A holds those triangles that have at least one angle not equal to 60. Whatever lies outside A has no angle different from 60, so all three angles are 60. That is exactly the family of equilateral triangles.

5.7 Miscellaneous exercise: the two proofs

Question 5 asks: if A is a subset of B, show that C - B is a subset of C - A. Take x in C - B. Then x is in C and not in B. Since A lies inside B, any member of A is also a member of B. If x were in A it would be in B, which contradicts x not in B. So x is not in A, and therefore x lies in C - A. The two inclusions are not needed here because the direction is fixed by the choice of x.

Question 3 asks: if A union B = A union C and A intersection B = A intersection C, show that B = C. Take x in B and argue by cases. If x is in A as well, then x lies in A intersection B, hence in A intersection C, so x is in C. If x is outside A, then x lies in A union B, hence in A union C; since x is not in A, x must be in C. So B is contained in C, and the same argument with the sets swapped gives C contained in B. Therefore B = C.

5.8 Mixed practice from the exercises

Three quick items cover the shapes that remain.

1. The set of integers whose square is at most 4 is {-2, -1, 0, 1, 2}, because -2 and 2 give 4 while -3 already gives 9.

2. The distinct letters of TRIGONOMETRY are T, R, I, G, O, N, M, E and Y, with the repeated letters written once.

3. The interval (6, 12] means real x with 6 < x <= 12: 12 is in and 6 is out, so set-builder form uses a strict left inequality and a weak right one.