Set Operations Laws and Theorems are an essential part of Class 11 CBSE Mathematics that help simplify and solve problems involving sets. These laws provide systematic rules for performing operations like union, intersection, and complement. Among these, De Morgan’s Laws are especially important for transforming complex set expressions into simpler forms, making problem-solving more efficient in competitive exams.
Laws of Operations On Sets
Laws of operations on sets are fundamental algebraic properties (commutative, associative, distributive, identity, complement, and De Morgan’s governing how sets are combined using union, intersection, and complementation. These laws ensure consistency in set theory, similar to algebraic laws in arithmetic, helping simplify complex set expressions.
THEOREM 1 (Idempotent laws)
For any set $A$, prove that:
1.1 $A \cup A = A$
1.2 $A \cap A = A$
PROOF
THEOREM 1.1
We have
$$
A \cup A = \{x : x \in A \text{ or } x \in A\} = \{x : x \in A\} = A.
$$
THEOREM 1.2
$$
A \cap A = \{x : x \in A \text{ and } x \in A\} = \{x : x \in A\} = A.
$$
THEOREM 2 (Identity laws)
For any set $A$, prove that:
2.1. $A \cup \emptyset = A$
2.2 $A \cap U = A$
(where $U$ is the universal set)
PROOF
THEOREM 2.1
We have
$$
A \cup \emptyset = \{x : x \in A \text{ or } x \in \emptyset\} = \{x : x \in A\} = A.
$$
THEOREM 2.2
$$
A \cap U = \{x : x \in A \text{ and } x \in U\} = \{x : x \in A\} = A.
$$
| Key Point |
|---|
| $\emptyset$ and $U$ are the identity elements for union and intersection of sets respectively. |
THEOREM 3 (Commutative laws)
For any two sets $A$ and $B$, prove that:
3.1. $A \cup B = B \cup A$ [Commutative law for union of sets]
3.2. $A \cap B = B \cap A$ [Commutative law for intersection of sets]
PROOF
THEOREM 3.1
Let $x$ be an arbitrary element of $A \cup B$. Then,
$$
x \in A \cup B \Rightarrow x \in A \text{ or } x \in B
\Rightarrow x \in B \text{ or } x \in A
\Rightarrow x \in B \cup A.
$$
$\therefore A \cup B \subseteq B \cup A$ …(i)
Again, let $y$ be an arbitrary element of $B \cup A$. Then,
$$
y \in B \cup A \Rightarrow y \in B \text{ or } y \in A
\Rightarrow y \in A \text{ or } y \in B
\Rightarrow y \in A \cup B.
$$
$\therefore B \cup A \subseteq A \cup B$ …(ii)
From (i) and (ii), we get $A \cup B = B \cup A$.
THEOREM 3.2
Let $x$ be an arbitrary element of $A \cap B$. Then,
$$
x \in A \cap B \Rightarrow x \in A \text{ and } x \in B
\Rightarrow x \in B \text{ and } x \in A
\Rightarrow x \in B \cap A.
$$
$\therefore A \cap B \subseteq B \cap A$ …(iii)
Again, let $y$ be an arbitrary element of $B \cap A$. Then,
$$
y \in B \cap A \Rightarrow y \in B \text{ and } y \in A
\Rightarrow y \in A \text{ and } y \in B
\Rightarrow y \in A \cap B.
$$
$\therefore B \cap A \subseteq A \cap B$ …(iv)
From (iii) and (iv), we get $A \cap B = B \cap A$.
THEOREM 4 (Associative laws)
For any sets $A, B, C$, prove that:
4.1. $(A \cup B) \cup C = A \cup (B \cup C)$ [Associative law for union of sets]
4.2. $(A \cap B) \cap C = A \cap (B \cap C)$ [Associative law for intersection of sets]
PROOF
THEOREM 4.1
Let $x$ be an arbitrary element of $(A \cup B) \cup C$. Then,
$ x \in (A \cup B) \cup C $
$\Rightarrow x \in (A \cup B) \text{ or } x \in C $
$\Rightarrow (x \in A \text{ or } x \in B) \text{ or } x \in C $
$\Rightarrow x \in A \text{ or } (x \in B \text{ or } x \in C) $
$\Rightarrow x \in A \cup (B \cup C).$
$\therefore (A \cup B) \cup C \subseteq A \cup (B \cup C)$. …(i)
Again, let $y$ be an arbitrary element of $A \cup (B \cup C)$. Then,
$y \in A \cup (B \cup C) \Rightarrow y \in A$ or $y \in (B \cup C)$
$\Rightarrow y \in A$ or $(y \in B$ or $y \in C)$
$\Rightarrow (y \in A$ or $y \in B)$ or $y \in C$
$\Rightarrow y \in (A \cup B)$ or $y \in C$
$\Rightarrow y \in (A \cup B) \cup C$.
$\therefore A \cup (B \cup C) \subseteq A \cup (B \cup C)$. …(ii)
From (i) and (ii), we get $A \cup B \cup C = A \cup (B \cup C)$.
THEOREM 4.2
Let $x$ be an arbitrary element of $(A \cap B) \cap C$. Then,
$x \in (A \cap B) \cap C \Rightarrow x \in (A \cap B)$ and $x \in C$
$\Rightarrow x \in A$ and $x \in B$ and $x \in C$
$\Rightarrow x \in A$ and $(x \in B$ and $x \in C)$
$\Rightarrow x \in A$ and $x \in (B \cap C)$
$\Rightarrow x \in A \cap (B \cap C)$.
$\therefore (A \cap B) \cap C \subseteq A \cap (B \cap C)$. …(iii)
Again, let $y$ be an arbitrary element of $A \cap (B \cap C)$. Then,
$y \in A \cap (B \cap C) \Rightarrow y \in A$ and $y \in (B \cap C)$
$\Rightarrow y \in A$ and $(y \in B$ and $y \in C)$
$\Rightarrow (y \in A$ and $y \in B)$ and $y \in C$
$\Rightarrow y \in (A \cap B)$ and $y \in C$
$\Rightarrow y \in (A \cap B) \cap C$.
$\therefore A \cap (B \cap C) \subseteq (A \cap B) \cap C$. …(iv)
From (iii) and (iv), we get $(A \cap B) \cap C = A \cap (B \cap C)$.
THEOREM 5 (Distributive laws)
For any three sets $A$, $B$, $C$ prove that:
5.1. $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$ [Distributive law of union over intersection]
5.2. $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ [Distributive law of intersection over union]
PROOF
THEOREM 5.1
Let $x$ be an arbitrary element of $A \cup (B \cap C)$. Then,
$x \in A \cup (B \cap C) \Rightarrow x \in A$ or $x \in (B \cap C)$
$\Rightarrow x \in A$ or $(x \in B$ and $x \in C)$
$\Rightarrow (x \in A$ or $x \in B)$ and $(x \in A$ or $x \in C)$
[‘or’ distributes ‘and’]
$\Rightarrow x \in (A \cup B)$ and $x \in (A \cup C)$.
$\Rightarrow x \in (A \cup B) \cap (A \cup C)$.
$\therefore A \cup (B \cap C) \subseteq (A \cup B) \cap (A \cup C)$. …(i)
Again, let $y$ be an arbitrary element of $(A \cup B) \cap (A \cup C)$. Then,
$y \in (A \cup B) \cap (A \cup C) \Rightarrow y \in (A \cup B)$ and $y \in (A \cup C)$
$\Rightarrow (y \in A$ or $y \in B)$ and $(y \in A$ or $y \in C)$
$\Rightarrow y \in A$ or $(y \in B$ and $y \in C)$
[‘or’ distributes ‘and’]
$\Rightarrow y \in A$ or $y \in (B \cap C)$
$\Rightarrow y \in A \cup (B \cap C)$.
$\therefore (A \cup B) \cap (A \cup C) \subseteq A \cup (B \cap C)$. …(ii)
From (i) and (ii), we get $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$.
THEOREM 5.2
Let $x$ be an arbitrary element of $A \cap (B \cup C)$. Then,
$x \in A \cap (B \cup C) \Rightarrow x \in A$ and $x \in (B \cup C)$
$\Rightarrow x \in A$ and $(x \in B$ or $x \in C)$
$\Rightarrow (x \in A$ and $x \in B)$ or $(x \in A$ and $x \in C)$
[‘and’ distributes ‘or’]
$\Rightarrow x \in (A \cap B)$ or $x \in (A \cap C)$
$\Rightarrow x \in (A \cap B) \cup (A \cap C)$.
$\therefore A \cap (B \cup C) \subseteq (A \cap B) \cup (A \cap C)$. …(iii)
Again, let $y$ be an arbitrary element of $(A \cap B) \cup (A \cap C)$. Then,
$y \in (A \cap B) \cup (A \cap C) \Rightarrow y \in (A \cap B)$ or $y \in (A \cap C)$
$\Rightarrow (y \in A$ and $y \in B)$ or $(y \in A$ and $y \in C)$
$\Rightarrow y \in A$ and $(y \in B$ or $y \in C)$
[‘and’ distributes ‘or’]
$\Rightarrow y \in A$ and $y \in (B \cup C)$
$\Rightarrow y \in A \cap (B \cup C)$.
$\therefore (A \cap B) \cup (A \cap C) \subseteq A \cap (B \cup C)$. …(iv)
From (iii) and (iv), we get $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$.
THEOREM 6 (De Morgan’s laws)
For any two sets $A$ and $B$, prove that:
6.1. $(A \cup B)’ = A’ \cap B’$
6.2. $(A \cap B)’ = A’ \cup B’$
PROOF
THEOREM 6.1
Let $x$ be an arbitrary element of $(A \cup B)’$. Then,
$x \in (A \cup B)’ \Rightarrow x \notin A \cup B$
$\Rightarrow x \notin A$ and $x \notin B$
$\Rightarrow x \in A’$ and $x \in B’$
$\Rightarrow x \in A’ \cap B’$.
$\therefore (A \cup B)’ \subseteq A’ \cap B’$. …(i)
Again, let $y$ be an arbitrary element of $A’ \cap B’$. Then,
$y \in A’ \cap B’ \Rightarrow y \in A’$ and $y \in B’$
$\Rightarrow y \notin A$ and $y \notin B$
$\Rightarrow y \notin (A \cup B)$
$\Rightarrow y \in (A \cup B)’$.
$\therefore A’ \cap B’ \subseteq (A \cup B)’$. …(ii)
From (i) and (ii), we get $(A \cup B)’ = A’ \cap B’$.
THEOREM 6.2
Let $x$ be an arbitrary element of $(A \cap B)’$. Then,
$x \in (A \cap B)’ \Rightarrow x \notin A \cap B$
$\Rightarrow x \notin A$ or $x \notin B$
$\Rightarrow x \in A’$ or $x \in B’$
$\Rightarrow x \in A’ \cup B’$.
$\therefore (A \cap B)’ \subseteq A’ \cup B’$. …(iii)
Again, let $y$ be an arbitrary element of $A’ \cup B’$. Then,
$y \in A’ \cup B’ \Rightarrow y \in A’$ or $y \in B’$
$\Rightarrow y \notin A$ or $y \notin B$
$\Rightarrow y \notin (A \cap B)$
$\Rightarrow y \in (A \cap B)’$.
$\therefore A’ \cup B’ \subseteq (A \cap B)’$. …(iv)
From (iii) and (iv), we get $(A \cap B)’ = A’ \cup B’$.
SOME MORE RESULTS
THEOREM 7
For any two sets $A$ and $B$, prove that:
I. $A \subseteq B \Rightarrow B’ \subseteq A’$
II. $A – B = A \cap B’$
III. $(A – B) \cap B = \emptyset$
IV. $(A – B) \cup (A \cap B) = A$
V. $(A – B) \cap (A \cap B) = \emptyset$
PROOF
I.
Let $A \subseteq B$ be given and let $x$ be an arbitrary element of $B’$. Then,
$x \in B’ \Rightarrow x \notin B$
$\Rightarrow x \notin A$
[∵ $A \subseteq B$]
$\Rightarrow x \in A’$.
$\therefore B’ \subseteq A’$.
Hence, $A \subseteq B \Rightarrow B’ \subseteq A’$.
II.
Let $x$ be an arbitrary element of $(A – B)$. Then,
$x \in (A – B) \Rightarrow x \in A$ and $x \notin B$
$\Rightarrow x \in A$ and $x \in B’$
$\Rightarrow x \in A \cap B’$.
$\therefore (A – B) \subseteq A \cap B’$. …(i)
Again, let $y$ be an arbitrary element of $(A \cap B’)$. Then,
$y \in (A \cap B’) \Rightarrow y \in A$ and $y \in B’$
$\Rightarrow y \in A$ and $y \notin B$
$\Rightarrow y \in (A – B)$.
$\therefore A \cap B’ \subseteq (A – B)$. …(ii)
Hence, from (i) and (ii), we get $A – B = A \cap B’$.
III.
If possible, let $(A – B) \cap B \neq \emptyset$ and let
$x \in (A – B) \cap B$. Then,
$x \in (A – B) \cap B \Rightarrow x \in (A – B)$ and $x \in B$
⇒ $(x \in A$ and $x \in B)$ and $x \in B$
⇒ $x \in A$ and $(x \in B$ and $x \in B)$.
But, $x \notin B$ and $x \in B$ can never hold simultaneously.
Thus, we arrive at a contradiction.
Since the contradiction arises by assuming that $(A – B) \cap B \ne \phi$
and hence $(A – B) \cap B = \phi$.
IV.
Let $x$ be an arbitrary element of $(A – B) \cup (B – A)$. Then,
$x \in (A – B) \cup (B – A)$ ⇒ $x \in (A – B)$ or $x \in (B – A)$
⇒ $(x \in A$ and $x \notin B)$ or $(x \notin A$ and $x \in B)$
⇒ $(x \in A$ or $x \in B)$ and $(x \in A$ or $x \notin B)$ [note it]
⇒ $x \in (A \cup B)$ and $x \in (A \cap B)$.
∴ $(A – B) \cup (B – A) = (A \cup B) – (A \cap B)$. … (iii)
Again, let $y$ be an arbitrary element of $(A \cup B) – (A \cap B)$. Then,
$y \in (A \cup B) – (A \cap B)$ ⇒ $y \in (A \cup B)$ and $y \notin (A \cap B)$
⇒ $(y \in A$ or $y \in B)$ and $(y \notin A$ or $y \notin B)$ [note it]
⇒ $(y \in A$ and $y \notin B)$ or $(y \notin A$ and $y \in B)$ [note it]
⇒ $y \in (A – B)$ or $y \in (B – A)$
⇒ $y \in (A – B) \cup (B – A)$.
∴ $(A \cup B) – (A \cap B) = (A – B) \cup (B – A)$. … (iv)
From (iii) and (iv), we get $(A – B) \cup (B – A) = (A \cup B) – (A \cap B)$.
V.
Let $(A – B) \ne \phi$ be given and we have to show that $A \cap B = \phi$.
If possible, let $A \cap B \ne \phi$ and let $x \in A \cap B$. Then,
$x \in A \cap B$ ⇒ $x \in A$ and $x \in B$
⇒ $(x \in A – B)$ and $x \in B$ [∵ $A = (A – B) \cup (A \cap B)$ (given)]
⇒ $(x \in A – B)$ and $x \in B$
But, $x \in B$ and $x \notin B$ both can never hold simultaneously.
Thus, we arrive at a contradiction.
Since the contradiction arises by assuming that $A \cap B \ne \phi$, so
$A \cap B = \phi$.
Thus, $(A – B) \ne \phi$ ⇒ $A \cap B = \phi$.
Again, let $(A – B) = \phi$ and we have to show that $(A – B) = A$.
Now, $x \in (A – B)$ ⇒ $x \in A$ and $x \notin B$
⇒ $x \in A$ (surely).
∴ $(A – B) \subseteq A$. … (v)
Again, $y \in A$ ⇒ $y \in B$ [∵ $A \cap B = \phi$]
⇒ $y \in A$ and $y \notin B$
⇒ $y \in (A – B)$.
∴ $A \subseteq (A – B)$. … (vi)
Thus, from (v) and (vi), we get $A = (A – B)$.
∴ $A \cap B = \phi$ ⇒ $(A – B) = A$.
Hence, $(A – B) \ne \phi$ ⇔ $(A – B) = A$ ⇔ $(A \cap B) = \phi$.
THEOREM 8
If $(A \cup B) = (A \cap B)$ then
prove that $A = B$.
PROOF
Let $(A \cup B) = (A \cap B)$ be given.
Let $x$ be an arbitrary element of $A$. Then,
$x \in A$ ⇒ $x \in A \cup B$
[∵ $A \subseteq A \cup B$]
⇒ $x \in A \cap B$
[∵ $A \cup B = A \cap B$]
⇒ $x \in A$ and $x \in B$
⇒ $x \in B$ (surely).
∴ $A \subseteq B$. … (i)
Again, let $y \in B$. Then,
$y \in B$ ⇒ $y \in A \cup B$
[∵ $B \subseteq A \cup B$]
⇒ $y \in A \cap B$
[∵ $A \cup B = A \cap B$]
⇒ $y \in A$ and $y \in B$
⇒ $y \in A$ (surely).
∴ $B \subseteq A$. … (ii)
Thus, from (i) and (ii), we get $A = B$.
THEOREM 9
If $A \subseteq B$ then for any set $C$,
prove that $(C – B) \subseteq (C – A)$.
PROOF
Let $A \subseteq B$ be given.
Let $x \in (C – B)$. Then,
$x \in (C – B)$ ⇒ $x \in C$ and $x \notin B$
⇒ $x \in C$ and $x \in A$ [∵ $A \subseteq B$]
⇒ $x \in (C – A)$.
∴ $(C – B) \subseteq (C – A)$.
Hence, $A \subseteq B$ ⇒ $(C – B) \subseteq (C – A)$.
THEOREM 10
For any sets $A$ and $B$, prove that:
(i) $A \cup (A \cap B) = A$ (ii) $A \cap (A \cup B) = A$
PROOF
(i) Since $(A \cap B) \subseteq A$, we have $A \cup (A \cap B) = A$
[∵ $X \subseteq Y \Rightarrow X \cup Y = Y$].
(ii) Since $A \subseteq (A \cup B)$, we have $A \cap (A \cup B) = A$
[∵ $X \subseteq Y \Rightarrow X \cap Y = X$].
THEOREM 11
For any sets $A$ and $B$, prove that:
(i) $(A \cap B) \cup (A – B) = A$
(ii) $A \cup (B – A) = A \cup B$
PROOF
(i) We have
$(A \cap B) \cup (A – B) = (A \cap B) \cup (A \cap B’)$ [∵ $(A – B) = (A \cap B’)$]
= $A \cap (B \cup B’)$ [by distributive law]
= $A \cap U$ [∵ $B \cup B’ = U$]
= $A$
Hence, $(A \cap B) \cup (A – B) = A$.
(ii) We have
$A \cup (B – A) = A \cup (B \cap A’)$ [∵ $(B – A) = (B \cap A’)$]
= $(A \cup B) \cap (A \cup A’)$ [by distributive law]
= $(A \cup B) \cap U$ [∵ $A \cup A’ = U$]
= $(A \cup B)$.
Hence, $A \cup (B – A) = (A \cup B)$.
THEOREM 12
If $A \cap B’ = \phi$ then prove that $A = A \cap B$ and hence show that $A \subseteq B$.
PROOF
Let $A \cap B’ = \phi$ be given. Then,
$A = A \cap U$, where $U$ is the universal set
= $A \cap (B \cup B’)$
[∵ $B \cup B’ = U$]
= $(A \cap B) \cup (A \cap B’)$
= $(A \cap B) \cup \phi$
[∵ $A \cap B’ = \phi$]
= $(A \cap B)$.
Hence, $A = A \cap B$.
Further, let $A \cap B = A$ and let $x \in A$. Then,
$x \in A$ ⇒ $x \in A \cap B$
[∵ $A \cap B = A$]
⇒ $x \in A$ and $x \in B$
⇒ $x \in B$ (surely).
∴ $A \subseteq B$.
THEOREM 13
If $A$, $B$ and $C$ be the sets such that $A \cup B = A \cup C$ and
$A \cap B = A \cap C$ then prove that $B = C$.
PROOF
Let $A \cup B = A \cup C$ and $A \cap B = A \cap C$ be given. Then,
$A \cup B = A \cup C$
⇒ $B = B \cap (A \cup B) = B \cap (A \cup C) = (B \cap A) \cup (B \cap C)$
and $C = C \cap (A \cup C) = C \cap (A \cup B) = (C \cap A) \cup (C \cap B)$
[∵ $A \cap B = A \cap C$ and $B \cap C = C \cap B$]
⇒ $B = (A \cap B) \cup (B \cap C)$ and $C = (A \cap B) \cup (B \cap C)$
Hence, $B = C$.
THEOREM 14
For any sets $A$, $B$ and $C$, prove that:
(i) $A – (B \cup C) = (A – B) \cap (A – C)$
(ii) $A – (B \cap C) = (A – B) \cup (A – C)$
(iii) $(A \cup B) – C = (A – C) \cup (B – C)$
(iv) $(A \cap B) – C = (A – C) \cap (B – C)$
PROOF
(i) We have
$A – (B \cup C) = A \cap (B \cup C)’$ [∵ $X – Y = X \cap Y’$]
= $A \cap (B’ \cap C’)$ [by De Morgan’s law]
= $(A \cap B’) \cap (A \cap C’)$
= $(A – B) \cap (A – C)$.
Hence, $A – (B \cup C) = (A – B) \cap (A – C)$.
(ii) We have
$A – (B \cap C) = A \cap (B \cap C)’$ [∵ $X – Y = X \cap Y’$]
= $A \cap (B’ \cup C’)$ [by De Morgan’s law]
= $(A \cap B’) \cup (A \cap C’)$ [by distributive law]
= $(A – B) \cup (A – C)$.
Hence, $A – (B \cap C) = (A – B) \cup (A – C)$.
(iii) $(A \cup B) – C = (A \cup B) \cap C’$ [∵ $X – Y = X \cap Y’$]
= $(A \cap C’) \cup (B \cap C’)$ [by distributive law]
= $(A – C) \cup (B – C)$.
Hence, $(A \cup B) – C = (A – C) \cup (B – C)$.
(iv) $(A \cap B) – C = (A \cap B) \cap C’$ [∵ $X – Y = X \cap Y’$]
= $(A \cap C’) \cap (B \cap C’)$ [note]
= $(A – C) \cap (B – C)$.
Hence, $(A \cap B) – C = (A – C) \cap (B – C)$.
THEOREM 15
Let $A$ and $B$ be sets. If $A \cap X = B \cap X$ and $A \cup X = B \cup X$ for some set $X$, show that $A = B$.
PROOF
$A \cap X = B \cap X$ [given]
⇒ $A = A \cup (A \cap X) = A \cup (B \cap X)$ [by distributive law]
⇒ $B \cap X \subseteq A$ … (i)
Again, $A \cup X = B \cup X$ [given]
⇒ $B = B \cup (B \cap X) = B \cup (A \cap X)$ [by distributive law]
⇒ $A \cap X \subseteq B$ … (ii)
From (i) and (ii), we get $A = B$.
THEOREM 16
Show that the following four conditions are equivalent:
(i) $A \subseteq B$ (ii) $A – B = \phi$ (iii) $A \cup B = B$ (iv) $A \cap B = A$
PROOF
In order to prove the required result, we will show that:
(i) ⇒ (ii) ⇒ (iii) ⇒ (iv) ⇒ (i)
Now, (i) ⇒ (ii):
Let $A \subseteq B$ be given.
Then, there is no element of $A$ which is not in $B$.
∴ $A – B = {x : x \in A \text{ and } x \notin B} = \phi$
[∵ there is no element of $A$ which is not in $B$].
Hence, $A \subseteq B \Rightarrow A – B = \phi$ and therefore, (i) ⇒ (ii).
(ii) ⇒ (iii):
Let $A – B = \phi$ be given. Then,
$A – B = \phi \Rightarrow$ every element of $A$ is in $B$
$\Rightarrow A \subseteq B$
$\Rightarrow A \cup B = B$.
Thus, $A – B = \phi \Rightarrow A \cup B = B$ and therefore, (ii) ⇒ (iii).
(iii) ⇒ (iv):
Let $A \cup B = B$ be given. Then,
$A \cup B = B \Rightarrow A \subseteq B \Rightarrow A \cap B = A$.
Thus, $A \cup B = B \Rightarrow A \cap B = A$ and therefore, (iii) ⇒ (iv).
(iv) ⇒ (i):
Let $A \cap B = A$ be given. Then,
$x \in A \Rightarrow x \in A \cap B$
[∵ $A \cap B = A$]
$\Rightarrow x \in A$ and $x \in B$
$\Rightarrow x \in B$ (surely).
∴ $A \subseteq B$
Thus, $A \cap B = A \Rightarrow A \subseteq B$ and therefore, (iv) ⇒ (i).
∴ (i) ⇒ (ii) ⇒ (iii) ⇒ (iv) ⇒ (i).
Hence, the given four conditions are equivalent.
THEOREM 17
For any sets $A$ and $B$, prove that
$$P(A \cap B) = P(A) \cap P(B).$$
PROOF
Let $X \in P(A \cap B)$. Then,
$X \in P(A \cap B) \Rightarrow X \subseteq A \cap B$
$\Rightarrow X \subseteq A$ and $X \subseteq B$
$\Rightarrow X \in P(A)$ and $X \in P(B)$
$\Rightarrow X \in P(A) \cap P(B)$
∴ $P(A \cap B) \subseteq P(A) \cap P(B)$. … (i)
Again, let $Y \in P(A) \cap P(B)$. Then,
$Y \in P(A) \cap P(B) \Rightarrow Y \in P(A)$ and $Y \in P(B)$
$\Rightarrow Y \subseteq A$ and $Y \subseteq B$
$\Rightarrow Y \subseteq A \cap B$
$\Rightarrow Y \in P(A \cap B)$
∴ $P(A) \cap P(B) \subseteq P(A \cap B)$ … (ii)
From (i) and (ii), we get $P(A \cap B) = P(A) \cap P(B)$.
THEOREM 18
For any two sets $A$ and $B$, prove that
$$P(A) \cup P(B) \subseteq P(A \cup B).$$
But, $P(A \cup B)$ is not necessarily a subset of $P(A) \cup P(B)$.
PROOF
Let $X$ be an arbitrary element of $P(A) \cup P(B)$. Then,
$X \in P(A) \cup P(B) \Rightarrow X \in P(A)$ or $X \in P(B)$
$\Rightarrow X \subseteq A$ or $X \subseteq B$
$\Rightarrow X \subseteq A \cup B$
$\Rightarrow X \in P(A \cup B)$
∴ $P(A) \cup P(B) \subseteq P(A \cup B)$.
However, $P(A \cup B) \subseteq P(A) \cup P(B)$ is not always true.
For example, let $A = {1}$ and $B = {2}$. Then, $A \cup B = {1, 2}$.
∴ $P(A) = {\phi, {1}}$, $P(B) = {\phi, {2}}$
and $P(A \cup B) = {\phi, {1}, {2}, {1, 2}}$.
Also, $P(A) \cup P(B) = {\phi, {1}, {2}}$.
∴ $P(A \cup B) \not\subseteq P(A) \cup P(B)$.
Hence, in general, $P(A \cup B) \neq P(A) \cup P(B)$.
THEOREM 19
If $P(A) = P(B)$, prove that $A = B$.
PROOF
Let $P(A) = P(B)$. Then,
$A \in P(A)$
$\Rightarrow A \in P(B)$
[∵ $P(A) = P(B)$]
$\Rightarrow A \subseteq B$ … (i)
Again, $B \in P(B)$
$\Rightarrow B \in P(A)$
[∵ $P(B) = P(A)$]
$\Rightarrow B \subseteq A$ … (ii)
From (i) and (ii), we get $A \subseteq B$ and $B \subseteq A$.
Hence, $A = B$.
FAQs (Very Short Questions and Answers) Based on Set Operations
What are the laws of set operations?
Set operation laws are standard rules used to simplify expressions involving sets, such as union, intersection, and complement. These include commutative, associative, distributive, identity, domination, and complement laws.
What is the commutative law in sets?
The commutative law states that the order of sets does not affect the result:
$$
A \cup B = B \cup A, \quad A \cap B = B \cap A
$$
What is the associative law in sets?
The associative law states that grouping of sets does not affect the result:
$$
(A \cup B) \cup C = A \cup (B \cup C)
$$
$$
(A \cap B) \cap C = A \cap (B \cap C)
$$
What is the distributive law in sets?
This law connects union and intersection:
$$
A \cup (B \cap C) = (A \cup B) \cap (A \cup C)
$$
$$
A \cap (B \cup C) = (A \cap B) \cup (A \cap C)
$$
What are identity laws in sets?
Identity laws involve the universal set $U$ and empty set $\emptyset$:
$$
A \cup \emptyset = A, \quad A \cap U = A
$$
What are domination laws in sets?
Domination laws give fixed results:
$$
A \cup U = U, \quad A \cap \emptyset = \emptyset
$$
What are complement laws in sets?
These laws involve complements of sets:
$$
A \cup A’ = U, \quad A \cap A’ = \emptyset
$$
What is De Morgan’s Law?
De Morgan’s Laws relate union and intersection through complements:
$$
(A \cup B)’ = A’ \cap B’
$$
$$
(A \cap B)’ = A’ \cup B’
$$
Why are De Morgan’s Laws important?
They help simplify complex set expressions and are widely used in solving problems involving complements and Venn diagrams.
Can set laws be verified using Venn diagrams?
Yes, Venn diagrams are a visual method to verify and understand all set operation laws and theorems.
PRACTICE EXERCISE
- If $A = {a, b, c, d, e}$, $B = {a, c, e, g}$ and $C = {b, e, f, g}$, verify that:
(i) $A \cup B = B \cup A$
(ii) $A \cup C = C \cup A$
(iii) $B \cup C = C \cup B$
(iv) $A \cap B = B \cap A$
(v) $B \cap C = C \cap B$
(vi) $A \cap C = C \cap A$
(vii) $(A \cup B) \cup C = A \cup (B \cup C)$
(viii) $(A \cap B) \cap C = A \cap (B \cap C)$ - If $A = {a, b, c, d, e}$, $B = {a, c, e, g}$ and $C = {b, e, f, g}$, verify that:
(i) $(A \cup B) \cup C = A \cup (B \cup C)$
(ii) $(A \cap B) \cap C = A \cap (B \cap C)$ - If $A = {x : x \in N, x \leq 7}$, $B = {x : x \text{ is prime}, x < 8}$ and $C = {x : x \text{ is odd and } x < 10}$, verify that:
(i) $(A \cup B) \cap C = (A \cap C) \cup (B \cap C)$
(ii) $(A \cap B) \cup C = (A \cup C) \cap (B \cup C)$ - If $U = {1, 2, 3, 4, 5, 6, 7, 8, 9}$, $A = {2, 4, 6, 8}$ and $B = {2, 3, 5, 7}$, verify that:
(i) $(A – B)’ = A’ \cup B’$
(ii) $(A \cap B)’ = A’ \cup B’$ - Let $A = {a, b, c}$, $B = {b, c, d, e}$ and $C = {c, d, e, f}$ be subsets of $U = {a, b, c, d, e, f}$. Then, verify that:
(i) $A’ = A$
(ii) $(A \cup B)’ = A’ \cap B’$
(iii) $(A \cap B)’ = A’ \cup B’$ - Given an example of three sets $A$, $B$, $C$ such that
$A \cap B \neq \phi$, $B \cap C \neq \phi$, $A \cap C \neq \phi$ and $A \cap B \cap C = \phi$. - For any sets $A$ and $B$, prove that:
(i) $(A – B) \cup B = A$
(ii) $(A – B) \cap B = \phi$
(iii) $(A \cup B) – A = B – A$
(iv) $(A \cap B) – A = \phi$
(v) $(A – B) \cup (A – B’) = A$ - For any sets $A$ and $B$, prove that:
(i) $A \cap B’ = A – B$
(ii) $A’ \cup B = U \Rightarrow A \subseteq B$
HINTS TO SOME SELECTED QUESTIONS
- Take $A = {1, 2}$, $B = {2, 3}$ and $C = {1, 3}$.
- (i) $(A – B) \cup B = (A \cup B) – (B – B) = A \cup B – \phi = A \cup B$ wait, use the identities.
(ii) Use distributive law.
(iii) Use the definition.
(iv) Use De Morgan or direct.
(v) Use distributive law repeatedly. - (i) Direct from definition.
(ii) Use complement properties.
Important Chapter Linking
Set Operations Laws and Theorems is an important part of Class 11 CBSE Mathematics that focuses on simplifying and analyzing set expressions using standard rules. In this topic, you will learn fundamental laws such as commutative, associative, distributive, identity, domination, and complement laws, along with powerful results like De Morgan’s Laws. These concepts help in transforming complex set operations involving union, intersection, and complement into simpler forms. Venn diagrams are also used to visualize and verify these laws effectively. To strengthen conceptual clarity and problem-solving skills, this section includes a wide range of JEE Previous Year Questions (PYQs), MCQs, and practice problems, making it highly useful for competitive exams like JEE, NDA, and IMUCET.