why is $f(A \cap B)$ a proper subset of $f(A) \cap f(B)?$ why it is proper? 
THEOREM. If $f: X \to Y$ is a function and $A$ and $B$ are subsets of $X$, then
(i) $\ f(A \cup B) = f(A) \cup f(B)$
(ii) $\ f(A \cap B) \subset f(A) \cap f(B)$
(iii) $\ f(A) - f(B) \subset f(A - B)$

Q 1: Can anyone please tell me why ii) is a proper subset instead of a subset? Why cann't              them be equal?
Q 2 :and what is the proof for three of them ?
 A: $f(A\cup B)=f(A)\cup f(B)$
Proof.
Suppose $t\in f(A\cup B)$. This means that for some $s\in A\cup B$, we have $t=f(s)$. So, we then have that $t\in f(A)$ or $t\in f(B)$ because $s\in A$ or $s\in B$. That is, $t\in f(A)\cup f(B)$. Hence $f(A\cup B)\subset f(A)\cup f(B)$.
Suppose $t\in f(A)\cup f(B)$. This means $t\in f(A)$ or $t\in f(B)$. So, for some $s\in A$ we have $t=f(s)$ or for some $s\in B$, we have $t=f(s).$ Regardless, we have that $t=f(s)$ for some $s\in A\cup B$ and so $s\in f(A\cup B)$. Hence $f(A)\cup f(B)\subset f(A\cup B)$.
Since they are subsets of each other, this forces equality.
A: As already mentioned $\subset$ is the notation for a (in general not proper) subset.
For (ii) let $y\in f(A\cap B)$. Then by definition, $y=f(x)$ where $x\in A\cap B$, i.e. $x\in A$ and $x\in B$. So in particular, $y=f(x)\in f(A)$ and $y=f(x)\in f(B)$ which concludes the prove.
For (iii) let $y\in f(A)\setminus f(B)$. Then you know that there exists a $x\in X$ s.t. $f(x) = y$ and $x\in A$ but $x\notin B$. Hence, $x\in A\setminus B$. Thus,  $y=f(x)\in f(A\setminus B)$ which proves the claim.
