The Heine-Borel Theorem

We’ve talked about compact subspaces, particularly of compact spaces and Hausdorff spaces (and, of course, compact Hausdorff spaces). So how can we use this to understand the space of real numbers, or higher-dimensional versions like ?

First off, is Hausdorff, which should be straightforward to prove. Unfortunately, it’s not compact. To see this, consider the open sets of the form for all positive real numbers . Given any real number we can find an with , so . Therefore the collection of these open intervals covers . But if we take any finite number of them, one will be the biggest, and so we must miss some real numbers. This open cover does not have a finite subcover, and is not compact. We can similarly show that is Hausdorff, but not compact.

So, since is Hausdorff, any compact subset of must be closed. But not every closed subset is compact. What else does compactness imply? Well, we can take the proof that isn’t compact and adapt it to any subset . We take the collection of all open “cubes” consisting of -tuples of real numbers, each of which is between and , and we form open subsets of by the intersections . Now the only way for there to be a finite subcover of this open cover of is for there to be some so that . That is, every component of every point of has absolute value less than , and so we say that is “bounded”.

We see now that every compact subset of is closed and bounded. It turns out that being closed and bounded is not only necessary for compactness, but they’re also sufficient! To see this, we’ll show that the closed cube is compact. Then a bounded set is contained in some such cube, and a closed subset of a compact space is compact. This is the Heine-Borel theorem.

In the case, we just need to see that the interval is compact. Take an open cover of this interval, and define the set to consist of all so that a finite collection of the cover . Then define to be the least upper bound of . Basically, is as far along the interval as we can get with a finite number of sets, and we’re hoping to show that . Clearly it can’t go past , since . But can it be less than ?

In fact it can’t, because if it were, then we can find some open set from the cover that contains . As an open neighborhood of , the set contains some interval . Then must be in , and so there is some finite collection of the which covers . But then we can just add in to get a finite collection of the which covers , and this contradicts the fact that is the supremum of . Thus and there is a finite subcover of , making this closed interval compact!

Now Tychonoff’s Theorem tells us that products of closed intervals are also compact. In particular, the closed cube is compact. And since any closed and bounded set is contained in some such cube, it will be compact as a closed subspace of a compact space. Incidentally, since is finite, we don’t need to wave the Zorn talisman to get this invocation of the Tychonoff magic to work.

As a special case, we can look back at the one-dimensional case to see that a compact, connected space must be a closed interval . Then we know that the image of a connected space is connected, and that the image of a compact space is compact, so the image of a closed interval under a continuous function is another closed interval.

The fact that this image is an interval gave us the intermediate value theorem. The fact that it’s closed now gives us the extreme value theorem: a continuous, real-valued function on a closed interval attains a maximum and a minimum. That is, there is some so that for all , and similarly there is some so that for all .