Let $ F : x \Rightarrow y $ be a function. Then prove that
$i. A_1 \subset A_2 \Rightarrow F(A_1) \subset F(A_2)$
$ii. F(\cup_i A_i) = \cup_i F(A_i) $
$iii.F(\cap_i A_i) \subset \cap_i f(A_i)$
Solution
$ Let \ f: x \rightarrow y $ be a function.
i.
$Let\ y \in f(A_1)$
$Then\ \exists x \in A_1 : y = f(x) $.
$But\ A_1 \subset A_2 $
$we\ can\ say\ , x \in A_2 $
$Hence\ y = f(x) $
ii.
$Let\ y \in f ( \cup_i A_i) then\ \exists x \in \cup_i A_i : y = f(x)$
$ Since\ x \in (\cup_iA_i) \Rightarrow x \in A_i\ for\ some\ i $
$ so\ y = f(x) \in f(A_i)\ and\ hence\ y \in \cup_i(A_i) $
Conversely
$ Let\ y \in \cup_if(A_i)\ then\ y \in f(A_i) for\ some\ i $
$ so\ \exists x \in A_i : y = f(x) $
$ and\ x \in \Rightarrow x \in \cup_i(A_i) $
$ so\ that\ f(x) \in f(\cup_i A_i)$
iii.
$
Let\ y \in f(\cup_i A_i)\ then\ \exists x \in \cap_i A_i : y = f(x) $
$ Since\ x \in \cap_i A_i $
$ we\ can\ write\ x \in A_i\ \forall i $
$ so\ f(x) \in f(A_i)\ \forall i $
$ Hence\ f(x) \in \cap_if(A_i) $