Chapter 1 · Sets

1.2 Subsets, Intervals and the Universal Set

$A\subset B$ when every element of $A$ is also in $B$. Intervals such as $[a,b)$ are the most important subsets of $\mathbb{R}$, and a universal set $U$ fixes what we are talking about.

A cricket team names a squad of $18$ players for a tournament, and before every match it picks a playing XI from that squad. Every player in the XI is a member of the squad, but not the other way round. This "lies inside" relationship between two sets is so common that it has its own name and symbol: the XI is a subset of the squad.

Subsets are everywhere in mathematics. Every natural number is an integer, every integer is a rational number, every rational number is real — a chain of subsets you have been using since Class 9 without naming it. The solutions of an inequality such as $2\le x\lt5$ form a special kind of subset of the real line called an interval, which you will need in Chapter 5 and in every chapter on functions and calculus. And whenever we solve a problem we work inside some background set — the integers, the real numbers, the set of all triangles — called the universal set. This section introduces all three ideas.

Subsets

📘 Definition 1.6 — Subset

A set $A$ is a subset of a set $B$, written $A\subset B$, if every element of $A$ is also an element of $B$: $$A\subset B\quad\text{means}\quad a\in A\ \Rightarrow\ a\in B .$$ If at least one element of $A$ is not in $B$, then $A$ is not a subset of $B$, written $A\not\subset B$.

In this course, as in the CBSE syllabus, $A\subset B$ allows $A=B$. (Some books write $A\subseteq B$ for this and keep $\subset$ for a subset that is smaller.)

To show $A\subset B$ you must check every element of $A$. To show $A\not\subset B$ one counterexample is enough — a single element of $A$ that is missing from $B$.

✏️ Example 1 — Subset or not?

Decide whether $A\subset B$ in each case, giving a reason. (a) $A=\{2,3\}$, $B=\{1,2,3,4\}$. (b) $A=\{x : x \text{ is an even natural number}\}$, $B=\{x : x \text{ is a natural number divisible by } 4\}$. (c) $A$ = the set of vowels, $B$ = the set of letters of the English alphabet. (d) $A=\{x : x\in\mathbb{R},\ x^2=4\}$, $B=\mathbb{N}$.

💡 Show solution▶

(a) $2\in B$ and $3\in B$: every element of $A$ is in $B$, so $A\subset B$.

(b) $2\in A$ but $2$ is not divisible by $4$, so $2\notin B$. Hence $A\not\subset B$. (The other way round is true: a number divisible by $4$ is even, so $B\subset A$.)

(d) $A=\{-2,2\}$ and $-2\notin\mathbb{N}$, so $A\not\subset\mathbb{N}$.

Three facts follow directly from the definition. They are used constantly, so we record them as a theorem.

📐 Theorem 1.1 — Basic properties of subsets

For any sets $A$, $B$, $C$:

$A\subset A$ (every set is a subset of itself);
$\phi\subset A$ (the empty set is a subset of every set);
$A=B$ if and only if $A\subset B$ and $B\subset A$;
if $A\subset B$ and $B\subset C$, then $A\subset C$.

Proof. (1) Every element of $A$ is an element of $A$. (2) To have $\phi\not\subset A$ we would need an element of $\phi$ that is not in $A$; but $\phi$ has no elements at all, so this cannot happen. (3) "$A\subset B$ and $B\subset A$" says every element of $A$ is in $B$ and every element of $B$ is in $A$ — exactly the definition of $A=B$. (4) If $x\in A$ then $x\in B$ (as $A\subset B$), and then $x\in C$ (as $B\subset C$). $\blacksquare$

📘 Definition 1.7 — Proper subset, superset, singleton

If $A\subset B$ and $A\ne B$, then $A$ is a proper subset of $B$, and $B$ is called a superset of $A$. A set with exactly one element, such as $\{5\}$, is called a singleton set.

✏️ Example 2 — Listing all the subsets

Write down all the subsets of (a) $\{a,b\}$ and (b) $\{1,2,3\}$. Which of them are proper subsets?

💡 Show solution▶

Method List the subsets by size — with $0$ elements, then $1$, then $2$, … — so that none is missed. Remember Theorem 1.1: $\phi$ and the set itself are always on the list.

(a) $\phi,\ \{a\},\ \{b\},\ \{a,b\}$: four subsets. All except $\{a,b\}$ are proper subsets.

(b) size 0: $\phi$; size 1: $\{1\},\{2\},\{3\}$; size 2: $\{1,2\},\{1,3\},\{2,3\}$; size 3: $\{1,2,3\}$. Eight subsets, of which seven are proper.

Check Building a subset of $\{1,2,3\}$ means deciding, for each of the three elements, "in" or "out": that is $2\times2\times2=8$ choices, matching our list. ✓ (You will meet this counting idea properly in Chapter 6.)

✏️ Example 3 — Proving two sets equal by two inclusions

Let $A=\{x : x=6k,\ k\in\mathbb{Z}\}$ and $B=\{x : x\in\mathbb{Z},\ x \text{ is divisible by both } 2 \text{ and } 3\}$. Prove that $A=B$.

💡 Show solution▶

Plan By Theorem 1.1(3), it is enough to show $A\subset B$ and $B\subset A$.

$A\subset B$ Let $x\in A$, so $x=6k=2(3k)=3(2k)$. Then $x$ is divisible by $2$ and by $3$, so $x\in B$.

$B\subset A$ Let $x\in B$. Since $2$ divides $x$, write $x=2m$. Now $3$ divides $2m$; since $3$ is a prime that does not divide $2$, it must divide $m$ (by the Fundamental Theorem of Arithmetic, Class 10). So $m=3k$ and $x=6k$, i.e. $x\in A$.

Conclude $A\subset B$ and $B\subset A$, hence $A=B$. $\blacksquare$

The inspector visits every element of $A$ and asks "is it in $B$?", then visits every element of $B$ and asks "is it in $A$?". One "no" is enough to break a subset relation. The picture places each element in its region inside the universal set $U$: when $A\subset B$, the circle of $A$ sits inside the circle of $B$. Try the pair with $A=\phi$ — there is nothing to check, so nothing can fail.

Pair:

now checking—

$A\subset B$?—

$B\subset A$?—

conclusion—

⚠️ "Belongs to" ($\in$) is not "is a subset of" ($\subset$)

$\in$ connects an element to a set; $\subset$ connects a set to a set. So $1\in\{1,2\}$ and $\{1\}\subset\{1,2\}$ are both correct, while $1\subset\{1,2\}$ and $\{1\}\in\{1,2\}$ are both wrong.

Things get interesting when a set has other sets as elements. If $A=\{1,\{2,3\}\}$, then $\{2,3\}\in A$ (it is one of the two elements), but $\{2,3\}\not\subset A$ (because $2\notin A$). Example 4 practises exactly this.

Also: $\phi\subset A$ is always true, but $\phi\in A$ is true only when $\phi$ is literally listed as an element of $A$.

✏️ Example 4 — Sets whose elements are sets

Let $A=\{1,\{2,3\},4\}$. Which of the following are true? (i) $\{2,3\}\in A$ (ii) $\{2,3\}\subset A$ (iii) $\{\{2,3\}\}\subset A$ (iv) $\{1,4\}\subset A$ (v) $2\in A$ (vi) $\phi\subset A$ (vii) $\phi\in A$

💡 Show solution▶

List $A$ has exactly three elements: the number $1$, the set $\{2,3\}$, and the number $4$. So $n(A)=3$.

(i)–(iii) (i) True: $\{2,3\}$ is one of the three elements. (ii) False: for $\{2,3\}\subset A$ we would need $2\in A$, and it is not. (iii) True: the only element of $\{\{2,3\}\}$ is $\{2,3\}$, which belongs to $A$.

(iv)–(v) (iv) True: $1\in A$ and $4\in A$. (v) False: $2$ is inside an element of $A$, not itself an element.

(vi)–(vii) (vi) True for every set (Theorem 1.1). (vii) False: $\phi$ is not one of the three listed elements.

Subsets of the Real Numbers

The number systems you know fit inside one another: $$\mathbb{N}\subset\mathbb{Z}\subset\mathbb{Q}\subset\mathbb{R}.$$ Each inclusion is proper: $0\in\mathbb{Z}$ but $0\notin\mathbb{N}$; $\tfrac12\in\mathbb{Q}$ but $\tfrac12\notin\mathbb{Z}$; $\sqrt2\in\mathbb{R}$ but $\sqrt2\notin\mathbb{Q}$. The real numbers that are not rational form the set of irrational numbers, $T=\{x : x\in\mathbb{R},\ x\notin\mathbb{Q}\}$ — for example $\sqrt2$, $\sqrt3$, $\pi$ (see Class 10 Mathematics, Section 1.3 Revisiting Irrational Numbers). Every real number is rational or irrational, never both, so $\mathbb{Q}\not\subset T$, $T\not\subset\mathbb{Q}$ and $T\subset\mathbb{R}$.

Intervals

When we solve $2\le x\lt5$ the answer is not a short list of numbers — it is a whole unbroken stretch of the number line. Such stretches are called intervals, and they have a compact notation in which a square bracket means "the end point is included" and a round bracket means "the end point is excluded".

📘 Definition 1.8 — Intervals

Let $a,b\in\mathbb{R}$ with $a\lt b$.

Name	Notation	Set-builder form	On the number line
open interval	$(a,b)$	$\{x : x\in\mathbb{R},\ a\lt x\lt b\}$	both end points hollow ○
closed interval	$[a,b]$	$\{x : x\in\mathbb{R},\ a\le x\le b\}$	both end points filled ●
closed–open (half-open)	$[a,b)$	$\{x : x\in\mathbb{R},\ a\le x\lt b\}$	● at $a$, ○ at $b$
open–closed (half-open)	$(a,b]$	$\{x : x\in\mathbb{R},\ a\lt x\le b\}$	○ at $a$, ● at $b$

The number $b-a$ is called the length of each of these intervals. Unbounded intervals use the symbols $\infty$ and $-\infty$: $(a,\infty)=\{x : x\gt a\}$, $[a,\infty)=\{x : x\ge a\}$, $(-\infty,b)=\{x : x\lt b\}$, $(-\infty,b]=\{x : x\le b\}$, and $(-\infty,\infty)=\mathbb{R}$.

⚠️ Common mistakes with intervals

1. $\infty$ is not a number. It only says "no end on this side", so it always takes a round bracket: $[2,\infty)$, never $[2,\infty]$.

2. The smaller number comes first. $(5,2)$ is meaningless; write $(2,5)$.

3. $(2,5)$ is not $\{3,4\}$. An interval contains all real numbers between its ends, such as $2.001$ and $\pi$, not just the integers.

4. The same notation $(a,b)$ also means an ordered pair (Chapter 2). The context — "the interval $(a,b)$" or "the point $(a,b)$" — tells you which.

✏️ Example 5 — Working with intervals

(a) Write in interval notation: (i) $\{x : x\in\mathbb{R},\ -4\lt x\le6\}$ (ii) $\{x : x\in\mathbb{R},\ 0\le x\lt7\}$ (iii) $\{x : x\in\mathbb{R},\ x\ge2\}$.
(b) Write $[-3,5)$ in set-builder form.
(c) Find the length of $(-2.5,3]$ and the number of integers it contains.
(d) Is $(2,5)\subset[2,5]$? Is $[2,5]\subset(2,5)$?

💡 Show solution▶

(a) (i) $-4$ excluded, $6$ included: $(-4,6]$. (ii) $[0,7)$. (iii) $[2,\infty)$.

(b) $\{x : x\in\mathbb{R},\ -3\le x\lt5\}$.

(d) Every $x$ with $2\lt x\lt5$ also satisfies $2\le x\le5$, so $(2,5)\subset[2,5]$ — a proper subset, since $2\in[2,5]$ but $2\notin(2,5)$. For the same reason $[2,5]\not\subset(2,5)$.

Choose the type of interval and move the end points. The shaded stretch is the interval; a filled dot means the end point belongs to it, a hollow dot means it does not. The test point $x$ sweeps along the line — watch the membership verdict flip exactly at the end points, and look closely at what happens at a hollow end.

Type:$a$:−2$b$:3

interval notation—

set-builder form—

length—

integers in it—

test point $x$—

The Universal Set

Whether $x^2=2$ "has a solution" depends on where you are looking: among rational numbers it has none, among real numbers it has two. Every discussion takes place inside some background set, and it is good practice to say which one.

📘 Definition 1.9 — Universal set

In a particular discussion, a set that contains all the sets under consideration as subsets is called the universal set for that discussion, and is denoted by $U$. It is usually drawn as a rectangle, with the other sets as regions inside it.

The universal set is not fixed once and for all — it depends on the context. In number theory we might take $U=\mathbb{Z}$; when studying triangles, $U$ could be the set of all triangles in a plane; for the human population census, $U$ could be the set of all people living in India. In Section 1.3 the rectangle becomes the frame of every Venn diagram.

✏️ Example 6 — Choosing a universal set

(a) Let $A=\{1,3,5\}$, $B=\{2,4,6\}$, $C=\{0,2,4,6,8\}$. Which of the following can be taken as a universal set for all three? (i) $\{0,1,2,3,4,5,6\}$ (ii) $\phi$ (iii) $\{0,1,2,3,\dots,10\}$ (iv) $\{1,2,3,4,5,6,7,8\}$
(b) Find the solution set of $x^2=2$ when the universal set is (i) $\mathbb{Q}$, (ii) $\mathbb{R}$.

💡 Show solution▶

Test A universal set must contain every element of $A$, $B$ and $C$, that is, every element of $\{0,1,2,3,4,5,6,8\}$.

(a) (i) No — $8\in C$ is missing. (ii) No — $\phi$ contains nothing. (iii) Yes. (iv) No — $0\in C$ is missing.

(b) The solutions of $x^2=2$ are $x=\pm\sqrt2$, which are irrational. (i) With $U=\mathbb{Q}$ the solution set is $\phi$. (ii) With $U=\mathbb{R}$ it is $\{-\sqrt2,\sqrt2\}$.

🚀 Beyond the board syllabus (JEE Main)

In JEE Main 2026, which lists the power set by name; not in the CBSE year-end exam, and not named in JEE Advanced 2026.

The set of all subsets of $A$ is called the power set of $A$, written $P(A)$. Example 2 showed $P(\{a,b\})=\{\phi,\{a\},\{b\},\{a,b\}\}$. Since each element of $A$ is either in or out of a subset, independently, a set with $n$ elements has $$n\big(P(A)\big)=2^n$$ subsets, of which $2^n-1$ are proper subsets and $2^n-2$ are non-empty proper subsets.

Worked example. Find $n\big(P(P(\phi))\big)$, and the number of elements of $A$ if $A$ has $63$ proper subsets.
Solution. $P(\phi)=\{\phi\}$ has one element, so $P(P(\phi))=\{\phi,\{\phi\}\}$ has $2^1=2$ elements. For the second part, $2^n-1=63$ gives $2^n=64$, so $n(A)=6$.

Exercises

Subsets. In parts 1–6 insert $\subset$ or $\not\subset$.

1 $\{2,4\}\ \square\ \{1,2,3,4,5\}$
2 $\{a,b,c\}\ \square\ \{b,c,d\}$
3 $\{x : x \text{ is a student of Class 11 of your school}\}\ \square\ \{x : x \text{ is a student of your school}\}$
4 $\{x : x \text{ is a square in a plane}\}\ \square\ \{x : x \text{ is a rectangle in the same plane}\}$
5 $\{x : x \text{ is a rectangle in a plane}\}\ \square\ \{x : x \text{ is a square in the same plane}\}$
6 $\{x : x\in\mathbb{N},\ x \text{ is a multiple of } 6\}\ \square\ \{x : x\in\mathbb{N},\ x \text{ is a multiple of } 3\}$
7 Let $A=\{1,2,\{3,4\},5\}$. Which are true? (a) $\{3,4\}\subset A$ (b) $\{3,4\}\in A$ (c) $\{\{3,4\}\}\subset A$ (d) $1\in A$ (e) $\{1,2,5\}\subset A$ (f) $\phi\in A$ (g) $\phi\subset A$
8 True or false: every set has at least two subsets.
9 If $A\subset B$ and $B\subset A$, what can you conclude? Which result justifies it?

💡 Show solutions▶

1 $\subset$: $2$ and $4$ are both in the larger set.

2 $\not\subset$: $a\notin\{b,c,d\}$.

3 $\subset$: every Class 11 student of your school is a student of your school.

4 $\subset$: a square is a rectangle with equal sides.

5 $\not\subset$: a $2\times3$ rectangle is not a square.

6 $\subset$: $6k=3(2k)$, so every multiple of $6$ is a multiple of $3$.

7 The elements of $A$ are $1$, $2$, $\{3,4\}$ and $5$. (a) False ($3\notin A$). (b) True. (c) True. (d) True. (e) True. (f) False. (g) True.

8 False: $\phi$ has only one subset, $\phi$ itself. (Every non-empty set has at least two: $\phi$ and itself.)

9 $A=B$, by Theorem 1.1(3).

Listing subsets.

1 Write all the subsets of $\{a\}$.
2 Write all the subsets of $\{1,2\}$.
3 Write all the subsets of $\{x,y,z\}$.
4 Write all the proper subsets of $\{-1,0,1\}$.
5 Write all the subsets of $\phi$.
6 Find all sets $X$ such that $\{1\}\subset X\subset\{1,2,3\}$.
7 Is $\{0\}\subset\mathbb{N}$? Is $\{0\}\subset\mathbb{Z}$?

💡 Show solutions▶

1 $\phi,\ \{a\}$.

2 $\phi,\ \{1\},\ \{2\},\ \{1,2\}$.

3 $\phi$; $\{x\},\{y\},\{z\}$; $\{x,y\},\{x,z\},\{y,z\}$; $\{x,y,z\}$ — eight subsets.

4 All subsets except the set itself: $\phi,\ \{-1\},\{0\},\{1\},\ \{-1,0\},\{-1,1\},\{0,1\}$ — seven proper subsets.

5 Only $\phi$ itself.

6 $X$ must contain $1$, and may or may not contain $2$ and $3$: $\{1\},\ \{1,2\},\ \{1,3\},\ \{1,2,3\}$ — four sets.

7 $\{0\}\not\subset\mathbb{N}$ because $0\notin\mathbb{N}$; $\ \{0\}\subset\mathbb{Z}$ because $0\in\mathbb{Z}$.

III

Intervals. In parts 1–5 write the set as an interval; in parts 6–7 write the interval in set-builder form.

1 $\{x : x\in\mathbb{R},\ -3\lt x\lt0\}$
2 $\{x : x\in\mathbb{R},\ 6\le x\le12\}$
3 $\{x : x\in\mathbb{R},\ 6\lt x\le12\}$
4 $\{x : x\in\mathbb{R},\ -23\le x\lt5\}$
5 $\{x : x\in\mathbb{R},\ x\lt1\}$
6 $(-2,7]$
7 $[-1,\infty)$
8 Find the length of $[3,7.5)$ and list the integers in it.
9 How many integers lie in $(-3.2,4.8)$?
10 Is $[1,4]\subset(0,5)$? Is $(0,5)\subset[1,4]$?
11 Medicine. A vaccine must be stored at a temperature $T$ from $2\,^\circ\text{C}$ to $8\,^\circ\text{C}$, both included. Write the safe range as an interval. Is $8\,^\circ\text{C}$ safe? Is $1.5\,^\circ\text{C}$?
12 Grading. A school awards grade A for marks $m$ with $80\le m\lt90$. Write this as an interval and decide whether a student with $90$ marks gets grade A.

Check: see the interval on the number line, with the integers it contains

Part:

interval—

length—

integers in it—

💡 Show solutions▶

1 $(-3,0)$.

2 $[6,12]$.

3 $(6,12]$.

4 $[-23,5)$.

5 $(-\infty,1)$.

6 $\{x : x\in\mathbb{R},\ -2\lt x\le7\}$.

7 $\{x : x\in\mathbb{R},\ x\ge-1\}$.

8 Length $7.5-3=4.5$; integers $3,4,5,6,7$ (five of them).

9 $-3,-2,\dots,4$: eight integers ($-3\gt-3.2$ is inside; $5\gt4.8$ is not).

10 If $1\le x\le4$ then $0\lt x\lt5$, so $[1,4]\subset(0,5)$. But $0.5\in(0,5)$ and $0.5\notin[1,4]$, so $(0,5)\not\subset[1,4]$.

11 $T\in[2,8]$ (in °C). $8\,^\circ\text{C}$ is safe (closed end); $1.5\,^\circ\text{C}$ is not, since $1.5\lt2$.

12 $m\in[80,90)$. A student with $90$ marks does not get grade A — $90$ is excluded and starts the next grade band.

Universal sets and proofs.

1 Let $A=\{2,3\}$, $B=\{3,5,7\}$, $C=\{1,9\}$. Which of these can be a universal set for $A$, $B$ and $C$? (a) $\{1,2,3,5,7\}$ (b) $\{1,2,3,\dots,9\}$ (c) $\mathbb{N}$ (d) the set of prime numbers
2 Find the solution set of $x^2-2=0$ when the universal set is (a) $\mathbb{Q}$, (b) $\mathbb{R}$.
3 Find the solution set of $2x+3=0$ when the universal set is (a) $\mathbb{N}$, (b) $\mathbb{Z}$, (c) $\mathbb{Q}$.
4 Prove that if $A\subset\phi$ then $A=\phi$.
5 Let $A=\{x : x=4n,\ n\in\mathbb{Z}\}$ and $B=\{x : x=2n,\ n\in\mathbb{Z}\}$. Show that $A$ is a proper subset of $B$.
6 Challenge. Give an example of sets $A$ and $B$ with both $A\in B$ and $A\subset B$.
7 Challenge. Let $A=\{x\in\mathbb{R} : x^2-5x+6=0\}$. Find all sets $B$ with $B\subset A$, and decide whether $[2,3]\subset A$.

💡 Show solutions▶

1 We need $\{1,2,3,5,7,9\}$ inside $U$. (a) No ($9$ missing). (b) Yes. (c) Yes. (d) No ($1$ and $9$ are not prime).

2 $x=\pm\sqrt2$, irrational. (a) $\phi$. (b) $\{-\sqrt2,\sqrt2\}$.

3 $x=-\tfrac32$. (a) $\phi$. (b) $\phi$. (c) $\left\{-\tfrac32\right\}$.

4 By Theorem 1.1(2), $\phi\subset A$. Together with $A\subset\phi$, Theorem 1.1(3) gives $A=\phi$. (Directly: an element of $A$ would have to lie in $\phi$, which has none.)

5 If $x=4n$ then $x=2(2n)$ with $2n\in\mathbb{Z}$, so $A\subset B$. But $2\in B$ ($n=1$) and $2=4n$ would need $n=\tfrac12\notin\mathbb{Z}$, so $2\notin A$ and $A\ne B$. Hence $A$ is a proper subset of $B$.

6 $A=\{1\}$, $B=\{1,\{1\}\}$: $\{1\}$ is listed as an element of $B$, so $A\in B$; and $1\in B$, so $A\subset B$. (Another: $A=\phi$, $B=\{\phi\}$.)

7 $A=\{2,3\}$. Its subsets are $\phi,\ \{2\},\ \{3\},\ \{2,3\}$. $[2,3]\not\subset A$, since for example $2.5\in[2,3]$ but $2.5\notin A$ — do not confuse the interval $[2,3]$ with the set $\{2,3\}$.

🎯 Exam corner Board · 1 markBoard · 2 marksJEE MainJEE Advanced

How this is examined: Board papers ask 1-mark questions on $\in$ versus $\subset$ and on the empty set, and 2-mark questions converting between set-builder form and intervals; JEE counts subsets (using $2^n$ from the 🚀 box) and sets traps with sets whose elements are themselves sets.

E1. Board · 1 mark Which of the following is not a subset of $\{1,2,3\}$? (A) $\phi$ (B) $\{1,3\}$ (C) $\{1,2,3\}$ (D) $\{0,1\}$

💡 Show solution▶

Idea $0\notin\{1,2,3\}$, so $\{0,1\}\not\subset\{1,2,3\}$. The others are subsets: $\phi$ is a subset of every set, and every set is a subset of itself.

Trap Students often pick (A) or (C), forgetting Theorem 1.1(1)–(2).

Answer (D).

E2. Board · 2 marks Write as intervals and state how many integers each contains: (i) $\{x : x\in\mathbb{R},\ -5\lt x\le2\}$ (ii) $\{x : x\in\mathbb{R},\ 3\le x\lt8\}$.

💡 Show solution▶

(i) $(-5,2]$; integers $-4,-3,\dots,2$: seven. (1 mark)

(ii) $[3,8)$; integers $3,4,5,6,7$: five. (1 mark)

Answer $(-5,2]$ with $7$ integers; $[3,8)$ with $5$ integers.

E3. Board · 1 mark Assertion (A): $\phi\subset\{\phi\}$. Reason (R): The empty set is a subset of every set. Choose: (A) both A and R are true and R is the correct explanation of A; (B) both 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.

💡 Show solution▶

Idea R is Theorem 1.1(2), true. Applying it to the set $\{\phi\}$ gives exactly A. (Separately, $\phi\in\{\phi\}$ is also true, but that is not what A says.)

Answer (A).

E4. JEE Main The number of sets $X$ such that $\{a,b\}\subset X\subset\{a,b,c,d,e\}$ is (A) $4$ (B) $8$ (C) $16$ (D) $32$

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Idea $X$ must contain $a$ and $b$. Each of $c$, $d$, $e$ may independently be in or out.

Count $2\times2\times2=8$ choices.

Trap (D) $=2^5$ counts all subsets of $\{a,b,c,d,e\}$, ignoring the condition $\{a,b\}\subset X$.

Answer (B). ⏱ Fast route: $2^{(\text{elements of the big set})-(\text{elements forced in})}=2^{5-2}$.

E5. JEE Main Numerical value. A set $A$ has $256$ subsets. Find the number of non-empty proper subsets of $A$.

💡 Show solution▶

Idea $2^n=256$ gives $n(A)=8$.

Remove From the $256$ subsets remove $\phi$ (empty) and $A$ itself (not proper): $256-2$.

Answer $254$.

E6. JEE Advanced One or more options correct. Let $A=\{1,\{2\},\{1,2\}\}$. Which of the following are true? (A) $\{2\}\in A$ (B) $\{1,2\}\subset A$ (C) $\{1,\{2\}\}\subset A$ (D) $A$ has exactly $6$ subsets

💡 Show solution▶

List The three elements of $A$ are $1$, $\{2\}$ and $\{1,2\}$; the number $2$ is not an element.

Options (A) True. (B) False: $2\notin A$ (although $\{1,2\}\in A$). (C) True: $1\in A$ and $\{2\}\in A$. (D) False: $n(A)=3$, so $A$ has $2^3=8$ subsets.

Answer (A), (C).

🔬 Research–Teaching NexusMini project · about 45 minutes

Research question. How are real-world "bands" — air-quality categories, BMI classes, tax slabs, exam grades — written as intervals, and do the published bands overlap or leave gaps?

Why it matters. A band system only works if every possible value falls in exactly one band. That requires the intervals to fit end to end, with each boundary value assigned to one side by a careful choice of $[$ or $($. Badly written bands ("0–50, 50–100") leave a reading of exactly $50$ in two categories at once — a real source of confusion in official tables.

What to do.

Find a published band table: India's Air Quality Index categories, the WHO adult BMI classes, the slabs of an income-tax regime, or your school's grading scheme.
Rewrite every band in interval notation, deciding each end point from the wording ("up to", "above", "and over").
Draw all the bands on one number line. Check: is every value covered? Is any value in two bands?
If you find a gap or an overlap, rewrite the table so that the bands are of the form $[a_1,a_2),[a_2,a_3),[a_3,a_4),\dots$ and explain why this form never fails.

Tools: Phone browser for the table; paper number line or a drawing app

Hand in: The table in interval notation, the number-line picture, and a two-sentence verdict on whether the original wording was unambiguous.

Going further. Measured values are rounded (an AQI of $100.4$ may be reported as $100$). How does rounding interact with an open or closed end point? Discuss with a partner.

Instructor note: many published tables use integer ranges such as "51–100", which are unambiguous only for whole-number readings; students should notice that for real-valued readings the gap $(50,51)$ appears. The half-open form $[a_k,a_{k+1})$ is the standard fix.

🧪 Self-check quiz

0/5 correct

Q1. Which statement is true?

Q2. $\{x : x\in\mathbb{R},\ -1\le x\lt4\}$ in interval notation is

Q3. How many subsets does $\{p,q\}$ have?

Q4. The length of the interval $(-2,5]$ is

Q5. Which can be a universal set for $A=\{1,2\}$, $B=\{2,3\}$ and $C=\{5\}$?

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