Editing Heine–Borel theorem (section)

== Proof ==

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

Let <math>S</math> be a subset of <math>\mathbb{R}^n</math>.  Observe first the following: if <math>a</math> is a [[limit point]] of <math>S</math>, then any finite collection <math>C</math> of open sets, such that each open set <math>U\in C</math> is disjoint from some [[Neighbourhood (mathematics)#Definitions|neighborhood]] <math>V_U</math> of <math>a</math>, fails to be a cover of <math>S</math>. Indeed, the intersection of the finite family of sets <math>V_U</math> is a neighborhood <math>W</math> of <math>a</math> in <math>\mathbb{R}^n</math>. Since <math>a</math> is a limit point of <math>S</math>, <math>W</math> must contain a point <math>x</math> in <math>S</math>. This <math>x\in S</math> is not covered by the family <math>C</math>, because every <math>U</math> in <math>C</math> is disjoint from <math>V_U</math> and hence disjoint from <math>W</math>, which contains <math>x</math>.

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

The proof above applies with almost no change to showing that any compact subset <math>S</math> of a [[Hausdorff space|Hausdorff]] topological space <math>X</math> is closed in <math>X</math>.

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

Let <math>S</math> be a compact set in <math>\mathbb{R}^n</math>, and <math>U_x</math> a ball of radius 1 centered at <math>x\in\mathbb{R}^n</math>. Then the set of all such balls centered at <math>x\in S</math> is clearly an open cover of <math>S</math>, since <math>\cup_{x\in S} U_x</math> contains all of <math>S</math>. Since <math>S</math> is 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 let <math>M</math> be the maximum of the distances between them. Then if <math>C_p</math> and <math>C_q</math> are the centers (respectively) of unit balls containing arbitrary <math>p,q\in S</math>, the triangle inequality says:

<math display="block">
d(p, q)\le d(p, C_p) + d(C_p, C_q) + d(C_q, q)\le 1 + M + 1 = M + 2.
</math>

So the diameter of <math>S</math> is bounded by <math>M+2</math>.

'''Lemma: A closed subset of a compact set is compact.'''

Let <math>K</math> be a closed subset of a compact set <math>T</math> in <math>\mathbb{R}^n</math> and let <math>C_K</math> be an open cover of <math>K</math>. Then <math>U=\mathbb{R}^n\setminus K</math> is an open set and
 
<math display="block"> C_T = C_K \cup \{U\} </math>

is an open cover of <math>T</math>. Since <math>T</math> is compact, then <math>C_T</math> has a finite subcover <math> C_T'</math>, that also covers the smaller set <math>K</math>.  Since <math>U</math> does not contain any point of <math>K</math>, the set <math>K</math> is already covered by <math> C_K' = C_T' \setminus \{U\} </math>, that is a finite subcollection of the original collection <math>C_K</math>.  It is thus possible to extract from any open cover <math>C_K</math> of <math>K</math> a finite subcover.

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

If a set <math>S</math> in <math>\mathbb{R}^n</math> is bounded, then it can be enclosed within an <math>n</math>-box

<math display="block"> T_0 = [-a, a]^n</math>

where <math>a>0</math>. By the lemma above, it is enough to show that <math>T_0</math> is compact.

Assume, by way of contradiction, that <math>T_0</math> is not compact. Then there exists an infinite open cover <math>C</math> of <math>T_0</math> that does not admit any finite subcover. Through bisection of each of the sides of <math>T_0</math>, the box <math>T_0</math> can be broken up into <math>2^n</math> sub <math>n</math>-boxes, each of which has diameter equal to half the diameter of <math>T_0</math>. Then at least one of the <math>2^n</math> sections of <math>T_0</math> must require an infinite subcover of <math>C</math>, otherwise <math>C</math> itself would have a finite subcover, by uniting together the finite covers of the sections. Call this section <math>T_1</math>.

Likewise, the sides of <math>T_1</math> can be bisected, yielding <math>2^n</math> sections of <math>T_1</math>, at least one of which must require an infinite subcover of <math>C</math>. Continuing in like manner yields a decreasing sequence of nested <math>n</math>-boxes:

<math display="block"> T_0 \supset T_1 \supset T_2 \supset \ldots \supset T_k \supset \ldots </math>

where the side length of <math>T_k</math> is <math>(2a)/2^k</math>, which tends to 0 as <math>k</math> tends to infinity. Let us define a sequence <math>(x_k)</math> such that each <math>x_k</math> is in <math>T_k</math>. This sequence is [[Cauchy_sequence | Cauchy]], so it must converge to some limit <math>L</math>. Since each <math>T_k</math> is closed, and for each <math>k</math> the sequence <math>(x_k)</math> is eventually always inside <math>T_k</math>, we see that <math>L\in T_k</math> for each <math>k</math>.

Since <math>C</math> covers <math>T_0</math>, then it has some member <math>U\in C</math> such that <math>L\in U</math>. Since <math>U</math> is open, there is an <math>n</math>-ball <math>B(L)\subseteq U</math>. For large enough <math>k</math>, one has <math>T_k\subseteq B(L)\subseteq U</math>, but then the infinite number of members of <math>C</math> needed to cover <math>T_k</math> can be replaced by just one: <math>U</math>, a contradiction.

Thus, <math>T_0</math> is compact. Since <math>S</math> is closed and a subset of the compact set <math>T_0</math>, then <math>S</math> is also compact (see the lemma above).