[proofplan] We compute the boundary using the identity $\partial A = \overline{A} \setminus A^\circ$, where closure and interior are taken in the ambient [topological space](/page/Topological%20Space) $X$. In an indiscrete space, [citetheorem:8869] gives the complete list of possible closures and interiors of subsets. A nonempty proper subset has closure $X$ and empty interior, while the two extreme subsets $\varnothing$ and $X$ have equal closure and interior, giving empty boundary. [/proofplan] [step:Use the closure and interior formula for the boundary] For every subset $B \subset X$, its boundary relative to $X$ is \begin{align*} \partial B = \overline{B} \setminus B^\circ. \end{align*} Here $\overline{B}$ denotes the closure of $B$ in $X$, and $B^\circ$ denotes the interior of $B$ in $X$. We apply this identity to the subset $A \subset X$. [guided] The boundary is being computed relative to the ambient space $X$. Thus the closure $\overline{B}$, the interior $B^\circ$, and the boundary $\partial B$ all refer to the topology on $X$, not to any larger space. For any subset $B \subset X$, the boundary is defined by \begin{align*} \partial B = \overline{B} \setminus B^\circ. \end{align*} This formula says that a point lies on the boundary exactly when it lies in the closure of $B$ but is not an interior point of $B$. We will use this formula with $B = A$ after computing $\overline{A}$ and $A^\circ$ in the [indiscrete topology](/page/Indiscrete%20Topology). [/guided] [/step] [step:Compute the boundary of a nonempty proper subset] Assume $\varnothing \subsetneq A \subsetneq X$. Since $A \neq \varnothing$, [citetheorem:8869] gives \begin{align*} \overline{A} = X. \end{align*} Since $A \neq X$, the same result gives \begin{align*} A^\circ = \varnothing. \end{align*} Therefore \begin{align*} \partial A = \overline{A} \setminus A^\circ = X \setminus \varnothing = X. \end{align*} [/step] [step:Compute the boundary of the empty subset] Assume $A = \varnothing$. By [citetheorem:8869], \begin{align*} \overline{A} = \overline{\varnothing} = \varnothing. \end{align*} Also $A^\circ = \varnothing^\circ = \varnothing$, since $\varnothing$ is open in every topology. Hence \begin{align*} \partial A = \overline{A} \setminus A^\circ = \varnothing \setminus \varnothing = \varnothing. \end{align*} [/step] [step:Compute the boundary of the whole space] Assume $A = X$. By [citetheorem:8869], \begin{align*} A^\circ = X^\circ = X. \end{align*} Also $\overline{A} = \overline{X} = X$, since $X$ is closed in every topology. Hence \begin{align*} \partial A = \overline{A} \setminus A^\circ = X \setminus X = \varnothing. \end{align*} This proves both asserted cases. [/step]