MathsCBSE Class 11 · Sets

Subsets, Intervals and Set-Builder Notation (Class 11)

Q: What is the difference between ⊂ and ⊆?

In NCERT, ⊂ means 'is a subset of' and allows equality, so A ⊂ A is true. A proper subset must also have A ≠ B, written 'A ⊂ B and A ≠ B' (or A ⊊ B). Some books write ⊆ for 'subset', with the same meaning as NCERT's ⊂.

Q: How do I write the set of real numbers greater than 3?

(3, ∞), with a round bracket at both ends.

Q: What is the difference between (a, b) as an interval and as an ordered pair?

They look the same, but the context tells you: in a set it is an interval, in coordinate geometry it is a point.

Q: Do I need the power set?

No, the NCERT section on the power set has been removed from the 2026-27 syllabus.

Updated 17 October 2026 · Fact-checked

A set can be written in roster form or set-builder form. A is a subset of B if every element of A is in B; it is proper if A ≠ B. Intervals are subsets of R: (a, b) excludes the end points, [a, b] includes them. |x − a| < r gives (a − r, a + r). Unions and intersections of intervals are found on the number line, taking care over open and closed ends.

Hollow dots leave the end point out (round bracket); filled dots include it (square bracket); |x−a|<r is an interval centred at a.

Illustration drawn for QuestionBuddy.

Understand Subsets, Intervals and Set-Builder Notation

Representations. *Roster form* lists the elements: ${- 1, 0, 1, 2, 3, 4}$ . *Set-builder form* gives the rule: ${x \in Z : - 2 < x \leq 4}$ . To convert, solve the condition in the stated domain ( $N$ , $Z$ , $R$ ) and list the values. The domain matters: ${x \in N : x < 4} = {1, 2, 3}$ but ${x \in Z : 0 \leq x < 4} = {0, 1, 2, 3}$ . Patterns can be written with a rule: ${1, 4, 9, 16, 25} = {n^{2} : n \in N, n \leq 5}$ .

Subsets. In NCERT, $A \subset B$ means every element of $A$ is in $B$ (equality is allowed). It is a proper subset if also $A \neq = B$ ; write this as ' $A \subset B$ and $A \neq = B$ ' (or $A ⊊ B$ ). The empty set $\emptyset$ is a subset of every set, and every set is a subset of itself. To decide a relation, list both sets and check element by element. For $A = {x : x^{2} - 5 x + 6 = 0} = {2, 3}$ and $B = {x \in N : x < 5} = {1, 2, 3, 4}$ , $A \subset B$ and $A \neq = B$ , so $A$ is a proper subset of $B$ .

Intervals. For real $a < b$ :

open $(a, b) = {x : a < x < b}$ ; closed $[a, b] = {x : a \leq x \leq b}$ ;
half-open $[a, b)$ and $(a, b]$ ;
rays $(a, \infty)$ , $[a, \infty)$ , $(- \infty, b)$ , $(- \infty, b]$ ; and $R = (- \infty, \infty)$ .

Square brackets include the end point; round brackets exclude it, and $\pm \infty$ always take round brackets.

Modulus conditions as intervals. $∣ x - a ∣ < r ⟺ a - r < x < a + r$ , i.e. $(a - r, a + r)$ . $∣ x - a ∣ \geq r$ gives $(- \infty, a - r] \cup [a + r, \infty)$ . The condition $1 \leq ∣ x ∣ \leq 3$ gives $[- 3, - 1] \cup [1, 3]$ .

Operations on intervals. Draw a number line, shade each set, and read the result. Remember the end points: with $A = [- 2, 3)$ and $B = (0, 5]$ ,

$A \cap B = (0, 3)$ (the point $3 \in / A$ , the point $0 \in / B$ );
$A \cup B = [- 2, 5]$ ;
$A - B = [- 2, 0]$ (here $0 \in A$ and $0 \in / B$ , so $0$ is kept);
$B - A = [3, 5]$ ( $3 \in B$ , $3 \in / A$ ).

Intervals joined by "or" with a gap stay as a union: $(- \infty, 2) \cup (3, 4)$ . The complement of an interval in $R$ is read off the same way.

Subsets, Intervals and Set-Builder Notation: formulas to remember

Open interval: $(a, b) = x : a < x < b$; Both ends excluded.
Closed interval: $[a, b] = x : a \leq x \leq b$; Both ends included.
Modulus inside: $∣ x - a ∣ < r g i v es (a - r, a + r)$; Distance less than r from a.
Modulus outside: $∣ x - a ∣ \geq r g i v es x \leq a - r or x \geq a + r$; Union of two rays.
Subset test: $A i s a s u b se t o f B i f x in A im pl i es x in B$; Check every element.

How to solve Subsets, Intervals and Set-Builder Notation questions

Use these steps for set-builder conversions, subset questions and intervals.

1Note the domain (N, Z or R) in a set-builder description.
2Solve the condition and list or shade the solution.
3For subsets, list both sets and check each element of $A$ in $B$ ; use "proper" if $A \neq = B$ .
4For intervals, draw a number line and mark each set with open or closed end points.
5For a union or intersection, read off the combined interval and test each end point separately.
6Convert modulus conditions to double inequalities (or two rays) first.

Subsets, Intervals and Set-Builder Notation in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Closed intervals as allowed ranges

Suppose a lift in a building can carry a load of at least 20 kg and at most 400 kg for it to start.

Write the set of valid loads as an interval and say whether 18 kg and 400 kg are valid.

Try it first, then show the solutionSolution

Valid loads $= {x \in R : 20 \leq x \leq 400} = [20, 400]$ .
$18 \in / [20, 400]$ : not valid.
$400 \in [20, 400]$ since the end point is included: valid.

Answer: $[20, 400]$ ; 18 kg is not valid, 400 kg is valid.

Memory hook: Square bracket: the end point is allowed.

Complement of an interval

Suppose a greenhouse must be kept between $1 8^{\circ}$ C and $2 8^{\circ}$ C (both included). An alarm rings whenever the temperature is outside this range.

Describe the alarm temperatures as a union of intervals.

Try it first, then show the solutionSolution

Safe set $= [18, 28]$ .
Alarm set $= R - [18, 28] = (- \infty, 18) \cup (28, \infty)$ .
Neither $18$ nor $28$ rings the alarm because they belong to the safe set.

Answer: $(- \infty, 18) \cup (28, \infty)$

Memory hook: The complement flips every bracket at the end points.

Tricks to remember Subsets, Intervals and Set-Builder Notation

Short memory aids. Say each one aloud once, then use it on a question the same day.

Square includes, round excludes
The bracket shows whether the end point belongs to the set.
Use it when: Reading and writing intervals.
Infinity gets a round bracket
You can never reach infinity, so ±∞ always take round brackets.
Use it when: Rays.
Distance picture
|x − a| < r reads: the distance of x from a is less than r.
Use it when: Modulus intervals.
Test the end points
After a union, intersection or difference, test each end point separately.
Use it when: Interval operations.

Subsets, Intervals and Set-Builder Notation: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Roster and set-builder forms · 2 marks · Very short answerBoard level

Write ${x \in N : x is a prime number less than 20}$ in roster form and ${3, 6, 9, 12, 15}$ in set-builder form.

Try it first, then show the solutionSolution

Primes below 20: ${2, 3, 5, 7, 11, 13, 17, 19}$ (eight elements). (1 mark)
${3, 6, 9, 12, 15} = {3 n : n \in N, n \leq 5}$ . (1 mark)

Answer: ${2, 3, 5, 7, 11, 13, 17, 19}$ ; ${3 n : n \in N, n \leq 5}$

Example 2: Operations on intervals · 3 marks · Short answerBoard level

Let $A = [- 2, 3)$ and $B = (0, 5]$ . Find $A \cap B$ , $A \cup B$ , $A - B$ and $B - A$ .

Try it first, then show the solutionSolution

$A \cap B = (0, 3)$ : common points are greater than $0$ (as $0 \in / B$ ) and less than $3$ (as $3 \in / A$ ). (1 mark)
$A \cup B = [- 2, 5]$ . (½ mark)
$A - B$ : points of $A$ not in $B$ are $x \leq 0$ in $A$ , so $[- 2, 0]$ . (1 mark)
$B - A$ : points of $B$ that are $\geq 3$ , so $[3, 5]$ . (½ mark)

Answer: $(0, 3)$ ; $[- 2, 5]$ ; $[- 2, 0]$ ; $[3, 5]$

Example 3: Deciding subset relations · 3 marks · Short answerHarder board level

Let $A = {x : x^{2} - 5 x + 6 = 0}$ , $B = {x \in N : x < 5}$ and $C = {x : x^{2} - 1 = 0}$ . List the sets and decide which are subsets of which.

Try it first, then show the solutionSolution

$A = {2, 3}$ , $B = {1, 2, 3, 4}$ , $C = {- 1, 1}$ . (1 mark)
Every element of $A$ is in $B$ and $A \neq = B$ , so $A \subset B$ and $A \neq = B$ ( $A$ is a proper subset of $B$ ). (1 mark)
$- 1 \in / B$ , so $C \neq \subset B$ ; also $A \neq \subset C$ and $C \neq \subset A$ . (1 mark)

Answer: $A \subset B$ (proper, as $A \neq = B$ ); $C \neq \subset B$

Example 4: Modulus conditions · 3 marks · Short answerHarder board level

Write as intervals (i) ${x \in R : ∣ x - 2∣ < 3}$ , (ii) ${x \in R : 1 \leq ∣ x ∣ \leq 3}$ .

Try it first, then show the solutionSolution

(i) $- 3 < x - 2 < 3$ , so $- 1 < x < 5$ : $(- 1, 5)$ . (1 mark)
(ii) $∣ x ∣ \geq 1$ gives $x \leq - 1$ or $x \geq 1$ ; $∣ x ∣ \leq 3$ gives $- 3 \leq x \leq 3$ . (1 mark)
Intersection: $[- 3, - 1] \cup [1, 3]$ . (1 mark)

Answer: (i) $(- 1, 5)$ (ii) $[- 3, - 1] \cup [1, 3]$

Example 5: Combining several interval sets · 5 marks · Long answerHOTS / competency

Let $A = {x \in R : x^{2} - 5 x + 6 \leq 0}$ , $B = {x \in R : ∣ x - 3∣ < 2}$ and $C = {x \in R : x \geq 4}$ . Find $A$ , $B$ , $A \cap B$ , $B - A$ , $(A \cup C)^{'}$ (complement in $R$ ) and $B \cap C$ .

Try it first, then show the solutionSolution

$A$ : $(x - 2) (x - 3) \leq 0$ gives $[2, 3]$ . $B$ : $- 2 < x - 3 < 2$ gives $(1, 5)$ . (1 mark)
$A \cap B = [2, 3]$ since $[2, 3] \subset (1, 5)$ . (1 mark)
$B - A = (1, 2) \cup (3, 5)$ . (1 mark)
$A \cup C = [2, 3] \cup [4, \infty)$ , so $(A \cup C)^{'} = (- \infty, 2) \cup (3, 4)$ . (1 mark)
$B \cap C = [4, 5)$ (4 is included from $C$ , 5 is excluded from $B$ ). (1 mark)

Answer: $A = [2, 3]$ ; $B = (1, 5)$ ; $A \cap B = [2, 3]$ ; $B - A = (1, 2) \cup (3, 5)$ ; $(A \cup C)^{'} = (- \infty, 2) \cup (3, 4)$ ; $B \cap C = [4, 5)$

Where students lose marks in Subsets, Intervals and Set-Builder Notation

The slips examiners see most often, and what to write instead.

Mistake: Wrong bracket at an end point
The inclusion of an end point is not checked.
Wrong working
$[- 2, 3) \cap (0, 5] = [0, 3]$ .
Right working
$(0, 3)$ .
Do this: Test each end point in both sets.
Usually costs: 1 mark.
Mistake: Ignoring the domain of a set-builder set
N, Z and R give different sets.
Wrong working
${x \in N : x < 4} = {0, 1, 2, 3}$ .
Right working
${1, 2, 3}$ .
Do this: Underline the domain.
Usually costs: The listing mark.
Mistake: Calling every subset proper
A set is a subset of itself.
Wrong working
Saying ' $A$ is a proper subset of $A$ ', or treating every $A \subset B$ as proper.
Right working
$A \subset A$ is true, but $A$ is not a proper subset of itself because $A = A$ . A proper subset needs $A \subset B$ and $A \neq = B$ .
Do this: Check whether the two sets are equal.
Usually costs: The reasoning mark.
Mistake: Writing |x| ≥ 1 as one interval
It has two separate pieces.
Wrong working
$∣ x ∣ \geq 1 \Rightarrow - 1 \leq x \leq 1$ (this is the answer for $∣ x ∣ \leq 1$ , not $∣ x ∣ \geq 1$ ).
Right working
$x \leq - 1$ or $x \geq 1$ .
Do this: Greater than means outside: use a union.
Usually costs: The interval mark.

Exam tips

Underline the domain in a set-builder description before listing.
Draw the number line for every interval question and test the end points.
Use "proper subset" only when the sets are different.
Write modulus conditions as a double inequality (or "or") before using intervals.
Show a membership check for each end point.

Practice questions on Subsets, Intervals and Set-Builder Notation

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
$[1, 4] \cap (3, 6)$ equals:
(a) $(3, 4]$
(b) $[3, 4]$
(c) $(3, 4)$
(d) $[1, 6)$
Hint
Check whether 3 and 4 belong to both sets.
Answer
(a). $3 \in / (3, 6)$ and $4 \in [1, 4]$ , $4 \in (3, 6)$ .
Q2 · 1 mark · MCQHarder board level
The set ${x \in R : ∣2 x - 1∣ \leq 3}$ is:
(a) $[- 1, 2]$
(b) $(- 1, 2)$
(c) $[- 2, 1]$
(d) $[- 1, 3]$
Hint
$- 3 \leq 2 x - 1 \leq 3$ .
Answer
(a). $- 2 \leq 2 x \leq 4$ , so $- 1 \leq x \leq 2$ .
Q3 · 3 marks · Short answerHarder board level
Let $A = {x \in N : x \leq 10}$ and $B = {x \in Z : - 2 < x < 5}$ . Write $A \cap B$ , $A \cup B$ and $B - A$ in roster form.
Hint
List $B$ first: integers greater than $- 2$ and less than $5$ .
Answer
$B = {- 1, 0, 1, 2, 3, 4}$ , $A = {1, \dots, 10}$ . $A \cap B = {1, 2, 3, 4}$ ; $A \cup B = {- 1, 0, 1, 2, \dots, 10}$ (12 elements); $B - A = {- 1, 0}$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): $(- \infty, 2) \cup (2, \infty) = R$ .
Reason (R): The number 2 belongs to neither of the two intervals.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Is 2 in the union?
Answer
(d). The union is $R - {2}$ , so A is false; R is true.
Q5 · 4 marks · Short answerHOTS / competency
Let $A = {x \in Z : x = 2 n, n \in Z}$ and $B = {x \in Z : x = 3 n, n \in Z}$ . Show that $A \cap B = {6 k : k \in Z}$ , and find three elements of $A - B$ and one integer in neither $A$ nor $B$ .
Hint
An integer divisible by 2 and by 3 is divisible by 6.
Answer
$x \in A \cap B$ means $x$ is divisible by both 2 and 3, hence by 6 (the lcm), so $x = 6 k$ ; conversely every $6 k$ is divisible by 2 and by 3. $A - B$ : even numbers not divisible by 3, e.g. $2, 4, 8$ . The integer $1$ is in neither set.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What does [a, b) mean?Show answer
a ≤ x < b.
2.Which bracket is used with infinity?Show answer
A round bracket.
3.What is |x − a| < r as an interval?Show answer
(a − r, a + r).
4.Is the empty set a subset of every set?Show answer
Yes.
5.What is {x in N : x < 4}?Show answer
{1, 2, 3}.

Subsets, Intervals and Set-Builder Notation: frequently asked questions

What is the difference between ⊂ and ⊆?

In NCERT, ⊂ means 'is a subset of' and allows equality, so A ⊂ A is true. A proper subset must also have A ≠ B, written 'A ⊂ B and A ≠ B' (or A ⊊ B). Some books write ⊆ for 'subset', with the same meaning as NCERT's ⊂.

How do I write the set of real numbers greater than 3?

(3, ∞), with a round bracket at both ends.

What is the difference between (a, b) as an interval and as an ordered pair?

They look the same, but the context tells you: in a set it is an interval, in coordinate geometry it is a point.

Do I need the power set?

No, the NCERT section on the power set has been removed from the 2026-27 syllabus.