If A = {a, b, c, d, e}, B = {a, c, e, g}, verify that:

 (i) LHS = A ∪ B

$=\{a, b, c, d, e\} \cup\{a, c, e, g\}$

$=\{a, b, c, d, e, g\}$

$=\{a, c, e, g\} \cup\{a, b, c, d, e\}$

$=B \cup A$

$=R H S$

Hence proved.

(ii) To prove: A ∪ C = C ∪ A

Since the element of set C is not provided,

let x be any element of C.

$\mathrm{LHS}=\mathrm{A} \cup \mathrm{C}$

$=\{a, b, c, d, e\} \cup\left\{x \mid x^{\in} C\right\}$

$=\{a, b, c, d, e, x\}$

$=\{x, a, b, c, d, e\}$

$=\left\{x \mid x^{\in} C\right\} \cup\{a, b, c, d, e\}$

$=C \cup A$

= RHS

Hence proved

(iii) To prove: B ∪ C = C ∪ B

Since the element of set C is not provided

let x be any element of C.

LHS = B ∪ C

$=\{a, c, e, g\} \cup\left\{x \mid x^{\in} C\right\}$

$=\{a, c, e, g, x\}$

$=\{x, a, c, e, g\}$

$=\left\{x \mid x^{\epsilon} C\right\} \cup\{a, c, e, g\}$

= C ∪ B

= RHS

Hence proved.

(iv) LHS = A ∩ B

$=\{a, b, c, d, e\} \cup\{a, c, e, g\}$

$=\{a, c, e\}$

RHS = B ∩ A

$=\{a, c, e, g\} \cap\{a, b, c, d, e\}$

$=\{a, c, e\}$

$\therefore \mathrm{A} \cap \mathrm{B}=\mathrm{B} \cap \mathrm{A}$

(v) Let x be an element of B ∩ C

$\Rightarrow \mathrm{x}^{\mathrm{E}} \mathrm{B} \cap \mathrm{C}$

$\Rightarrow \mathrm{x}^{\in} \mathrm{B}$ and $\mathrm{x}^{\in} \mathrm{C}$

$\Rightarrow \mathrm{x}^{\in} \mathrm{C}$ and $\mathrm{x}^{\in}{ }_{\mathrm{B}}$ [by definition of intersection]

$\Rightarrow x^{\in} C \cap B$

$\Rightarrow \mathrm{B} \cap \mathrm{C} \subset \mathrm{C} \cap \mathrm{B} \ldots .$ (i)

Now let x be an element of C ∩ B

Then, $x \in C \cap B$

$\Rightarrow \mathrm{x}^{\in} \mathrm{C}$ and $\mathrm{x}^{\in} \mathrm{B}$

$\Rightarrow \mathrm{x}^{\in} \mathrm{B}$ and $\mathrm{x}^{\in} \mathrm{C}$ [by definition of intersection]

$\Rightarrow \mathrm{x}^{\in} \mathrm{B} \cap \mathrm{C}$

$\Rightarrow \mathrm{C} \cap \mathrm{B}^{\subset} \mathrm{B} \cap \mathrm{C} \ldots$ (ii)

From (i) and (ii) we have,

B ∩ C = C ∩ B [ every set is a subset of itself]

Hence proved.

(vi) Let $x$ be an element of $A \cap C$

$\Rightarrow \mathrm{x}^{\in} \mathrm{A} \cap \mathrm{C}$

$\Rightarrow \mathrm{x}^{\in} \mathrm{A}$ and $\mathrm{x}^{\in} \mathrm{C}$

$\Rightarrow \mathrm{x}^{\in} \mathrm{C}$ and $\mathrm{x}^{\in} \mathrm{A}$ [by definition of intersection]

$\Rightarrow \mathrm{x}^{\in} \mathrm{C} \cap \mathrm{A}$

$\Rightarrow \mathrm{A} \cap \mathrm{C}^{\subset} \mathrm{C} \cap \mathrm{A} \ldots .$ (i)

Now let $x$ be an element of $C \cap A$

Then, $x \in C \cap A$

$\Rightarrow \mathrm{x}^{\in} \mathrm{C}$ and $\mathrm{x}^{\in} \mathrm{A}$

$\Rightarrow \mathrm{x}^{\in} \mathrm{A}$ and $\mathrm{x}^{\in} \mathrm{C}$ [by definition of intersection]

$\Rightarrow \mathrm{x}^{\in} \mathrm{A} \cap \mathrm{C}$

$\Rightarrow \mathrm{C} \cap \mathrm{A} \subset \mathrm{A} \cap \mathrm{C} \ldots$ (ii)

From (i) and (ii) we have,

$A \cap C=C \cap A$ [ every set is a subset of itself]

Hence proved.

(vii) Let $x$ be any element of $(A \cup B) \cup C$

$\Rightarrow{ }_{\mathrm{x}} \in_{(\mathrm{A} \cup \mathrm{B}) \text { or } \mathrm{x} \in_{\mathrm{C}}}$

$\Rightarrow{ }_{\mathrm{x}} \in_{\mathrm{A} \text { or } \mathrm{x}} \in_{\mathrm{B} \text { or } \mathrm{x}} \in_{\mathrm{C}}$

$\Rightarrow{ }_{\mathrm{x}} \in_{\mathrm{A} \text { or } \mathrm{x}} \in_{(\mathrm{B} \cup \mathrm{C})}$

$\Rightarrow{ }_{\mathrm{x}} \in_{\mathrm{A} \cup(\mathrm{B} \cup \mathrm{C})}$

$\Rightarrow(\mathrm{A} \cup \mathrm{B}) \cup \mathrm{C}^{\subset} \mathrm{A} \cup(\mathrm{B} \cup \mathrm{C}) \ldots \ldots(\mathrm{i})$

Now, let $\mathrm{x}$ be an element of $\mathrm{A} \cup(\mathrm{B} \cup \mathrm{C})$

Then, $x \in A$ or $(B \cup C)$

$\Rightarrow x^{\in} A$ or $x^{E} B$ or $x^{\in} C$

$\Rightarrow x^{\in}(A \cup B)$ or $x^{\in} C$

$\Rightarrow x^{\in}(A \cup B) \cup C$

$\Rightarrow A \cup(B \cup C) \subset(A \cup B) \cup C \ldots \ldots$ (ii)

From $i$ and $i i,(A \cup B) \cup C=A \cup(B \cup C)$

[ every set is a subset of itself]

Hence , proved.

(viii) Let $x$ be any element of $(A \cap B) \cap C$

$\Rightarrow x \in(A \cap B)$ and $x \in C$

$\Rightarrow x^{\in} A$ and $x \in_{B}$ and $x \in C$

$\Rightarrow x_{x} \in_{A}$ and $x^{\in}(B \cap C)$

$\Rightarrow{ }_{x} \in_{A} \cap(B \cap C)$

$\Rightarrow(A \cap B) \cap C^{\subset} A \cap(B \cap C) \ldots . .(\mathrm{i})$

Now, let $x$ be an element of $A \cap(B \cap C)$

Then, $x \in{ }_{A}$ and $(B \cap C)$

$\Rightarrow{ }_{x} \in{ }_{A}$ and $x \in{ }_{B}$ and $x \in_{C}$

$\Rightarrow{ }_{x} \in_{(A \cap B) \text { and } x} \in_{C}$

$\Rightarrow{ }_{x} \in_{(A \cap B) \cap C}$

${\Rightarrow{A} \cap(B \cap C)} \subset_{(A \cap B) \cap C \ldots \ldots(\text { ii })}$

From $\mathrm{i}$ and $\mathrm{ii},(\mathrm{A} \cap \mathrm{B}) \cap \mathrm{C}=\mathrm{A} \cap(\mathrm{B} \cap \mathrm{C})$

[every set is a subset of itself]

Hence, proved.