For any two sets A and B, show that the following statements are equivalent:
(i) $A \subset B$
(ii) $A - B = ϕ$
(iii) $A \cup B = B$
(iv) $A \cap B = A .$

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Solution

We have that the following statements are equivalent:
(i) $A \subset B$
(ii) $A - B = ϕ$
(iii) $A \cup B = B$
(iv) $A \cap B = A$

Proof:
$Let A \subset B Let x be an arbitary element of (A - B) . Now, x \in (A - B) \Rightarrow x \in A & x \notin B (Which is contradictory) Also, ∵ A \subset B \Rightarrow A - B \subseteq ϕ . . . (1) We know that null sets are the subsets of every set . ∴ ϕ \subseteq A - B . . . (2) From (1) & (2), we get, (A - B) = ϕ ∴ (i) = (ii) Now, we have, (A - B) = ϕ That means that there is no element in A that does not belong to B . Now, A \cup B = B ∴ (ii) = (iii) We have, A \cup B = B \Rightarrow A \subset B \Rightarrow A \cap B = A ∴ (iii) = (iv) We have, A \cap B = A It should be possible if A \subset B . Now, A \subset B ∴ (iv) = (i) We have, (i) = (ii) = (iii) = (iv) Therefore, we can say that all statements are equivalent .$

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