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 funxtion.

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.

$Let\ y \in f(\cup_i A_i)\ then\ \exists x \in \cap_i A_i : y = f(x)$ $Since\ x \in \cap_i A_i$ $we\ can\ write\ x \in A_i\ for\ all\ i$ $so\ f(x) \in f(A_i)\ for\ all\ i$ $Hence\ f(x) \in \cap_if(A_i)$

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