[proofplan] We prove $\overline{B}^A = \overline{B} \cap A$ by showing each side is contained in the other, using the [Neighbourhood Characterisation of Closure](/theorems/1005) in both the ambient space $X$ and the subspace $A$. The key link between the two is that the open sets of the subspace topology on $A$ are precisely the sets $U \cap A$ for $U \in \tau$. [/proofplan] [step:Show $\overline{B}^A \subset \overline{B} \cap A$] Let $x \in \overline{B}^A$. Since $\overline{B}^A \subset A$ by definition of closure in $A$, we have $x \in A$. We show $x \in \overline{B}$ using the [Neighbourhood Characterisation of Closure](/theorems/1005) in $X$. Let $U \in \tau$ with $x \in U$. Then $U \cap A$ is open in the subspace topology on $A$ and contains $x$. Since $x \in \overline{B}^A$, the [Neighbourhood Characterisation of Closure](/theorems/1005) applied in $A$ gives $(U \cap A) \cap B \neq \varnothing$. Since $B \subset A$, this simplifies to $U \cap B \neq \varnothing$. As $U$ was an arbitrary open neighbourhood of $x$ in $X$, the [Neighbourhood Characterisation of Closure](/theorems/1005) applied in $X$ gives $x \in \overline{B}$. [guided] Let $x \in \overline{B}^A$. The closure $\overline{B}^A$ is taken inside the topological space $(A, \tau_A)$, so $\overline{B}^A \subset A$, giving $x \in A$. To show $x \in \overline{B}$, we verify the neighbourhood condition in $X$. Let $U \in \tau$ with $x \in U$. The subspace topology on $A$ consists of all sets $V \cap A$ where $V \in \tau$. In particular, $U \cap A$ is open in $A$ and contains $x$ (since $x \in U$ and $x \in A$). Applying the [Neighbourhood Characterisation of Closure](/theorems/1005) in the subspace $(A, \tau_A)$, since $x \in \overline{B}^A$, every open neighbourhood of $x$ in $A$ meets $B$. So $(U \cap A) \cap B \neq \varnothing$. Since $B \subset A$, we have $(U \cap A) \cap B = U \cap B$. Thus $U \cap B \neq \varnothing$. Since this holds for every $U \in \tau$ containing $x$, the [Neighbourhood Characterisation of Closure](/theorems/1005) in $X$ gives $x \in \overline{B}$. [/guided] [/step] [step:Show $\overline{B} \cap A \subset \overline{B}^A$] Let $x \in \overline{B} \cap A$. We show $x \in \overline{B}^A$ by the [Neighbourhood Characterisation of Closure](/theorems/1005) applied in $A$. Let $W$ be open in $A$. By definition of the subspace topology, $W = U \cap A$ for some $U \in \tau$. Suppose $x \in W$, so $x \in U$ and $x \in A$. Since $x \in \overline{B}$ and $U \in \tau$ contains $x$, the [Neighbourhood Characterisation of Closure](/theorems/1005) in $X$ gives $U \cap B \neq \varnothing$. Since $B \subset A$, we have $U \cap B = (U \cap A) \cap B = W \cap B$, so $W \cap B \neq \varnothing$. Since $W$ was an arbitrary open neighbourhood of $x$ in $A$, we conclude $x \in \overline{B}^A$. [guided] Let $x \in \overline{B} \cap A$. We verify the neighbourhood condition in $A$. Let $W$ be an open neighbourhood of $x$ in $A$. By the definition of the subspace topology, $W = U \cap A$ for some $U \in \tau$. Since $x \in W$, we have $x \in U$. Now $x \in \overline{B}$ and $U$ is an open neighbourhood of $x$ in $X$, so by the [Neighbourhood Characterisation of Closure](/theorems/1005) applied in $X$, $U \cap B \neq \varnothing$. Choose $b \in U \cap B$. Since $b \in B \subset A$, we get $b \in U \cap A = W$, so $b \in W \cap B$. Hence $W \cap B \neq \varnothing$. Since $W$ was an arbitrary open neighbourhood of $x$ in $A$, the [Neighbourhood Characterisation of Closure](/theorems/1005) in $(A, \tau_A)$ gives $x \in \overline{B}^A$. [/guided] [/step]