What is definition of order topology?

Question

I am doing self study of topology, the book I am following has used term order topology in problems and few theorems. What is order topology? Kindly give explanation with some examples that will be very helpful for me.

Answer 1

If a set $X$ has a linear order $<$ (with the convention that $x < x$ never holds, we use $\le$ for those linear orders) we can define a topology on $X$ by the defining a subbase $\mathcal{S}=\{L(x), U(x): x \in X\}$ where we define the lower set $L(x)=\{y \in X: y < x\}$ and the upper set $U(x) = \{y \in X: y > x\}$.

In other words, by definition the order topology on $X$ is the smallest (in inclusion) topology in which all sets of the form $L(x)$ and $U(x)$ are open. This is mostly inspired by how the topology on standard ordered sets like $\Bbb R, \Bbb Q$ is defined. Note that any open interval $(x,y)=\{z \in Z: x < z \land z < y\}$ is also open in $X$, as the intersection of two open sets $L(y) \cap U(x)$. But if $X$ has a maximal element, like $1$ in $[0,1]$, it will never lie in any open interval (there is no right-hand endpoint that is strictly bigger than it) but it will be in open sets of the form $(a,1]=U(a)$ and these sets then form its local base of neighbourhoods.

So the standard base for the order topology is all open intervals $(x,y), x,y \in X$ plus all sets $[m,x)=L(x), x \in X$ when $m$ has a minimum $m$, plus all sets $(x,M], x \in X$ when $X$ has a maximum $M$. (If we have neither we don't need such sets in our topology). All open sets are then the unions of basic open sets, as always.

Munkres (2nd ed.) introduces this topology in §14, p.85 and Engelking introduces the topology in Exercise 1.7.4. In papers they're often abbreviated as LOTS (linearly ordered topological space). There is quite a rich theory surrounding them, and many special theorems that hold for them.

Answer 2

The order topology is the natural generalisation of the usual topology on $\mathbb{R}$.

Given a partially ordered set $(X,\leq)$ with more than one element you can define a topology on $X$ having as sub-basis sets of the form $$ (-\infty,x_1)=\{x\in X|\; x<x_1\}\\ (x_1,\infty)=\{x\in X|\; x_1<x\}\\ $$

Examples include, as stated, $\mathbb{R}$ (and $\mathbb{N}$ and $\mathbb{Q},$ I guess) in the usual topology.

Given any set $Y,$ let $X=\mathcal{P}(Y)$ denote its power set. Then, there is a natural ordering on $X$ given by set inclusion and so, you get a topology on the power set, where a set $U\subseteq X$ is open if and only if for every $A\in U,$ there exists $A_0$ and $A_1$ such that $A_0\subsetneq U\subsetneq A_1$ and for any set $A_0\subsetneq B\subsetneq A_1,$ we have $B$ in $U$.

A last example that Munkres likes to use is $\mathbb{R}^2$ in the lexicographical ordering. Here, we say that $(x_1,y_1)\leq (x_2,y_2)$ if $y_2>y_1$ or $y_2=y_1$ and $x_2\geq x_1$ (we first compare the second coordinates and if they agree, we compare the first ones).

In the order topology here, it's true that a sequence $(z_n,w_n)_{n\in\mathbb{N}}$ converges if and only if $w_n$ is eventually constant, and $z_n$ converges in $\mathbb{R}$. If you want something to get started, try proving this last fact.