Show that for any sets A and B, A = (A ∩ B) ∪ (A - B) and A ∪ (B - A) = (A ∪ B)

Show that for any sets A and B, A = (A ∩ B) ∪ (A - B) and A ∪ (B -…

Question 43 of 47advanced🔍 AnalyzeLong Answer5 marks

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

Correct Answer

(i) A = (A ∩ B) ∪ (A – B)

(ii) A ∪ (B – A) = (A ∪ B)

Exercise: Miscellaneous Exercise on Chapter 1 | Q: 6 | (Page 21)

For More Understanding

Explanation

This question requires proving two set identities using the definition of set difference and the method of showing mutual subset relations.

The textbook defines $A - B$ as 'the set of elements which belong to $A$ but not to $B$ '.

Solution Steps

Step 1: To prove $A = (A \cap B) \cup (A - B)$ , we show both sets are subsets of each other.
Step 2: Let $x \in (A \cap B) \cup (A - B)$ . Then $x \in (A \cap B)$ or $x \in (A - B)$ . If $x \in (A \cap B)$ , then $x \in A$ . If $x \in (A - B)$ , then $x \in A$ . Hence $(A \cap B) \cup (A - B) \subseteq A$ .
Step 3: Conversely, let $x \in A$ . Then either $x \in B$ or $x \notin B$ . If $x \in B$ , then $x \in (A \cap B)$ . If $x \notin B$ , then $x \in (A - B)$ . In both cases, $x \in (A \cap B) \cup (A - B)$ . Hence $A \subseteq (A \cap B) \cup (A - B)$ .
Step 4: Since both sets are subsets of each other, $A = (A \cap B) \cup (A - B)$ .
Step 5: To prove $A \cup (B - A) = (A \cup B)$ , let $x \in A \cup (B - A)$ . Then $x \in A$ or $x \in (B - A)$ . If $x \in A$ , then $x \in (A \cup B)$ . If $x \in (B - A)$ , then $x \in B$ , so $x \in (A \cup B)$ . Hence $A \cup (B - A) \subseteq (A \cup B)$ .
Step 6: Conversely, let $x \in (A \cup B)$ . Then $x \in A$ or $x \in B$ . If $x \in A$ , then $x \in A \cup (B - A)$ . If $x \in B$ and $x \notin A$ , then $x \in (B - A)$ , so $x \in A \cup (B - A)$ . If $x \in B$ and $x \in A$ , then $x \in A$ . Hence $(A \cup B) \subseteq A \cup (B - A)$ .
Step 7: Since both sets are subsets of each other, $A \cup (B - A) = (A \cup B)$ .

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