Question
Show that for any sets A and B, A = ( A $\cap$B ) $\cup$( A – B ) and A $\cup$( B – A ) = ( A $\cup$B )

Answer

We have to prove: $A=(A \cap B) \cup(A-B)$
Proof: Let $x \in A$
Now, we need to show that $x \in(A \cap B) \cup(A-B)$
In Case I,
$x \in(A \cap B)$
$\Rightarrow X \in(A \cap B) \subset(A \cup B) \cup(A-B)$
In Case II,
$X \notin A \cap B$
$\Rightarrow x \notin B \ or \ x \notin A$
$\Rightarrow X \notin B(X \notin A)$
$\Rightarrow X \notin A-B \subset(A \cup B) \cup(A-B)$
$\therefore A \subset(A \cap B) \cup(A-B)(i)$
It can be concluded that, $A \cap B \subset A \ and \ (A-B) \subset A$
Therefore, (A $\cap$B) $\cap$(A - B) $\subset $A (ii)
Equating (i) and (ii),we get
$A=(A \cap B) \cup(A-B)$
Now, we need to show, $A \cup(B-A) \subset A \cup B$
Suppose that,
$X \in A \cup(B-A)$
$X \in A \ or \ X \in(B-A)$
$\Rightarrow X \in A \ or \ (X \in B \text { and } X \notin A)$
$\Rightarrow(X \in A \text { or } X \in B) \ and \ (X \in A \text { and } X \notin A)$
$\Rightarrow X \in(B \cup A)$
$\therefore A \cup(B-A) \subset(A \cup B)$(iii)
Now, to prove: $(A \cup B) \subset A \cup(B-A)$
Let y $\in A\cup B$
$\mathrm{y} \in \mathrm{A}\ or \ \mathrm{y} \in \mathrm{B}$
$(y \in A \text { or } y \in B)\ and \ (X \in A \text { and } X \notin A)$
$\Rightarrow y \in A \ or \ (y \in B \text { and } y \notin A)$
$\Rightarrow y \in A \cup(B-A)$
Therefore,$A \cup B \subset A \cup(B-A)$(iv)
$\therefore$ Using (iii) and (iv), we obtain
$A \cup(B-A)=A \cup B$

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