Let me just start by saying that I'm basically trying to prove this: How to prove every closed interval in R is compact?
Except that I need to do it in a very strange way... I'm teaching an Inquiry-Based course (based on the University of Chicago IBL courses) and I have a bunch of material that has been passed down to me from previous instructors. So far it's been good, but I'm now running in circles trying to follow a particular string of theorems from this material. Let me explain a little more:
The students have been building "the Continuum" from scratch in a set-theoretic/point-set topological way. They have 4 axioms:
Axioms:
(1) $\mathcal{C}$ is nonempty.
(2) $\mathcal{C}$ has an ordering $<$.
(3) $\mathcal{C}$ has no first or last point.
(4) $\mathcal{C}$ is connected.
and from these alone they have proved many important facts about the real numbers, but never appealing to their arithmetic structure. In particular, they have never worked with addition or multiplication of reals, and instead of referring to them as numbers, the materials provided simply refer to elements of $\mathcal{C}$ as "points".
This means that when talking about regions in $\mathcal{C}$, the students only have the definition $(a,b) = \{x\in \mathcal{C} \mid a<x<b\}$ to work with. No $\varepsilon$-balls or metric space structure to work with here.
We are in the section of the course now that deals with compactness. They already have shown that compact implies closed and bounded, and that a closed subset of a compact set is also compact. But they now must prove the reverse direction. As per usual, the method the provided material comes with has them first show that the closed interval $[a,b] = \{x\in\mathcal{C} \mid a\leq x \leq b \}$ is compact. But it has a few lemmas leading up to it that I'm having trouble with.
They begin with a strange definition:
A simple chain of regions from $a$ to $b$ is a finite collection of regions $R_1, \ldots, R_n$ satisfying:
(1) $a\in R_i \iff i=1$
(2) $b\in R_i \iff i=n$
(3) $R_i\cap R_j \neq \emptyset \iff j=i+1$
From here they are asked to prove the following two lemmas:
Lemma 1: Let $a < x < b$, and let $R_1, \ldots, R_n$ form a simple chain of regions from $a$ to $b$. Then there exists $k$ such that $1 \leq k \leq n$ and $R_1, \ldots, R_k$ form a simple chain of regions from $a$ to $x$.
Lemma 2: If $\mathscr{U}$ is a collection of regions covering $[a, b]$, then $\mathscr{U}$ has a finite subset that is a simple chain of regions from $a$ to $b$.
The first lemma isn't bad, and in particular the students should take away from it that a simple chain of regions forms an open cover of $[a,b]$.
The second lemma however is the one giving me a lit of trouble, and I'm struggling to see how Lemma 1 even comes into play. In the spirit of this solution I tried looking at the set $$ X = \{ x\in[a,b] \mid \text{$\exists\mathscr{U}'\subseteq \mathscr{U}$, where $\mathscr{U}'$ is a simple chain of regions from $a$ to $x$} \}, $$ and proving that $b=\sup X$. Clearly $\sup X \leq b$, but I'm having trouble finalizing that contradiction when $\sup X <b$. Here's what I have so far:
- Let $s= \sup X$, and suppose $s<b$. Clearly $s\in[a,b]$ and $\mathscr{U}$ is an open cover of $[a,b]$, hence there is a region $\tilde{R}\in \mathscr{U}$ with $s\in\tilde{R}$. Explicitly $\tilde{R}=(u,v)$ and $u<s<v$.
- Since $u\neq \sup X$, then there must be a point $x\in X$ between $u$ and $s$. Namely, $u<x\leq s < v$.
- Therefore, there exists a simple chain of regions from $a$ to $x$, say $R_1, \ldots, R_n\in \mathscr{U}$.
From here, we can see that $R_1,\ldots,R_n,\tilde{R}$ are an open cover of $[a,s]$. However, to reach the proper contradiction, I would need to show that this collection is also a simple chain of regions from $a$ to $s$. Then I would have some point $q\in\tilde{R}$, with $u<s<q<b$ and so $q\in X$, contradicting the fact that $s = \sup X$.
But it always comes back to this awkward definition of a "simple chain". This is where I get stuck. Can someone see why this open cover should actually be a simple chain of regions? Or am I missing some simpler argument, perhaps by appealing to Lemma 1? I appreciate any help you can offer.
