Let \(A \subset X\). If \(d\) is a metric for the topology of \(X\), show that \(d \mid A \times A\) is a metric for the subspace topology on \(A\).
Solution
The restriction \(d_A=d|_{A\times A}\) is a metric on \(A\), since the metric axioms for \(d\) hold in particular for points of \(A\). For \(a\in A\) and \(r>0\), its open ball is \[ B_{d_A}(a,r)=\{x\in A:d(a,x)<r\}=A\cap B_d(a,r). \] Thus every \(d_A\)-ball is open in the subspace topology. Conversely, if \(U=A\cap V\) is subspace-open and \(a\in U\), choose \(r>0\) with \[ B_d(a,r)\subseteq V. \] Then \[ B_{d_A}(a,r)=A\cap B_d(a,r)\subseteq U. \] Hence the metric topology of \(d_A\) is exactly the subspace topology.