NCERT Class 11 Mathematics — Sets, Exercise 1.3. All 8 questions solved.

This exercise turns on one distinction that catches almost everybody once: the difference between being an element of a set and being a subset of it.

• $A \subset B$ means every element of $A$ is also an element of $B$. It relates two sets.
• $a \in B$ means $a$ is one of the things listed inside $B$. It relates an object to a set.

Two conventions follow from the definition and are used throughout: $\varphi \subset A$ for every set $A$ (there is no element of $\varphi$ to fail the test), and $A \subset A$. A set with $n$ elements has exactly $2^n$ subsets.

The second half of the exercise introduces intervals, which are just subsets of $\mathbf{R}$ with a compact notation: a round bracket excludes the endpoint, a square bracket includes it.

Key insight. When a set contains another set as one of its members — as $A = \{1, 2, \{3,4\}, 5\}$ does — you must ask of every statement at what level it is operating. $\{3,4\} \in A$ is true because $\{3,4\}$ is literally one of the four things listed; $\{3,4\} \subset A$ is false because $3$ and $4$ are not themselves in $A$. Question 3 is nothing but this test, eleven times.

Question 1

Make correct statements by filling in the symbols $\subset$ or $\not\subset$ in the blank spaces:

(i) $\{2, 3, 4\} \ldots \{1, 2, 3, 4, 5\}$ (ii) $\{a, b, c\} \ldots \{b, c, d\}$ (iii) $\{x : x \text{ is a student of Class XI of your school}\} \ldots \{x : x \text{ is a student of your school}\}$ (iv) $\{x : x \text{ is a circle in the plane}\} \ldots \{x : x \text{ is a circle in the same plane with radius } 1 \text{ unit}\}$ (v) $\{x : x \text{ is a triangle in a plane}\} \ldots \{x : x \text{ is a rectangle in the plane}\}$ (vi) $\{x : x \text{ is an equilateral triangle in a plane}\} \ldots \{x : x \text{ is a triangle in the same plane}\}$ (vii) $\{x : x \text{ is an even natural number}\} \ldots \{x : x \text{ is an integer}\}$

Solution. In each part, ask whether every member of the left-hand collection is forced to be a member of the right-hand one. A single counterexample gives $\not\subset$.

(i) $\subset$. $2$, $3$ and $4$ all appear on the right.

(ii) $\not\subset$. $a$ is on the left but not on the right.

(iii) $\subset$. Every Class XI student of the school is a student of the school.

(iv) $\not\subset$. The containment runs the other way. Circles of radius $1$ are among all circles, but a circle of radius $5$ is not among the unit circles.

(v) $\not\subset$. A triangle has three sides and a rectangle four; the two collections have nothing in common at all.

(vi) $\subset$. Every equilateral triangle is a triangle.

(vii) $\subset$. Every even natural number — $2, 4, 6, \ldots$ — is an integer.

Question 2

Examine whether the following statements are true or false:

(i) $\{a, b\} \not\subset \{b, c, a\}$    (ii) $\{a, e\} \subset \{x : x \text{ is a vowel in the English alphabet}\}$ (iii) $\{1, 2, 3\} \subset \{1, 3, 5\}$    (iv) $\{a\} \subset \{a, b, c\}$ (v) $\{a\} \in \{a, b, c\}$    (vi) $\{x : x \text{ is an even natural number less than } 6\} \subset \{x : x \text{ is a natural number which divides } 36\}$

Solution.

(i) False. Both $a$ and $b$ do appear in $\{b, c, a\}$, so $\{a,b\}$ is a subset. The statement asserts the opposite, so it is false. Read the symbol carefully — the question is not whether $\{a,b\} \subset \{b,c,a\}$, but whether the printed statement is correct.

(ii) True. $a$ and $e$ are both vowels.

(iii) False. $2 \in \{1,2,3\}$ but $2 \notin \{1,3,5\}$.

(iv) True. $a$ is the only element on the left, and it appears on the right.

(v) False. The elements of $\{a, b, c\}$ are the letters $a$, $b$ and $c$ — not the set $\{a\}$. The correct statement is $\{a\} \subset \{a, b, c\}$, which is part (iv). Putting (iv) and (v) side by side is the whole point of the question.

(vi) True. The left set is $\{2, 4\}$. The divisors of $36$ are $1, 2, 3, 4, 6, 9, 12, 18, 36$, which contains both $2$ and $4$.

Question 3

Let $A = \{1, 2, \{3, 4\}, 5\}$. Which of the following statements are incorrect and why?

(i) $\{3,4\} \subset A$   (ii) $\{3,4\} \in A$   (iii) $\{\{3,4\}\} \subset A$ (iv) $1 \in A$   (v) $1 \subset A$   (vi) $\{1, 2, 5\} \subset A$ (vii) $\{1, 2, 5\} \in A$   (viii) $\{1, 2, 3\} \subset A$   (ix) $\varphi \in A$ (x) $\varphi \subset A$   (xi) $\{\varphi\} \subset A$

Solution. First, be completely clear about what $A$ contains. It has four elements: the numbers $1$, $2$ and $5$, and the set $\{3, 4\}$. The numbers $3$ and $4$ are not elements of $A$ — they are elements of an element of $A$.

(i) $\{3,4\} \subset A$ — incorrect. For this, $3$ and $4$ would each have to belong to $A$. Neither does.

(ii) $\{3,4\} \in A$ — correct. $\{3,4\}$ is one of the four listed members.

(iii) $\{\{3,4\}\} \subset A$ — correct. The single element of the left-hand set is $\{3,4\}$, and by (ii) that does belong to $A$.

(iv) $1 \in A$ — correct.

(v) $1 \subset A$ — incorrect. $\subset$ compares two sets, and $1$ is a number, not a set. The statement is not merely false, it is ill-formed. ($\{1\} \subset A$ would be correct.)

(vi) $\{1,2,5\} \subset A$ — correct. Each of $1$, $2$, $5$ belongs to $A$.

(vii) $\{1,2,5\} \in A$ — incorrect. $A$’s members are $1$, $2$, $\{3,4\}$ and $5$; the set $\{1,2,5\}$ is not among them. Compare with (vi): the same three numbers, but subset and element are not interchangeable.

(viii) $\{1,2,3\} \subset A$ — incorrect. $3 \notin A$.

(ix) $\varphi \in A$ — incorrect. The empty set is not one of the four listed members.

(x) $\varphi \subset A$ — correct. The empty set is a subset of every set, vacuously: there is no element of $\varphi$ that could fail to lie in $A$.

(xi) $\{\varphi\} \subset A$ — incorrect. The only element of $\{\varphi\}$ is $\varphi$, and by (ix) $\varphi \notin A$. Note this is a different reason from (ix)’s — here it is the containment that fails because of what $\{\varphi\}$ holds, not a claim about $\varphi$ being a member in its own right.

A note on the printed key. NCERT justifies both (ix) and (xi) with the same clause, “as $\varphi \subset A$”. That is the right reason for (ix) — $\varphi$ is a subset, not a member — but (xi) fails for the separate reason given above. The list of incorrect statements in the key is correct; only the shared justification is careless.

Question 4

Write down all the subsets of the following sets:

(i) $\{a\}$    (ii) $\{a, b\}$    (iii) $\{1, 2, 3\}$    (iv) $\varphi$

Solution. A subset is formed by deciding, for each element independently, whether to include it. With $n$ elements that is $2^n$ choices, which is exactly why an $n$-element set has $2^n$ subsets — and a useful check that none has been missed.

(i) $2^1 = 2$ subsets: $\varphi,\ \{a\}$

(ii) $2^2 = 4$ subsets: $\varphi,\ \{a\},\ \{b\},\ \{a, b\}$

(iii) $2^3 = 8$ subsets. Listing them by size keeps the count honest:

$$\varphi;\quad \{1\}, \{2\}, \{3\};\quad \{1,2\}, \{1,3\}, \{2,3\};\quad \{1,2,3\}$$

That is $1 + 3 + 3 + 1 = 8$ ✓.

(iv) $2^0 = 1$ subset: $\varphi$ itself. The empty set has exactly one subset, not none — a distinction worth pausing on.

Question 5

Write the following as intervals:

(i) $\{x : x \in \mathbf{R},\ -4 < x \le 6\}$    (ii) $\{x : x \in \mathbf{R},\ -12 < x < -10\}$ (iii) $\{x : x \in \mathbf{R},\ 0 \le x < 7\}$    (iv) $\{x : x \in \mathbf{R},\ 3 \le x \le 4\}$

Solution. Translate each inequality symbol into the matching bracket: strict $<$ becomes a round bracket (endpoint excluded), $\le$ becomes a square bracket (endpoint included). The two ends are decided independently, which is why mixed brackets like $(-4, 6]$ are perfectly ordinary.

Question 6

Write the following intervals in set-builder form:

(i) $(-3, 0)$    (ii) $[6, 12]$    (iii) $(6, 12]$    (iv) $[-23, 5)$

Solution. The same translation, run backwards. Note that (ii) and (iii) differ only at the left endpoint, and that difference has to survive into the inequality.

Question 7

What universal set(s) would you propose for each of the following:

(i) The set of right triangles.    (ii) The set of isosceles triangles.

Solution. A universal set is any set large enough to contain the set under discussion, chosen so that the discussion stays relevant. It is not unique — the skill is picking one that is neither too small to contain everything nor so large that it stops being informative.

Both collections are collections of triangles, so the natural choice for each is

$$U = \text{the set of all triangles in a plane}$$

The set of all polygons, or of all plane figures, would also work, but each is looser than it needs to be: nothing in either question ever concerns a quadrilateral, so widening $U$ that far only adds irrelevant material.

Question 8

Given the sets $A = \{1, 3, 5\}$, $B = \{2, 4, 6\}$ and $C = \{0, 2, 4, 6, 8\}$, which of the following may be considered as universal set(s) for all the three sets $A$, $B$ and $C$?

(i) $\{0, 1, 2, 3, 4, 5, 6\}$    (ii) $\varphi$ (iii) $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$    (iv) $\{1, 2, 3, 4, 5, 6, 7, 8\}$

Solution. A candidate qualifies only if it contains $A \cup B \cup C$ entirely. Collecting everything that must be inside:

$$A \cup B \cup C = \{0, 1, 2, 3, 4, 5, 6, 8\}$$

Now test each candidate against that list.

(i) Stops at $6$, so it misses $8 \in C$. Rejected.

(ii) $\varphi$ has no elements at all, so it contains none of $A$, $B$, $C$. Rejected.

(iii) Contains $0$ through $10$, which includes every one of $0, 1, 2, 3, 4, 5, 6, 8$. Accepted.

(iv) Starts at $1$, so it misses $0 \in C$. Rejected. This is the near miss worth checking twice — it reaches up to $8$ correctly and fails only at the bottom.

Common mistakes

• Question 2(i), answering the wrong question. The statement printed is $\{a,b\} \not\subset \{b,c,a\}$. Since $\{a,b\}$ is a subset, the statement is false. Students who work out the containment correctly and then write “True” have done the mathematics and lost the mark on the reading.
• Question 3, treating $\{3,4\}$ as though $3$ and $4$ were in $A$. They are not. $A$ has four elements, one of which happens to be a set. Writing out the four members explicitly before touching any part of the question prevents nearly every error here.
• Question 3(v), accepting $1 \subset A$. A number is not a set, so it cannot be a subset of anything. The correct form is $\{1\} \subset A$.
• Question 3(ix) and (x), treating $\varphi \in A$ and $\varphi \subset A$ as the same claim. The second is true for every set; the first is true only if somebody has deliberately put $\varphi$ inside the list.
• Question 4(iv), saying $\varphi$ has no subsets. It has exactly one, namely $\varphi$ — consistent with $2^0 = 1$.
• Question 5, using a square bracket at an infinite end or on a strict inequality. $(-4, 6]$ means $-4$ is out and $6$ is in; writing $[-4, 6]$ silently adds an element the set does not contain.
• Question 8, checking only the top of the range. Candidate (iv) reaches $8$ and looks convincing, but omits $0$. Form $A \cup B \cup C$ once, in full, and test every candidate against that single list.

Practise next

• Exercise 1.4 — union, intersection and difference. Question 8’s $A \cup B \cup C$ is the first appearance of the operation the next exercise is built on.
• Exercise 1.2 — worth revising if the equality arguments in question 3 felt uncertain, since $A = B$ is exactly $A \subset B$ together with $B \subset A$.