If x belongs to the domain then does f(x) belong to the range?

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.

$Let\ y \in f(\cup_i A_i)\ then\ \exists x \in \cap_i A_i : y = f(x)$

$Since\ x \in \cap_i A_i$

$we\ can\ write\ x \in A_i\ \forall i$

$so\ f(x) \in f(A_i)\ \forall i$

$Hence\ f(x) \in \cap_if(A_i)$

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