[proofplan] We prove each of the three stability properties independently. Part (i) uses monotonicity of closure and interior. Part (ii) proceeds by induction, with the base case $n = 2$ established using the [equivalent characterisation](/theorems/1084) (iii) of nowhere dense sets — the "local avoidance" formulation composes cleanly with unions because we can dodge one set at a time. Part (iii) follows from the idempotence of closure: $\overline{\overline{A}} = \overline{A}$, so $\operatorname{int}(\overline{\overline{A}}) = \operatorname{int}(\overline{A}) = \varnothing$. [/proofplan] [step:Prove (i): every subset of a nowhere dense set is nowhere dense] Let $A \subset X$ be nowhere dense and let $B \subset A$. Since $B \subset A$, monotonicity of closure gives $\overline{B} \subset \overline{A}$. By [monotonicity of the interior operator](/theorems/1013) (property 5), $\operatorname{int}(\overline{B}) \subset \operatorname{int}(\overline{A})$. Since $A$ is nowhere dense, $\operatorname{int}(\overline{A}) = \varnothing$. The only subset of $\varnothing$ is $\varnothing$ itself, so $\operatorname{int}(\overline{B}) = \varnothing$, and $B$ is nowhere dense. [guided] Let $A \subset X$ be nowhere dense and let $B \subset A$. We must show $\operatorname{int}(\overline{B}) = \varnothing$. The argument chains two monotonicity properties. First, since $B \subset A$, the closure operator is monotone: if $S \subset T$ then $\overline{S} \subset \overline{T}$ (every closed set containing $T$ also contains $S$, so the smallest closed set containing $S$ is contained in the smallest closed set containing $T$). This gives $\overline{B} \subset \overline{A}$. Second, [monotonicity of the interior operator](/theorems/1013) (property 5) states: if $S \subset T$ then $\operatorname{int}(S) \subset \operatorname{int}(T)$. Applying this with $S = \overline{B}$ and $T = \overline{A}$: \begin{align*} \operatorname{int}(\overline{B}) \subset \operatorname{int}(\overline{A}) = \varnothing, \end{align*} where the last equality is the nowhere dense hypothesis on $A$. Since $\operatorname{int}(\overline{B}) \subset \varnothing$, we conclude $\operatorname{int}(\overline{B}) = \varnothing$. [/guided] [/step] [step:Prove (ii) for two sets: $A_1 \cup A_2$ is nowhere dense when $A_1$ and $A_2$ are] Let $A_1, A_2 \subset X$ be nowhere dense. We use the [equivalent characterisation](/theorems/1084), specifically (i) $\Leftrightarrow$ (iii): $A$ is nowhere dense if and only if every nonempty open set contains a nonempty open subset disjoint from $A$. Let $G \in \tau$ be nonempty. Since $A_1$ is nowhere dense, characterisation (iii) provides a nonempty open $H_1 \subset G$ with $H_1 \cap A_1 = \varnothing$. Since $A_2$ is nowhere dense and $H_1$ is nonempty open, applying (iii) again provides a nonempty open $H_2 \subset H_1$ with $H_2 \cap A_2 = \varnothing$. The set $H_2$ satisfies: $H_2$ is nonempty, $H_2$ is open, $H_2 \subset H_1 \subset G$, and \begin{align*} H_2 \cap (A_1 \cup A_2) = (H_2 \cap A_1) \cup (H_2 \cap A_2). \end{align*} The second term is empty by construction. For the first term: $H_2 \subset H_1$ and $H_1 \cap A_1 = \varnothing$, so $H_2 \cap A_1 = \varnothing$. Therefore $H_2 \cap (A_1 \cup A_2) = \varnothing$, and characterisation (iii) confirms that $A_1 \cup A_2$ is nowhere dense. [guided] Let $A_1, A_2 \subset X$ be nowhere dense. A direct approach via the definition $\operatorname{int}(\overline{A_1 \cup A_2}) = \varnothing$ would require computing $\overline{A_1 \cup A_2} = \overline{A_1} \cup \overline{A_2}$ and then showing $\operatorname{int}(\overline{A_1} \cup \overline{A_2}) = \varnothing$. However, the interior of a union does not decompose as simply as the interior of an intersection (we have $\operatorname{int}(S_1 \cap S_2) = \operatorname{int}(S_1) \cap \operatorname{int}(S_2)$ by [Properties of the Interior Operator](/theorems/1013) property 4, but no analogous formula for unions). So the direct approach is awkward. Instead, we use the [equivalent characterisation](/theorems/1084) (iii): $A$ is nowhere dense if and only if every nonempty open set contains a nonempty open subset disjoint from $A$. This formulation composes well with unions because we can "dodge" one set at a time. Let $G \in \tau$ be nonempty. Since $A_1$ is nowhere dense, (iii) provides a nonempty open $H_1 \subset G$ with $H_1 \cap A_1 = \varnothing$. Now $H_1$ is itself a nonempty open set, so since $A_2$ is nowhere dense, (iii) applied to $H_1$ provides a nonempty open $H_2 \subset H_1$ with $H_2 \cap A_2 = \varnothing$. We verify the conclusion. First, $H_2 \subset H_1 \subset G$, so $H_2$ is a nonempty open subset of $G$. Second: \begin{align*} H_2 \cap (A_1 \cup A_2) = (H_2 \cap A_1) \cup (H_2 \cap A_2). \end{align*} Since $H_2 \subset H_1$ and $H_1 \cap A_1 = \varnothing$, we have $H_2 \cap A_1 = \varnothing$. By construction, $H_2 \cap A_2 = \varnothing$. So $H_2 \cap (A_1 \cup A_2) = \varnothing$. By (iii), $A_1 \cup A_2$ is nowhere dense. [/guided] [/step] [step:Extend (ii) to finite unions by induction] The general case follows by induction on the number of sets. The base case $n = 1$ is immediate. The inductive step: if $A_1 \cup \cdots \cup A_{n-1}$ is nowhere dense (by the inductive hypothesis) and $A_n$ is nowhere dense, then the previous step applied to the two nowhere dense sets $A_1 \cup \cdots \cup A_{n-1}$ and $A_n$ shows that $A_1 \cup \cdots \cup A_n$ is nowhere dense. [/step] [step:Prove (iii): the closure of a nowhere dense set is nowhere dense] Let $A \subset X$ be nowhere dense, so $\operatorname{int}(\overline{A}) = \varnothing$. We must show $\operatorname{int}(\overline{\overline{A}}) = \varnothing$. By idempotence of closure, $\overline{\overline{A}} = \overline{A}$ (since $\overline{A}$ is closed, and the closure of a closed set is itself), so \begin{align*} \operatorname{int}(\overline{\overline{A}}) = \operatorname{int}(\overline{A}) = \varnothing. \end{align*} Therefore $\overline{A}$ is nowhere dense. [guided] Let $A \subset X$ be nowhere dense, meaning $\operatorname{int}(\overline{A}) = \varnothing$. We must verify $\operatorname{int}(\overline{\overline{A}}) = \varnothing$. The closure operator is idempotent: for any subset $S \subset X$, $\overline{\overline{S}} = \overline{S}$. This holds because $\overline{S}$ is closed, and the closure of a closed set is itself. Applying this with $S = A$: \begin{align*} \operatorname{int}(\overline{\overline{A}}) = \operatorname{int}(\overline{A}) = \varnothing, \end{align*} where the final equality is the nowhere dense hypothesis on $A$. This result is worth pausing on: it says that passing from $A$ to $\overline{A}$ — potentially a much larger set — does not destroy the nowhere dense property. This is specific to nowhere dense sets and fails for related notions. For instance, a set can be meagre (a countable union of nowhere dense sets) while its closure is not meagre — in $\mathbb{R}$, the set $\mathbb{Q}$ is meagre but $\overline{\mathbb{Q}} = \mathbb{R}$ is not meagre by the [Baire Category Theorem](/theorems/1072). [/guided] [/step]