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I am having some difficulty grasping the concept of an ordered topology in $\mathbb R \times\mathbb R $. The definition I was given is that this is the topology where $(a,b) < (c,d)$ if $a<c$ or $a= c$ and $b <d$. Would the set $(0, 1)\times(0,1]$ be open in this topology? Or how should I be thinking about this?

I read this post, as well as many others, and am still confused.

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The set $U:=(0,1)\times(0,1]$ is not open.

Proof.

Any neighborhood of the point $\langle 1/2, 1\rangle\in U$ contains some point $\langle 1/2,y\rangle$ with $y>1$ which is therefore not in $U$. So not all points of $U$ are inner points and $U$ cannot be open.


You can think about the basic open sets $\{(x,y)\in\Bbb R^2\mid (x_1,y_1)<(x,y)<(x_2,y_2)\}$ of your space as some uncountable generalization of the following constellation:

The colored region should represent the "open" set. Note that the first coordinate is the top-down-direction, the second one the left-right-direction (might be flipped compared to usual mathematical plots). These open sets have (at most) two incompletely filled "lines", the upper and the lower one. All "lines" in between are completely filled. The first and the last line have open ends, something which is not representable in this discrete visualization.

In this sense, you see that open sets (that span several lines) include "whole lines". The basic open sets (that span several lines) can be decomposed as follows:

\begin{align} \{(x,y)\in\Bbb R^2\mid &(x_1,y_1)<(x,y)<(x_2,y_2)\} \\[0.5em]&= \color{red}{\underbrace{\{x_1\} \times (y_1,\infty)}_{\text{upper half line}} } \;\;\cup \color{blue}{\underbrace{(x_1,x_2)\times \Bbb R}_{\text{whole lines in between}} } \cup\;\;\color{green}{\underbrace{ \{x_2\}\times (-\infty, y_2)}_{\text{bottom half line}}} \end{align}

as highlighted in the following discrete visualization:

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  • $\begingroup$ To clarify whether I understand this, would $(0, 1) \times (1, 2]$ be open then? Or does it have to be $(0, 1) \times (1, 2)$ (in which case how do you distinguish open sets on this topology from those in the standard topology)? $\endgroup$ – Sveinn Feb 16 '18 at 16:00
  • $\begingroup$ @quixotic I removed some misleading statment from my answer. $(0,1)\times (1,2]$ is still not open for the same reason as $U$. However, $(0,1)\times (1,2)$ is open as it is the union of the open sets $\{x\}\times(1,2)$ for all $x\in(0,1)$. My answer focuses on the open sets spanning several lines and therefore containing whole lines. But an open set can start and end on the same line, like e.g. $\{x\}\times(1,2)$. In this case it looks like an open interval embedded into $\Bbb R^2$. I will add a final note to my post to answer your second question. $\endgroup$ – M. Winter Feb 16 '18 at 16:25
  • $\begingroup$ So would the set {3} $\times (1,2)$ be an example of a set open in the order topology over $\mathbb R \times \mathbb R$ but closed in the standard topology over $\mathbb R \times \mathbb R$? Or am I still confused? $\endgroup$ – Sveinn Feb 16 '18 at 16:28
  • $\begingroup$ @quixotic Yes and No. $\{3\}\times (1,2)$ is open in your topology, and not open in the standard one. But it is also not closed in the standard one. An example which would be closed is $\{3\}\times\Bbb R$. Also, the order topology is finer. Everything which is open in the standard topology is also open in your order topology. The converse does not hold has you have seen already. $\endgroup$ – M. Winter Feb 16 '18 at 16:31
  • $\begingroup$ Ah right, it slipped my mind that a set that in not open is not necessarily closed. Thank you, this has been very helpful. $\endgroup$ – Sveinn Feb 16 '18 at 16:36
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The topology that you have described is called the dictionary order topology on $\mathbb R \times\mathbb R$.

Assume that you are a librarian, any you have two books in your hand with inner library code $C3$ and $E2$.

Now, what first you do is that you compare the alphabets of the library codes of the books, namely you determine whether $C <, >, = E$ in the natural sense. In this case, since $C < E$, you do know that in a shelf starting from left and going to right, you put the first book, with the code $C3$, to the left of the second book, with the code $E2$, in that shelf.

Now, consider the books $D4$ and $D2$.To determine the order of the book in the shelf, if you compare the alphabets, they are equal, so what you, you check the numbers, since $4 > 2$, the first book, with the code $D4$, should be in the right of the second book, with the code $D2$, in the shelf.

Now, coming to you question, in the dictionary order topology $\mathbb R \times\mathbb R$, you do the same thing, given two elements $(a,b)$ and $(c,d)$, you first compare the first inputs, or say components, of the elements, namely you check whether $a <, >, = c$ in the usual sense, and if $a = c$, you check the second component of the elements, and determine which element is "less" than other element.

Now, in this way, you order all the elements of $\mathbb R \times\mathbb R$, and in an order topology, the open sets of that topology are the intervals, i.e (e,f), where $e,f \in \mathbb R \times\mathbb R$.

In particular, in your example, the set $$(0, 1)\times(0,1] = \{(x,y) \in \mathbb R \times\mathbb R| 0 <x<1, 0<y \leq 1 \} \\ = \cup_{x \in (0,1)} \{(x,y) | 0 < y <1\} \cup_{x \in (0,1)} \{(x,1)\}$$

and since the set contanining only the points of the form $(x,1)$ where $x \in (0,1)$, is not open, its union with some open set is not open.

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