A set is simply a well-defined collection of distinct objects. "Well-defined" means there is no ambiguity — you can always tell whether something belongs to the collection or not.
Real-life examples that ARE sets
The collection of all vowels in the English alphabet: {a, e, i, o, u} — well-defined, no ambiguity.
The collection of all prime numbers less than 10: {2, 3, 5, 7}.
The collection of months of the year starting with 'J': {January, June, July}.
The collection of students in your class whose roll numbers are below 5.
These are NOT sets — because they are NOT well-defined
"The collection of good students" — what does "good" mean? Different people decide differently.
"The collection of tall people" — "tall" is subjective, not precise.
"The collection of beautiful flowers" — beauty is subjective.
The key test: Can any two people independently, without discussion, agree on whether an object belongs? If yes, it's a set.
Notation basics
Sets are denoted by capital letters: A, B, C, X, Y, Z...
Elements (objects in the set) are denoted by small letters: a, b, x, y...
$x \in A$ means "$x$ belongs to set $A$" (x is an element of A).
$x
otin A$ means "$x$ does not belong to set $A$".
Example: If $A = \{1, 2, 3, 4, 5\}$, then $3 \in A$ but $7
otin A$.
2. How to Write Sets
Method 1: Roster (Tabular) Form
List all elements inside curly braces {}, separated by commas. Each element appears only once — repetitions are ignored. Order doesn't matter.
$A = \{1, 2, 3, 4, 5\}$ — set of natural numbers less than 6.
$B = \{a, e, i, o, u\}$ — set of vowels.
$C = \{2, 4, 6, 8, ...\}$ — even natural numbers (dots mean the pattern continues infinitely).
The word SCHOOL in roster form: $\{S, C, H, O, L\}$ — repeated letters written only once.
Method 2: Set-Builder (Rule/Property) Form
Write the rule that an element must satisfy. Format: $\{x : x ext{ satisfies some property}\}$ — read as "the set of all x such that x satisfies..."
$A = \{x : x ext{ is a natural number and } x < 6\} = \{1, 2, 3, 4, 5\}$
$B = \{x : x = 2n, n \in \mathbb{N}\} = \{2, 4, 6, 8, ...\}$ — even natural numbers.
$C = \{x : x \in \mathbb{R} ext{ and } x^2 = 4\} = \{-2, 2\}$
$D = \{x : x \in \mathbb{N}, 1 \le x \le 5\} = \{1, 2, 3, 4, 5\}$
Tip: The colon ":" or "|" means "such that". Both are acceptable.
A set with NO elements. Written as $\phi$ or $\{\}$. Example: Set of even prime numbers greater than 10.
Singleton Set
A set with exactly ONE element. Example: $\{0\}$, $\{5\}$, $\{\phi\}$ (set containing the empty set).
Finite Set
A set with a countable (limited) number of elements. Example: Set of days in a week = $\{Mon, Tue, ..., Sun\}$, $n = 7$.
Infinite Set
A set with unlimited elements, cannot be fully listed. Example: $\mathbb{N}, \mathbb{Z}, \mathbb{R}$, set of all stars in the universe.
Equal Sets
Two sets A and B are equal ($A = B$) if they have exactly the same elements. Order doesn't matter. $\{1,2,3\} = \{3,1,2\}$.
Equivalent Sets
Sets with the same NUMBER of elements (same cardinality), but not necessarily the same elements. $\{1,2,3\}$ and $\{a,b,c\}$ are equivalent.
Disjoint Sets
Two sets with NO common elements. $A \cap B = \phi$. Example: $\{1,3,5\}$ and $\{2,4,6\}$.
Intervals (Subsets of $\mathbb{R}$)
$[a,b]$ = closed (both included). $(a,b)$ = open (both excluded). $[a,b)$ and $(a,b]$ = half-open.
Critical Note: Empty Set Confusion
$\phi$ = the empty set (has 0 elements).
$\{\phi\}$ = a set containing the empty set (has 1 element — NOT empty!).
$\{0\}$ = a set containing zero (has 1 element — NOT empty!).
The empty set $\phi$ is a subset of every set.
4. Subsets, Proper Subsets & Power Set
Subset ($A \subseteq B$)
$A$ is a subset of $B$ if every element of $A$ is also an element of $B$. In other words, $A$ is "contained inside" $B$.
Formally: $A \subseteq B \iff orall x, (x \in A \implies x \in B)$
Understanding subsets with an example
Let $B = \{1, 2, 3\}$. All subsets of $B$:
$\phi$ (empty set — always a subset of any set)
$\{1\}, \{2\}, \{3\}$ (single-element subsets)
$\{1,2\}, \{1,3\}, \{2,3\}$ (two-element subsets)
$\{1,2,3\}$ (the set itself — every set is a subset of itself)
Total: 8 subsets. Since $|B| = 3$, total subsets $= 2^3 = 8$. ✓
Proper Subset ($A \subset B$ or $A \subsetneq B$)
$A$ is a proper subset of $B$ if $A \subseteq B$ AND $A
e B$. In other words, $A$ is a subset but not equal to $B$ — at least one element of $B$ is not in $A$.
Power Set, Subset Count Formulas
$$ ext{If } |A| = n ext{ (n elements in set A):}$$
$$ ext{Total subsets} = 2^n$$
$$ ext{Total proper subsets} = 2^n - 1$$
$$ ext{Total non-empty proper subsets} = 2^n - 2$$
$$ ext{Power Set } P(A) = ext{set of ALL subsets of } A, \quad |P(A)| = 2^n$$
Subset vs Element — the most common mistake in exams
Let $A = \{1, \{2, 3\}, 4\}$
$1 \in A$ ✓ (1 is an element of A)
$\{1\} \subset A$ ✓ ({1} is a subset of A)
$\{2, 3\} \in A$ ✓ ({2,3} is an element of A — it's listed as a single element)
$\{2, 3\} \subset A$ ✗ (2 is not directly in A; {2,3} as a whole is an element)
$\{\{2, 3\}\} \subset A$ ✓ (the set containing {2,3} is a subset of A)
5. Universal Set & Venn Diagrams
Universal Set ($U$)
The universal set is the "master set" — it contains all elements relevant to the problem at hand. Every other set in the discussion is a subset of $U$.
If we study natural numbers: $U = \mathbb{N}$
If we study students in a school: $U$ = set of all students in the school
If we study triangles: $U$ = set of all triangles
Venn Diagrams
Venn diagrams use overlapping circles inside a rectangle to visualize sets:
The rectangle represents the Universal Set $U$.
Each circle represents a set (A, B, C...).
Overlapping regions represent elements in multiple sets.
Elements outside all circles but inside the rectangle are in $U$ only.
Venn diagrams are extremely useful for proving set identities and solving word problems — always draw one for survey-type questions!
6. Set Operations
Union: $A \cup B$
All elements that are in $A$, or $B$, or both. Think of it as combining both sets (removing duplicates).
$A \cup B = \{x : x \in A ext{ or } x \in B\}$
Example: $\{1,2,3\} \cup \{3,4,5\} = \{1,2,3,4,5\}$
Intersection: $A \cap B$
Only the elements that are in BOTH $A$ AND $B$. The "overlap".
$A \cap B = \{x : x \in A ext{ and } x \in B\}$
Example: $\{1,2,3\} \cap \{3,4,5\} = \{3\}$
Difference: $A - B$ (or $A \setminus B$)
Elements that are in $A$ but NOT in $B$. "Remove from A everything that's also in B."
$A - B = \{x : x \in A ext{ and } x
otin B\}$
Example: $\{1,2,3,4\} - \{3,4,5\} = \{1,2\}$
Note: $A - B
e B - A$ in general!
Complement: $A'$ (or $A^c$)
All elements of the Universal Set that are NOT in $A$.
$A' = U - A = \{x : x \in U ext{ and } x
otin A\}$
Example: If $U = \{1,...,10\}$ and $A = \{1,3,5,7,9\}$, then $A' = \{2,4,6,8,10\}$
Properties of Set Operations
$$ ext{Commutativity: } A \cup B = B \cup A \quad ext{and} \quad A \cap B = B \cap A$$
$$ ext{Associativity: } (A \cup B) \cup C = A \cup (B \cup C)$$
$$ ext{Distributivity: } A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$$
$$ ext{Identity: } A \cup \phi = A \quad ext{and} \quad A \cap U = A$$
$$ ext{Idempotent: } A \cup A = A \quad ext{and} \quad A \cap A = A$$
$$A \cup A' = U \quad ext{and} \quad A \cap A' = \phi$$
$$(A')' = A \quad ext{(Double complement)}$$
$$\phi' = U \quad ext{and} \quad U' = \phi$$
7. De Morgan's Laws
De Morgan's Laws — Two Laws, Critical for Exams
Law 1: $(A \cup B)' = A' \cap B'$
The complement of a union = intersection of the complements.
Law 2: $(A \cap B)' = A' \cup B'$
The complement of an intersection = union of the complements.
Memory trick: When you "distribute" the complement inside, the $\cup$ flips to $\cap$ and vice versa.
8. Cardinality Formulas
Number of elements formulas — used in every survey word problem
$$n(A \cup B) = n(A) + n(B) - n(A \cap B)$$
$$n(A - B) = n(A) - n(A \cap B) = n(A \cup B) - n(B)$$
$$n(A \cap B) = n(A) + n(B) - n(A \cup B)$$
$$ extbf{For three sets:}$$
$$n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(A \cap C) + n(A \cap B \cap C)$$
$$ ext{Only in A} = n(A) - n(A \cap B) - n(A \cap C) + n(A \cap B \cap C)$$
$$ ext{Exactly two sets} = n(A \cap B) + n(B \cap C) + n(A \cap C) - 3 \cdot n(A \cap B \cap C)$$
$$ ext{Exactly one set} = n(A \cup B \cup C) - [ ext{exactly two}] - [ ext{all three}]$$
9. Worked Examples — Easy to Hard
Example 1: Write in Roster FormEasy
Write the set $A = \{x : x$ is a natural number and $x^2 < 25\}$ in Roster form.
Show Solution
We need natural numbers where $x^2 < 25$, i.e., $x < 5$ (since $x$ is positive).
Natural numbers less than 5: $x = 1, 2, 3, 4$. Check: $1^2=1, 2^2=4, 3^2=9, 4^2=16$ — all $< 25$. And $5^2 = 25$, not $< 25$.
So every element is a multiple of 3, and there are 5 elements (multiples 1 through 5).
$B = \{x : x = 3n,\ n \in \mathbb{N},\ 1 \le n \le 5\}$
Example 3: Which of these are sets?Easy
State which of the following are sets. Justify.
(i) The collection of all months of a year beginning with the letter J
(ii) The collection of ten most talented writers of India
(iii) The collection of all even integers
Show Solution
(i) YES — it is a set. Months starting with J are: January, June, July. These can be listed without ambiguity.
(ii) NO — not a set. "Most talented" is subjective. Different people would choose different writers. Not well-defined.
(iii) YES — it is a set. Even integers are perfectly well-defined: $\{..., -4, -2, 0, 2, 4, ...\}$. There is no ambiguity about whether a number is even.
Example 4: Counting SubsetsEasy
Let $A = \{a, b, c, d\}$. Find: (i) Number of subsets (ii) Number of proper subsets (iii) Write the power set $P(A)$... just count it.
(i) $A \cup B$ = all elements in $A$ or $B$ = $\{1, 2, 3, 4, 5, 6, 8\}$
(ii) $A \cap B$ = elements in both $A$ and $B$ = $\{2, 4, 6\}$
(iii) $A - B$ = elements in $A$ but not $B$ = $\{1, 3, 5\}$ (remove 2, 4, 6 from A)
(iv) $B - A$ = elements in $B$ but not $A$ = $\{8\}$ (remove 2, 4, 6 from B — only 8 remains)
Notice: $A - B
e B - A$. The difference operation is not commutative!
Example 7: Complement & De Morgan's VerificationMedium
Let $U = \{1,2,3,4,5,6,7,8,9,10\}$, $A = \{1,3,5,7,9\}$, $B = \{2,4,6,8,10\}$.
Verify: $(A \cup B)' = A' \cap B'$
Show Solution
LHS: $A \cup B = \{1,2,3,4,5,6,7,8,9,10\} = U$, so $(A \cup B)' = U' = \phi$
RHS: $A' = U - A = \{2,4,6,8,10\}$ and $B' = U - B = \{1,3,5,7,9\}$
$A' \cap B'$ = elements common to both $A'$ and $B'$ = $\phi$ (they share no elements)
LHS $= \phi =$ RHS $\checkmark$ — De Morgan's Law verified!
Example 8: Nested Set ProblemMedium
Let $A = \{1, 2, \{3, 4\}, 5\}$. State True or False:
(i) $\{3,4\} \in A$ (ii) $\{3,4\} \subset A$ (iii) $\{\{3,4\}\} \subset A$ (iv) $1 \in A$ (v) $\{1, 2, 5\} \subset A$
Show Solution
Think of $A$ as having 4 elements: $1$, $2$, $\{3,4\}$ (the whole set is one element), and $5$.
(i) $\{3,4\} \in A$ — TRUE. The set $\{3,4\}$ is listed as one of the 4 elements of $A$.
(ii) $\{3,4\} \subset A$ — FALSE. For this to be true, $3 \in A$ and $4 \in A$ must both hold. But $3$ is not a direct element of $A$ (only $\{3,4\}$ is).
(iii) $\{\{3,4\}\} \subset A$ — TRUE. This set has one element: $\{3,4\}$. Since $\{3,4\} \in A$, the set $\{\{3,4\}\}$ is a subset of $A$.
(iv) $1 \in A$ — TRUE. $1$ is directly listed as an element.
(v) $\{1, 2, 5\} \subset A$ — TRUE. All three ($1, 2, 5$) are elements of $A$.
Example 9: Using Cardinality Formula (2 sets)Medium
In a class of 40 students, 25 play cricket, 20 play football, and 15 play both. How many students play neither sport?
Show Solution
Let $C$ = cricket players, $F$ = football players. Given: $n(C) = 25$, $n(F) = 20$, $n(C \cap F) = 15$, total $= 40$.
Students playing at least one sport: $n(C \cup F) = n(C) + n(F) - n(C \cap F) = 25 + 20 - 15 = 30$
Students playing neither: $40 - n(C \cup F) = 40 - 30 = \mathbf{10}$
Example 10: Find $n(A \cap B)$ when union is givenMedium
If $n(A) = 14$, $n(B) = 11$, $n(A \cup B) = 21$, find $n(A \cap B)$.
Show Solution
Using: $n(A \cup B) = n(A) + n(B) - n(A \cap B)$
$21 = 14 + 11 - n(A \cap B)$
$n(A \cap B) = 14 + 11 - 21 = \mathbf{4}$
Example 11: Prove a Set IdentityHard
Prove that $A - B = A \cap B'$
Show Solution
To prove A - B = A ∩ B', we show each is a subset of the other.
Step 1 (LHS ⊆ RHS): Let $x \in A - B$. By definition, $x \in A$ and $x
otin B$. Since $x
otin B$, we have $x \in B'$. So $x \in A$ and $x \in B'$, which means $x \in A \cap B'$.
Step 2 (RHS ⊆ LHS): Let $x \in A \cap B'$. Then $x \in A$ and $x \in B'$. Since $x \in B'$, $x
otin B$. So $x \in A$ and $x
otin B$, which means $x \in A - B$.
Since $A - B \subseteq A \cap B'$ and $A \cap B' \subseteq A - B$, we have $A - B = A \cap B'$. (Proved)
Example 12: Classic 3-Set Survey (NCERT type)Hard
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, and 3 read all three. Find:
(i) How many read at least one newspaper?
(ii) How many read exactly one newspaper?
(iii) How many read exactly two newspapers?
(i) At least one newspaper:
$n(H \cup T \cup I) = 25+26+26 - (11+8+9) + 3 = 77 - 28 + 3 = \mathbf{52}$
(ii) Exactly two (meaning two only, not three):
- Only H and T (not I) $= n(H \cap T) - n(H \cap T \cap I) = 11 - 3 = 8$
- Only T and I (not H) $= 8 - 3 = 5$
- Only H and I (not T) $= 9 - 3 = 6$
Total exactly two $= 8 + 5 + 6 = \mathbf{19}$
If $n(A) = 3$, $n(B) = 6$ and $A \subset B$, find $n(A \cup B)$ and $n(A \cap B)$.
Show Solution
Since $A \subset B$ (A is a proper subset of B), every element of A is already in B.
Therefore $A \cap B = A$, so $n(A \cap B) = n(A) = \mathbf{3}$
And $A \cup B = B$, so $n(A \cup B) = n(B) = \mathbf{6}$
Example 14: Power Set of Power SetHard
If $A = \{a, b\}$, find $P(P(A))$.
Show Solution
Step 1: Find $P(A)$. Since $|A| = 2$, $P(A)$ has $2^2 = 4$ elements.
$P(A) = \{\phi, \{a\}, \{b\}, \{a,b\}\}$
Step 2: Find $P(P(A))$. Now $|P(A)| = 4$, so $P(P(A))$ has $2^4 = 16$ elements. It is the set of all subsets of $P(A)$, which includes $\phi$, all singletons, all pairs, all triples, and $P(A)$ itself.
Example 15: Real-life 3-Set Problem (NCERT Ex 1.6)Hard
In a group of 65 people, 40 like cricket, 10 like both cricket and tennis. How many like tennis only and not cricket? (Total tennis lovers = 35)
Q3(iii): Write the set $C = \{17, 26, 35, 44, 53, 62, 71, 80\}$ in Set-Builder form.
Pattern: $17 = 9(1)+8,\ 26=9(2)+8,\ ...$. But simpler: these are numbers of the form $9n+8$ for $n=1,...,8$, or equivalently numbers where digits add to 8 and they are two-digit numbers: $C = \{x : x = 9n+8,\ n \in \mathbb{N},\ 1 \le n \le 8\}$
Q4(i): Write the set in Set-Builder form: $\{3, 6, 9, 12\}$
$\{x : x = 3n,\ n \in \mathbb{N},\ 1 \le n \le 4\}$
Q5: Write as intervals: (i) $\{x \in \mathbb{R} : -4 < x \le 6\}$ (ii) $\{x \in \mathbb{R} : -12 < x < -10\}$
The set $\{a, b, \{a,b\}\}$ has 3 elements: $a$, $b$, and the set $\{a,b\}$. Since $\{a,b\}$ is listed as an element: YES, $\{a,b\} \in \{a, b, \{a,b\}\}$. TRUE.
Q4: How many elements does $P(A)$ have if $A = \phi$?
$|A| = 0$, so $|P(A)| = 2^0 = 1$. $P(\phi) = \{\phi\}$ — one element (the empty set itself).
Exercise 1.6 — Key Questions (Cardinality)
Q2: If $n(A) = 5$, $n(B) = 7$, $n(A \cup B) = 10$, find $n(A \cap B)$.
$n(A \cap B) = n(A) + n(B) - n(A \cup B) = 5 + 7 - 10 = 2$
Q8: In a committee, 50 people speak French, 20 speak Spanish, 10 speak both. How many speak at least one of these?
$n(F \cup S) = 50 + 20 - 10 = 60$ people speak at least one language.
Common Mistakes to Avoid
Trap 1: $\phi
e \{\phi\}$
$\phi$ is the empty set with zero elements. $\{\phi\}$ is a set containing the empty set — it has ONE element. They are completely different.
Trap 2: $\in$ vs $\subset$ — never mix these up
$x \in A$ says $x$ is an ELEMENT. $\{x\} \subset A$ says the set $\{x\}$ is a SUBSET. You can never write $x \subset A$ for a non-set element — it's meaningless.
Trap 3: $A - B
e B - A$
Set difference is not commutative. Always check the order — "A minus B" removes B's elements from A.
Trap 4: Equal sets vs Equivalent sets
Equal sets have the same elements. Equivalent sets just have the same number of elements. $\{1,2,3\}
e \{4,5,6\}$ but they are equivalent (both have 3 elements).
Quick Self-Check
Q1: Is $0 \in \phi$? (Is zero in the empty set?)
Answer
No. The empty set has NO elements. Nothing belongs to $\phi$.
Q2: If $n(A) = 5$, how many non-empty proper subsets does $A$ have?
Answer
$2^5 - 2 = 32 - 2 = 30$. (Subtract 1 for $\phi$ and 1 for $A$ itself.)