Heine–Borel theorem

In the topology of metric spaces 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
*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.

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 were published only in 1904. 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. Lebesgue (1898) and Schoenflies (1900) generalized it to arbitrary covers.

Proof

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

Let a set "S" be compact. A finite collection "C" of open sets which do not intersect at least one neighborhood of an accumulation point $vec\; a$ cannot be an open cover of "S", because the intersection of those neighborhoods not met forms an open set, which contains a point in "S" not covered by "C".

Consider a collection "C’" consisting of a neighborhood $N(vec\; x)$ for each $vec\; x\; in\; S$. If "S" is not closed then it has an accumulation point not in "S", call it $vec\; a,\text{'}$. Then choose (when constructing "C’"), for each $vec\; x\; in\; S$, an $N(vec\; x)\; in\; C\text{'}$ which is small enough to not intersect some neighborhood of $vec\; a,\text{'}$. Then any subcover of "C"' has the form of "C" discussed previously, an thus cannot be an open subcover of "S". This contradicts the compactness of "S". Thus every accumulation point of "S" is in "S", so "S" is closed.

"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.

"If a set is closed and bounded, then it is compact."

If a set "S" in $mathbb\{R\}^n$ is bounded, then it can be enclosed within an "n"-box:$[a\_1,\; b\_1]\; imes\; [a\_2,\; b\_2]\; imes\; ...\; imes\; [a\_n,\; b\_n]$where ; $a\_k,\; b\_k\; in\; mathbb\{R\}$ for $k\; in\; 1,\; ...,\; n$. Call this "n"-box "T"₀.

Through bisection of each of the sides of "T"₀, "T"₀ can be broken up into $2^n$ sub "n"-boxes, each of which has $1/2^n$ the size of "T"₀.

Assume, by way of contradiction, that "T"₀ is not compact. Then, given an infinite open cover "C" of "T"₀, 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 "T"₁ 's sides 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::$T\_0\; supset\; T\_1\; supset\; T\_2\; supset\; ...\; supset\; T\_k\; supset\; ...$whose length, when projected onto a given "x_j"-axis is $(b\_j\; -\; a\_j)/2^k$, which tends to 0 as $k\; o\; infty$.

Then:$T\_0\; cap\; T\_1\; cap\; T\_2\; cap\; ...\; cap\; T\_k\; cap\; ...\; e\; varnothing$, (Cantor’s Intersection Theorem)but instead contains some point $vec\; p\; in\; T\_0$. Since "C" covers "T"₀, then it has some member $U\; in\; C$ such that $vec\; p\; in\; U$. Since "U" is open, there is an "n"-ball $B(vec\; p)\; subseteq\; U$. For large enough "n", $T\_n\; subseteq\; B(vec\; p)\; subseteq\; U$, but then the infinite number of members of "C" needed to cover "T_n" can be replaced by just one: "U", a contradiction.

Thus, "T" is compact. Since "S" is closed and a subset of the compact set "T", then by the lemma (below) it is also compact.

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

Let "K" be a closed subset of a compact set "T" in $mathbb\{R\}^n$. Let $C\_K$ be an infinite open cover of "K". If $C\_K$ also covers "T", then since "T" is compact, then $C\_K$ has a finite subcover, and we are done.

Otherwise is an open set containing points in "T" not covered by $C\_K$. Let :be an open cover of "T". Since "T" is compact, then $C\_T$ has a finite subcover $C\_T\text{'}$. Since covers points in "T" not covered by $C\_K$, then . Thus where $C\_K\text{'}$ must be a finite subcover of $C\_K$. Since does not cover $K\; subset\; T$, then $C\_K\text{'}\; e\; varnothing$, so $C\_K$ has a finite subcover.

Generalizations

The theorem does not hold as stated for general metric spaces. A metric space (or topological vector space) is said to have the Heine-Borel property if every closed and bounded subset is 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) fails to have the Heine-Borel property. Complete metric spaces may also fail to have the property. For instance, no infinite-dimensional Banach space has the Heine-Borel property.

The theorem can be generalized to arbitrary metric spaces by strengthening the conditions required for compactness::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:
# It is obvious that any compact set "E" is totally bounded.
# 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"_"n""k") of "E", hence the intersection of the "F"_"n""k" 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").
# 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:
# 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.
# 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.
# Let $I\_0$ be an index such that $mbox\{\; \}\_\{U\_\{I\_0$ contains "a"; since ("x"_"n") converges to "a" and $mbox\{\; \}\_\{U\_\{I\_0$ is open, there is a large "n" such that the ball "B"_"n" is a subset of $mbox\{\; \}\_\{U\_\{I\_0$ - 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 [ [http://www.math.vanderbilt.edu/~schectex/ccc/excerpts/equivuf2.gifUF24] , p. 506. ] , a strong form of the axiom of choice. (Already, in general metric spaces, the "<=" direction requires the Axiom of dependent choice.)

See also

* Bolzano-Weierstrass theorem

Notes

References

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External links

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