[proofplan]
We prove the three claims in order. Part (1) follows from the definitions: the boundary $\partial A = \overline{A} \cap \overline{X \setminus A}$ consists of those closure points that are not interior, giving $\partial A = \overline{A} \setminus A^\circ$. Part (2) is an immediate rearrangement. Part (3) partitions $X$ into three sets by classifying each point $x$ according to whether $x$ is interior to $A$, interior to $X \setminus A$, or in neither interior — the last case being exactly the boundary.
[/proofplan]
[step:Show $\partial A = \overline{A} \setminus A^\circ$]
By definition, $\partial A = \overline{A} \cap \overline{X \setminus A}$. By the [Interior–Closure Duality](/theorems/1014), $A^\circ = X \setminus \overline{X \setminus A}$, which rearranges to $\overline{X \setminus A} = X \setminus A^\circ$. Substituting:
\begin{align*}
\partial A = \overline{A} \cap (X \setminus A^\circ) = \overline{A} \setminus A^\circ.
\end{align*}
[guided]
Recall that the boundary is defined as $\partial A := \overline{A} \cap \overline{X \setminus A}$ — the set of points in the closure of both $A$ and its complement. We need to re-express $\overline{X \setminus A}$ in terms of the interior. The [Interior–Closure Duality](/theorems/1014) gives $A^\circ = X \setminus \overline{X \setminus A}$. Solving for the closure of the complement: $\overline{X \setminus A} = X \setminus A^\circ$. That is, a point lies in $\overline{X \setminus A}$ precisely when it is not an interior point of $A$. Intersecting with $\overline{A}$:
\begin{align*}
\partial A = \overline{A} \cap (X \setminus A^\circ) = \overline{A} \setminus A^\circ.
\end{align*}
This says: the boundary of $A$ consists of those points that are in the closure of $A$ but not in its interior.
[/guided]
[/step]
[step:Decompose $\overline{A}$ as the disjoint union $A^\circ \cup \partial A$]
From part (1), $\partial A = \overline{A} \setminus A^\circ$. Since $A^\circ \subset \overline{A}$ (because $A^\circ \subset A \subset \overline{A}$ by the [Properties of the Interior Operator](/theorems/1013), property (1), and extensivity of closure), the sets $A^\circ$ and $\overline{A} \setminus A^\circ$ partition $\overline{A}$:
\begin{align*}
\overline{A} = A^\circ \cup (\overline{A} \setminus A^\circ) = A^\circ \cup \partial A.
\end{align*}
These two sets are disjoint by construction: $A^\circ \cap (\overline{A} \setminus A^\circ) = \varnothing$.
[/step]
[step:Partition $X$ into $A^\circ$, $\partial A$, and $(X \setminus A)^\circ$]
We show $X = A^\circ \cup \partial A \cup (X \setminus A)^\circ$ and that the three sets are pairwise disjoint.
**Exhaustion.** Let $x \in X$. By part (2), if $x \in \overline{A}$, then $x \in A^\circ \cup \partial A$. If $x \notin \overline{A}$, then $x \in X \setminus \overline{A}$. Since $X \setminus \overline{A}$ is open (as the complement of the closed set $\overline{A}$) and $X \setminus \overline{A} \subset X \setminus A$, it follows that $x \in (X \setminus A)^\circ$. Hence every $x \in X$ belongs to at least one of the three sets.
**Disjointness.** We verify pairwise:
- $A^\circ \cap \partial A = \varnothing$: this holds because $\partial A = \overline{A} \setminus A^\circ$.
- $A^\circ \cap (X \setminus A)^\circ = \varnothing$: by property (1) of the [Properties of the Interior Operator](/theorems/1013), $A^\circ \subset A$ and $(X \setminus A)^\circ \subset X \setminus A$, and $A \cap (X \setminus A) = \varnothing$.
- $\partial A \cap (X \setminus A)^\circ = \varnothing$: suppose for contradiction that $x \in \partial A \cap (X \setminus A)^\circ$. Since $x \in (X \setminus A)^\circ$, there exists $U \in \tau$ with $x \in U \subset X \setminus A$. Since $x \in \partial A \subset \overline{A}$ (by part (1)), the [Neighbourhood Characterisation of Closure](/theorems/1005) gives $U \cap A \neq \varnothing$. This contradicts $U \subset X \setminus A$.
[guided]
The exhaustion step deserves emphasis. We split $X$ into the cases $x \in \overline{A}$ and $x \notin \overline{A}$. If $x \in \overline{A}$, part (2) places $x$ in $A^\circ \cup \partial A$. If $x \notin \overline{A}$, we need $x \in (X \setminus A)^\circ$. Why is this true? The set $X \setminus \overline{A}$ is open because $\overline{A}$ is closed. Moreover, $\overline{A} \supset A$ implies $X \setminus \overline{A} \subset X \setminus A$. So $X \setminus \overline{A}$ is an open subset of $X \setminus A$ containing $x$, which witnesses $x \in (X \setminus A)^\circ$.
For the disjointness of $\partial A$ and $(X \setminus A)^\circ$: the key is that boundary points of $A$ are adherent to $A$ (they lie in $\overline{A}$), while interior points of $X \setminus A$ have an open neighbourhood entirely missing $A$. These two conditions are incompatible.
[/guided]
[/step]
