Let $(X,d)$ be a metric space, and let $B_r(x)$ be the $x$ -centered open ball of radius $r$ . If $d(x,y)\ge r+s$ , then the balls $B_r(x)$ and $B_s(y)$ are separated.

To prove this, suppose that $B_r(x)$ and $B_s(y)$ are not separated. Then there exists a $z\in X$ such that either $$ d(x,z)<r, \quad d(y,z)\le s, $$ or $$ d(x,z)\le r, \quad d(y,z)< s. $$ In either case, $$ d(x,y)\le d(x,z)+d(z,y)<r+s. $$