Show that the dictionary order topology on the set \(\mathbb{R} \times \mathbb{R}\) is the same as the product topology \(\mathbb{R}_d \times \mathbb{R}\), where \(\mathbb{R}_d\) denotes \(\mathbb{R}\) in the discrete topology. Compare this topology with the standard topology on \(\mathbb{R}^2\).
Solution
In the dictionary order, for every \(x\in\mathbb R\) and \(a<b\), \[ \{x\}\times(a,b)=((x,a),(x,b)), \] so each basic set for \(\mathbb R_d\times\mathbb R\) is dictionary-order open. Hence the product topology is contained in the dictionary-order topology.
Conversely, a dictionary-order interval between \((a,b)\) and \((c,d)\) is a union of vertical open pieces: if \(a<c\), it is \[ \{a\}\times(b,\infty)\;\cup\! \bigcup_{a<x<c}\!\{x\}\times\mathbb R \;\cup\;\{c\}\times(-\infty,d), \] and when \(a=c\) it is simply \(\{a\}\times(b,d)\). These sets are open in \(\mathbb R_d\times\mathbb R\). The two topologies therefore coincide.
This topology is strictly finer than the standard topology on \(\mathbb R^2\). Every ordinary open rectangle is a union of sets \(\{x\}\times(c,d)\), hence is dictionary-open, while for example \[ \{0\}\times\mathbb R \] is dictionary-open but not standard-open.