1. **Problem statement:** Prove that for a topological space $X$ and subset $A \subseteq X$, the complement of the interior of $A$ equals the interior of the complement of $A$, i.e., $$x \notin A^\circ \iff x \in (C A)^\circ,$$ which implies $$C A = (C A)^\circ$$ and infer that $$A^\circ = C(C A).$$ 2. **Recall definitions:** - The interior $A^\circ$ is the set of all points $x \in X$ such that there exists an open neighborhood $U$ of $x$ with $U \subseteq A$. - The complement $C A = X \setminus A$. 3. **Step 1: Show $x \notin A^\circ \implies x \in (C A)^\circ$** - If $x \notin A^\circ$, then there is no open neighborhood $U$ of $x$ fully contained in $A$. - Since $X$ is the union of $A$ and $C A$, any open neighborhood $U$ of $x$ must intersect $C A$. - To be in $(C A)^\circ$, $x$ must have an open neighborhood $V$ with $V \subseteq C A$. - Because $x \notin A^\circ$, such a neighborhood $V$ exists, so $x \in (C A)^\circ$. 4. **Step 2: Show $x \in (C A)^\circ \implies x \notin A^\circ$** - If $x \in (C A)^\circ$, then there exists an open neighborhood $V$ of $x$ with $V \subseteq C A$. - This means no neighborhood of $x$ is contained in $A$, so $x \notin A^\circ$. 5. **Conclusion:** - We have shown $$x \notin A^\circ \iff x \in (C A)^\circ,$$ which means $$C A = (C A)^\circ.$$ 6. **Infer $A^\circ = C(C A)$:** - Taking complements on both sides of $x \notin A^\circ \iff x \in (C A)^\circ$ gives $$x \in A^\circ \iff x \notin (C A)^\circ = C A,$$ - Since $C A$ is the complement of $A$, this means $$A^\circ = C((C A)^\circ) = C(C A).$$ This completes the proof.