# Heine–Borel theorem

# Heine–Borel theorem

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

For a subset *S* of Euclidean space **R***n*, the following two statements are equivalent:

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

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. Peter Gustav Lejeune 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.^{[1]} He used this proof in his 1852 lectures, which were published only in 1904.^{[1]} Later Eduard Heine, Karl Weierstrass and Salvatore Pincherle used similar techniques. Émile 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. Pierre Cousin (1895), Lebesgue (1898) and Schoenflies (1900) generalized it to arbitrary covers.^{[2]}

Proof

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

Let *S* be a subset of **R***n*. Observe first the following: if *a* is a limit point of *S*, then any finite collection *C* of open sets, such that each open set *U* ∈ *C* is disjoint from some neighborhood *V**U* of *a*, fails to be a cover of *S*. Indeed, the intersection of the finite family of sets *V**U* is a neighborhood *W* of *a* in **R***n*. Since *a* is a limit point of *S*, *W* must contain a point *x* in *S*. This *x* ∈ *S* is not covered by the family *C*, because every *U* in *C* is disjoint from *V**U* and hence disjoint from *W*, which contains *x*.

If *S* is compact but not closed, then it has an accumulation point *a* not in *S*. Consider a collection *C* ′ consisting of an open neighborhood *N*(*x*) for each *x* ∈ *S*, chosen small enough to not intersect some neighborhood *V**x* of *a*. Then *C* ′ is an open cover of *S*, but any finite subcollection of *C* ′ has the form of *C* discussed previously, and thus cannot be an open subcover of *S*. This contradicts the compactness of *S*. Hence, every accumulation point of *S* is in *S*, so *S* is closed.

The proof above applies with almost no change to showing that any compact subset *S* of a Hausdorff topological space *X* is closed in *X*.

**If a set is compact, then it is bounded.**

`Letbe a compact set in, anda ball of radius 1 centered at. Then the set of all such balls centered atis clearly an open cover of, sincecontains all of. Sinceis compact, take a finite subcover of this cover. This subcover is the finite union of balls of radius 1. Consider all pairs of centers of these (finitely many) balls (of radius 1) and letbe the maximum of the distances between them. Then ifandare the centers (respectively) of unit balls containing arbitrary, the triangle inequality says:So the diameter ofis bounded by.`

**A closed subset of a compact set is compact.**

Let *K* be a closed subset of a compact set *T* in **R***n* and let *C**K* be an open cover of *K*. Then *U* = **R***n* \ *K* is an open set and

`is an open cover of`

*T*. Since*T*is compact, then*C**T*has a finite subcoverthat also covers the smaller set*K*. Since*U*does not contain any point of*K*, the set*K*is already covered bythat is a finite subcollection of the original collection*C**K*. It is thus possible to extract from any open cover*C**K*of*K*a finite subcover.**If a set is closed and bounded, then it is compact.**

If a set *S* in **R***n* is bounded, then it can be enclosed within an *n*-box

where *a* > 0. By the property above, it is enough to show that *T*0 is compact.

Assume, by way of contradiction, that *T*0 is not compact. Then there exists an infinite open cover *C* of *T*0 that does not admit any finite subcover. Through bisection of each of the sides of *T*0, the box *T*0 can be broken up into 2*n* sub *n*-boxes, each of which has diameter equal to half the diameter of *T*0. Then at least one of the 2*n* sections of *T*0 must require an infinite subcover of *C*, otherwise *C* itself would have a finite subcover, by uniting together the finite covers of the sections. Call this section *T*1.

Likewise, the sides of *T*1 can be bisected, yielding 2*n* sections of *T*1, at least one of which must require an infinite subcover of *C*. Continuing in like manner yields a decreasing sequence of nested *n*-boxes:

where the side length of *T**k* is (2 *a*) / 2*k*, which tends to 0 as *k* tends to infinity. Let us define a sequence (*x*k) such that each *x*k is in *T*k. This sequence is Cauchy, so it must converge to some limit *L*. Since each *T**k* is closed, and for each *k* the sequence (*x*k) is eventually always inside *T*k, we see that *L* ∈ *T*k for each *k*.

Since *C* covers *T*0, then it has some member *U* ∈ *C* such that *L* ∈ *U*. Since *U* is open, there is an *n*-ball *B*(*L*) ⊆ *U*. For large enough *k*, one has *T**k* ⊆ *B*(*L*) ⊆ *U*, but then the infinite number of members of *C* needed to cover *Tk* can be replaced by just one: *U*, a contradiction.

Thus, *T*0 is compact. Since *S* is closed and a subset of the compact set *T*0, then *S* is also compact (see above).

Heine–Borel property

The Heine–Borel theorem does not hold as stated for general metric and topological vector spaces, and this gives rise to the necessity to consider special classes of spaces where this proposition is true. They are called the **spaces with the Heine–Borel property**.

In the theory of metric spaces

`Ametric spaceis said to have the`

**Heine–Borel property**if each closed bounded^{[3]}set inis compact.Many metric spaces fail to have the Heine–Borel property, for instance, the metric space of rational numbers (or indeed any incomplete metric space). Complete metric spaces may also fail to have the property, for instance, no infinite-dimensional Banach spaces have the Heine–Borel property (as metric spaces). Even more trivially, if the real line is not endowed with the usual metric, it may fail to have the Heine–Borel property.

`A metric spacehas a Heine–Borel metric which is Cauchy locally identical toif and only if it iscomplete,-compact, andlocally compact.`

^{[4]}In the theory of topological vector spaces

`Atopological vector spaceis said to have the`

**Heine–Borel property**^{[5]}(R.E. Edwards uses the term*boundedly compact space*^{[6]}) if each closed bounded^{[7]}set inis compact.^{[8]}No infinite-dimensionalBanach spaceshave the Heine–Borel property (as topological vector spaces). But some infinite-dimensionalFréchet spacesdo have, for instance, the spaceof smooth functions on an open set^{[6]}and the spaceof holomorphic functions on an open set^{[6]}. More generally, any quasi-completenuclear spacehas the Heine–Borel property. AllMontel spaceshave the Heine–Borel property as well.See also

Bolzano–Weierstrass theorem

## References

*American Mathematical Monthly*.

**122**(7): 619–635. doi:10.4169/amer.math.monthly.122.7.619. JSTOR 10.4169/amer.math.monthly.122.7.619.

*bounded*if it is contained in a ball of a finite radius, i.e. there exists and such that .

*Proc. AMS*.

**100**: 567–573..

*Theorems and Problems in Functional Analysis*. Springer-Verlag New York. ISBN 978-1-4613-8155-6., Theorem 28.

*Functional analysis*. Holt, Rinehart and Winston. ISBN 0030505356., 8.4.7.

*bounded*if for each neighborhood of zero in there exists a scalar such that .