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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