**Problem statement:** We are given a function $f: E \to F$, two subsets $A, B$ and subsets $A_1, A_2 \subseteq E$, $B_1, B_2 \subseteq F$. We need to prove: 1. If $A_1 \subset A_2$, then $f(A_1) \subset f(A_2)$. 2. $f(A_1 \cap A_2) \subset f(A_1) \cap f(A_2)$. 3. $f(A_1 \cup A_2) = f(A_1) \cup f(A_2)$. 4. If $B_1 \subset B_2$, then $f^{-1}(B_1) \subset f^{-1}(B_2)$. 5. $f^{-1}(B_1 \cap B_2) = f^{-1}(B_1) \cap f^{-1}(B_2)$. --- **Steps:** 1. To prove $A_1 \subset A_2 \Rightarrow f(A_1) \subset f(A_2)$: - Let $y \in f(A_1)$. Then by definition of direct image, $\exists x \in A_1$ such that $f(x) = y$. - Since $A_1 \subset A_2$, $x \in A_2$. - Hence $y = f(x) \in f(A_2)$. - So $f(A_1) \subset f(A_2)$. 2. To prove $f(A_1 \cap A_2) \subset f(A_1) \cap f(A_2)$: - Let $y \in f(A_1 \cap A_2)$. - Then $\exists x \in A_1 \cap A_2$ with $f(x)=y$. - Since $x \in A_1$ and $x \in A_2$, $y = f(x)$ is in both $f(A_1)$ and $f(A_2)$. - Therefore, $y \in f(A_1) \cap f(A_2)$. - Hence $f(A_1 \cap A_2) \subset f(A_1) \cap f(A_2)$. 3. To prove $f(A_1 \cup A_2) = f(A_1) \cup f(A_2)$: - For $\subseteq$ direction: - Let $y \in f(A_1 \cup A_2)$. - Then $\exists x \in A_1 \cup A_2$ with $f(x)=y$. - So $x \in A_1$ or $x \in A_2$. - Then $y \in f(A_1)$ or $y \in f(A_2)$, so $y \in f(A_1) \cup f(A_2)$. - For $\supseteq$ direction: - Let $y \in f(A_1) \cup f(A_2)$. - So $y \in f(A_1)$ or $y \in f(A_2)$. - In first case, $\exists x \in A_1$ with $f(x)=y$; similarly for second case with $A_2$. - Then $x \in A_1 \cup A_2$ and $y = f(x) \in f(A_1 \cup A_2)$. - Therefore, equality holds. 4. To prove $B_1 \subset B_2 \Rightarrow f^{-1}(B_1) \subset f^{-1}(B_2)$: - Let $x \in f^{-1}(B_1)$. - By definition of reciprocal image, $f(x) \in B_1$. - Since $B_1 \subset B_2$, $f(x) \in B_2$. - Therefore, $x \in f^{-1}(B_2)$. 5. To prove $f^{-1}(B_1 \cap B_2) = f^{-1}(B_1) \cap f^{-1}(B_2)$: - For $\subseteq$: - Let $x \in f^{-1}(B_1 \cap B_2)$. - Then $f(x) \in B_1 \cap B_2$ i.e. $f(x) \in B_1$ and $f(x) \in B_2$. - Hence $x \in f^{-1}(B_1)$ and $x \in f^{-1}(B_2)$. - So $x \in f^{-1}(B_1) \cap f^{-1}(B_2)$. - For $\supseteq$: - Let $x \in f^{-1}(B_1) \cap f^{-1}(B_2)$. - Then $f(x) \in B_1$ and $f(x) \in B_2$. - So $f(x) \in B_1 \cap B_2$. - Thus $x \in f^{-1}(B_1 \cap B_2)$. Therefore, equality holds. --- **Final answer:** All 5 set inclusion and equality properties hold as proved.