Heine–Borel theorem

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In mathematical analysis, the Heine–Borel theorem, named after Eduard Heine and Émile Borel, states:

For a subset S of Euclidean space Rn, the following two statements are equivalent:

In the context of real analysis, the former property is sometimes used as the defining property of compactness. However, the two definitions cease to be equivalent when we consider subsets of more general metric spaces and in this generality only the latter property is used to define compactness. In fact, the Heine–Borel theorem for arbitrary metric spaces reads:

A subset of a metric space is compact if and only if it is complete and totally bounded.

Contents

[edit] History and motivation

The history of what today is called the Heine–Borel theorem starts in the 19th century, with the search for solid foundations of real analysis. Central to the theory was the concept of uniform continuity and the theorem stating that every continuous function on a closed interval is uniformly continuous. Dirichlet was the first to prove this and implicitly he used the existence of a finite subcover of a given open cover of a closed interval in his proof. He used this proof in his 1862 lectures, which got published only in 1904. Later Heine, Weierstrass and Pincherle used similar techniques. Emile Borel in 1895 was the first to state and prove a form of what is now called the Heine–Borel theorem. His formulation restricted to countable covers. Lebesgue (1898) and Schoenflies (1900) generalized it to arbitrary covers.

[edit] Discussion of the theorem

If a set is not closed, then it cannot be compact.

If a set is not closed, then it is either an open set, or it is partially open: part of its boundary is open, by which is meant that that part of the boundary does not belong to the set.

Then it is possible to come up with an infinite cover whose elements (which are all, by definition, open) are all subsets of the given open set, but whose boundaries are never tangent to the open boundary of the given set. Non-tangency implies that the elements in the cover will have to approach the boundary by decreasing both their diameter and their distance to the boundary asymptotically to zero.

Therefore points infinitesimally close to the boundary of the given (open) set can only be covered by infinite subcovers of the infinite cover. Why infinite subcovers? Pick a point on the boundary of the given set, then pick a point P1 in the given set at a distance less than ε from the chosen boundary point. Then, due to the requirement of non-tangency, pick a ball inside the open set which is not tangent to the boundary. Call it C1. C1 will cover P1, but then one can pick another point P2 even closer to B than P1 but which is not in C1. Then pick an open ball C2 whose boundary is not tangent to the boundary of the given set, but which includes P2... This process can go on for P3, C3, P4, C4, P5, C5, etc. without end. This means that if a cover of an open boundary point has elements which are all subsets of the given set and whose boundaries are never tangent to the boundary of the given set, then this cover can not be finite, and so any such infinite cover cannot have a finite subcover.

If a set is unbounded, then it cannot be compact.

Why? Because one can always come up with an infinite cover, whose elements have an upper finite bound to their size, i.e. the elements of the cover are not allowed to grow in size without bound.

But there is no finite cover of an unbounded set such that its elements do not grow in size without bounds: if one adds up a finite set of n numbers whose upper limit is m, then their sum can be no greater than n \times m. A similar case holds for unbounded sets with finite covers: the elements of the finite cover could not possibly be bounded in size, otherwise the union of all the elements of the finite cover would itself be bounded, and could not cover an unbounded set.

So if an unbounded set is covered with an infinite cover whose elements have an upper finite bound to their size, then this infinite cover of the given set will have no finite subcover, because any subcover will be made up of elements of the cover, but the elements of the cover have an upper bound to their size, so the elements of the subcover will also have an upper bound to their size, and there is no finite cover of an unbounded set such that its elements have an upper bound to their diameters.

Therefore, given an unbounded set, there exists at least one infinite cover which has no finite subcover: namely, an infinite cover whose elements have diameters all of which have the same finite upper bound. Thus, an unbounded set cannot be compact.

[edit] Generalizations

The proper generalization to arbitrary metric spaces is:

A subset of a metric space is compact if and only if it is complete and totally bounded.

This generalisation also applies to topological vector spaces and, more generally, to uniform spaces.

Here is a sketch of the "⇒"-part of the proof, in the context of a general metric space, according to Jean Dieudonné:

  1. It is obvious that any compact set E is totally bounded.
  2. Let (xn) be an arbitrary Cauchy sequence in E; let Fn be the closure of the set { xk : k ≥ n } in E and Un := E − Fn. If the intersection of all Fn would be empty, (Un) would be an open cover of E, hence there would be a finite subcover (Unk) of E, hence the intersection of the Fnk would be empty; this implies that Fn is empty for all n larger than any of the nk, which is a contradiction. Hence, the intersection of all Fn is not empty, and any point in this intersection is an accumulation point of the sequence (xn).
  3. Any accumulation point of a Cauchy sequence is a limit point (xn); hence any Cauchy sequence in E converges in E, in other words: E is complete.

A proof of the "<="-part can be sketched as follows:

  1. If E were not compact, there would exist a cover (Ul)l of E having no finite subcover of E. Use the total boundedness of E to define inductively a sequence of balls (Bn) in E with
    • the radius of Bn is 2−n;
    • there is no finite subcover (Ul∩Bn)l of Bn;
    • Bn+1 ∩ Bn is not empty.
  2. Let xn be the center point of Bn and let yn be any point in Bn+1 ∩ Bn; hence we have d(xn+1, xn) ≤ d(xn+1, yn) + d(yn, xn) ≤ 2−n−1 + 2−n ≤ 2−n+1. It follows for n ≤ p < q: d(xp, xq) ≤ d(xp, xp+1) + ... + d(xq−1, xq) ≤ 2−p+1 + ... + 2−q+2 ≤ 2−n+2. Therefore, (xn) is a Cauchy sequence in E, converging to some limit point a in E, because E is complete.
  3. Let I0 be an index such that \mbox{ }_{U_{I_0}} contains a; since (xn) converges to a and \mbox{ }_{U_{I_0}} is open, there is a large n such that the ball Bn is a subset of \mbox{ }_{U_{I_0}} - in contradiction to the construction of Bn.

The proof of the "=>" part easily generalises to arbitrary uniform spaces, but the proof of the "<=" part is more complicated and is equivalent to the ultrafilter principle [1], a strong form of the axiom of choice. (Already, in general metric spaces, the "<=" direction requires the Axiom of dependent choice.)

[edit] References

  1. ^ UF24 in Eric Schechter's Handbook of Analysis and its Foundations.
  • P. Dugac (1989). "Sur la correspondance de Borel et le théorème de Dirichlet-Heine-Weierstrass-Borel-Schoenflies-Lebesgue". Arch. Internat. Hist. Sci. 39: 69-110. 
  • proof of Heine-Borel theorem on PlanetMath
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