Proof of Decomposition of a Topological Space by the Boundary of a Subset

By the definition of the boundary we have $\partial_X A=\operatorname{cl}_X(A)\setminus\operatorname{int}_X(A)$. The claims are proved in the order 1, 2, 3, 5, 4.

Claim 1. Since $\operatorname{int}_X(A)\subseteq X$, removing $\operatorname{int}_X(A)$ from a subset of $X$ is the same as intersecting with $X\setminus\operatorname{int}_X(A)$, so

By claim 1 of Duality Between Interior and Closure Under Complementation we have $X\setminus\operatorname{int}_X(A)=\operatorname{cl}_X(X\setminus A)$, and substituting gives $\partial_X A=\operatorname{cl}_X(A)\cap\operatorname{cl}_X(X\setminus A)$.

Claim 2. By claim 2 of The Closure is the Smallest Closed Superset, applied to $A$ and to $X\setminus A$, both $\operatorname{cl}_X(A)$ and $\operatorname{cl}_X(X\setminus A)$ are closed in $(X,\mathcal{T})$. Index these two sets by the set $I=\{1,2\}$, setting $C_1=\operatorname{cl}_X(A)$ and $C_2=\operatorname{cl}_X(X\setminus A)$. The set of all $x\in X$ lying in $C_a$ for every $a\in I$ is exactly $C_1\cap C_2$, and it is closed by claim 3 of Complements, Unions and Intersections of Closed Sets in a Topological Space. By claim 1 this set is $\partial_X A$, which is therefore closed.

Claim 3. Applying claim 1 with $X\setminus A$ in place of $A$ gives

Since $A\subseteq X$ we have $X\setminus(X\setminus A)=A$, so the right-hand side equals $\operatorname{cl}_X(X\setminus A)\cap\operatorname{cl}_X(A)$, which is $\partial_X A$ by claim 1.

Claim 5. By claim 1 of The Interior is the Largest Open Subset we have $\operatorname{int}_X(A)\subseteq A$, and by claim 1 of The Closure is the Smallest Closed Superset we have $A\subseteq\operatorname{cl}_X(A)$; hence $\operatorname{int}_X(A)\subseteq\operatorname{cl}_X(A)$. Since $\partial_X A=\operatorname{cl}_X(A)\setminus\operatorname{int}_X(A)$, no point of $\partial_X A$ lies in $\operatorname{int}_X(A)$, so the two sets are disjoint; and their union is

where the equality uses $\operatorname{int}_X(A)\subseteq\operatorname{cl}_X(A)$.

Claim 4. By claim 2 of Duality Between Interior and Closure Under Complementation we have $\operatorname{int}_X(X\setminus A)=X\setminus\operatorname{cl}_X(A)$. In particular $\operatorname{int}_X(X\setminus A)$ is disjoint from $\operatorname{cl}_X(A)$, and by claim 5 both $\operatorname{int}_X(A)$ and $\partial_X A$ are contained in $\operatorname{cl}_X(A)$; hence $\operatorname{int}_X(X\setminus A)$ is disjoint from each of $\operatorname{int}_X(A)$ and $\partial_X A$. Together with the disjointness of $\operatorname{int}_X(A)$ and $\partial_X A$ from claim 5, the three sets are pairwise disjoint. Finally, by claim 5 the union of the first two is $\operatorname{cl}_X(A)$, so the union of all three is

using $\operatorname{cl}_X(A)\subseteq X$.