The Unapologetic Mathematician

Mathematics for the interested outsider

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 \mathbb{R} of real numbers, or higher-dimensional versions like \mathbb{R}^n?

First off, \mathbb{R} is Hausdorff, which should be straightforward to prove. Unfortunately, it’s not compact. To see this, consider the open sets of the form (-x,x) for all positive real numbers x. Given any real number y we can find an x with |y|<x, so y\in(-x,x). Therefore the collection of these open intervals covers \mathbb{R}. 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 \mathbb{R} is not compact. We can similarly show that \mathbb{R}^n is Hausdorff, but not compact.

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

We see now that every compact subset of \mathbb{R}^n 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 \left[-x,x\right]^n is compact. Then a bounded set A is contained in some such cube, and a closed subset of a compact space is compact. This is the Heine-Borel theorem.

In the n=1 case, we just need to see that the interval \left[-x,x\right] is compact. Take an open cover \{U_i\} of this interval, and define the set S to consist of all y\in\left[-x,x\right] so that a finite collection of the U_i cover \left[-x,y\right]. Then define t to be the least upper bound of S. Basically, t is as far along the interval as we can get with a finite number of sets, and we’re hoping to show that t=x. Clearly it can’t go past x, since S\subseteq\left[-x,x\right]. But can it be less than x?

In fact it can’t, because if it were, then we can find some open set U from the cover that contains t. As an open neighborhood of t, the set U contains some interval (t-\epsilon,t+\epsilon). Then t-\epsilon must be in S, and so there is some finite collection of the U_i which covers \left[-x,t-\epsilon\right]. But then we can just add in U to get a finite collection of the U_i which covers \left[-x,t+\frac{\epsilon}{2}\right], and this contradicts the fact that t is the supremum of S. Thus t=x and there is a finite subcover of \left[-x,x\right], making this closed interval compact!

Now Tychonoff’s Theorem tells us that products of closed intervals are also compact. In particular, the closed cube \left[-x,x\right]^n\subseteq\mathbb{R}^n 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 n 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 \left[a,b\right]. 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 f:\mathbb{R}\rightarrow\mathbb{R} 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 f on a closed interval \left[a,b\right] attains a maximum and a minimum. That is, there is some c\in\left[a,b\right] so that f(c)\geq f(x) for all x\in\left[a,b\right], and similarly there is some d\in\left[a,b\right] so that f(d)\leq f(x) for all x\in\left[a,b\right].

January 18, 2008 - Posted by | Analysis, Calculus, Point-Set Topology, Topology


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