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. 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.

    Like

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