Part of Let $X$ be a simply ordered set with the least upper bound property with respect to the order topology. Then each closed interval is compact.

Question

From my topology notes...

Let $X$ be a simply ordered set with the least upper bound property with respect to the order topology. Then each closed interval is compact.

Proof. First step toward showing $[a, b]$ is compact: Assuming $a < b$, let $A$ be a covering of $[a, b]$ by open sets in the relative topology. If $x \in [a, b]$ with $x \neq b$, then there should exist a point $y > x$ in $[a, b]$ such that $[x, y]$ can be covered by at most two elements of $A$.

This statement is confusing me. I am guessing the author means the intervals $(a,x], (x,y]$, but why "at most 2". Insights appreciated.

Answer 1

The number $2$ is irrelevant, all that matters is that it's finite.

This can be seen from the remainder of the proof where the set $C$ is defined of all $y > a$ such that $[a,y]$ can be covered by finitely many cover elements.

This first step shows that $C \neq \emptyset$, so $\sup C$ exists and the final steps are $\sup C \in C$ and $\sup C = b$, all of which only use the finitely many property.

Answer 2

Let $\mathscr C$ be a cover of $[a,b]$. If $x\in [a,b],$ there is an open set $U\in \mathscr C$ which covers $x$. If $x$ has no successor, then there is a $x<y\in U$ and so $x$ and $y$ are both covered by $U$, so that so is $[x,y]$. On the other hand, if $x$ has a successor $y$ , then as $y$ us covered by some $V\in \mathscr C$, the set $[x,y]$ is covered $U\cup V$.

Answer 3

Assume X is the reals. Cover X with
A = { [a,x), (x,b] : a < x < b }.
More than two sets of A will cover [x,y].

Could he be trying to prove [a,b] compact by using
Alexander's subbase theorem, the easiest method?

A space is compact iff every covering by
subbase sets has a finite subcovering.

A subbase for [a,b] is { [a,x), (x,b] : a < x < b }.
Given a subbase covering, show there is a two set subcover.
Give an example showing this method fails for [a,b] $\cap$ Q.