Past exam of the mathematics course of the University of Cambridge / 2002 / ib / Paper 2 / 13G / b / Solution

The lexicographic order is a total order. Each point $p=(a,b)$ lies in the order interval between $(a,b-1)$ and $(a,b+1)$, so the order intervals cover ${\mathbb{R}}^2$.

If $p$ lies in both $⟨x_1,y_1⟩$ and $⟨x_2,y_2⟩$, let $x=\max (x_1,x_2)$ and $y=\min (y_1,y_2)$ in that total order. Then $x<p<y$ andIn particular an order interval contains $p$ and is contained in the intersection. Both conditions of part (a) hold, proving that these intervals form a basis of a topology. The resulting topology is the lexicographic order topology on the real plane.