1. Chapter at a Glance
Almost every branch of modern mathematics is written in the language of sets. Developed by the German mathematician Georg Cantor in the late 19th century, set theory gives precise words for "a collection of objects", "belongs to", "is part of", "together" and "in common". These ideas are used immediately in the next chapter (relations and functions), and later in probability, calculus and beyond. This chapter builds the vocabulary, the operations on sets, Venn diagrams and the counting formula for unions — all frequent in CBSE exams and in JEE Main.
- Sets and their representation: roster and set-builder form.
- Empty, finite, infinite and equal sets.
- Subsets, proper subsets, power set; intervals as subsets of ℝ.
- Universal set and Venn diagrams.
- Union, intersection, difference, complement; their laws.
- De Morgan's laws.
- Counting: n(A ∪ B) = n(A) + n(B) − n(A ∩ B) and its 3-set version.
2. Sets and Their Representation
A set is a well-defined collection of distinct objects — for any object we can decide clearly whether it belongs. "The five most talented writers of India" is not a set (not well-defined); "the vowels of the English alphabet" is.
- If a belongs to A, we write a ∈ A; otherwise a ∉ A.
- Standard sets: ℕ (natural numbers), ℤ (integers), ℚ (rationals), ℝ (reals), ℤ⁺ (positive integers).
| Form | How | Example |
|---|---|---|
| Roster (tabular) | List the elements in braces | {1, 2, 3, 6} |
| Set-builder | State the property | {x : x is a positive divisor of 6} |
In roster form, order does not matter and repeated elements are written once: {1, 2, 2, 3} = {1, 2, 3}.
Write {x : x is an integer and −3 ≤ x < 7} in roster form, and {1/2, 2/3, 3/4, 4/5, 5/6, 6/7} in set-builder form.
Solution: {−3, −2, −1, 0, 1, 2, 3, 4, 5, 6}. {x : x = n/(n + 1), n ∈ ℕ, 1 ≤ n ≤ 6}.
3. Types of Sets
| Type | Meaning | Example |
|---|---|---|
| Empty (null) set ∅ or { } | No elements | {x : x ∈ ℕ, 1 < x < 2} |
| Singleton | Exactly one element | {0} |
| Finite | Countable number of elements; n(A) = number of elements | Days of the week, n = 7 |
| Infinite | Not finite | ℕ, points on a line |
| Equal sets | Exactly the same elements | {1, 2} = {2, 1} |
{0} is not empty — it contains 0. And {∅} is not ∅ — it is a set containing one element (the empty set).
4. Subsets and Power Set
- A ⊆ B (A is a subset of B) if every element of A is in B.
- A ⊂ B (proper subset) if A ⊆ B and A ≠ B.
- A = B ⇔ A ⊆ B and B ⊆ A.
- ∅ is a subset of every set; every set is a subset of itself.
- The power set P(A) is the set of all subsets of A. If n(A) = m, then n[P(A)] = 2ᵐ.
Example: A = {1, 2}: P(A) = {∅, {1}, {2}, {1, 2}} — 2² = 4 elements.
ℕ ⊂ ℤ ⊂ ℚ ⊂ ℝ, and the irrationals T ⊂ ℝ with ℚ ∩ T = ∅.
Intervals (subsets of ℝ)
| Notation | Set | Type |
|---|---|---|
| (a, b) | {x : a < x < b} | Open |
| [a, b] | {x : a ≤ x ≤ b} | Closed |
| [a, b) | {x : a ≤ x < b} | Half-open |
| (a, b] | {x : a < x ≤ b} | Half-open |
| (a, ∞) | {x : x > a} | Ray |
5. Operations on Sets
The universal set U contains all elements under discussion. In Venn diagrams (after John Venn), U is a rectangle and sets are circles inside it.
| Operation | Definition | Example (A = {1,2,3,4}, B = {3,4,5}, U = {1,…,6}) |
|---|---|---|
| Union A ∪ B | {x : x ∈ A or x ∈ B} | {1, 2, 3, 4, 5} |
| Intersection A ∩ B | {x : x ∈ A and x ∈ B} | {3, 4} |
| Difference A − B | {x : x ∈ A, x ∉ B} | {1, 2} |
| Complement A′ | U − A | {5, 6} |
A and B are disjoint if A ∩ B = ∅.
- Commutative: A ∪ B = B ∪ A; A ∩ B = B ∩ A.
- Associative: (A ∪ B) ∪ C = A ∪ (B ∪ C); same for ∩.
- Distributive: A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C); A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C).
- Identity: A ∪ ∅ = A; A ∩ U = A. Idempotent: A ∪ A = A; A ∩ A = A.
- Complement: A ∪ A′ = U; A ∩ A′ = ∅; (A′)′ = A; U′ = ∅.
- De Morgan's laws: (A ∪ B)′ = A′ ∩ B′; (A ∩ B)′ = A′ ∪ B′.
Also A − B = A ∩ B′.
6. Counting with Sets
- n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
- If A, B disjoint: n(A ∪ B) = n(A) + n(B).
- n(A − B) = n(A) − n(A ∩ B).
- n(A ∪ B ∪ C) = n(A) + n(B) + n(C) − n(A∩B) − n(B∩C) − n(C∩A) + n(A∩B∩C).
In a group of 400 people, 250 speak Hindi and 200 speak English. How many speak both (everyone speaks at least one)?
Solution: 400 = 250 + 200 − n(H ∩ E) → n(H ∩ E) = 50.
In a survey of 60 people: 25 read newspaper H, 26 read T, 26 read I, 9 read H and I, 11 read H and T, 8 read T and I, 3 read all three. Find how many read none, and how many read exactly one.
Solution: n(H ∪ T ∪ I) = 25 + 26 + 26 − 11 − 8 − 9 + 3 = 52. None = 8. Exactly two = (11 − 3) + (8 − 3) + (9 − 3) = 19. Exactly one = 52 − 19 − 3 = 30.
7. Solved Problems (Board Pattern)
Show that A ∩ (A ∪ B) = A.
Solution: If x ∈ A then x ∈ A ∪ B, so x ∈ A ∩ (A ∪ B): A ⊆ A ∩ (A ∪ B). Conversely, x ∈ A ∩ (A ∪ B) ⇒ x ∈ A. Hence equal. ∎
U = {1, 2, …, 9}, A = {2, 4, 6, 8}, B = {2, 3, 5, 7}. Verify (A ∪ B)′ = A′ ∩ B′.
Solution: A ∪ B = {2, 3, 4, 5, 6, 7, 8} → complement {1, 9}. A′ = {1, 3, 5, 7, 9}, B′ = {1, 4, 6, 8, 9} → A′ ∩ B′ = {1, 9} ✓.
How many elements has P(P(∅))?
Solution: P(∅) = {∅} has 1 element → P(P(∅)) has 2¹ = 2 elements: {∅, {∅}}.
8. Master Formula Table
| Result | Formula |
|---|---|
| Power set size | n[P(A)] = 2ⁿ⁽ᴬ⁾ |
| Number of proper subsets | 2ⁿ − 1 |
| Union of two sets | n(A ∪ B) = n(A) + n(B) − n(A ∩ B) |
| Difference | n(A − B) = n(A) − n(A ∩ B) |
| Exactly one of A, B | n(A) + n(B) − 2n(A ∩ B) |
| Three sets | ΣA − Σ(pairs) + n(A ∩ B ∩ C) |
| De Morgan | (A ∪ B)′ = A′ ∩ B′; (A ∩ B)′ = A′ ∪ B′ |
| Difference via complement | A − B = A ∩ B′ |
9. Master Summary Table
| Concept | Key point | Exam focus |
|---|---|---|
| Well-defined | Membership must be clear | Which are sets? |
| Roster / set-builder | Convert between forms | 1-mark questions |
| Subsets, power set | 2ⁿ subsets | Counting subsets |
| Intervals | Open/closed ends | Write in set notation |
| Operations, laws | Venn diagrams | Prove identities |
| Counting | Inclusion–exclusion | Survey problems |
Chart — number of subsets as n(A) grows:
10. Mind Map
- Roster
- Set-builder
- Empty, finite, infinite
- Subsets, P(A)
- ∪, ∩, −, ′
- De Morgan
- n(A ∪ B)
- 3-set formula
11. Tips and Tricks
- For survey problems, fill the Venn diagram from the centre (all three) outward.
- To prove set identities, show each side is a subset of the other (element method).
- Remember: ∈ is for elements, ⊂ is for sets: 1 ∈ {1, 2} but {1} ⊂ {1, 2}.
- Writing {1} ∈ {1, 2} (wrong) instead of {1} ⊂ {1, 2}.
- Forgetting ∅ and the set itself when listing subsets.
- Subtracting the triple intersection instead of adding it in the 3-set formula.
12. Important One-liners
- A set must be well-defined.
- The empty set is a subset of every set.
- A set with n elements has 2ⁿ subsets.
- A − B, A ∩ B and B − A are pairwise disjoint.
- De Morgan: complement of a union is the intersection of complements.
- Set theory was founded by Georg Cantor.
13. Exam-pattern Questions with Detailed Answers
Which of these is the empty set?
- {0}
- {x : x ∈ ℕ, x² = 4}
- {x : x is an even prime greater than 2}
- {∅}
Show answer
The only even prime is 2, so there is none greater than 2. (A) and (D) have one element; (B) = {2}.
Answer: (C)
Write all subsets of {a, b, c}. How many are proper?
Show answer
∅, {a}, {b}, {c}, {a,b}, {b,c}, {a,c}, {a,b,c} — 8 subsets; 7 proper.
Answer: 8 subsets, 7 proper
Write (−3, 0] in set-builder form.
Show answer
{x : x ∈ ℝ, −3 < x ≤ 0}.
Answer: {x ∈ ℝ : −3 < x ≤ 0}
If A = {x : x is a natural number less than 6} and B = {x : x is a prime less than 10}, find A ∪ B, A ∩ B, A − B, B − A.
Show answer
A = {1, 2, 3, 4, 5}, B = {2, 3, 5, 7}. A ∪ B = {1, 2, 3, 4, 5, 7}; A ∩ B = {2, 3, 5}; A − B = {1, 4}; B − A = {7}.
Answer: As above
If n(U) = 50, n(A) = 28, n(B) = 32 and n(A ∩ B) = 15, find n(A′ ∩ B′).
Show answer
n(A ∪ B) = 45. A′ ∩ B′ = (A ∪ B)′ → 50 − 45 = 5.
Answer: 5
In a class of 35 students, 24 like cricket and 16 like football; each likes at least one. How many like both? How many like only cricket?
Show answer
Both = 24 + 16 − 35 = 5. Only cricket = 24 − 5 = 19.
Answer: 5 and 19
Is it true that for any sets A and B, P(A) ∪ P(B) = P(A ∪ B)? Justify.
Show answer
False. A = {1}, B = {2}: {1, 2} ∈ P(A ∪ B) but not in P(A) ∪ P(B) = {∅, {1}, {2}}.
Answer: False (counterexample)
Show that A ∪ B = A ∩ B implies A = B.
Show answer
x ∈ A ⇒ x ∈ A ∪ B = A ∩ B ⇒ x ∈ B, so A ⊆ B. Similarly B ⊆ A. Hence A = B. ∎
Answer: Proved
Prove that (A − B) ∪ (A ∩ B) = A.
Show answer
(A ∩ B′) ∪ (A ∩ B) = A ∩ (B′ ∪ B) = A ∩ U = A (distributive law). ∎
Answer: Proved
In a survey of 400 students, 100 drink apple juice, 150 orange juice and 75 both. How many drink neither?
Show answer
n(A ∪ O) = 100 + 150 − 75 = 175. Neither = 400 − 175 = 225.
Answer: 225
A college awarded 38 medals in football, 15 in basketball and 20 in cricket. These went to 58 students, and only 3 students got medals in all three. How many received medals in exactly two sports?
Show answer
Total medals 73 = (exactly one) + 2(exactly two) + 3(all three). Students 58 = (one) + (two) + 3. Subtract: 73 − 58 = (two) + 2(3) → two = 15 − 6 = 9.
Answer: 9
Show that for any sets A and B: A = (A ∩ B) ∪ (A − B), and A ∪ (B − A) = A ∪ B.
Show answer
First: (A ∩ B) ∪ (A ∩ B′) = A ∩ (B ∪ B′) = A ∩ U = A. Second: A ∪ (B ∩ A′) = (A ∪ B) ∩ (A ∪ A′) = (A ∪ B) ∩ U = A ∪ B. ∎
Answer: Proved using distributive law
14. Quick Revision Summary
- Set = well-defined collection; roster and set-builder forms.
- ∅ ⊆ every set; n(P(A)) = 2ⁿ.
- Operations ∪, ∩, −, ′; laws incl. distributive and De Morgan.
- n(A ∪ B) = n(A) + n(B) − n(A ∩ B); extend to three sets.
- Intervals (a, b), [a, b], etc. are subsets of ℝ.