# Creeping Along

In my ongoing love affair with compactness I am constantly revisiting a particular proof of the Heine-Borel theorem, a characterization of compactness in $\mathbb{R}$. There are two proofs that I know of: the standard “subdivision” proof and the “creeping along” proof. I am going to focus on the creeping along proof.

Heine-Borel Theorem. A subset $A \subseteq \mathbb{R}$ is compact if and only if it is closed and bounded.

To do some creeping we need to collect some useful facts.

Fact 1. A subset $\mathcal{A} \subseteq \mathbb{R}$ is bounded if and only if is contained in some closed interval $[a,b]$

Fact 2. The set $\mathbb{R}$ is complete (as a linear order) because every non-empty set $A \subseteq \mathbb{R}$ with an upper bound has a least upper bound, called $\sup A$.

Fact 3. Closed subsets of compact subsets of $\mathbb{R}$ are in fact themselves compact. With fact 1 this means that it is enough to show that closed and bounded intervals in $\mathbb{R}$ are compact. (In general closed subsets of compact $T_2$ spaces are compact.)

So now let us creep:

Creeping along proof of the Heine-Borel theorem. For notational clarity, let $[0,1]$ be the closed interval we look at. I won’t use anything about 0 or 1 other than that they are the endpoints of the interval, so this proof will go through if we decided to instead look at $[a,b]$. Let $\mathcal{U}$ be any open cover of $[0,1]$. Consider $\displaystyle T := \{t \in [0,1]: [0,t] \textrm{ is contained in a finite subcover of }\mathcal{U}]\}$
This $T$ is a measure of how true the theorem is. If $1 \in T$ then the theorem will be proved. We first note that $0 \in T$, so $T$ is non-empty, and $T$ is clearly bounded above by 1. So by the completeness of $\mathbb{R}$ we know that $T$ has a least upper bound called $\sup T \leq 1$.

Here is where the creeping happens: Suppose that $\sup T < 1$. Since $\mathcal{U}$ is an open cover of $[0,1]$ there is an open set $U \in \mathcal{U}$ so that $\sup T \in U$. Now $U$ catches some element of $t \in T$ to the left of $\sup T$ as $\sup T$ is the least upper bound of $T$. So then $[0,t] \cup [t, \sup T]$ has a finite open subcover. (Whatever the finite subcover for $[0,t]$ was plus $U$.) Now also since $\sup T < 1$, we know that $U$ contains something to the right of $\sup T$. So now we know that $\sup T$ really isn’t the least upper bound! So in fact we had to have $\sup T = 1$, and since we have already established that $\sup T \in T$ this tells us that $[0,1]$ has a finite open subcover! [QED]

Great! We see also that this technique really cares about the linear order on $\mathbb{R}$ and not really about “compactness”. (The subdivision proof, which is valid in $\mathbb{R}^n$ does not use orders, but uses that compactness is equivalent to closed subsets with the finite intersection property always have a common point.) So let’s see what this proof does for other linear orders.

First, we distill the essential properties from the proof to get an ordered Heine-Borel Theorem:

Ordered Heine-Borel Theorem. Let $(X,<)$ be a complete linear order. If $A\subseteq X$ is closed and contained in a closed interval $[a,b]$ then $A$ is compact.

We need to be a little more careful at the creeping stage here, because we want to use some sort of connectedness of $X$, but it isn’t too tricky. (In fact complete linear orders have a very nice connectedness property which you should try to discover if you don’t know it already.)

In particular this says:

Corollary 1. For any ordinals $\alpha < \beta$ the interval $[\alpha, \beta]$ is compact.

Corollary 2. Any ordinal with countable cofinality is the countable union of compact sets (i.e. $\sigma$-compact). In particular, $\aleph_\omega$ is $\sigma$-compact.

Now we can adapt this proof slightly to get a result about another type of compactness:

Example. The Sorgenfrey line, $\R_l$, ( $\mathbb{R}$ with the topology given by the basis of sets $[a,b)$) is a Lindelöf space. That is every open cover of $\R_l$ has a countable subcover. For this all we need to show is that $[0,1]$ or equivalently $[0,1)$ is Lindelöf. Try the usual creeping along proof and see where you get stuck. Then try adapting the set $T$ in the Heine-Borel proof for this situation.

Next week I will use this method to show compactness of another linearly ordered space that both looks like it should be compact and looks like it should not be compact.

## 3 thoughts on “Creeping Along”

1. Peter Krautzberger says:

Good post. I really like the image of creeping forward on the line and transferring the idea to more complicated settings.

Just a thought: if you have the time, you could add tags and categories to the post.

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