Subspace topology and order topology

From Munkres, p.90 example 3:

Let $I=[0,1]$. The dictionary order on $I\times I$ is just the restriction to $I\times I$ of the dictionary order on the plane $\mathbb{R}\times\mathbb{R}$. However, the dictionary order topology on $I\times I$ is not the same as the subspace topology on $I\times I$ obtained from the dictionary topology on $\mathbb{R}\times\mathbb{R}$! For example, the set $\{1/2\}\times (1/2,1]$ is open in $I\times I$ in the subspace topology but not in the order topology.

Could you explain how can $\{1/2\}\times (1/2,1]$ be open in the subspace topology? We need to find an $A\times B$ open in $\mathbb{R}\times\mathbb{R}$ such that $([0,1]\cap A)\times ([0,1]\cap B)=\{1/2\}\times(1/2,1]$. How can the intersection of a real open interval with $[0,1]$ be a singleton?

Solution 1:

You wrote:

We need to find an $A\times B$ open in $\mathbb{R}\times\mathbb{R}$ such that

This would be true if we were working with the product topology on $\mathbb R\times\mathbb R$. But in this example we are working with the order topology coming from the lexicographic order.

In this topology on $\mathbb R\times\mathbb R$ the set $\{1/2\}\times(1/2,3/2)$ is open, since it is precisely the set of all points which are between $(1/2,1/2)$ and $(1/2,3/2)$ (w.r.t. the linear order which we are working with).

The set $\{1/2\}\times(1/2,1]$ is not open in order topology (from the lexicographic order on $I\times I$) since every neighborhood of the point $(1/2,1)$ contains some points with $x$-coordinate greater than $1/2$. (Since the point $(1/2,1)$ does not have an immediate successor nor is it the greatest element in this order.)