Chapter 3 — Sets

We use collections all the time: a cricket team, the vowels of the alphabet, the rivers of Telangana, the even numbers. In mathematics a collection is called a set only when it is well defined, which means that for any object whatsoever we can decide definitely whether it belongs to the collection or not. There must be no room for opinion.

The collection of vowels in the English alphabet is a set, because every letter is either a vowel or not: a, e, i, o, u are in, every other letter is out. The collection of even natural numbers less than 20 is a set. The collection of months of the year beginning with J is a set: January, June, July. The collection of students in your class who are taller than 150 cm is a set, because a measurement decides membership.

On the other hand, the collection of intelligent students in your class is not a set, because intelligence is a matter of opinion: one teacher might include a student that another leaves out. The collection of beautiful flowers, of difficult chapters in this book, of good cricketers, of tasty dishes: none of these is a set. Recognising this distinction is the first exercise of the chapter, and examination questions ask it directly.

The objects in a set are called its elements or members. Sets are usually named by capital letters A, B, C, ... and elements by small letters. The symbol ∈ means is an element of and ∉ means is not an element of. If V is the set of vowels, then a ∈ V and b ∉ V. If N is the set of natural numbers, 5 ∈ N but −5 ∉ N and 1/2 ∉ N.

Some standard sets have fixed names: N for the natural numbers {1, 2, 3, ...}, W for the whole numbers {0, 1, 2, 3, ...}, Z for the integers {..., −2, −1, 0, 1, 2, ...}, Q for the rational numbers and R for the real numbers. These letters are reserved and should not be used for other sets.

Two facts about elements matter. First, order does not matter: {1, 2, 3} and {3, 1, 2} are the same set. Second, repetition does not matter: the set of letters in the word MISSISSIPPI is {M, I, S, P}, with each letter listed once. A set is determined entirely by which objects belong to it, not by how they are arranged or how often they are mentioned.

There are two standard ways of writing a set.

Roster form (also called list form or tabular form) lists all the elements inside curly brackets, separated by commas. The set of odd natural numbers less than 10 is {1, 3, 5, 7, 9}. The set of letters in the word MATHEMATICS is {M, A, T, H, E, I, C, S}. When a set is large but its pattern is clear, we may use three dots: the set of natural numbers up to 100 is {1, 2, 3, ..., 100}, and the set of all even natural numbers is {2, 4, 6, ...}. The dots must only be used when the pattern is unmistakable.

Set-builder form describes the elements by a property they share, instead of listing them. It is written {x : x has the property} or {x | x has the property}, read as the set of all x such that x has the property. The set {1, 3, 5, 7, 9} in set-builder form is {x : x is an odd natural number less than 10} or, more compactly, {x : x = 2n − 1, n ∈ N, n ≤ 5}. The set of vowels is {x : x is a vowel of the English alphabet}. The set of real numbers between 0 and 1 cannot be listed at all, so set-builder form {x : x ∈ R, 0 < x < 1} is the only way to write it.

Converting from set-builder to roster. Take the property, test each candidate, and list those that pass. {x : x is a natural number and 3 < x < 10} = {4, 5, 6, 7, 8, 9}. {x : x is a prime number that divides 60} = {2, 3, 5}. {x : x is a two-digit number whose digits add to 8} = {17, 26, 35, 44, 53, 62, 71, 80}. {x : x ∈ Z, x² = 25} = {−5, 5}. {x : x ∈ N, x² = 25} = {5}. Notice that the same property gives different sets depending on the stated domain, so read the domain carefully.

Converting from roster to set-builder. Look for the rule. {1, 4, 9, 16, 25} = {x : x = n², n ∈ N, n ≤ 5}. {1/2, 2/3, 3/4, 4/5} = {x : x = n/(n + 1), n ∈ N, n ≤ 4}. {2, 3, 5, 7, 11} = {x : x is a prime number less than 12}. {Monday, Tuesday, ..., Sunday} = {x : x is a day of the week}. {0, 5, 10, 15, ...} = {x : x is a multiple of 5 together with 0} = {5n : n ∈ W}.

Each form has its use. Roster form shows the elements at a glance and is best for small sets; set-builder form captures infinite sets and shows the reason the elements belong together. Examination questions ask for both conversions, and the one-mark questions frequently ask for the roster form of a set given by a slightly tricky rule, such as {x : x is a letter in the word COMMITTEE}, whose answer {C, O, M, I, T, E} lists each letter once.

The empty set. A set with no elements at all is called the empty set or null set. It is written ∅ or { }. Examples: the set of natural numbers less than 1; the set of odd numbers divisible by 2; the set of triangles with four sides; {x : x ∈ N, x + 5 = 3}; the set of months with 32 days. Each of these descriptions is perfectly well defined, and in each case nothing satisfies it. There is only one empty set, whatever description produced it.

A frequent error is to write {0} or {∅} for the empty set. The set {0} has one element, the number 0, so it is not empty. The set {∅} has one element, the empty set itself, so it is not empty either. Only ∅ or { } denotes a set with nothing in it.

Finite sets. A set is finite if we can count its elements and the counting comes to an end. The set of letters of the English alphabet is finite (26 elements). The set of districts of Telangana is finite. The set {x : x ∈ N, x < 10⁹} is finite even though it is very large. The empty set is finite, with 0 elements. The number of elements in a finite set A is called its cardinal number, written n(A). So if A = {2, 4, 6, 8}, then n(A) = 4, and n(∅) = 0.

Infinite sets. A set is infinite if the counting never ends. N, W, Z, Q and R are all infinite. The set of multiples of 7 is infinite. The set of points on a line segment is infinite. The set of rational numbers between 0 and 1 is infinite, even though the interval is short. A set given in roster form with three dots and no final element, such as {5, 10, 15, ...}, is infinite.

Deciding finite versus infinite sometimes needs thought. {x : x ∈ N, x is a multiple of 5 and x < 100} is finite (19 elements). {x : x ∈ N, x is a multiple of 5} is infinite. {x : x ∈ R, 0 ≤ x ≤ 1} is infinite. {x : x ∈ Z, −100 ≤ x ≤ 100} is finite (201 elements). The set of prime numbers is infinite, a fact proved by Euclid.

Equivalent sets. Two finite sets are equivalent if they have the same number of elements, even if the elements are different. {a, b, c} and {1, 2, 3} are equivalent because both have 3 elements. This is different from equal sets, which must have the very same elements, as the next section explains.

Equal sets. Two sets A and B are equal, written A = B, if they have exactly the same elements: every element of A is in B and every element of B is in A. Since order and repetition do not matter, {1, 2, 3} = {3, 2, 1} = {1, 1, 2, 3}. The set of letters in FOLLOW and the set of letters in WOLF are equal: both are {F, O, L, W}. The set {x : x is a prime number less than 10} and the set {2, 3, 5, 7} are equal. But {1, 2, 3} and {1, 2, 4} are not equal, and {a, b, c} and {1, 2, 3} are not equal though they are equivalent.

Subsets. A set A is a subset of a set B, written A ⊂ B, if every element of A is also an element of B. The set of vowels {a, e, i, o, u} is a subset of the set of all letters. {2, 4} ⊂ {1, 2, 3, 4}. N ⊂ W ⊂ Z ⊂ Q ⊂ R. If A is not a subset of B, we write A ⊄ B; for example {1, 5} ⊄ {1, 2, 3, 4} because 5 is missing from the second set.

Two consequences follow from the definition. First, every set is a subset of itself, since every element of A is certainly in A. Second, the empty set is a subset of every set, because there is no element of ∅ that could fail to be in B. So for any set A, both ∅ ⊂ A and A ⊂ A.

Equality through subsets. A = B if and only if A ⊂ B and B ⊂ A. This is the standard way to prove two sets equal: show each contains the other.

Listing all subsets. The subsets of {1, 2} are ∅, {1}, {2}, {1, 2}: four subsets. The subsets of {a, b, c} are ∅, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}: eight subsets. In general a set with n elements has 2ⁿ subsets, because each element is either included or not, giving two choices per element. A set with 4 elements has 16 subsets and a set with 5 has 32. A subset other than the set itself is called a proper subset; a set with n elements has 2ⁿ − 1 proper subsets.

Care with element versus subset. If A = {1, {2}, 3}, then 1 ∈ A, {2} ∈ A, 2 ∉ A, {1} ⊂ A, {{2}} ⊂ A, and {2} ⊄ A (because 2 is not an element of A). Such questions test whether you distinguish the object 2 from the set {2}. Read each symbol slowly.

Universal set. In any discussion, all the sets under consideration are subsets of one fixed larger set, called the universal set and denoted μ (in this syllabus) or U. If we are talking about even numbers and prime numbers, the universal set might be N. If we are talking about the boys and girls of a class, the universal set is the set of all students of that class. The choice of universal set depends on the problem, and it must be stated or understood before complements can be found.

Venn diagrams. A Venn diagram is a picture in which the universal set is drawn as a rectangle and each set inside it as a closed curve, usually a circle. Elements may be written inside the appropriate regions. Venn diagrams make relationships between sets visible at a glance and are used throughout the rest of the chapter.

To draw the Venn diagram of A = {1, 2, 3} inside μ = {1, 2, 3, 4, 5, 6}: draw the rectangle, write μ at its corner, draw a circle labelled A, write 1, 2, 3 inside the circle and 4, 5, 6 inside the rectangle but outside the circle.

Subsets in a Venn diagram. If A ⊂ B, the circle for A is drawn entirely inside the circle for B. The picture of N ⊂ W ⊂ Z ⊂ Q ⊂ R is a nest of five curves, one inside the other.

Overlapping sets. If A and B share some elements but neither is inside the other, the circles are drawn overlapping. The region where they overlap holds the common elements. For A = {1, 2, 3, 4} and B = {3, 4, 5, 6}, the overlap contains 3 and 4; the crescent of A alone contains 1 and 2; the crescent of B alone contains 5 and 6.

Disjoint sets. If A and B have no element in common, the circles are drawn apart, not touching. The set of even numbers and the set of odd numbers are disjoint.

Reading a Venn diagram. A two-circle diagram inside a rectangle has four regions: inside A only, inside B only, inside both, and inside neither. Every element of the universal set lies in exactly one of these four regions. Counting problems later in the chapter come down to filling in the number of elements in each region, always starting from the overlap and working outwards.

When drawing for an examination, label the rectangle with μ, label each circle with its set name, and if elements are given, write every element of the universal set somewhere in the picture, including those outside all the circles. A Venn diagram that omits the elements outside the circles is incomplete and loses a mark.

The union of two sets A and B, written A ∪ B and read A union B, is the set of all elements that are in A, or in B, or in both. In set-builder form, A ∪ B = {x : x ∈ A or x ∈ B}. The word or here is inclusive: an element in both sets belongs to the union, but it is listed only once.

Examples. If A = {2, 5, 6, 8} and B = {5, 7, 9, 1}, then A ∪ B = {1, 2, 5, 6, 7, 8, 9}. The element 5 appears in both but is written once. If A = {a, e, i, o, u} and B = {a, i, u}, then A ∪ B = {a, e, i, o, u} = A, because B is a subset of A. If A is the set of even natural numbers and B is the set of odd natural numbers, A ∪ B = N.

Venn diagram. In a diagram of two overlapping circles, A ∪ B is the entire region covered by either circle, shaded as one connected shape. If the sets are disjoint, the union is the two separate circles together. If A ⊂ B, the union is just the larger circle B.

Properties. The union is commutative: A ∪ B = B ∪ A, because or works in either order. It is associative: (A ∪ B) ∪ C = A ∪ (B ∪ C), so A ∪ B ∪ C is unambiguous. A ∪ A = A. A ∪ ∅ = A, since the empty set adds nothing. A ∪ μ = μ, since everything is already in the universal set. If A ⊂ B, then A ∪ B = B. Both A and B are subsets of A ∪ B.

Union of three sets. If A = {1, 2, 3}, B = {3, 4, 5} and C = {5, 6}, then A ∪ B ∪ C = {1, 2, 3, 4, 5, 6}. Work in stages: A ∪ B = {1, 2, 3, 4, 5}, then join C.

Word interpretation. If A is the set of students who play cricket and B the set who play football, then A ∪ B is the set of students who play at least one of the two games. Whenever a question says at least one, either, or, or any of, the union is meant.

Counting the union. The number of elements in the union is not simply n(A) + n(B), because the common elements would be counted twice. In the first example n(A) = 4, n(B) = 4, but n(A ∪ B) = 7, not 8, because 5 is shared. The exact rule is developed later in the chapter after intersection is defined.

A common slip is to list a repeated element twice, writing {1, 2, 5, 5, 6, 7, 8, 9}. Since a set has no repeated elements, this is wrong, and in an examination the answer would be marked incorrect.

The intersection of two sets A and B, written A ∩ B and read A intersection B, is the set of all elements that are in both A and B. In set-builder form, A ∩ B = {x : x ∈ A and x ∈ B}. The word and demands membership of both sets.

Examples. If A = {5, 6, 7, 8} and B = {7, 8, 9, 10}, then A ∩ B = {7, 8}. If A = {1, 2, 3, 4, 5} and B = {2, 4, 6, 8}, then A ∩ B = {2, 4}. If A is the set of multiples of 4 and B the set of multiples of 6, then A ∩ B is the set of multiples of 12, the common multiples. If A = {a, b, c} and B = {d, e}, then A ∩ B = ∅: the sets have nothing in common.

Venn diagram. A ∩ B is the lens-shaped region where the two circles overlap; only this region is shaded. If the sets are disjoint, there is no overlap and nothing is shaded. If A ⊂ B, the intersection is the smaller circle A.

Properties. Intersection is commutative, A ∩ B = B ∩ A, and associative, (A ∩ B) ∩ C = A ∩ (B ∩ C). A ∩ A = A. A ∩ ∅ = ∅, since nothing can be in the empty set. A ∩ μ = A, since every element of A is in the universal set. If A ⊂ B, then A ∩ B = A. The intersection A ∩ B is a subset of both A and of B.

Comparing the two operations. The union collects, the intersection filters. A ∩ B is always a subset of A ∪ B. For the sets A = {5, 6, 7, 8} and B = {7, 8, 9, 10}, A ∩ B = {7, 8} while A ∪ B = {5, 6, 7, 8, 9, 10}. When the sets are equal, union and intersection coincide with the set itself.

Distributive laws. Intersection distributes over union and union over intersection: A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) and A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C). These can be verified with a Venn diagram or with specific sets. Take A = {1, 2, 3, 4}, B = {2, 4, 6}, C = {3, 4, 5}. Then B ∪ C = {2, 3, 4, 5, 6} and A ∩ (B ∪ C) = {2, 3, 4}. On the other side, A ∩ B = {2, 4}, A ∩ C = {3, 4}, and their union is {2, 3, 4}. The two sides agree.

Word interpretation. If A is the set of students who like tea and B those who like coffee, A ∩ B is the set who like both. Questions that say both, common, and, or as well as are asking for the intersection.

Intersection of three sets. A ∩ B ∩ C is the set of elements common to all three; in a three-circle Venn diagram it is the small central region where all three circles overlap.

Two sets are disjoint if they have no element in common, that is, if A ∩ B = ∅. The set of even numbers and the set of odd numbers are disjoint. The set of vowels and the set of consonants are disjoint. The set of boys and the set of girls in a class are disjoint. The sets {1, 3, 5} and {2, 4, 6} are disjoint. The sets {x : x ∈ N, x < 5} and {x : x ∈ N, x > 5} are disjoint, and note that neither contains 5 itself.

In a Venn diagram, disjoint sets are drawn as two circles that do not overlap. This is the only case in which the circles are separated; if there is any possibility of common elements, the circles are drawn overlapping even if the overlap turns out to be empty in a particular case.

Deciding disjointness. Test whether any element belongs to both. A = {2, 3, 5, 7} and B = {4, 6, 8, 9}: no common element, so disjoint. A = {x : x is a multiple of 3} and B = {x : x is a multiple of 5}: 15 belongs to both, so not disjoint. A = {x : x is a prime} and B = {x : x is even}: 2 belongs to both, so not disjoint, even though they look unrelated. A = {x : x² = 4} and B = {x : x² = 9}: A = {−2, 2}, B = {−3, 3}, disjoint.

Union of disjoint sets. When A and B are disjoint, the union is easy to count: n(A ∪ B) = n(A) + n(B), because no element is double counted. In a class of 40 with 22 boys and 18 girls, the sets of boys and girls are disjoint and 22 + 18 = 40. This is the special case of the general formula in the last section of the chapter.

Partitions. If several sets are pairwise disjoint and together make up the universal set, they are said to partition it. The sets of even and odd natural numbers partition N. The sets of numbers of the form 3q, 3q + 1 and 3q + 2 partition the positive integers, which is the idea behind Euclid's division lemma. A set and its complement, defined in the next section, always partition the universal set.

Disjoint does not mean unrelated. Disjoint sets can be closely related. The set of natural numbers less than 100 and the set of natural numbers greater than 100 are disjoint, yet together with {100} they make up N. Disjointness is only a statement about shared elements.

Questions on disjoint sets usually appear as one-mark items: state whether the given pair is disjoint, with a reason, or give an example of two disjoint sets. Always justify with the intersection: A ∩ B = ∅, therefore disjoint; or name a common element, therefore not disjoint.

The difference of two sets A and B, written A − B and read A minus B, is the set of elements that are in A but not in B. In set-builder form, A − B = {x : x ∈ A and x ∉ B}. It removes from A everything that B contains.

Examples. If A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7}, then A − B = {1, 2, 3}, the elements of A that are not in B. And B − A = {6, 7}, the elements of B not in A. Notice that A − B and B − A are different sets. If A = {a, b, c, d} and B = {b, d}, then A − B = {a, c} and B − A = ∅. If A and B are disjoint, A − B = A and B − A = B, since there is nothing to remove.

Venn diagram. A − B is the crescent of A that lies outside B: the part of circle A not covered by circle B. B − A is the other crescent. The three regions A − B, A ∩ B and B − A are pairwise disjoint and together make up A ∪ B.

Properties. Difference is not commutative: in general A − B ≠ B − A; they are equal only when both are empty, which happens when A = B. A − A = ∅. A − ∅ = A. ∅ − A = ∅. A − B = A exactly when A and B are disjoint. A − B = ∅ exactly when A ⊂ B. Also A − B, A ∩ B and B − A are pairwise disjoint, and (A − B) ∪ (A ∩ B) = A.

Difference with a subset. If B ⊂ A, then A − B is the part of A left after removing B. With A = {1, 2, ..., 10} and B = {2, 4, 6, 8, 10}, A − B = {1, 3, 5, 7, 9}. This is the pattern for complements in the next section.

Word interpretation. If A is the set of students who play cricket and B those who play football, A − B is the set who play cricket but not football, and B − A is the set who play football but not cricket. The phrases only, but not, and except signal a difference.

Computing with three sets. With A = {1, 2, 3, 4, 5, 6}, B = {2, 4, 6, 8} and C = {3, 4, 5, 6}: A − (B ∪ C): B ∪ C = {2, 3, 4, 5, 6, 8}, so A − (B ∪ C) = {1}. (A − B) ∩ (A − C): A − B = {1, 3, 5}, A − C = {1, 2}, intersection {1}. The two agree, illustrating the law A − (B ∪ C) = (A − B) ∩ (A − C).

The examination often asks for A − B and B − A from given sets, and for a Venn diagram of each. Shade only the crescent; a common error is to shade the whole of A.

Once a universal set μ is fixed, every set A inside it has a complement: the set of all elements of μ that are not in A. It is written A' (read A dash or A complement) and sometimes A^c. In set-builder form, A' = {x : x ∈ μ and x ∉ A}, which is the same as μ − A.

Examples. If μ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and A = {2, 4, 6, 8, 10}, then A' = {1, 3, 5, 7, 9}. If μ is the set of letters of the alphabet and V is the set of vowels, then V' is the set of consonants. If μ = N and E is the set of even natural numbers, then E' is the set of odd natural numbers. If μ is the set of students of a school and A is the set of Class 10 students, A' is the set of students not in Class 10.

The complement depends entirely on the universal set. The complement of {1, 2} in μ = {1, 2, 3} is {3}, but in μ = {1, 2, 3, 4, 5} it is {3, 4, 5}. A question that asks for a complement without stating μ is incomplete; in the examination, μ is always given.

Venn diagram. A' is everything inside the rectangle but outside the circle A. To shade A', shade the rectangle and leave the circle blank.

Properties. A ∪ A' = μ: every element is either in A or not. A ∩ A' = ∅: nothing is both in A and not in A. (A')' = A: the complement of the complement is the original set. μ' = ∅ and ∅' = μ. If A ⊂ B, then B' ⊂ A' (the inclusion reverses). And n(A') = n(μ) − n(A) for finite sets.

De Morgan's laws. (A ∪ B)' = A' ∩ B' and (A ∩ B)' = A' ∪ B'. In words: not (A or B) is the same as (not A) and (not B); not (A and B) is the same as (not A) or (not B). Verify with μ = {1, 2, 3, 4, 5, 6}, A = {1, 2, 3}, B = {3, 4}. A ∪ B = {1, 2, 3, 4}, so (A ∪ B)' = {5, 6}. A' = {4, 5, 6}, B' = {1, 2, 5, 6}, so A' ∩ B' = {5, 6}. Agreement. Similarly A ∩ B = {3}, (A ∩ B)' = {1, 2, 4, 5, 6}, and A' ∪ B' = {1, 2, 4, 5, 6}.

Word interpretation. If A is the set of students who passed, A' is the set who did not pass. Neither ... nor corresponds to (A ∪ B)', the students who did neither activity. Not both corresponds to (A ∩ B)'.

Difference and complement together. A − B = A ∩ B': the elements of A that are not in B are exactly the elements in both A and B'. This identity lets every difference be rewritten using intersection and complement, which is useful in proofs.

For finite sets there is a simple relation between the number of elements in the union and the numbers in each set. Look at the Venn diagram of two overlapping sets. A ∪ B consists of three disjoint regions: A only, both, and B only. Now n(A) counts A only plus both, and n(B) counts B only plus both. So n(A) + n(B) counts the overlap twice. Subtracting one copy of the overlap gives the correct count:

n(A ∪ B) = n(A) + n(B) − n(A ∩ B).

If A and B are disjoint, n(A ∩ B) = 0 and the formula reduces to n(A ∪ B) = n(A) + n(B).

Verification. A = {1, 2, 3, 4, 5}, B = {4, 5, 6, 7}. n(A) = 5, n(B) = 4, A ∩ B = {4, 5} so n(A ∩ B) = 2, and A ∪ B = {1, 2, 3, 4, 5, 6, 7} so n(A ∪ B) = 7. Check: 5 + 4 − 2 = 7.

Finding a missing quantity. Any one of the four numbers can be found from the other three. If n(A) = 12, n(B) = 15 and n(A ∪ B) = 20, then n(A ∩ B) = 12 + 15 − 20 = 7. If n(A ∩ B) = 5, n(A ∪ B) = 30 and n(A) = 18, then n(B) = 30 − 18 + 5 = 17.

Word problem 1. In a class of 50 students, 30 play cricket, 25 play football and 10 play both. How many play at least one game? n(C ∪ F) = 30 + 25 − 10 = 45. How many play neither? 50 − 45 = 5. How many play only cricket? 30 − 10 = 20. How many play only football? 25 − 10 = 15. Check: 20 + 10 + 15 + 5 = 50.

Word problem 2. In a group of 60 people, 27 like tea, 42 like coffee and each person likes at least one. How many like both? Here n(T ∪ C) = 60, so n(T ∩ C) = 27 + 42 − 60 = 9. How many like only tea? 27 − 9 = 18. How many like only coffee? 42 − 9 = 33.

Word problem 3. In a survey of 100 people, 70 read newspaper X, 60 read newspaper Y and 20 read neither. How many read both? Those who read at least one number 100 − 20 = 80, so n(X ∩ Y) = 70 + 60 − 80 = 50.

The Venn diagram method. Draw two overlapping circles, put the number for both in the overlap first, then fill each crescent by subtraction, then find the outside region by subtracting the total inside from n(μ). Filling from the middle outward avoids double counting. This method answers any question about the situation, not just the one asked.

A frequent error is to add n(A) and n(B) and forget the subtraction, or to subtract the overlap from only one of them. Always write the formula before substituting.

Examination problems often give sets by rules rather than lists, and combine several operations. The safe method is: convert every set to roster form first, then perform the operations one at a time, then draw the Venn diagram if asked.

Problem 1. μ = {x : x ∈ N, x ≤ 12}, A = {x : x is a multiple of 2}, B = {x : x is a multiple of 3}. Find A ∪ B, A ∩ B, A − B, B − A, A' and (A ∩ B)'. First list: μ = {1, ..., 12}, A = {2, 4, 6, 8, 10, 12}, B = {3, 6, 9, 12}. Then A ∪ B = {2, 3, 4, 6, 8, 9, 10, 12}; A ∩ B = {6, 12}; A − B = {2, 4, 8, 10}; B − A = {3, 9}; A' = {1, 3, 5, 7, 9, 11}; (A ∩ B)' = {1, 2, 3, 4, 5, 7, 8, 9, 10, 11}.

Problem 2. A = {x : x is a letter in the word ASSASSINATION}, B = {x : x is a letter in the word STATION}. A = {A, S, I, N, T, O}, B = {S, T, A, I, O, N}. Every element of A is in B and every element of B is in A, so A = B. This kind of question tests the rule that repeated letters count once.

Problem 3. If A = {1, 2, 3, 4}, B = {2, 4, 6, 8} and C = {3, 4, 5, 6}, verify that A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C). Left: B ∩ C = {4, 6}, so A ∪ (B ∩ C) = {1, 2, 3, 4, 6}. Right: A ∪ B = {1, 2, 3, 4, 6, 8}, A ∪ C = {1, 2, 3, 4, 5, 6}, intersection {1, 2, 3, 4, 6}. Equal.

Problem 4. If A ⊂ B, show that A ∩ B = A and A ∪ B = B, with an example. Take A = {1, 2}, B = {1, 2, 3, 4}. A ∩ B = {1, 2} = A and A ∪ B = {1, 2, 3, 4} = B. In general, every element of A is in B, so the common elements are all of A, and joining B to A adds nothing new to B.

Problem 5. Is {x : x ∈ N, x² − 5x + 6 = 0} equal to {2, 3}? Solve: (x − 2)(x − 3) = 0, x = 2 or 3, both natural. Yes, the sets are equal. Is {x : x ∈ N, x² + 1 = 0} empty? x² = −1 has no real solution, so yes, it is ∅.

Problem 6. Write {x : x ∈ Z, −3 < x < 4} in roster form and find its number of subsets. The set is {−2, −1, 0, 1, 2, 3}, 6 elements, so it has 2⁶ = 64 subsets.

Problem 7. State which of the following are true for A = {1, 2, 3} and B = {3, 2, 1, 1}: A = B; A ⊂ B; A and B are disjoint; n(A) = n(B). Since B = {1, 2, 3} after removing repetition, A = B is true, A ⊂ B is true, disjoint is false (they share everything), and n(A) = n(B) = 3 is true.

Keep a checklist for these problems: convert to roster, perform one operation per line, list elements in increasing order without repeats, and for a complement confirm the universal set. Marks are given for the working as well as the final sets.

This chapter introduced the vocabulary of sets and the four operations on them. Here is what to remember and how the SSC paper tests it.

Definitions to know exactly. A set is a well-defined collection. Roster form lists; set-builder form gives the rule. The empty set ∅ has no elements; {0} is not empty. A set is finite if its elements can be counted to an end, otherwise infinite. A = B means the same elements; A ⊂ B means every element of A is in B; a set of n elements has 2ⁿ subsets. The universal set μ holds everything under discussion.

Operations. A ∪ B: in A or B or both. A ∩ B: in both. A − B: in A, not in B. A': in μ, not in A. Disjoint: A ∩ B = ∅.

Laws. Commutative and associative laws for union and intersection; A ∪ ∅ = A, A ∩ ∅ = ∅, A ∪ μ = μ, A ∩ μ = A; distributive laws; De Morgan's laws; A − B = A ∩ B'; A ∪ A' = μ, A ∩ A' = ∅.

Counting. n(A ∪ B) = n(A) + n(B) − n(A ∩ B); n(A') = n(μ) − n(A).

One-mark questions. Write a given set in roster or set-builder form. Is the given collection a set? Is ∅ a subset of {1, 2}? Write all subsets of {a, b}. Give an example of two disjoint sets. State whether {x : x ∈ N, x < 1} is empty. These need one line with a precise reason.

Two-mark questions. Given A and B, find A ∪ B and A ∩ B, or A − B and B − A. Draw the Venn diagram of A ∩ B. If A = {1, 2, 3} and B = {3, 4}, verify that A − B ≠ B − A. Show that if A ⊂ B then A ∪ B = B.

Four-mark questions. Given μ and two or three sets by rules, find several operations and draw the Venn diagram with every element placed. Verify a distributive law or a De Morgan law for given sets. Solve a counting word problem with a Venn diagram showing all four regions.

Common mistakes. Repeating an element in a set. Confusing ∈ with ⊂. Writing {0} for the empty set. Shading the whole of A instead of the crescent for A − B. Forgetting the elements outside the circles in a Venn diagram. Forgetting to subtract the overlap in the counting formula. Not stating the universal set when finding a complement.

Revision routine. Redraw the four basic Venn diagrams (union, intersection, difference, complement) from memory. Recite the counting formula and solve one word problem. Convert three sets between roster and set-builder form. Verify one De Morgan law with sets of your own choosing.

Looking ahead. The language of sets returns in the chapters on polynomials (the set of zeroes), coordinate geometry (the set of points on a line), probability (the sample space and events, where an event is a subset of the sample space) and statistics, and in Intermediate mathematics it becomes the basis of relations and functions.