Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
I was also puzzled by this before, note that $I\times I$ is not convex in the order topology. – Ziqian Xie Feb 13 at 15:55
up vote 9 down vote accepted

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.)

share|cite|improve this answer
I still don't understand. The book says that a set is open in a subset $Y$ of $\mathbb{R}$ if it is an intersection of $Y$ with some open subset of $\mathbb{R}$. So what is this open subset of $\mathbb{R}$? – Xena Apr 27 '13 at 13:15
It is $\{1/2\} \times (1/2, 3/2)$, see second paragraph. Perhaps you are slightly confused because the point $(1/2,3/2)$ looks like the interval $(1/2,3/2)$ in this notation. – Rudy the Reindeer Apr 27 '13 at 13:17
Now everything is clear! Thank you. – Xena Apr 27 '13 at 13:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.