1. **Problem:** Prove that for a function $f : X \to Y$ between topological spaces, the following conditions are equivalent: (a) $f$ is continuous; (d) $f(\overline{A}) \subset \overline{f(A)}$ for every subset $A$ of $X$. 2. **Recall the criterion:** From topology, a function $f$ is continuous if and only if for every closed set $C$ in $Y$, the preimage $f^{-1}(C)$ is closed in $X$ (criterion (c) of 3.4.5). 3. **Step to show (a) implies (d):** - Assume $f$ is continuous. - Let $A \subset X$ be any subset. - The closure $\overline{A}$ is the smallest closed set containing $A$. - Since $f$ is continuous, $f^{-1}(\overline{f(A)})$ is closed and contains $A$. - Therefore, $\overline{A} \subset f^{-1}(\overline{f(A)})$. - Applying $f$ to both sides, $f(\overline{A}) \subset f(f^{-1}(\overline{f(A)})) \subset \overline{f(A)}$. 4. **Step to show (d) implies (a):** - Assume $f(\overline{A}) \subset \overline{f(A)}$ for every $A \subset X$. - Let $C$ be a closed subset of $Y$. - Then $f^{-1}(C)$ is a subset of $X$. - Since $C$ is closed, $\overline{C} = C$. - Using the assumption with $A = f^{-1}(C)$, we get: $$f(\overline{f^{-1}(C)}) \subset \overline{f(f^{-1}(C))} \subset \overline{C} = C.$$ - This implies $\overline{f^{-1}(C)} \subset f^{-1}(C)$ because if $x \in \overline{f^{-1}(C)}$, then $f(x) \in C$. - Hence, $f^{-1}(C)$ is closed. - By the closed set criterion, $f$ is continuous. **Final conclusion:** Conditions (a) and (d) are equivalent.