Uniqueness of endpoints of non-empty open bounded intervals in a linear order

Question

This is a follow-up to my previous question on uniqueness of endpoints of closed bounded intervals. Suppose $(L, \leq)$ is a linear order, and suppose $a,b,c,d$ are elements of $L$ such that $a < b$ and $c <d$. Suppose further that the open intervals $(a,b)$ and $(c,d)$ are both non-empty. If $(a,b)=(c,d)$, must $a=c$ and $b=d$? I would be very surprised if there was a counterexample in a linear order. I know there are counterexamples in general partial orders, but what about linear orders?

Answer 1

It's not true if the open intervals are empty. For example, take $a=1,\ b=2,\ c=3,\ d=4$ in $\Bbb N$.

On the other hand, if we suppose that $(a,b)=(c,d)$ is nonempty, the result is true.

First, by definition of intervals, both $b$ and $d$ must be strict upper bounds of the nonempty set $(a,b)=(c,d)$, similarly both $a$ and $c$ are strict lower bounds, hence in particular $c<x< b$ for any $x\in(a,b)=(c,d)$, so $c<b$ follows. Similarly, $a<d$.

In a linear order we either have $a<c,\ a>c$ or $a=c$.
In the first case we would have $c\in (a,b)$ whereas $c\notin (c,d)$, and similarly in the second case we would have $a\in (c,d)$ whereas $a\notin (a,b)$. So we must have $a=c$.
Analogously, $b=d$ also follows.