[proofplan] We show $X = A^\circ \cup \partial A \cup (X \setminus A)^\circ$ by proving two things: (i) every point of $X$ belongs to at least one of the three sets, and (ii) the three sets are pairwise disjoint. Part (i) follows from the [Neighbourhood Characterisation of Closure](/theorems/1005): for any $x \in X$, either some open neighbourhood of $x$ is contained in $A$, or some open neighbourhood is contained in $X \setminus A$, or every open neighbourhood meets both $A$ and $X \setminus A$. Part (ii) follows from the observation that membership in $A^\circ$ or $(X \setminus A)^\circ$ provides a neighbourhood avoiding one side, contradicting the boundary condition. [/proofplan] [step:Show every point of $X$ belongs to at least one of $A^\circ$, $\partial A$, $(X \setminus A)^\circ$] Let $x \in X$. Consider two cases. **Case 1:** There exists an open set $U \in \tau$ with $x \in U \subset A$. Then $x \in A^\circ$ by definition of the interior. **Case 2:** No such open set exists. Then either there exists an open set $V \in \tau$ with $x \in V \subset X \setminus A$, in which case $x \in (X \setminus A)^\circ$, or no open neighbourhood of $x$ is contained entirely in $A$ or entirely in $X \setminus A$. In the latter sub-case, every open neighbourhood $U$ of $x$ satisfies both $U \cap A \neq \varnothing$ (since $U \not\subset X \setminus A$) and $U \cap (X \setminus A) \neq \varnothing$ (since $U \not\subset A$), so $x \in \partial A$. [guided] Let $x \in X$. We perform a case analysis on the behaviour of open neighbourhoods of $x$. **Case 1: $x \in A^\circ$.** This occurs when there exists $U \in \tau$ with $x \in U \subset A$. **Case 2: $x \notin A^\circ$.** Then no open set $U$ containing $x$ is contained in $A$; equivalently, every open neighbourhood of $x$ meets $X \setminus A$. We split further. **Case 2a: $x \in (X \setminus A)^\circ$.** There exists $V \in \tau$ with $x \in V \subset X \setminus A$. **Case 2b: $x \notin (X \setminus A)^\circ$.** Then no open neighbourhood of $x$ is contained in $X \setminus A$, so every open neighbourhood of $x$ meets $A$. Combined with Case 2 (every open neighbourhood meets $X \setminus A$), every open neighbourhood of $x$ meets both $A$ and $X \setminus A$, which is the definition of $x \in \partial A$. In all cases, $x$ belongs to at least one of the three sets. [/guided] [/step] [step:Show the three sets are pairwise disjoint] **$A^\circ \cap (X \setminus A)^\circ = \varnothing$:** If $x \in A^\circ$, then $x \in A$. If $x \in (X \setminus A)^\circ$, then $x \in X \setminus A$. Since $A$ and $X \setminus A$ are disjoint, so are $A^\circ$ and $(X \setminus A)^\circ$. **$A^\circ \cap \partial A = \varnothing$:** If $x \in A^\circ$, there exists $U \in \tau$ with $x \in U \subset A$, so $U \cap (X \setminus A) = \varnothing$. A boundary point $x \in \partial A$ requires every open neighbourhood to meet $X \setminus A$, which $U$ does not. Hence $x \notin \partial A$. **$(X \setminus A)^\circ \cap \partial A = \varnothing$:** If $x \in (X \setminus A)^\circ$, there exists $V \in \tau$ with $x \in V \subset X \setminus A$, so $V \cap A = \varnothing$. A boundary point requires every open neighbourhood to meet $A$, which $V$ does not. Hence $x \notin \partial A$. [guided] We verify that no point can belong to two of the three sets simultaneously. **$A^\circ$ and $(X \setminus A)^\circ$ are disjoint:** $A^\circ \subset A$ and $(X \setminus A)^\circ \subset X \setminus A$, and $A \cap (X \setminus A) = \varnothing$. **$A^\circ$ and $\partial A$ are disjoint:** Suppose $x \in A^\circ$. Then there exists $U \in \tau$ with $x \in U \subset A$. This neighbourhood satisfies $U \cap (X \setminus A) = \varnothing$, so $x$ fails the condition for $\partial A$ (which requires every open neighbourhood to meet $X \setminus A$). **$(X \setminus A)^\circ$ and $\partial A$ are disjoint:** Suppose $x \in (X \setminus A)^\circ$. Then there exists $V \in \tau$ with $x \in V \subset X \setminus A$. This neighbourhood satisfies $V \cap A = \varnothing$, so $x$ fails the condition for $\partial A$ (which requires every open neighbourhood to meet $A$). Since the three sets cover $X$ and are pairwise disjoint, $X = A^\circ \sqcup \partial A \sqcup (X \setminus A)^\circ$. [/guided] [/step]