If $(X,\preccurlyeq)$ is a totally ordered set equipped with the order topology then which is the closure of $(x_1,x_2)$ for $x_1\prec x_2$?

Question

Let be $(X,\preccurlyeq)$ is a totally ordered set equipped with the order topology and thus for $x_1,x_2\in X$ such that $x_1\prec x_2$ we let try to determine the closure of $(x_1, x_2)$. So I know that the set $$ \mathcal S:=\{(x,\to):x\in X\}\cup\{(\leftarrow,x):x\in X\} $$ is a prebasis for $X$ and thus let be $\xi_1,\xi,\xi_2\in X$ such that $$ \xi_1<x_1<\xi<x_2<\xi _2 $$ provided that $(x_1,x_2)$ is not empty. So we observe that $$ [\xi_1,\to)\cap(x_1,x_2)=(x_1,x_2)\quad(\xi_1,\xi)\cap(x_1,x_2)=(x_1,\xi)\\(\xi,\xi_2)\cap(x_1,x_2)=(\xi,x_2)\quad (\leftarrow,\xi_2)\cap(x_1,x_2)=(x_1,x_2) $$ so that if $(x_1,x_2)$ is not empty then $x_1$ and $x_2$ are adherent to it, that is $$ [x_1,x_2]\subseteq\operatorname{cl}\big((x_1,x_2)\big) $$ However I do not able to prove or disprove if the reverse inclusion hols so that I thought to put a specific question. So could someone help me, pelase?

Answer 1

Consider a well ordered set, for example. Then $x_2$ is in the closure of $(x_1,x_2)$ iff $x_2$ is a limit ordinal, that is, $x_2$ is not isolated. Thus in the ordinals, $(4,7)$ is closed, but $(4,\omega)$ has closure $(4,\omega]$. In a well ordered set there are no downward sequences, so the left endpoint will never be in the closure.

For a general totally ordered set, the closure of $(x_1,x_2)$ could be any of $$ (x_1,x_2),\quad(x_1,x_2],\quad[x_1,x_2),\quad[x_1,x_2]. $$