basis of ordered square

Question

In ordered topology, $(a × b, a × c)$ for b < d is basis element.

so $(0 × 0.5, 0 × 2) ∩ [0,1]×[0,1] = (0 × 0.5, 0×1)$ is open in ordered square.

Is this right?

Answer 1

The intersection is actually $S=(0\times0.5,0\times1\color{red}]$, which is infact open in the unit square with subspace topology induced by dictionary order topology in $\Bbb R^2$. But this is not open in the unit square with dictionary order topology, often called the ordered square, whose basis elements are of the form $\{[0\times0,a\times b),(a\times b,c\times d),(c\times d,1\times1]:a\times b,c\times d\in I\}$. This is because any basis element $(a\times b,c\times d)$ which contains the point $0\times1$ is not a subset of $S$.