proof of Heine-Borel theorem

To shorten the proofs, we will be using certain concepts and facts from general point-set topology, such as the very first assertion below. Even if you are not familiar with them, these assertions are all easily proven from the definitions (left as an exercise).

Compact implies closed and bounded.

Since n is a Hausdorff space, any compact subset A has to be closed as well. That A is also boundedPlanetmathPlanetmathPlanetmath is very easily seen by taking the open cover {(-k,k)n}k. Passing to a finite cover, we see that A is contained in a bounded set. ∎

Reduction for closed and bounded implies compact.

To prove that if A is closed and bounded implies it is compact, it is only necessary to prove that the rectangle [a,b]n is compact. For general A, since A is bounded, it is contained in some such compact rectangle [a,b]n. Since A is closed and contained in a compact set, A is also compact. ∎

The case n=1: the closed interval is compact..

Let 𝒞 be an arbitrary cover of [a,b] by open sets in 1. Define

S={x[a,b]:some finite subcollection of 𝒞 covers [a,x]}.

Set t=supS; then atb<.

We first note that the supremum t is attained: that is, there is a finite subcollection of 𝒞 that covers [a,t]. Obviously we can choose one open set U from 𝒞 that covers the point t. This U must also cover the open interval (t-ϵ,t+ϵ) for some ϵ>0. But by the definition of t, the interval [a,t-ϵ] has a finite subcover. When this finite subcover is put together with U, we obtain a finite subcover for [a,t].

It is easy to see that t cannot be less than b. For the same open set U above covers the interval (t,t+ϵ), so we have a finite subcover for [a,t+ϵ/2]. If t is the supremum then it has go to be at the right endpoint of the interval [a,b], i.e. t=b. ∎

For the case n>1, apply TychonoffPlanetmathPlanetmath’s Theorem, in the finite case (, which states:

If X1,,Xn are each compact, then X1××Xn in the product topology is compact.

Here we set Xi to be the closed intervals; then the general closed rectangle in n is compact. It is not hard to see that the product topology for the closed rectangle is the same as its subspace topology in the product topology n=××. And it is again a standard exercise to show that the product topology on n is the same as the norm topology on n as a vector spaceMathworldPlanetmath.

This completesPlanetmathPlanetmath the proof of the Heine-Borel theorem.

Proof by a bisection argument

There is another proof of the Heine-Borel theorem for n without resorting to Tychonoff’s Theorem. It goes by bisecting the rectangle along each of its sides. At the first stage, we divide up the rectangle A into 2n subrectangles. Suppose the open cover 𝒞 of A has no finite subcover. Then one of the subrectangles — call it A1 — must have no finite subcover by 𝒞. We can subdivide A1 into 2n pieces; since A1 has no finite subcover, one of the new subrectangles of A1 also has no finite subcover. And continue dividing to get a nested sequence of rectangles AA1A2 whose side lengths approach zero, and possessing no finite subcover.

By the nested interval theorem, the “limit rectangle” i=1Ai must consist of a sole point x, and this obviously has a finite subcover by an open set U𝒞. But U must contain a small rectangle with centre x, which for i large enough, contradicts Ai having no finite subcover.

Of course, in both proofs of the Heine-Borel theorem, the completeness of the reals (the least upper bound property) enters in an essential way.

Title proof of Heine-Borel theorem
Canonical name ProofOfHeineBorelTheorem
Date of creation 2013-03-22 12:57:40
Last modified on 2013-03-22 12:57:40
Owner stevecheng (10074)
Last modified by stevecheng (10074)
Numerical id 23
Author stevecheng (10074)
Entry type Proof
Classification msc 54D30