(1)Let $\displaystyle f: X \to Y $ be a function and $\displaystyle A_{1}, A_{2} \in \mathcal{P}(X) $. (i) Prove that $\displaystyle A_{1} \subseteq A_{2} \Rightarrow \overrightarrow{f}(A_{1}) \subseteq \overrightarrow{f}(A_{2}) $. Prove that the converse is not universally true. Give a simple condition on $\displaystyle f $ which is equivalent to the converse. (ii) Prove that $\displaystyle \overrightarrow{f}(A_{1} \cap A_{2}) \subseteq \overrightarrow{f}(A_{1}) \cap \overrightarrow{f}(A_{2}) $. Prove that equality is not universally true. (iii) Prove that $\displaystyle \overrightarrow{f}(A_{1} \cup A_{2}) = \overrightarrow{f}(A_{1}) \cup \overrightarrow{f}(A_{2}) $.

(i) So $\displaystyle x \in A_1 \Rightarrow x \in A_2 $. Then $\displaystyle \overrightarrow{f}(A_{1}) = \{f(x) | x \in A_1 \} \Rightarrow \overrightarrow{f}(A_{2}) = \{f(x) | x \in A_2 \} $. Thus $\displaystyle \overrightarrow{f}(A_1) \subseteq \overrightarrow{f}(A_2) $. The converse is $\displaystyle \overrightarrow{f}(A_1) \subseteq \overrightarrow{f}(A_2) \Rightarrow A_1 \subseteq A_2 $. Suppose that $\displaystyle A_{1} = \{c \} $ and $\displaystyle A_{2} = \{d \} $. Let $\displaystyle f(a) = a $. Then $\displaystyle \overrightarrow{f}(A_1) = \overrightarrow{f}(A_2) = \emptyset $ but $\displaystyle A_1 \not \subseteq A_2 $. Is this correct? What would be the condition on $\displaystyle f $ which is equivalent to its converse?

(ii) $\displaystyle \overrightarrow{f}(A_{1} \cap A_{2}) = \{ f(x) | x \in A_{1} \cap A_2 \} $ which is a subset of $\displaystyle \overrightarrow{f}(A_{1}) \cap \overrightarrow{f}(A_{2}) $. What would be a counterexample to show inequality?

(iii) This is basically the same as part (ii) except with an $\displaystyle \text{or} $?

Thanks