The collection of balls $B_{d'}(a,r)\subset A$ centered at points $a\in A$ in the metric $d'=d|A\times A$ is a basis for the metric topology on $A$ . $A$ as a subspace has a basis element $B'(x,r)=A\cap B_{d}(x,r)=\{y\in A|d(x,y)<r\}$ for all $x\in X$ . Clearly, $B_{d'}(a,r)=A\cap B_{d}(a,r)=B'(a,r)$ so that the subspace topology on $A$ is finer than the metric topology on $A$ . Further, consider any basis element $B'(x,r)$ for the subspace topology. If $a\in B'(x,r)\subset A$ , let $r'=r-d(x,a)$ . Then $a\in B_{d'}(a,r')=$ $B_{d}(a,r')\cap A\subset$ $B_{d}(x,r)\cap A=B'(x,r)$ . Therefore, $B'(x,r)$ is open in the metric topology on $A$ , and the metric topology on $A$ is finer than the subspace topology on $A$ .