Any interval in the ordered square is a linear continuum

Question

How to prove that any interval in the ordered square is a linear continuum?

I can see why the second condition holds, but I don't understand why any interval must have a least upper bound property.

Answer 1

If $(a,b)$ is any interval with $a,b \in [0,1]^2$, let $B \subseteq (a,b)$ be a bounded above subset of the ordered square, with upper bound $c \in (a,b)$. Then $\sup(B)$ exists in $[0,1]^2$ (example 1 on page 151 shows this) and as $\sup(B) \le c$, $\sup(B) \in (a,b)$. So $(a,b)$ also has the least upper bound property. In general, an interval (open or closed) in a linear continuum (in Munkres' definition) is again a linear continuum.

Answer 2

The lexicographic square is complete and densely ordered, hence compact and connected. Hence every closed interval is compact and connected as well.