Let \(I = [0, 1]\). Compare the product topology on \(I \times I\), the dictionary order topology on \(I \times I\), and the topology \(I \times I\) inherits as a subspace of \(\mathbb{R} \times \mathbb{R}\) in the dictionary order topology.
Solution
Let \(\mathcal P\) be the product topology on \(I^2\), \(\mathcal D\) the dictionary order topology on \(I^2\), and \(\mathcal S\) the topology inherited from the dictionary order topology on \(\mathbb R^2\).
The product and dictionary-order topologies are incomparable. The set \[ I\times(1/2,1] \] is product-open (since \((1/2,1]\) is open in the subspace \(I\)), but it is not dictionary-order open at \((0,1)\): every dictionary-order neighborhood of \((0,1)\) contains points whose first coordinate is positive. Conversely, \[ \{0\}\times(0,1) \] is dictionary-order open, but is not product-open.
The subspace topology \(\mathcal S\) is finer than the product topology because, by E-QMZO5, the ambient dictionary topology on \(\mathbb R^2\) is \(\mathbb R_d\times\mathbb R\); restricting to \(I^2\) gives the product of discrete \(I\) in the first coordinate with the usual subspace topology in the second, which contains the ordinary product topology.
It is also finer than the dictionary order topology \(\mathcal D\): for \(a\in I^2\), the open rays \(\{y\in I^2:y<a\}\) and \(\{y\in I^2:y>a\}\) of the ordered square are the intersections of \(I^2\) with the corresponding open rays of \(\mathbb R^2\), so every \(\mathcal D\)-open set is \(\mathcal S\)-open. Both inclusions are strict. Strictness over \(\mathcal P\) follows from the open set \(\{0\}\times(0,1)\). Strictness over \(\mathcal D\) follows, for example, from \[ \{1/2\}\times(1/2,1], \] which is open in \(\mathcal S\) but not in the dictionary order topology on the ordered square. Hence \[ \mathcal P\not\subseteq\mathcal D,\qquad \mathcal D\not\subseteq\mathcal P, \qquad \mathcal P,\mathcal D\subsetneq\mathcal S. \]