[proofplan] The proof uses only the defining feature of the [indiscrete topology](/page/Indiscrete%20Topology): the only open subsets of $X$ are $\varnothing$ and $X$. Taking complements shows that the only closed subsets are also $\varnothing$ and $X$. The closure is determined by the closed sets containing $A$, while the interior is determined by the open sets contained in $A$. [/proofplan] [step:Identify the closed subsets of the indiscrete space] Since $\tau=\{\varnothing,X\}$, the open subsets of $X$ are exactly $\varnothing$ and $X$. A subset $F\subset X$ is closed exactly when $X\setminus F$ is open. Hence the only closed subsets of $X$ are $\varnothing$ and $X$, because $X\setminus \varnothing=X$ and $X\setminus X=\varnothing$. [/step] [step:Compute the closure from the closed supersets of $A$] By definition, $\overline{A}$ is the intersection of all closed subsets of $X$ that contain $A$. If $A=\varnothing$, then both $\varnothing$ and $X$ are closed subsets containing $A$. Since $\varnothing\subset X$, their intersection is $\varnothing$, so $\overline{A}=\varnothing$. If $A\neq\varnothing$, then $\varnothing$ does not contain $A$. The only closed subset of $X$ containing $A$ is therefore $X$, so $\overline{A}=X$. [guided] We compute the closure using the definition: $\overline{A}$ is the smallest closed subset of $X$ containing $A$, equivalently the intersection of all closed subsets of $X$ that contain $A$. From the previous step, the only closed subsets available are $\varnothing$ and $X$. First suppose $A=\varnothing$. Then $\varnothing$ is closed and contains $A$, and $X$ is also closed and contains $A$. The intersection of all closed supersets of $A$ is therefore the intersection of $\varnothing$ and $X$, which is $\varnothing$. Hence $\overline{A}=\varnothing$. Now suppose $A\neq\varnothing$. Then $\varnothing$ cannot contain $A$, because a nonempty set has an element and the empty set has none. Thus the only closed subset of $X$ containing $A$ is $X$. The intersection of the single closed superset $X$ is $X$, so $\overline{A}=X$. [/guided] [/step] [step:Compute the interior from the open subsets contained in $A$] By definition, $A^\circ$ is the union of all open subsets of $X$ that are contained in $A$. If $A=X$, then $X$ is open and contained in $A$, so the union of all open subsets contained in $A$ is $X$. Hence $A^\circ=X$. If $A\neq X$, then $X$ is not contained in $A$. Since the only open subsets of $X$ are $\varnothing$ and $X$, the only open subset contained in $A$ is $\varnothing$. Therefore $A^\circ=\varnothing$. [/step]