**Question:**

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) $A \cap B=A$.

**Solution:**

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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