Exam 2024 (B)

5

\begin{matrix} Functions f, g : P (N) \to P (N) are called nice to each other if \\ \forall A, B \in P (N) : \begin{matrix} A \subseteq B ⟹ f (B) \subseteq f (A) \\ A \subseteq B ⟹ g (B) \subseteq g (A) \end{matrix} \\ And \forall A \in P (N) : \begin{matrix} A \subseteq f (g (A)) \\ A \subseteq g (f (A)) \end{matrix} \end{matrix}

\begin{matrix} Find two functions that are nice to each other \\ Solution: \\ Let \forall A \in P (N) : f (A) = g (A) = N \\ Let A \subseteq B \in P (N) \\ f (B) = f (A) = N ⟹ f (A) \subseteq f (B) \\ g (A) = g (B) = N ⟹ g (A) \subseteq g (B) \\ A \subseteq N = f (g (A)) = g (f (A)) \end{matrix}

\begin{matrix} Prove: f, g are nice to each other ⟺ \forall A, B \in P (N) : A \subseteq f (B) ⟺ B \subseteq g (A) \\ Proof: \\ Let f, g be nice to each other \\ Let A, B \in P (N) \\ Let A \subseteq f (B) \\ ⟹ g (f (B)) \subseteq g (A) ⟹ B \subseteq g (A) \\ Let B \subseteq g (A) \\ ⟹ f (g (A)) \subseteq f (B) ⟹ A \subseteq f (B) \\ Let \forall A, B \in P (N) : A \subseteq f (B) ⟺ B \subseteq g (A) \\ Let A, B \in P (N) \\ \underset{^{'} B^{'}}{\underset{⏟}{g (A)}} \subseteq g (A) ⟹ A \subseteq f (\underset{^{'} B^{'}}{\underset{⏟}{g (A)}}) \\ \underset{^{'} A^{'}}{\underset{⏟}{f (B)}} \subseteq f (B) ⟹ B \subseteq g (\underset{^{'} A^{'}}{\underset{⏟}{f (B)}}) \\ Let A \subseteq B \\ B \subseteq f (g (B)) ⟹ A \subseteq f (g (B)) ⟹ g (B) \subseteq g (A) \\ B \subseteq g (f (B)) ⟹ A \subseteq g (f (B)) ⟹ f (B) \subseteq f (A) \\ ⟹ f, g are nice to each other \end{matrix}

\begin{matrix} Let f, g be nice to each other \\ Prove: f = f \circ g \circ f \\ Proof: \\ D o m (f \circ g \circ f) = P (N) = D o m (f) \\ R a n g e (f \circ g \circ f) = P (N) = R a n g e (f) \\ Let A \in P (N) \\ A \subseteq g (f (A)) ⟹ f (g (f (A))) \subseteq f (A) \\ g (f (A)) \subseteq g (f (A)) ⟹ f (A) \subseteq f (g (f (A))) \\ ⟹ f (A) = f (g (f (A))) ⟹ f = f \circ g \circ f \end{matrix}

\begin{matrix} Let f, g be nice to each other \\ Prove: \forall A, B \in P (N) : f (A \cup B) = f (A) \cap f (B) \\ Proof: \\ Let A, B \in P (N) \\ x \in f (A \cup B) ⟺ {x} \subseteq f (A \cup B) ⟺ A \cup B \subseteq g ({x}) \\ ⟺ A \subseteq g ({x}) \land B \subseteq g ({x}) ⟺ {x} \subseteq f (A) \land {x} \subseteq f (B) \\ ⟺ {x} \subseteq f (A) \cap f (B) ⟺ x \in f (A) \cap f (B) \\ ⟹ f (A \cup B) = f (A) \cap f (B) \end{matrix}