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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:

• S is closed and bounded
• every open cover of S has a finite subcover, that is, S is compact.

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.

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 Eduard Heine, Karl Weierstrass and Salvatore 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 was restricted to countable covers. Lebesgue (1898) and Schoenflies (1900) generalized it to arbitrary covers.

If a set is compact, then it must be closed.

Let a set be compact. A finite set of sets which do not intersect at least one neighborhood of an accumulation point cannot be an open cover, because the intersection of those neighborhoods not met forms an open set, which contains a point within the set not contained within that finite set of sets. Consider the neighborhoods of every element of the set minus the set containing the accumulation point, such that each neighborhood does not intersect at least one neighborhood of the accumulation point. All subsets of this set of neighborhoods are thus of the form discussed earlier, and thus cannot be an open subcover. Thus this original set of such neighborhoods not intersecting the neighborhood of the accumulation point does not form an open cover either, even though it does not include the accumulation point but includes all other elements that are within the set. Thus, the accumulation point must be within the set.

If a set is compact, then it is bounded.

Why? Consider the open balls centered upon a common point, with any radius. This can cover any set, because all points in the set are some distance away from that point. Any finite subcover of this cover must be bounded, because it will bound it within the largest open ball within that subcover. Therefore, any set covered by this subcover must also be bounded.

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 (x_n) be an arbitrary Cauchy sequence in E; let F_n be the closure of the set { x_k : k ≥ n } in E and U_n := E − F_n. If the intersection of all F_n were empty, (U_n) would be an open cover of E, hence there would be a finite subcover (U_nk) of E, hence the intersection of the F_nk would be empty; this implies that F_n is empty for all n larger than any of the n_k, which is a contradiction. Hence, the intersection of all F_n is not empty, and any point in this intersection is an accumulation point of the sequence (x_n).
3. Any accumulation point of a Cauchy sequence is a limit point (x_n); 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 (U_l)_l of E having no finite subcover of E. Use the total boundedness of E to define inductively a sequence of balls (B_n) in E with
   • the radius of B_n is 2⁻ⁿ;
   • there is no finite subcover (U_l∩B_n)_l of B_n;
   • B_n+1 ∩ B_n is not empty.
2. Let x_n be the center point of B_n and let y_n be any point in B_n+1 ∩ B_n; hence we have d(x_n+1, x_n) ≤ d(x_n+1, y_n) + d(y_n, x_n) ≤ 2⁻ⁿ⁻¹ + 2⁻ⁿ ≤ 2⁻ⁿ⁺¹. It follows for n ≤ p < q: d(x_p, x_q) ≤ d(x_p, x_p+1) + ... + d(x_q−1, x_q) ≤ 2^−p+1 + ... + 2^−q+2 ≤ 2⁻ⁿ⁺². Therefore, (x_n) 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 contains a; since (x_n) converges to a and is open, there is a large n such that the ball B_n is a subset of - in contradiction to the construction of B_n.

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.)

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