Munkres Example 16.3

Question

The example says:

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 order 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, as you can check.

I have been reading responses about this same problem for a while and they have been very helpful. However, I want to check if my own explanation is correct.

1. The set $\{1/2\} \times (1/2,1]$ is open in $I \times I$ in the subspace topology. This is because we can get it as a result of $I \times I \cap \{ 1/2 \} \times (1/2,3/2)$ where $\{ 1/2 \} \times (1/2,3/2)$ is open in the dictionary order on the plane $\mathbb{R} \times \mathbb{R}$.

2. (NOTE: Corrected notation in the end) Now for the order topology on $I \times I$, the sets are of the form:

   • $(a,b) \times (c,d)$ where $a < c$ and $b < d$
   • $[0,b) \times [0,d)$ where $b < d$
   • $(a,1] \times (c,1]$ where $a < c\\$

   The set $\{1/2\} \times (1/2,1]$ does not correspond to any of the forms shown for a basis in the order topology on $I \times I$. Then, the set $\{1/2\} \times (1/2,1]$ will be closed in the order topology on $I \times I$.

To make sure I am understanding, the set he set $\{1/2\} \times (1/2,1)$ should be open in both topologies, right?

Edit: As @Berci and @Henno Brandsma noted, I had problems using Munkres' notation. I will not delete my previous notation as it might be helpful for other people to identify the same mistake

2. (Corrected) Now for the order topology on $I \times I$, the sets are of the form:
   • $((a,b),(c,d))$ where $(a,b) < (c,d)$ on the lexicographic order
   • $[(0,0),(a,b))$ where $(0,0) \leq (a,b)$ on the lexicographic order
   • $((a,b),(1,1)]$ where $(a,b) \leq (1,1)$ on the lexicographic order

Then the set $\{1/2\} \times (1/2,1]$ on Munkres' notation will be $((1/2,1/2), (1/2,1)]$. So this set will be closed on the order topology on $I \times I$.

Answer 1

As to point 2. we only know a base element is of that form, except you need as the first to mention open intervals $((a,b), (c,d))$ where $(a,b) <_{l} (c,d)$, which is different from what you wrote (the lexicographic order is not a product order!), and this is enough to refute $\{\frac{1}{2}\} \times (\frac{1}{2}, 1]$ being open, as $(\frac{1}{2},1 )$ is not an interior point (no base element containing it sits inside the set). This stems from the fact that open intervals in te restricted order on the square need to have both endpoints in the square, by definition!

$\{\frac{1}{2}\} \times (\frac{1}{2}, 1)$ is indeed open in both topologies as an open interval in the (restricted) lexicographic order: from the point $(\frac{1}{2}, \frac{1}{2})$ to the end point $(\frac{1}{2}, 1)$.