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Class 11 Maths Notes

Sets Class 11 Notes

Complete, exam-ready notes on sets: representation in roster and set-builder form, the types of sets, subsets, the power set and universal set, the operations of union, intersection and difference, complements and De Morgan's laws, and cardinality problems — written for CBSE, JEE and NEET revision.

Class11SubjectMathematicsCoversCBSE · JEE

Written byDeep Narayan· Science & Mathematics EducatorReviewed byPushpanjali

What is this chapter about in one line?

A set is a well-defined collection of distinct objects, and this chapter studies how they are represented, related (subsets) and combined (union, intersection, complement) using Venn diagrams and cardinality formulas.

Sets and Their Representation

Set (roster and set-builder)

A set is a well-defined collection of distinct (unique) objects called its elements. Roster form lists elements explicitly, e.g. {1, 2, 3}; set-builder form states a property, e.g. {x : x is a natural number less than 4}. Order does not matter and repeated elements are written once.

Membership symbol

a ∈ A means a belongs to set A; a ∉ A means a does not. The empty set ∅ has no elements, and the null symbol {} is the same set.

Types of Sets

  • Finite set: a fixed, countable number of elements, e.g. {1, 2, 3}; infinite set: never-ending, e.g. the natural numbers ℕ.
  • Empty (null) set: no elements at all, written ∅ or {}.
  • Singleton set: exactly one element, e.g. {5}.
  • Equal sets A = B have exactly the same elements; equivalent sets have the same number of elements but different members.

Empty set

A set with no elements. It is finite with cardinality 0, and it is a subset of every set. Note {0} is not empty — it is the singleton containing the number zero.

Subsets, Power Set and Universal Set

Subset and power set

A ⊂ B (A is a subset of B) means every element of A is also in B. The empty set and the set itself are always subsets. The power set P(A) is the collection of all subsets of A; if A has n elements, P(A) has 2ⁿ subsets.

∣P(A)∣=2n,n=∣A∣|P(A)| = 2^n,\qquad n = |A|
Number of subsets in the power set

Universal Set and Set Operations

Universal set U

The set that contains all elements relevant to a discussion. The complement A′ (or Aᶜ) is everything in U that is not in A. Union (A ∪ B) gathers all elements of A and B; intersection (A ∩ B) keeps only the elements common to both; difference (A − B) keeps elements in A not in B.

  • Union: A ∪ B = {x : x ∈ A or x ∈ B}.
  • Intersection: A ∩ B = {x : x ∈ A and x ∈ B}.
  • Complement: A′ = {x : x ∈ U and x ∉ A}.
  • Difference: A − B = {x : x ∈ A and x ∉ B}.

Venn Diagrams and De Morgan's Laws

De Morgan's laws

Venn diagrams represent sets as regions inside the universal set. They make union, intersection and complement relations visually obvious, and they are used to prove De Morgan's laws.

(A∪B)′=A′∩B′,(A∩B)′=A′∪B′(A \cup B)' = A' \cap B',\qquad (A \cap B)' = A' \cup B'
De Morgan's laws

Complement swaps the operator

When you take the complement of a union it becomes an intersection of complements, and the complement of an intersection becomes a union of complements — the AND/OR operator flips.

Cardinality and Counting Problems

Cardinality

The number of elements in a set, written n(A). For two sets, union size subtracts the overlap counted twice; for three sets, the triple-overlap must also be added back.

n(A∪B)=n(A)+n(B)−n(A∩B)n(A \cup B) = n(A) + n(B) - n(A \cap B)
Two-set inclusion–exclusion
n(A∪B∪C)=n(A)+n(B)+n(C)−n(A∩B)−n(B∩C)−n(C∩A)+n(A∩B∩C)n(A \cup B \cup C) = n(A)+n(B)+n(C) - n(A \cap B) - n(B \cap C) - n(C \cap A) + n(A \cap B \cap C)
Three-set inclusion–exclusion

Solved Examples

Example: If A = {1, 2, 3, 4} and B = {3, 4, 5, 6}, find A ∪ B, A ∩ B and A − B.

Solution: A ∪ B = {1, 2, 3, 4, 5, 6}; A ∩ B = {3, 4}; A − B = {1, 2} — the elements in A that are not in B.

Example: In a class of 60 students, 35 play cricket and 28 play football, and 12 play both. How many play neither?

Solution: n(C ∪ F) = 35 + 28 − 12 = 51. So the number playing neither = 60 − 51 = 9 students.

Revision

Key formulas at a glance

Memorise these before attempting numericals — most exam questions hinge on one of them.

Power set size

∣P(A)∣=2n|P(A)| = 2^n

Two-set union

n(A∪B)=n(A)+n(B)−n(A∩B)n(A \cup B) = n(A) + n(B) - n(A \cap B)

Three-set union

n(A∪B∪C)=n(A)+n(B)+n(C)−n(A∩B)−n(B∩C)−n(C∩A)+n(A∩B∩C)n(A \cup B \cup C) = n(A)+n(B)+n(C) - n(A\cap B) - n(B\cap C) - n(C\cap A) + n(A\cap B\cap C)

De Morgan first law

(A∪B)′=A′∩B′(A \cup B)' = A' \cap B'

De Morgan second law

(A∩B)′=A′∪B′(A \cap B)' = A' \cup B'

Subset meaning

A⊂B  ⟺  every element of A is in BA \subset B \iff \text{every element of } A \text{ is in } B

Set-builder form

{x:x has property P}=roster {x1,x2,… }\{x : x \text{ has property } P\} = \text{roster } \{x_1, x_2, \dots\}

Exam tips

How this chapter is asked

Where this topic appears in CBSE, JEE Main and NEET papers.

  • Roster form lists elements; set-builder states a defining property.
  • Empty set ∅ has 0 elements; {0} is a singleton, not empty.
  • A set with n elements has exactly 2ⁿ subsets.
  • De Morgan's laws flip the operator when complementing a union/intersection.
  • Union count subtracts the double-counted intersection once: n(A ∪ B) = n(A) + n(B) − n(A ∩ B).
  • For three sets, add back the triple intersection n(A ∩ B ∩ C).
  • Equal sets share all elements; equivalent sets only match in size.

FAQ

Common questions

What are roster and set-builder forms of a set?

Roster form lists every element inside braces, e.g. {1, 2, 3}; set-builder form states a property, e.g. {x : x ∈ ℕ, x < 4}. Both describe the same set.

What is the power set of a set?

The power set P(A) is the collection of all subsets of A. If A has n elements, P(A) contains 2ⁿ subsets, including the empty set and A itself.

What are De Morgan's laws?

They relate complements to unions and intersections: (A ∪ B)′ = A′ ∩ B′ and (A ∩ B)′ = A′ ∪ B′. Complementing a union gives an intersection of complements and vice versa.

How do you find n(A ∪ B)?

Add the sizes and subtract the overlap: n(A ∪ B) = n(A) + n(B) − n(A ∩ B). The intersection is counted twice, so it is subtracted once.

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