What this chapter is about
Sets form the foundation of modern mathematics. A set is a well-defined collection of distinct objects, called elements or members. In Class 11, you study sets because nearly every branch of mathematics—relations, functions, probability, sequences—uses the language of sets. Understanding sets now prepares you for all senior secondary mathematics.
This chapter teaches you how to describe a set, how to classify sets by size and content, how to compare sets, and how to combine or separate sets using operations like union, intersection and complement. You also learn to visualise these operations with Venn diagrams and to apply set results in counting problems.
After working through this chapter, you should be able to write any collection in set notation, identify subsets and power sets, perform operations on two or more sets, prove simple set identities, and solve word problems that ask how many elements lie in overlapping groups.
Key ideas
- A set is well-defined when, for any object, we can decide with certainty whether it belongs to the set or not. For example, the collection of all vowels in the English alphabet is well-defined; the collection of tall students in a class is not.
- Two ways to describe a set: roster (or tabular) form lists every element inside braces, such as {a, e, i, o, u}; set-builder form states a property, such as {x : x is a vowel in the English alphabet}.
- A set with no elements is the empty set (or null set), written ∅ or {}. A set with a finite number of elements is a finite set; otherwise it is infinite.
- If every element of set A is also in set B, then A is a subset of B, written A ⊆ B. Every set is a subset of itself, and ∅ is a subset of every set.
- Two sets are equal if and only if each is a subset of the other, that is, A = B when A ⊆ B and B ⊆ A.
- The power set of A, written P(A), is the set of all subsets of A. If A has n elements, P(A) has 2ⁿ elements.
- Union (A ∪ B) contains elements in A or B or both. Intersection (A ∩ B) contains elements in both A and B. Difference (A − B) contains elements in A but not in B. Complement (A′) contains elements in the universal set U but not in A.
- Venn diagrams represent sets as circles inside a rectangle (the universal set) and help visualise operations and relationships.
Formulas and facts to remember
- Formula / Rule: n(A ∪ B) = n(A) + n(B) − n(A ∩ B) · Meaning: Count of union equals sum of counts minus overlap.
- Formula / Rule: n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A ∩ B) − n(B ∩ C) − n(A ∩ C) + n(A ∩ B ∩ C) · Meaning: Inclusion-exclusion for three sets.
- Formula / Rule: n(P(A)) = 2ⁿ where n = n(A) · Meaning: Number of subsets of a set with n elements.
- Formula / Rule: (A′)′ = A · Meaning: Complement of complement returns the original set.
- Formula / Rule: A ∪ A′ = U and A ∩ A′ = ∅ · Meaning: A set and its complement together make U; they share nothing.
- Formula / Rule: De Morgan's laws: (A ∪ B)′ = A′ ∩ B′ and (A ∩ B)′ = A′ ∪ B′ · Meaning: Complement of union is intersection of complements, and vice versa.
- Formula / Rule: A − B = A ∩ B′ · Meaning: Difference can be written as intersection with complement.
Worked examples
Example 1. In a class of 40 students, 25 play cricket and 18 play football. If 8 students play both games, how many play neither?
Step 1: Let C = set of cricket players, F = set of football players. Given n(C) = 25, n(F) = 18, n(C ∩ F) = 8.
Step 2: Use inclusion-exclusion. n(C ∪ F) = 25 + 18 − 8 = 35.
Step 3: Students playing neither = total − n(C ∪ F) = 40 − 35 = 5.
Answer: 5 students play neither game.
Example 2. Write the set A = {x : x is a positive integer less than 10 and divisible by 3} in roster form and find its power set.
Step 1: Identify elements. Positive integers less than 10 divisible by 3 are 3, 6, 9.
Step 2: Roster form: A = {3, 6, 9}.
Step 3: Number of elements n(A) = 3, so n(P(A)) = 2³ = 8.
Step 4: List subsets: ∅, {3}, {6}, {9}, {3, 6}, {3, 9}, {6, 9}, {3, 6, 9}.
Answer: P(A) has 8 subsets as listed.
Example 3. Let U = {1, 2, 3, 4, 5, 6, 7, 8}, A = {1, 3, 5, 7}, B = {2, 3, 5, 8}. Find (A ∪ B)′ and verify De Morgan's law.
Step 1: A ∪ B = {1, 2, 3, 5, 7, 8}.
Step 2: (A ∪ B)′ = U − (A ∪ B) = {4, 6}.
Step 3: A′ = {2, 4, 6, 8}, B′ = {1, 4, 6, 7}.
Step 4: A′ ∩ B′ = {4, 6}.
Since (A ∪ B)′ = A′ ∩ B′ = {4, 6}, De Morgan's law is verified.
Common mistakes
- Writing the empty set as {∅} instead of ∅ or {} → {∅} is a set containing one element (the empty set), not the empty set itself.
- Confusing ∈ (element of) with ⊆ (subset of) → 3 ∈ {1, 2, 3} but {3} ⊆ {1, 2, 3}; an element is not the same as a singleton subset.
- Forgetting to subtract the intersection when counting a union → always use n(A ∪ B) = n(A) + n(B) − n(A ∩ B).
- Assuming A − B = B − A → set difference is not commutative; the two results usually differ.
- Listing duplicate elements in roster form → each element appears only once in a set; {2, 2, 3} should be {2, 3}.
Quick revision
- A set is well-defined; order and repetition do not matter.
- ∅ is a subset of every set; every set is a subset of itself.
- Power set of n elements has 2ⁿ subsets.
- Union adds, intersection keeps common, difference removes.
- De Morgan: complement of union = intersection of complements, and vice versa.
- Inclusion-exclusion prevents double-counting in union problems.