In my ongoing love affair with compactness I am constantly revisiting a particular proof of the Heine-Borel theorem, a characterization of compactness in . 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 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 is bounded if and only if is contained in some closed interval

**Fact 2.** The set is complete (as a linear order) because every non-empty set with an upper bound has a least upper bound, called .

**Fact 3.** Closed subsets of compact subsets of are in fact themselves compact. With fact 1 this means that it is enough to show that closed and bounded *intervals* in are compact. (In general closed subsets of compact spaces are compact.)

So now let us creep:

**Creeping along proof of the Heine-Borel theorem.** For notational clarity, let 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 . Let be any open cover of . Consider

This is a measure of how true the theorem is. If then the theorem will be proved. We first note that , so is non-empty, and is clearly bounded above by 1. So by the completeness of we know that has a least upper bound called .

Here is where the creeping happens: Suppose that . Since is an open cover of there is an open set so that .

Now catches some element of to the left of as is the *least* upper bound of . So then has a finite open subcover. (Whatever the finite subcover for was plus .) Now also since , we know that contains something to the right of . So now we know that really isn’t the least upper bound! So in fact we had to have , and since we have already established that this tells us that has a finite open subcover! [**QED**]

Great! We see also that this technique really cares about the linear order on and not really about “compactness”. (The subdivision proof, which is valid in 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 be a complete linear order. If is closed and contained in a closed interval then 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 , 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 the interval is compact.

**Corollary 2. **Any ordinal with countable cofinality is the countable union of compact sets (i.e. -compact). In particular, is -compact.

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

**Example.** The Sorgenfrey line, , ( with the topology given by the basis of sets ) is a Lindelöf space. That is every open cover of has a *countable* subcover. For this all we need to show is that or equivalently is Lindelöf. Try the usual creeping along proof and see where you get stuck. Then try adapting the set 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.

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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