4
$\begingroup$

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

  1. (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 $I \times I$.

$\endgroup$
5
  • 2
    $\begingroup$ For 2, the order topology has the open intervals as base, indeed including $\{1/2\}\times(1/2,\, 1)$, however your description seems to exclude it. Why take Cartesian product of two intervals in the first place? $\endgroup$
    – Berci
    Mar 9, 2018 at 23:04
  • 2
    $\begingroup$ The set $\{1/2\}\times(1/2,1]$ is not closed in the order topology on $I\times I$. Its closure is $\{1/2\}\times[1/2,1]$. Did you perhaps confuse "closed" with "not open"? $\endgroup$ Mar 10, 2018 at 3:11
  • $\begingroup$ @Berci You are right. There is no need for cartesian products. I had misunderstood and misused the notation. $\endgroup$
    – Amphiaraos
    Mar 11, 2018 at 3:23
  • $\begingroup$ @AndreasBlass The book has not yet introduced what closure and closed sets are. I have checked the book again. hat I quoted is the same $\endgroup$
    – Amphiaraos
    Mar 11, 2018 at 3:26
  • $\begingroup$ @Amphiarao Now it's clear, thanks $\endgroup$
    – Berci
    Mar 11, 2018 at 9:20

1 Answer 1

2
$\begingroup$

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 $(\frac{1}{2}, 1)$.

$\endgroup$
5
  • $\begingroup$ First, it's misleading that open intervals and pairs are both denoted by simple parenthesis. Second, I guess a line is missing from the list of possible forms of open sets, as $\{1/2\}\times(1/2,\,1)$ does not contain any full 'rectangle'. $\endgroup$
    – Berci
    Mar 9, 2018 at 22:43
  • 2
    $\begingroup$ @Berci the first item in 2 should be: an open interval $((a,b),(c,d))$ where $(a,b) < (c,d)$ where $<$ is the lexicographic order. And also, context makes clear where I mean a point in the square in an open interval in the coordinate. Munkres uses $a\times b$ for the point $(a,b)$ instead, which I don’t like myself. $\endgroup$ Mar 9, 2018 at 22:49
  • $\begingroup$ @Berci the OP uses a mixed notation, not pure Munkres. $\endgroup$ Mar 9, 2018 at 22:58
  • $\begingroup$ Ok, having the first item correctly, it would include the rest, wouldn't it? $\endgroup$
    – Berci
    Mar 9, 2018 at 23:09
  • $\begingroup$ @Berci no, the neighbourhoods at the minimum and maximum are not open intervals, but open segments. These form a subbase. $\endgroup$ Mar 10, 2018 at 6:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .