1. **Show that if $A_1 \subset A_2$, then $f(A_1) \subset f(A_2)$:** By definition, $f(A) = \{y \in F \mid \exists x \in A, f(x) = y\}$. If $A_1 \subset A_2$, then every element $x$ of $A_1$ is also in $A_2$. Therefore, any $y = f(x)$ with $x \in A_1$ is also an image of some $x$ in $A_2$, so $y \in f(A_2)$. Hence, $f(A_1) \subset f(A_2)$. 2. **Show that $f(A_1 \cap A_2) \subset f(A_1) \cap f(A_2)$:** Take any $y \in f(A_1 \cap A_2)$. Then there exists $x \in A_1 \cap A_2$ such that $f(x) = y$. Since $x$ is in both $A_1$ and $A_2$, $y$ is in both $f(A_1)$ and $f(A_2)$. Thus $y \in f(A_1) \cap f(A_2)$, which proves the inclusion. 3. **Show that $f(A_1 \cup A_2) = f(A_1) \cup f(A_2)$:** By definition, - For $y \in f(A_1 \cup A_2)$, there exists $x \in A_1 \cup A_2$ with $f(x) = y$. Hence, $x \in A_1$ or $x \in A_2$, so $y \in f(A_1)$ or $y \in f(A_2)$, thus $y \in f(A_1) \cup f(A_2)$. - Conversely, if $y \in f(A_1) \cup f(A_2)$, then $y$ is an image of some $x$ in $A_1$ or some $x$ in $A_2$, so $x \in A_1 \cup A_2$, which implies $y \in f(A_1 \cup A_2)$. Hence, $f(A_1 \cup A_2) = f(A_1) \cup f(A_2)$. 4. **Show that if $B_1 \subset B_2$, then $f^{-1}(B_1) \subset f^{-1}(B_2)$:** By definition, $f^{-1}(B) = \{x \in E \mid f(x) \in B\}$. If $B_1 \subset B_2$ and $x \in f^{-1}(B_1)$, then $f(x) \in B_1 \subset B_2$, so $x \in f^{-1}(B_2)$. Hence, $f^{-1}(B_1) \subset f^{-1}(B_2)$. 5. **Show that $f^{-1}(B_1 \cap B_2) = f^{-1}(B_1) \cap f^{-1}(B_2)$:** By the definition of inverse image: - If $x \in f^{-1}(B_1 \cap B_2)$, then $f(x) \in B_1 \cap B_2$. This means $f(x) \in B_1$ and $f(x) \in B_2$, so $x \in f^{-1}(B_1)$ and $x \in f^{-1}(B_2)$, implying $x \in f^{-1}(B_1) \cap f^{-1}(B_2)$. - Conversely, if $x \in f^{-1}(B_1) \cap f^{-1}(B_2)$, then $f(x) \in B_1$ and $f(x) \in B_2$, so $f(x) \in B_1 \cap B_2$, which means $x \in f^{-1}(B_1 \cap B_2)$. Therefore, $f^{-1}(B_1 \cap B_2) = f^{-1}(B_1) \cap f^{-1}(B_2)$. **Summary:** All the prescribed set relations for direct and reciprocal images under $f$ are demonstrated by applying the definitions clearly and using elementary set theory.