Every order topology is Hausdorff.

Question

We are asked to show that every set $X$ in the order topology is a Hausdorff space. In other words, for every distinct $x_1, x_2 \in X$, we need to find open sets $U,V$ that contain $x_1$ and $x_2$ respectively but are disjoint.

What if we suppose that $x_1$ is the smallest (with respect to the order) element in $X$ and $x_2$ is the immediate successor of $x_1$ (an example would be $0$,$1$ in the non-negative integers). Is it still possible to construct two open sets that contain $0$ and $1$ respectively but are disjoint?

Answer 1

For any $x\in X$ let $(\leftarrow,x)=\{y\in X:y<x\}$ and $(x,\to)=\{y\in X:y>x\}.$ And let $(\leftarrow,x]=\{y\in X:y\le x\}$ and $[x,\to)=\{y\in X:y\ge x\}.$ For any $x,y\in X$ let $(x,y)=\{z\in X: x<z<y\}.$

By definition of the order topology, $(\leftarrow,x)$ and $(x,\to)$ are open.

Let $x,y\in X$ with $x<y.$

$(i).$ If $(x,y)=\emptyset$ then $U=(\leftarrow, y)=(\leftarrow, x]$ and $V=(x,\to)=[y,\to)$ are disjoint open sets with $x\in U$ and $y\in V.$

$(ii).$ If $(x,y)\ne\emptyset$ then let $z$ be some (any) member of $(x,y).$ Then $U=(\leftarrow,z)$ and $V=(z,\to)$ are disjoint open sets with $x\in U$ and $y\in V.$

Answer 2

As to your example, $[0,1) = \{0\}$ is basic open (also equal to the subbasic open set $(-\infty,1)=\{x\mid x < 1\}$). And for $1$ (the immediate successor of $0$, the minimum, we can use the subbasic $(0,\infty)$ in general, but for $\Bbb N$ we use the basic open $(0,2)=\{1\}$, but in either case $1$ is in this open set and is disjoint from $\{0\}$, which is open, as we saw.

A point in an order topology with a direct predecessor and a direct successor is an isolated point (as we can form the open interval from the predecessor to the successor and get an open singleton).

But for all order topologies, the subbasic sets (open segments) $(-\infty,x)$ and $(y,\infty)$ are enough to separate all pairs of points.