Let $d_{d}((x,y),(a,b))=1$ if $x\neq a$ and $\overline{d}(y,b)=\min\{|y-b|,1\}$ otherwise (i.e. if $x=a$ ). This is a metric, as the only nontrivial property to check is the triangle inequality, but it is easy to check using the facts that the distance between any two points is at most $1$ , and $\overline{d}$ is a metric on $\mathbb{R}$ . Now, using Lemma 20.2, every ball of size $\epsilon$ under this metric is either a vertical interval open in the dictionary order topology or the whole space (if $\epsilon>1$ ). In either case it is open in the dictionary order topology on $\mathbb{R}\times\mathbb{R}$ . Vice versa, for every open set in the dictionary order topology and any its point there is an open vertical interval centered at the point and contained in the set. Therefore, there is an open ball contained in the set.

Alternatively, the dictionary order topology on $\mathbb{R}\times\mathbb{R}$ can be generated by the set of vertical intervals $\{x\}\times(y,z)$ as its basis, and even by the set of “small” vertical intervals $\{x\}\times(y,z)$ where $y<z\le y+2$ . But each such interval can be rewritten as $\{x\}\times(\tfrac{y+z}{2}-\tfrac{z-y}{2},\tfrac{y+z}{2}+\tfrac{z-y}{2})=B_{d_{d}}((x,\tfrac{y+z}{2}),\tfrac{z-y}{2})$ where $0<\tfrac{z-y}{2}\le1$ . And vice versa, for every $\epsilon\in(0,1]$ , $B_{d_{d}}((x,y),\epsilon)=\{x\}\times(y-\epsilon,y+\epsilon)$ where $y-\epsilon<y+\epsilon\le y-\epsilon+2$ . Hence, the dictionary order topology is generated by the basis consisting of all $\epsilon$ -balls, $\epsilon\le1$ , in the $d_{d}$ metric. But the same basis generates the topology induced by $d_{d}$ , because to generate a metric topology it is enough to consider only “small” balls in its basis (the first paragraph on page 122).