[proofplan] The identity $A^\circ = X \setminus \overline{X \setminus A}$ is proved by showing both inclusions. For $(\subset)$: if $x \in A^\circ$, then some open neighbourhood of $x$ misses $X \setminus A$, so $x \notin \overline{X \setminus A}$, hence $x \in X \setminus \overline{X \setminus A}$. For $(\supset)$: if $x \notin \overline{X \setminus A}$, then by the [Neighbourhood Characterisation of Closure](/theorems/1005), some open neighbourhood of $x$ is disjoint from $X \setminus A$, placing $x$ in $A^\circ$. The equivalent form $\overline{A} = X \setminus (X \setminus A)^\circ$ follows by substituting $X \setminus A$ for $A$ and taking complements. [/proofplan] [step:Show $A^\circ \subset X \setminus \overline{X \setminus A}$] Let $x \in A^\circ$. By definition, there exists $U \in \tau$ with $x \in U \subset A$. Since $U \subset A$, we have $U \cap (X \setminus A) = \varnothing$. The set $U$ is an open neighbourhood of $x$ that does not intersect $X \setminus A$, so by the [Neighbourhood Characterisation of Closure](/theorems/1005), $x \notin \overline{X \setminus A}$. Equivalently, $x \in X \setminus \overline{X \setminus A}$. [/step] [step:Show $X \setminus \overline{X \setminus A} \subset A^\circ$] Let $x \in X \setminus \overline{X \setminus A}$, so $x \notin \overline{X \setminus A}$. By the [Neighbourhood Characterisation of Closure](/theorems/1005), there exists $U \in \tau$ with $x \in U$ and $U \cap (X \setminus A) = \varnothing$. The condition $U \cap (X \setminus A) = \varnothing$ is equivalent to $U \subset A$. Since $U$ is open and $x \in U \subset A$, we conclude $x \in A^\circ$. [guided] The [Neighbourhood Characterisation of Closure](/theorems/1005) states: $x \in \overline{B}$ if and only if every open set containing $x$ meets $B$. Negating this, $x \notin \overline{B}$ if and only if there exists an open set $U$ containing $x$ with $U \cap B = \varnothing$. We apply the negated form with $B = X \setminus A$. The condition $U \cap (X \setminus A) = \varnothing$ means every point of $U$ lies in $A$, i.e., $U \subset A$. Since $U$ is open and contains $x$, this witnesses $x \in A^\circ$. [/guided] [/step] [step:Derive the equivalent formulation $\overline{A} = X \setminus (X \setminus A)^\circ$] Substituting $X \setminus A$ in place of $A$ in the identity $A^\circ = X \setminus \overline{X \setminus A}$ gives \begin{align*} (X \setminus A)^\circ = X \setminus \overline{X \setminus (X \setminus A)} = X \setminus \overline{A}. \end{align*} Taking the complement of both sides in $X$: \begin{align*} X \setminus (X \setminus A)^\circ = \overline{A}. \end{align*} [/step]