Show that \(\mathbb{R} \times \mathbb{R}\) in the dictionary order topology is metrizable.
Solution
By E-QMZO5, the dictionary order topology on \(\mathbb R\times\mathbb R\) is exactly the product topology \[ \mathbb R_d\times\mathbb R, \] where \(\mathbb R_d\) is discrete.
Let \[ \delta(x,x')=\begin{cases}0,&x=x',\\1,&x\ne x',\end{cases} \qquad \rho(y,y')=\min\{|y-y'|,1\}. \] Both are metrics inducing the discrete and usual topologies respectively. Then \[ D((x,y),(x',y'))=\delta(x,x')+\rho(y,y') \] is a metric on \(\mathbb R^2\). For \(0<r\le1\), \[ B_D((x,y),r)=\{x\}\times(y-r,y+r), \] since \(\delta(x,x')\ge1\) when \(x'\ne x\). These sets form a basis of \(\mathbb R_d\times\mathbb R\) at \((x,y)\), and every \(D\)-ball about \((x,y)\) contains one of them. Hence \(D\) metrizes the dictionary order topology.