For any two sets A and B, show that the following statements are equivalent:
(i) $A \subset B$
(ii) $A-B=\phi$
(iii) $A \cup B=B$
(iv) $A \cap B=A$.
We have that the following statements are equivalent:
(i) $A \subset B$
(ii) $A-B=\phi$
(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,
$x \in(A-B)$
$\Rightarrow x \in A \& x \notin B \quad$ (Which is contradictory)
Also,
$\because A \subset B$
$\Rightarrow A-B \subseteq \phi$ ...(1)
We know that null sets are the subsets of every set.
$\therefore \phi \subseteq A-B$ ...(2)
From $(1) \&(2)$, we get,
$(A-B)=\phi$
$\therefore(\mathrm{i})=(\mathrm{ii})$
Now, We have,
$(A-B)=\phi$
That means that there is no element in $A$ that does not belong to $B$.
Now,
$A \cup B=B$
$\therefore(\mathrm{ii})=(\mathrm{iii})$
WE have,
$A \cup B=B$
$\Rightarrow A \subset B$
$\Rightarrow A \cap B=A$
$\therefore($ iii $)=($ iv $)$
WE have,
$A \cap B=A$
It should be possible if $A \subset B$.
Now,
$A \subset B$
$\therefore(\mathrm{iv})=(\mathrm{i})$
We have,
$(\mathrm{i})=(\mathrm{ii})=(\mathrm{iii})=(\mathrm{iv})$
Therefore, we can say that all statements are equivalent.
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