[proofplan] We prove the four-way equivalence by a circular chain (i) $\Rightarrow$ (ii) $\Rightarrow$ (iv) $\Rightarrow$ (iii) $\Rightarrow$ (i). The key observation throughout is the interplay between interior and closure via [Interior–Closure Duality](/theorems/1014): $\operatorname{int}(S) = X \setminus \overline{X \setminus S}$. The implication (i) $\Rightarrow$ (ii) rewrites the nowhere dense condition as a density statement using this duality. The implication (ii) $\Rightarrow$ (iv) translates density of the complement into a local avoidance property. The step (iv) $\Rightarrow$ (iii) is a weakening (disjointness from $\overline{A}$ implies disjointness from $A$). Finally, (iii) $\Rightarrow$ (i) recovers the empty interior condition by contrapositive. [/proofplan] [step:Show (i) $\Rightarrow$ (ii) by applying interior–closure duality] Assume $\operatorname{int}(\overline{A}) = \varnothing$. By the [Interior–Closure Duality](/theorems/1014), for any subset $S \subset X$, \begin{align*} \operatorname{int}(S) = X \setminus \overline{X \setminus S}. \end{align*} Applying this with $S = \overline{A}$ yields \begin{align*} \operatorname{int}(\overline{A}) = X \setminus \overline{X \setminus \overline{A}}. \end{align*} Since $\operatorname{int}(\overline{A}) = \varnothing$, this gives $X \setminus \overline{X \setminus \overline{A}} = \varnothing$, which rearranges to $\overline{X \setminus \overline{A}} = X$. This is precisely the statement that the open set $X \setminus \overline{A}$ is dense in $X$. [guided] We want to show that the nowhere dense condition $\operatorname{int}(\overline{A}) = \varnothing$ forces the open set $X \setminus \overline{A}$ to be dense. The connection between interior and closure is provided by the [Interior–Closure Duality](/theorems/1014), which states that for any $S \subset X$, \begin{align*} \operatorname{int}(S) = X \setminus \overline{X \setminus S}. \end{align*} Why is this the right tool? Because the nowhere dense condition involves $\operatorname{int}(\overline{A})$, and the density condition involves $\overline{X \setminus \overline{A}}$ — these are related by exactly this duality identity. Substituting $S = \overline{A}$: \begin{align*} \operatorname{int}(\overline{A}) = X \setminus \overline{X \setminus \overline{A}}. \end{align*} The hypothesis gives $\operatorname{int}(\overline{A}) = \varnothing$, so $X \setminus \overline{X \setminus \overline{A}} = \varnothing$. Taking complements of both sides yields $\overline{X \setminus \overline{A}} = X$, which says that the closure of $X \setminus \overline{A}$ is all of $X$. By definition, this means $X \setminus \overline{A}$ is dense in $X$. [/guided] [/step] [step:Show (ii) $\Rightarrow$ (iv) by intersecting a dense open set with an arbitrary open set] Assume $\overline{X \setminus \overline{A}} = X$. Let $G \in \tau$ be nonempty. Since $X \setminus \overline{A}$ is dense in $X$, every nonempty open set meets $X \setminus \overline{A}$. Define $H := G \cap (X \setminus \overline{A})$. Since $G$ is open and $X \setminus \overline{A}$ is open (as the complement of the closed set $\overline{A}$), $H$ is open. Since $X \setminus \overline{A}$ is dense and $G$ is nonempty open, $H \neq \varnothing$. We verify the two required properties. First, $H = G \cap (X \setminus \overline{A}) \subset G$. Second, $H \subset X \setminus \overline{A}$, so $H \cap \overline{A} = \varnothing$. [guided] Assume $\overline{X \setminus \overline{A}} = X$, i.e., the open set $X \setminus \overline{A}$ is dense. We must show that every nonempty open set $G$ contains a nonempty open subset disjoint from $\overline{A}$. The natural candidate is $H := G \cap (X \setminus \overline{A})$. Why this choice? We need $H$ to be (a) open, (b) nonempty, (c) contained in $G$, and (d) disjoint from $\overline{A}$. The intersection of two open sets is open, so (a) holds. For (c), $H \subset G$ by construction. For (d), $H \subset X \setminus \overline{A}$ by construction, so $H \cap \overline{A} = \varnothing$. The only nontrivial point is (b): why is $H$ nonempty? A subset $D \subset X$ is dense if and only if $D \cap V \neq \varnothing$ for every nonempty open set $V$. Since $X \setminus \overline{A}$ is dense and $G$ is nonempty open, the intersection $H = G \cap (X \setminus \overline{A})$ is nonempty. [/guided] [/step] [step:Show (iv) $\Rightarrow$ (iii) by noting that $A \subset \overline{A}$] Assume (iv): for every nonempty $G \in \tau$, there exists a nonempty $H \in \tau$ with $H \subset G$ and $H \cap \overline{A} = \varnothing$. Since $A \subset \overline{A}$, the condition $H \cap \overline{A} = \varnothing$ implies $H \cap A = \varnothing$. The set $H$ therefore satisfies all requirements of (iii): $H$ is nonempty, $H$ is open, $H \subset G$, and $H \cap A = \varnothing$. [/step] [step:Show (iii) $\Rightarrow$ (i) by contrapositive] We prove the contrapositive: $\operatorname{int}(\overline{A}) \neq \varnothing$ implies the negation of (iii). Suppose $\operatorname{int}(\overline{A}) \neq \varnothing$. Then there exists a nonempty open set $G \subset \overline{A}$. We show that no nonempty open $H \subset G$ can satisfy $H \cap A = \varnothing$. Let $H \in \tau$ with $H \subset G$ and $H \neq \varnothing$. Since $G \subset \overline{A}$, we have $H \subset \overline{A}$. Pick any $x \in H$. Since $x \in \overline{A}$, the [Neighbourhood Characterisation of Closure](/theorems/1005) guarantees that every open set containing $x$ intersects $A$. The set $H$ is itself an open set containing $x$, so $H \cap A \neq \varnothing$. Therefore the open set $G = \operatorname{int}(\overline{A})$ witnesses the failure of (iii), completing the contrapositive argument. [guided] We prove the contrapositive: assume $\operatorname{int}(\overline{A}) \neq \varnothing$ and derive the failure of (iii). If $\operatorname{int}(\overline{A}) \neq \varnothing$, there exists a nonempty open set $G$ with $G \subset \overline{A}$. We claim that $G$ witnesses the negation of (iii): there is no nonempty open $H \subset G$ with $H \cap A = \varnothing$. To see this, let $H$ be any nonempty open subset of $G$. Then $H \subset G \subset \overline{A}$. Pick any $x \in H$. Since $x \in \overline{A}$, the [Neighbourhood Characterisation of Closure](/theorems/1005) guarantees that every open set containing $x$ intersects $A$. The set $H$ is itself an open set containing $x$ (it is open and $x \in H$), so $H \cap A \neq \varnothing$. This shows that for the specific nonempty open set $G = \operatorname{int}(\overline{A})$, every nonempty open subset meets $A$. Condition (iii) demands that *every* nonempty open set has a nonempty open subset disjoint from $A$, and $G$ violates this. Hence (iii) fails. Contrapositively, (iii) implies (i), completing the cycle (i) $\Rightarrow$ (ii) $\Rightarrow$ (iv) $\Rightarrow$ (iii) $\Rightarrow$ (i). [/guided] [/step]