Basis of order topology on $\mathbb{R} × \mathbb{R} $

Question

On James r Monkers Topology book in the second chapter the section of order topology, in the example 2 of order topology on $\mathbb{R} × \mathbb{R}$ he defined the order topology the one whose the collection of all open intervals $(a×b , c×d)$ is its basis, then he said that the subcollection of all intervals of the form $(a×b, c×d]$ can be also its basis. How to show that? In other word how to prove that the collection of all intervals of the form $(a×b, c×d)$ and the collection of all intervals of the form $(a×b,c×d]$ define the same topology.

Answer 1

This is not true. There's no way to generate a half-open interval $(a\times b, a\times d]$ from open intervals, necessarily of the form $(a\times b', a\times d')$. Note that the first coordinate is the same. Just like in $\mathbb{R}$ you can't generate $(b,d]$ from $(b',d')$.

But that's not what Munkres says. When Munkres describes the example, he talks about two kinds of intervals: the first one of the most general form $(a\times b, c\times d)$ and the second one of the form $(a\times b, a\times d)$. The second one has the same first coordiante. These generate the same topology.