Sets And Relations — Complete Guide with Solved Examples

Complete guide to sets and relations for Class 11. Solved examples, exam tips, PYQs.

CBSE JEE-MAIN 15 min read

What Are Sets and Why Do They Matter?

Sets form the language of modern mathematics. Every time we talk about “all solutions to this equation” or “which students passed both Physics and Chemistry,” we’re thinking in sets — we just don’t always call it that.

A set is a well-defined collection of distinct objects. The phrase “well-defined” is doing real work here: we need to be able to say, without ambiguity, whether any given object belongs to the collection or not. “All integers greater than 5” is a set. “All good songs” is not — because “good” is subjective.

For Class 11 CBSE and JEE Main, sets and relations together form the first chapter — and a surprisingly high-weightage one. JEE Main regularly pulls 1-2 questions from this chapter, often disguised inside a function problem. CBSE board exams typically have one 3-mark and one 5-mark question here.

A relation builds on sets: it’s a rule that pairs elements from one set with elements of another. When that rule becomes perfectly structured — every input maps to exactly one output — we get a function. So sets and relations are literally the foundation of the function chapter, which is one of the highest-scoring topics across all competitive exams.

Let’s get the definitions sharp, then work through techniques that will cover both board exams and MCQs comfortably.


Key Terms and Definitions

Sets

Roster (Tabular) Form: List all elements inside curly braces.

A={1,2,3,4,5}A = \{1, 2, 3, 4, 5\}

Set Builder Form: Describe the property elements must satisfy.

A={x:xN, x5}A = \{x : x \in \mathbb{N},\ x \leq 5\}

Empty Set (\emptyset or {}\{\}): A set with no elements. Note: {}\{\emptyset\} is not empty — it contains one element (the empty set itself).

Universal Set (UU): The “master set” for a given problem. Everything lives inside UU.

Subset: ABA \subseteq B means every element of AA is also in BB. Every set is a subset of itself. \emptyset is a subset of every set.

Power Set P(A)P(A): The collection of all subsets of AA. If A=n|A| = n, then P(A)=2n|P(A)| = 2^n.

A common source of free marks: if A=3|A| = 3, the number of subsets is 23=82^3 = 8, and the number of proper subsets is 231=72^3 - 1 = 7 (excluding AA itself). This distinction has appeared in multiple board papers.

Set Operations

AB={x:xA or xB}A \cup B = \{x : x \in A \text{ or } x \in B\} AB={x:xA and xB}A \cap B = \{x : x \in A \text{ and } x \in B\} AB={x:xA and xB}A - B = \{x : x \in A \text{ and } x \notin B\} A=UA(complement of A)A' = U - A \quad \text{(complement of } A\text{)}

Disjoint Sets: AB=A \cap B = \emptyset — no elements in common.

Relations

A relation RR from set AA to set BB is a subset of A×BA \times B (the Cartesian product).

The Cartesian product A×B={(a,b):aA, bB}A \times B = \{(a, b) : a \in A,\ b \in B\}.

If A=m|A| = m and B=n|B| = n, then A×B=mn|A \times B| = mn, and the number of possible relations from AA to BB is 2mn2^{mn}.

Domain of R: All first elements (inputs) actually used in RR. Range of R: All second elements (outputs) actually appearing in RR. Codomain: The full set BB — range \subseteq codomain always.


Types of Relations (on a Set)

When RR is a relation from AA to AA (i.e., RA×AR \subseteq A \times A), we classify it:

Reflexive

Every element is related to itself: (a,a)R(a, a) \in R for all aAa \in A.

Example: “is equal to” on R\mathbb{R} is reflexive. “is less than” is not (no number is less than itself).

Symmetric

If (a,b)R(a, b) \in R, then (b,a)R(b, a) \in R.

Example: “is a sibling of” is symmetric. “is the father of” is not.

Transitive

If (a,b)R(a, b) \in R and (b,c)R(b, c) \in R, then (a,c)R(a, c) \in R.

Example: “is less than” is transitive: 2<52 < 5 and 5<95 < 9 implies 2<92 < 9.

Equivalence Relation

A relation that is simultaneously reflexive, symmetric, and transitive.

JEE Main loves equivalence relations. A typical question gives you a relation and asks you to check each property. Work systematically: reflexive → symmetric → transitive. If any one fails, stop — it cannot be an equivalence relation.

Anti-Symmetric

If (a,b)R(a, b) \in R and (b,a)R(b, a) \in R, then a=ba = b.

Example: “is less than or equal to” (\leq) on integers: if aba \leq b and bab \leq a, then a=ba = b.


Venn Diagrams and Counting Formulas

For two overlapping sets:

AB=A+BAB|A \cup B| = |A| + |B| - |A \cap B|

For three sets:

ABC=A+B+CABBCAC+ABC|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |B \cap C| - |A \cap C| + |A \cap B \cap C|

The three-set formula is CBSE’s favourite 5-mark question structure. We’ll use it in Example 3 below.

De Morgan’s Laws

(AB)=AB(A \cup B)' = A' \cap B' (AB)=AB(A \cap B)' = A' \cup B'

Read these as: “complement of a union = intersection of complements” and vice versa.


Solved Examples

Example 1 — Power Set (CBSE Level)

Q: If A={1,2,3}A = \{1, 2, 3\}, write P(A)P(A) and find P(P(A))|P(P(A))|.

Solution:

P(A)={, {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3}}P(A) = \{\emptyset,\ \{1\},\ \{2\},\ \{3\},\ \{1,2\},\ \{1,3\},\ \{2,3\},\ \{1,2,3\}\}

So P(A)=23=8|P(A)| = 2^3 = 8.

Now, P(P(A))=28=256|P(P(A))| = 2^8 = 256.

The logic: P(A)P(A) has 8 elements, so its power set has 282^8 elements.


Example 2 — Checking Relation Properties (JEE Main Level)

Q: Let R={(a,b):a,bZ, ab is divisible by 5}R = \{(a, b) : a, b \in \mathbb{Z},\ a - b \text{ is divisible by } 5\}. Show RR is an equivalence relation.

Reflexive: For any aZa \in \mathbb{Z}: aa=0a - a = 0, and 505 \mid 0. So (a,a)R(a, a) \in R. ✓

Symmetric: Suppose (a,b)R(a, b) \in R, i.e., 5(ab)5 \mid (a - b). Then ba=(ab)b - a = -(a - b), and if 5(ab)5 \mid (a - b), then 5(ab)5 \mid -(a-b) too. So (b,a)R(b, a) \in R. ✓

Transitive: Suppose (a,b)R(a, b) \in R and (b,c)R(b, c) \in R. Then 5(ab)5 \mid (a - b) and 5(bc)5 \mid (b - c). Adding: 5(ab)+(bc)=ac5 \mid (a - b) + (b - c) = a - c. So (a,c)R(a, c) \in R. ✓

All three hold, so RR is an equivalence relation.

For divisibility relations aba - b divisible by nn, they are always equivalence relations. Recognising this pattern saves you proof-writing time in exams.


Example 3 — Three-Set Inclusion-Exclusion (CBSE 5-Mark)

Q: In a class of 100 students, 60 play cricket, 45 play football, 35 play basketball. 20 play both cricket and football, 15 play both football and basketball, 25 play both cricket and basketball. 8 play all three. Find how many play at least one sport.

Solution:

Let CC, FF, BB be the sets.

CFB=60+45+35201525+8|C \cup F \cup B| = 60 + 45 + 35 - 20 - 15 - 25 + 8 =14060+8=88= 140 - 60 + 8 = 88

88 students play at least one sport, so 10088=12100 - 88 = 12 play none.


Example 4 — Number of Relations (JEE Main Level)

Q: If A=3|A| = 3 and B=4|B| = 4, find the number of relations from AA to BB that contain a specific ordered pair (a1,b1)(a_1, b_1).

Solution:

Total ordered pairs in A×BA \times B: 3×4=123 \times 4 = 12.

A relation containing (a1,b1)(a_1, b_1) must include this pair. For the remaining 121=1112 - 1 = 11 pairs, each is either included or not — giving 2112^{11} choices.

Answer: 211=20482^{11} = 2048.


Example 5 — Equivalence Classes (JEE Advanced Level)

Q: On Z\mathbb{Z}, define aba \sim b iff aba - b is divisible by 3. Find the equivalence classes.

When we divide any integer by 3, the remainder is 0, 1, or 2. So:

  • Class [0]={,6,3,0,3,6,}[0] = \{\ldots, -6, -3, 0, 3, 6, \ldots\} — multiples of 3
  • Class [1]={,5,2,1,4,7,}[1] = \{\ldots, -5, -2, 1, 4, 7, \ldots\} — remainder 1
  • Class [2]={,4,1,2,5,8,}[2] = \{\ldots, -4, -1, 2, 5, 8, \ldots\} — remainder 2

These three classes partition Z\mathbb{Z} — every integer belongs to exactly one class. This is the key property of equivalence relations: they always create a partition.


Exam-Specific Tips

CBSE Class 11 Board Exam

The board paper typically has:

  • 1 MCQ (1 mark): Usually “find P(A)|P(A)|” or identify subset count
  • 1 short answer (3 marks): Venn diagram word problem or De Morgan’s law proof
  • 1 long answer (5 marks): Three-set inclusion-exclusion application

Marking scheme tip: In proofs (like showing reflexive/symmetric/transitive), each property carries equal marks. Even if you can’t prove transitivity, write down the definitions and show your attempt — partial credit is awarded.

JEE Main Strategy

JEE Main has shifted toward asking about equivalence relations in disguise — a function-based question where you must identify if the underlying relation is reflexive/symmetric/transitive. The chapter appears in the syllabus as “Sets, Relations and Functions” with a combined weightage of 3-4% (roughly 2-3 questions in a 90-question paper).

Focus areas for JEE Main:

  1. Equivalence relation verification — 1 direct question every 3-4 papers
  2. Cartesian product counting — often embedded in permutation questions
  3. Power set + subset counting — quick 1-mark questions if you know the formulas

SAT Strategy

SAT doesn’t test formal relations theory, but set operations (union, intersection, complement) appear regularly in data interpretation and Venn diagram problems. The three-set formula is especially useful.


Common Mistakes to Avoid

Mistake 1: Confusing \emptyset and {}\{\emptyset\}. The empty set \emptyset has 0 elements. The set {}\{\emptyset\} has 1 element (which happens to be the empty set). So P({})=21=2|P(\{\emptyset\})| = 2^1 = 2, not 20=12^0 = 1.

Mistake 2: Thinking “not symmetric” means “anti-symmetric.” A relation can be neither symmetric nor anti-symmetric. Anti-symmetric means: (a,b)R(a,b) \in R and (b,a)R(b,a) \in R together force a=ba = b. These are completely different conditions.

Mistake 3: Forgetting that \emptyset is a subset of every set. When listing subsets of A={1,2}A = \{1, 2\}, students write {1},{2},{1,2}\{1\}, \{2\}, \{1,2\} and miss \emptyset. This gives 3 instead of 4, losing marks on power set questions.

Mistake 4: Double-counting in three-set problems. When you add A+B+C|A| + |B| + |C|, elements in two-way intersections are counted twice and elements in the three-way intersection are counted three times. The formula corrects this — but students sometimes subtract all three pairwise intersections and forget to add back ABC|A \cap B \cap C|.

Mistake 5: Applying transitivity incorrectly. Transitivity requires: if (a,b)R(a,b) \in R and (b,c)R(b,c) \in R, then (a,c)R(a,c) \in R. The key word is “and” — you need both pairs in RR before the conclusion applies. If (a,b)R(a,b) \in R but there’s no (b,anything)(b, \text{anything}) in RR, transitivity isn’t violated.


Practice Questions

Q1. If A={x:x25x+6=0}A = \{x : x^2 - 5x + 6 = 0\} and B={x:x23x+2=0}B = \{x : x^2 - 3x + 2 = 0\}, find ABA \cup B, ABA \cap B, and ABA - B.

Solve each equation: A={2,3}A = \{2, 3\}, B={1,2}B = \{1, 2\}.

AB={1,2,3}A \cup B = \{1, 2, 3\}, AB={2}A \cap B = \{2\}, AB={3}A - B = \{3\}.


Q2. If AA has 4 elements, how many relations on AA are reflexive?

A×AA \times A has 42=164^2 = 16 pairs. A reflexive relation must contain all 4 diagonal pairs (a,a)(a,a). The remaining 164=1216 - 4 = 12 pairs can be freely included or not.

Answer: 212=40962^{12} = 4096.


Q3. In a survey of 200 people, 120 read newspaper A, 90 read B, and 70 read C. 40 read A and B, 30 read B and C, 50 read A and C, and 20 read all three. How many read none?

ABC=120+90+70403050+20=180|A \cup B \cup C| = 120 + 90 + 70 - 40 - 30 - 50 + 20 = 180

People reading none: 200180=20200 - 180 = 20.


Q4. Is the relation R={(a,b):ab2}R = \{(a,b) : |a - b| \leq 2\} on Z\mathbb{Z} an equivalence relation?

Reflexive: aa=02|a - a| = 0 \leq 2

Symmetric: If ab2|a - b| \leq 2, then ba=ab2|b - a| = |a - b| \leq 2

Transitive: Take a=1,b=2,c=3a = 1, b = 2, c = 3: (1,2)R(1,2) \in R and (2,3)R(2,3) \in R, and 13=22|1-3| = 2 \leq 2 ✓. But try a=1,b=3,c=4a = 1, b = 3, c = 4: (1,3)R(1,3) \in R, (3,4)R(3,4) \in R, but 14=3>2|1-4| = 3 > 2, so (1,4)R(1,4) \notin R. ✗

Not transitive, so not an equivalence relation.


Q5. Let A={1,2,3,4,5}A = \{1, 2, 3, 4, 5\}. Define R={(a,b):a+b=6}R = \{(a,b) : a + b = 6\}. Find RR, its domain, and range. Is RR a function from AA to AA?

R={(1,5),(2,4),(3,3),(4,2),(5,1)}R = \{(1,5), (2,4), (3,3), (4,2), (5,1)\}.

Domain ={1,2,3,4,5}=A= \{1, 2, 3, 4, 5\} = A, Range ={1,2,3,4,5}=A= \{1, 2, 3, 4, 5\} = A.

Yes, RR is a function — each element of AA maps to exactly one element of AA.


Q6. Using De Morgan’s law, if U={1,2,3,4,5,6,7,8}U = \{1,2,3,4,5,6,7,8\}, A={1,3,5,7}A = \{1,3,5,7\}, B={2,3,5,6}B = \{2,3,5,6\}, verify (AB)=AB(A \cup B)' = A' \cap B'.

AB={1,2,3,5,6,7}A \cup B = \{1,2,3,5,6,7\}, so (AB)={4,8}(A \cup B)' = \{4, 8\}.

A={2,4,6,8}A' = \{2,4,6,8\}, B={1,4,7,8}B' = \{1,4,7,8\}, so AB={4,8}A' \cap B' = \{4, 8\}. ✓ Verified.


Q7. How many elements does P(P())P(P(\emptyset)) have?

P()={}P(\emptyset) = \{\emptyset\}, which has 1 element.

P(P())=P({})={,{}}P(P(\emptyset)) = P(\{\emptyset\}) = \{\emptyset, \{\emptyset\}\}, which has 21=22^1 = 2 elements.


Q8. On the set of all lines in a plane, define RR as “is parallel to.” Is RR an equivalence relation? (Convention: a line is parallel to itself.)

Reflexive: Every line is parallel to itself (by convention). ✓

Symmetric: If l1l2l_1 \parallel l_2, then l2l1l_2 \parallel l_1. ✓

Transitive: If l1l2l_1 \parallel l_2 and l2l3l_2 \parallel l_3, then l1l3l_1 \parallel l_3. ✓

Yes, RR is an equivalence relation. The equivalence classes are bundles of parallel lines — each class is determined by the common slope.


FAQs

What is the difference between a relation and a function?

A relation from AA to BB is any subset of A×BA \times B — there are no restrictions. A function is a special relation where every element of AA (the domain) maps to exactly one element of BB. Every function is a relation, but not every relation is a function.

What does “well-defined” mean for a set?

It means we can always determine, without ambiguity, whether an object belongs to the set. “Even integers” is well-defined. “Famous mathematicians” is not — “famous” is a matter of opinion.

Can a relation be both symmetric and anti-symmetric?

Yes. The only relations that are both symmetric and anti-symmetric are subsets of the diagonal {(a,a):aA}\{(a,a) : a \in A\}. The “equality” relation on any set is the classic example.

How many equivalence relations can be defined on a 3-element set?

Five. They correspond to the five ways to partition {1,2,3}\{1,2,3\}: {{1},{2},{3}}\{\{1\},\{2\},\{3\}\}, {{1,2},{3}}\{\{1,2\},\{3\}\}, {{1,3},{2}}\{\{1,3\},\{2\}\}, {{2,3},{1}}\{\{2,3\},\{1\}\}, and {{1,2,3}}\{\{1,2,3\}\}.

What is the difference between range and codomain?

The codomain is the set BB that we’re mapping to — it’s declared upfront. The range (or image) is the set of values actually achieved. Range \subseteq codomain always, but they’re only equal when every element of BB is hit by some element of AA (surjective/onto function).

Why does the empty set have 20=12^0 = 1 subset?

The only subset of \emptyset is \emptyset itself. This is consistent with the general formula: A=0P(A)=20=1|A| = 0 \Rightarrow |P(A)| = 2^0 = 1.

In JEE, how do I quickly check if a relation is transitive?

Sketch a directed graph: draw nodes for each element, draw arrows for each pair in RR. Transitivity means: if there’s a path of length 2 from aa to cc (through some bb), there must also be a direct arrow from aa to cc. Look for “two-step paths” without direct arrows — these are your counterexamples.

What is an equivalence class and why does it matter?

Given an equivalence relation RR on AA, the equivalence class of aa is [a]={bA:(a,b)R}[a] = \{b \in A : (a,b) \in R\} — all elements related to aa. The key theorem: equivalence classes always form a partition of AA (they’re non-empty, pairwise disjoint, and their union is all of AA). This idea appears explicitly in JEE Advanced and is the conceptual foundation for modular arithmetic.

Practice Questions