NCERT Class 11 Mathematics — Sets, Miscellaneous Exercise on Chapter 1. All 10 questions solved.

This exercise is a change of register. The earlier exercises asked you to compute with sets; this one asks you to prove things about them, and to disprove things by producing a counterexample. Almost every proof here uses the same opening move:

To show $P \subset Q$, take an arbitrary $x \in P$ and argue that $x$ must lie in $Q$. To show $P = Q$, do that twice, once in each direction.

The algebraic identities used are the distributive laws and the two absorption laws:

$$A \cap (B \cup C) = (A \cap B) \cup (A \cap C), \qquad A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$$ $$A \cap (A \cup B) = A, \qquad A \cup (A \cap B) = A$$

Key insight. A statement of the form “if $\ldots$ then $\ldots$” is disproved by one counterexample and proved only by an argument covering every case. Question 2 alternates between the two, and knowing which job you are doing before you start is most of the work. When hunting a counterexample, the smallest sets that could possibly work — one or two elements, sometimes a set containing a set — are where to look first.

Question 1

Decide, among the following sets, which sets are subsets of one and another:

$$A = \{x : x \in \mathbf{R} \text{ and } x \text{ satisfies } x^2 – 8x + 12 = 0\},\quad B = \{2, 4, 6\},$$ $$C = \{2, 4, 6, 8, \ldots\},\quad D = \{6\}$$

Solution. The one set not given in roster form is $A$, so solve its equation first:

$$x^2 – 8x + 12 = (x – 2)(x – 6) = 0 \quad\Longrightarrow\quad x = 2 \text{ or } x = 6$$

$$A = \{2, 6\}$$

$C$ is the set of even natural numbers. Now compare each pair, smallest first:

• $D = \{6\}$: its only element $6$ lies in $A$, in $B$ and in $C$. So $D \subset A$, $D \subset B$, $D \subset C$.
• $A = \{2, 6\}$: both elements lie in $B$ and in $C$. So $A \subset B$, $A \subset C$.
• $B = \{2, 4, 6\}$: all three are even, so $B \subset C$.
• Nothing runs the other way: $4 \in B$ but $4 \notin A$, and $8 \in C$ but $8 \notin B$.

The four sets therefore form a chain, $D \subset A \subset B \subset C$, and every containment above follows from that single chain by transitivity — which is question 2(iii) put to work.

Question 2

In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example.

(i) If $x \in A$ and $A \in B$, then $x \in B$ (ii) If $A \subset B$ and $B \in C$, then $A \in C$ (iii) If $A \subset B$ and $B \subset C$, then $A \subset C$ (iv) If $A \not\subset B$ and $B \not\subset C$, then $A \not\subset C$ (v) If $x \in A$ and $A \not\subset B$, then $x \in B$ (vi) If $A \subset B$ and $x \notin B$, then $x \notin A$

Solution.

(i) False. Membership is not transitive, because $A \in B$ says $A$ is one of $B$’s objects, not that $A$’s contents are. Take

$$A = \{1\},\qquad B = \big\{\{1\},\ 2\big\}$$

Then $1 \in A$ and $A \in B$, but $B$’s elements are $\{1\}$ and $2$ — the bare number $1$ is not among them, so $1 \notin B$.

(ii) False. Mixing $\subset$ and $\in$ in one chain fails for the same reason. Take

$$A = \{1\},\qquad B = \{1, 2\},\qquad C = \big\{\{1, 2\},\ 3\big\}$$

Then $A \subset B$ and $B \in C$, but $C$’s elements are $\{1,2\}$ and $3$, and $\{1\}$ is neither. So $A \notin C$.

(iii) True. Let $x \in A$. Since $A \subset B$, we get $x \in B$; since $B \subset C$, we get $x \in C$. As $x$ was arbitrary, every element of $A$ lies in $C$, which is exactly $A \subset C$.

(iv) False. Two failures of containment say nothing about a third. Take

$$A = \{1\},\qquad B = \{2\},\qquad C = \{1, 3\}$$

Then $A \not\subset B$ (since $1 \notin B$) and $B \not\subset C$ (since $2 \notin C$), yet $A \subset C$, because $1 \in C$.

(v) False. $A \not\subset B$ means some element of $A$ is outside $B$ — it says nothing about the particular $x$ you happen to have chosen, and certainly does not force it into $B$. Take

$$A = \{1, 2\},\qquad B = \{1\}$$

Here $A \not\subset B$ because $2 \notin B$. Choosing $x = 2$ gives $x \in A$ but $x \notin B$.

(vi) True. This is the contrapositive of the definition of subset. Suppose, for contradiction, that $x \in A$. Since $A \subset B$, this would give $x \in B$, contradicting $x \notin B$. Hence $x \notin A$.

Question 3

Let $A$, $B$ and $C$ be sets such that $A \cup B = A \cup C$ and $A \cap B = A \cap C$. Show that $B = C$.

Solution. Neither hypothesis alone is enough — $A \cup B = A \cup C$ holds for $A = \{1\}$, $B = \{1\}$, $C = \varphi$, where $B \ne C$ — so the proof must use both. The trick is to start from the absorption law and substitute.

$$B = B \cap (A \cup B) \qquad \text{(absorption)}$$

Now replace $A \cup B$ by $A \cup C$, which the first hypothesis permits:

$$B = B \cap (A \cup C) = (B \cap A) \cup (B \cap C) \qquad \text{(distributive law)}$$

The second hypothesis says $B \cap A = A \cap B = A \cap C$, so

$$B = (A \cap C) \cup (B \cap C) = (A \cup B) \cap C \qquad \text{(distributive law, backwards)}$$

Applying the first hypothesis once more, $A \cup B = A \cup C$, and then absorption in the other direction:

$$B = (A \cup C) \cap C = C$$

Hence $B = C$. $\blacksquare$

Notice where each hypothesis entered: the union equality was used twice, at the start and the end, and the intersection equality once in the middle. Dropping either breaks the chain.

Question 4

Show that the following four conditions are equivalent:

(i) $A \subset B$    (ii) $A – B = \varphi$    (iii) $A \cup B = B$    (iv) $A \cap B = A$

Solution. Four statements are equivalent when any one implies any other. Proving all twelve implications would be wasteful; instead close a cycle

$$\text{(i)} \Rightarrow \text{(ii)} \Rightarrow \text{(iii)} \Rightarrow \text{(iv)} \Rightarrow \text{(i)}$$

after which any statement reaches any other by going round.

(i) $\Rightarrow$ (ii). Suppose $A \subset B$ and, for contradiction, that $A – B \ne \varphi$. Then some $x$ satisfies $x \in A$ and $x \notin B$. But $A \subset B$ forces $x \in B$ — a contradiction. So $A – B = \varphi$.

(ii) $\Rightarrow$ (iii). Suppose $A – B = \varphi$; that is, no element of $A$ lies outside $B$, so every element of $A$ lies in $B$. Take any $x \in A \cup B$. Either $x \in B$ already, or $x \in A$ and hence $x \in B$. Either way $x \in B$, so $A \cup B \subset B$. The reverse inclusion $B \subset A \cup B$ is automatic. Hence $A \cup B = B$.

(iii) $\Rightarrow$ (iv). Suppose $A \cup B = B$. Then

$$A \cap B = A \cap (A \cup B) = A$$

by the absorption law.

(iv) $\Rightarrow$ (i). Suppose $A \cap B = A$ and take any $x \in A$. Then $x \in A \cap B$, and every element of an intersection lies in both sets, so $x \in B$. Hence $A \subset B$.

The cycle closes, so all four conditions are equivalent. $\blacksquare$

Question 5

Show that if $A \subset B$, then $C – B \subset C – A$.

Solution. Take an arbitrary $x \in C – B$. By definition,

$$x \in C \quad\text{and}\quad x \notin B$$

We must show $x \in C – A$, that is, $x \in C$ and $x \notin A$. The first half is already given, so the whole question is why $x \notin A$.

Suppose instead that $x \in A$. Since $A \subset B$, that would give $x \in B$, contradicting $x \notin B$. Hence $x \notin A$, and therefore $x \in C – A$.

Since $x$ was arbitrary, $C – B \subset C – A$. $\blacksquare$

The containment reverses direction, and that is worth noticing rather than memorising: the bigger the set you subtract, the smaller what remains. $B$ is at least as big as $A$, so removing $B$ from $C$ leaves at most what removing $A$ leaves.

Question 6

Show that for any sets $A$ and $B$,

$$A = (A \cap B) \cup (A – B) \qquad\text{and}\qquad A \cup (B – A) = A \cup B$$

Solution.

First identity. Every element of $A$ either does or does not lie in $B$ — there is no third possibility. Those that do form $A \cap B$; those that do not form $A – B$. So the two pieces together account for all of $A$, and neither contains anything outside $A$.

Written as a two-way inclusion:

• ($\supset$) $A \cap B \subset A$ and $A – B \subset A$, so their union is contained in $A$.
• ($\subset$) Let $x \in A$. If $x \in B$ then $x \in A \cap B$; if $x \notin B$ then $x \in A – B$. Either way $x \in (A \cap B) \cup (A – B)$.

Hence $A = (A \cap B) \cup (A – B)$. The two pieces are disjoint, so this splits $A$ cleanly in two.

Second identity.

• ($\subset$) Let $x \in A \cup (B – A)$. If $x \in A$ then certainly $x \in A \cup B$. If $x \in B – A$ then $x \in B$, so again $x \in A \cup B$.
• ($\supset$) Let $x \in A \cup B$. If $x \in A$, we are done. If $x \notin A$, then $x \in B$ must hold, and together with $x \notin A$ that gives $x \in B – A$.

Hence $A \cup (B – A) = A \cup B$. $\blacksquare$

The second identity says the same thing as the first, seen from the other side: $A \cup B$ can be built as $A$ plus only the part of $B$ that $A$ does not already supply.

Question 7

Using properties of sets, show that

(i) $A \cup (A \cap B) = A$    (ii) $A \cap (A \cup B) = A$

Solution. “Using properties of sets” means an algebraic derivation from the distributive laws and the identities $A \cap U = A$, $A \cup \varphi = A$ — not an element-chase. Both proofs work by inserting the identity element for the other operation.

(i) Write the first $A$ as $A \cap U$:

$$A \cup (A \cap B) = (A \cap U) \cup (A \cap B)$$

Now factor out $A$ using the distributive law:

$$= A \cap (U \cup B) = A \cap U = A$$

since $U \cup B = U$ for any subset $B$ of $U$.

(ii) Symmetrically, write the first $A$ as $A \cup \varphi$:

$$A \cap (A \cup B) = (A \cup \varphi) \cap (A \cup B)$$

and factor out $A$ using the other distributive law:

$$= A \cup (\varphi \cap B) = A \cup \varphi = A$$

since $\varphi \cap B = \varphi$. $\blacksquare$

These are the absorption laws, and they are worth recognising by name — they are what made questions 3 and 4 short.

Question 8

Show that $A \cap B = A \cap C$ need not imply $B = C$.

Solution. “Need not imply” asks for a single counterexample. The mechanism to exploit is that $A$ can be too small to see the difference between $B$ and $C$ — whatever $B$ and $C$ disagree about, put it outside $A$.

Take

$$A = \{1, 2\},\qquad B = \{2, 3\},\qquad C = \{2, 4\}$$

Then

$$A \cap B = \{2\} \qquad\text{and}\qquad A \cap C = \{2\}$$

so $A \cap B = A \cap C$. But $3 \in B$ and $3 \notin C$, so $B \ne C$. $\blacksquare$

Contrast this with question 3: adding the second hypothesis $A \cup B = A \cup C$ would rule this example out, since here $A \cup B = \{1,2,3\}$ and $A \cup C = \{1,2,4\}$ differ. The two hypotheses together are exactly what is needed; either alone is not.

Question 9

Let $A$ and $B$ be sets. If $A \cap X = B \cap X = \varphi$ and $A \cup X = B \cup X$ for some set $X$, show that $A = B$.

(Hint: $A = A \cap (A \cup X)$, $B = B \cap (B \cup X)$ and use the distributive law.)

Solution. Follow the hint. Start from the absorption law and substitute the hypothesis:

$$A = A \cap (A \cup X) = A \cap (B \cup X)$$

Distribute:

$$A = (A \cap B) \cup (A \cap X) = (A \cap B) \cup \varphi = A \cap B$$

using $A \cap X = \varphi$. Now run the identical argument starting from $B$:

$$B = B \cap (B \cup X) = B \cap (A \cup X) = (B \cap A) \cup (B \cap X) = (A \cap B) \cup \varphi = A \cap B$$

Both $A$ and $B$ equal the same set $A \cap B$, so

$$A = B \qquad \blacksquare$$

The role of $X$ is worth naming: $A \cap X = \varphi$ says $X$ contributes nothing to $A$, and $A \cup X = B \cup X$ says $A$ and $B$ agree once $X$ is added. Since $X$ overlaps neither, it can be cancelled — which is precisely what the two lines above do formally.

Question 10

Find sets $A$, $B$ and $C$ such that $A \cap B$, $B \cap C$ and $A \cap C$ are non-empty sets and $A \cap B \cap C = \varphi$.

Solution. The requirement is that the three sets overlap in pairs but have no element common to all three. So give each pair its own private shared element, and make sure no element is shared by more than two.

Three sets of two elements each do the job:

$$A = \{1, 2\},\qquad B = \{1, 3\},\qquad C = \{2, 3\}$$

Checking:

$$A \cap B = \{1\} \ne \varphi, \qquad B \cap C = \{3\} \ne \varphi, \qquad A \cap C = \{2\} \ne \varphi$$

$$A \cap B \cap C = \{1\} \cap C = \varphi$$

since $1 \notin C$. All four conditions hold.

Drawn as a Venn diagram, this is three circles overlapping pairwise with an empty region at the centre — the arrangement that shows why pairwise intersection is a strictly weaker condition than triple intersection.

Common mistakes

• Question 2(i) and (ii), treating $\in$ as though it were transitive like $\subset$. $A \in B$ makes $A$ one of $B$’s objects; it says nothing about $A$’s own contents. Every counterexample here needs a set that contains a set, which is why the smallest working examples look artificial.
• Question 2(iv) and (v), proving a false statement by choosing a convenient example. An example can only disprove. If a statement seems to hold in the case you tried, that is not a proof — and here it is not even true.
• Question 3, using only one hypothesis. $A \cup B = A \cup C$ alone does not give $B = C$ (take $A = \{1\}$, $B = \{1\}$, $C = \varphi$). Any proof that never uses $A \cap B = A \cap C$ has gone wrong somewhere.
• Question 4, proving all twelve implications. A cycle of four is logically complete and a quarter of the work. Writing (i) $\Leftrightarrow$ (ii), (i) $\Leftrightarrow$ (iii), (i) $\Leftrightarrow$ (iv) separately is also valid but longer.
• Question 5, getting the direction backwards. The conclusion is $C – B \subset C – A$, not the reverse. Subtracting more leaves less — test it on $A = \{1\}$, $B = \{1,2\}$, $C = \{1,2,3\}$ if the direction ever feels uncertain.
• Question 7, “proving” absorption by drawing a Venn diagram. A diagram is a good check but not the derivation the question asks for. The step that carries the proof is inserting $U$ or $\varphi$ so the distributive law can be applied.
• Question 10, making the three sets overlap in a common element. If any element sits in all three sets, $A \cap B \cap C \ne \varphi$ and the last condition fails. Each shared element must belong to exactly two of the three.

Practise next

• Exercise 2.1 — the next chapter, Relations and Functions, opens with Cartesian products, which are built directly from the set operations proved here.
• Exercise 1.5 — worth revising alongside question 7, since De Morgan’s laws and the absorption laws are the two families of identity this chapter expects you to use without re-deriving.