Order topology neighborhood

Question

I would like to know how we can describe the neighborhood of $(0,0)$ and $(1,1)$ in respect of $Z$ where $Z:=[0,1]^2\subseteq \mathbb R^2$ with a lexicographic order $(x,y)\le(x',y') \iff x<x'$ or $x=x',y\le y'$. I have no idea how in general describe neighborhoods for points $(x,y)\in Z$ with respect to the order topology, may you could help me.

Answer 1

Let $p=\langle a,b\rangle,q=\langle c,d\rangle\in[0,1]^2$, and let $\preceq$ be the lexicographic order on $Z=[0,1]^2$. You know that $p\preceq q$ if either $a<c$, or $a=c$ and $b\le d$. Clearly $\langle 0,0\rangle\preceq p\preceq\langle 1,1\rangle$ for all $p\in Z$, so $\langle Z,\preceq\rangle$ is a linear order with endpoints $\langle 0,0\rangle$ and $\langle 1,1\rangle$.

In general a basic open set in $Z$ is going to have the form $(p,q)$ for some $p,q\in Z$ with $p\prec q$, just as in $[0,1]$ most basic open sets have the form $(x,y)$ for some $x,y\in[0,1]$ with $x<y$. But just as in $[0,1]$, the endpoints require separate handling. A basic open neighborhood of $0$ in $[0,1]$ has the form $[0,x)$ for some $x\in(0,1]$, and similarly a basic open nbhd of $1$ in $[0,1]$ has the form $(x,1]$ for some $x\in[0,1)$. Basic open nbhds of the endpoints $\langle 0,0\rangle$ and $\langle 1,1\rangle$ of $Z$ are formed analogously: for each $p=\langle a,b\rangle\in Z\setminus\{\langle 0,0\rangle\}$ the left-closed, right-open interval

$$\begin{align*} \Big[\langle 0,0\rangle,p\Big)&=\Big[\langle 0,0\rangle,\langle a,b\rangle\Big)\\ &=\big\{\langle x,y\rangle\in Z:\langle 0,0\rangle\preceq\langle x,y\rangle\prec\langle a,b\rangle\big\} \end{align*}\tag{1}$$

is a basic open nhbd of $\langle 0,0\rangle$. To see what this interval looks like, you should consider two cases separately, preferably making a sketch of each.

• If $a=0$, the point $p$ is on the $y$-axis, somewhere directly above the point $\langle 0,0\rangle$, and the set $(1)$ is a half-open vertical line segment lying on the $y$-axis: it’s $\{0\}\times[0,b)$.

• If $a>0$, the set $(1)$ consists of the entire rectangle $[0,a)\times[0,1]$ together with the half-open line segment $\{a\}\times[0,b)$ lying on the line $x=a$.

I’ll leave the basic open nbhds of $\langle 1,1\rangle$ to you.

Added: The situation at $\langle 1,1\rangle$ is the mirror image of that at $\langle 0,0\rangle$. If $p=\langle a,b\rangle\in Z\setminus\{\langle 1,1\rangle\}$, then the left-open, right-closed interval

$$\begin{align*} \Big(p,\langle 1,1\rangle\Big]&=\Big(\langle a,b\rangle,\langle 1,1\rangle\Big]\\ &=\big\{\langle x,y\rangle\in Z:\langle a,b\rangle\prec\langle x,y\rangle\preceq\langle 1,1\rangle\big\} \end{align*}\tag{2}$$

is a basic open nbhd of $\langle 1,1\rangle$. If $a=1$, the point $p$ is on the line $x=1$ somewhere directly below $\langle 1,1\rangle$, and the interval is $\{1\}\times(b,1]$. If $a<1$, it contains all of the rectangle $(a,1]\times[0,1]$ and every point $\langle a,y\rangle$ with $b<y\le 1$, so it’s $$\Big(\{a\}\times(b,1]\Big)\cup\Big((a,1]\times[0,1]\Big)\;.$$ Since $(a,1]=\varnothing$ if $a=1$, we could actually just write

$$\Big(p,\langle 1,1\rangle\Big]=\Big(\{a\}\times(b,1]\Big)\cup\Big((a,1]\times[0,1]\Big)\;,$$

taking care of both cases at once. (You might try to do the same thing for the other endpoint.)

You work out basic open nbhds of other points in the same way, except that now you have to worry about what happens on both sides of the point, just as you do when you’re defining open nbhds in $[0,1]$ of points other than $0$ and $1$. Let $p=\langle a,0\rangle$, for instance, where $0<a\le 1$. Basic open nbhds of $p$ have the form $(q,r)$, where $q\prec p\prec r$. Suppose that $q=\langle b,c\rangle$ and $r=\langle d,e\rangle$. Then we must have $b<a$ (in order to have $q\prec p$), and we must have $a\le d$. Moreover, if $a=d$, then we must have $e>0$. Then

$$\begin{align*} (q,r)&=\Big(\langle b,c\rangle,\langle d,e\rangle\Big)\\ &=\big\{\langle x,y\rangle:\langle b,c\rangle\prec\langle x,y\rangle\prec\langle d,e\rangle\big\}\\ &=\Big(\{b\}\times(c,1]\Big)\cup\Big((b,d)\times[0,1]\Big)\cup\Big(\{d\}\times[0,e)\Big)\;. \end{align*}\tag{3}$$

(Note that if $a<d$ and $e=0$, then $[0,e)=\varnothing$, and the last term in that union is empty.)

Given the interval in $(3)$, we can always find $q\,'=\langle b\,',0\rangle$ and $r\,'=\langle a,e\,'\rangle$ such that $b<b\,'<a$, and either $0<e\,'<e$, or $d>a$ and $e\,'=1$. I leave it to you to check that $p\in(q\,',r\,')\subseteq(q,r)$, and

$$(q\,',r\,')=\Big((b\,',a)\times[0,1]\Big)\cup\Big(\{a\}\times[0,e\,')\Big)\;.$$

Draw some pictures to illustrate the various possibilities; this is very helpful in understanding the structure of the order. All other cases can be analyzed similarly.

Answer 2

What is a basic open set in an order topology? When would such a set contain $(x,y)$?