A Study in Campi of Cities of Interior
3. Procedimentos metodológicos
4.3 Compact Sets
Compact sets are sets (usually infinite) which share many of the most important properties of finite sets. They play an important role in real analysis.
Definition 4.26 A set S ⊆ R is called compact if every sequence in S has a subsequence that converges to an element of S.
Theorem 4.27 (Heine-Borel) A set S ⊆ R is compact if and only if it is closed and bounded.
Proof: That a closed, bounded set has the property of compactness is the content of Corollary 4.25 and Proposition 4.12.
Now let S be a set that is compact. If S is not bounded, then there is an element s1 of S that has absolute value larger than 1. Also there must be an element s2 of S that has absolute value larger than 2. Continuing, we find elements sj ∈ S satisfying
|sj| > j
for each j. But then no subsequence of the sequence{sj} can be Cauchy. This contradiction shows that S must be bounded.
If S is compact but S is not closed, then there is a point x which is the limit of a sequence{sj} ⊆ S but which is not itself in S. But every sequence in S is, by definition of “compact,” supposed to have a subsequence converging to an element of S. For the sequence{sj} that we are considering, x is the only candidate for the limit of a subsequence. Thus it must be that x ∈ S. That contradiction establishes that S is closed.
In the abstract theory of topology (where there is no notion of distance), sequences cannot be used to characterize topological properties (instead there is a concept of nets, which we cannot treat here). Therefore a different definition of compactness is used. For interest’s sake, and for future use, we now show that the definition of compactness that we have been discussing is equivalent to the one used in topology theory. First we need a new definition.
Definition 4.28 Let S be a subset of the real numbers. A collection of open sets{Oα}α∈A(eachOαis an open set of real numbers) is called an open covering
of S if [
α∈A
Oα⊇ S . See Figure 4.13.
Example 4.29 The collection C = {(1/j, 1)}∞j=1 is an open covering of the interval I = (0, 1). No finite subcollection of the elements ofC covers I.
The collectionD = {(1/j, 1)}∞j=1∪ {(−1/5, 1/5), (4/5, 6/5)} is an open cov-ering of the interval J = [0, 1]. However, not all the elements D are actually needed to cover J. In fact
74 CHAPTER 4. BASIC TOPOLOGY
Figure 4.13: Open covers and compactness.
(−1/5, 1/5) , (1/6, 1) , (4/5, 6/5) cover the interval J.
It is the distinction displayed in this example that distinguishes compact sets from the point of view of topology. To understand the point, we need another definition:
Definition 4.30 If C is an open covering of a set S and if D is another open covering of S such that each element ofD is also an element of C then we call D a subcovering of C.
We callD a finite subcovering if D has just finitely many elements.
Example 4.31 The collection of intervals
C = {(j − 1, j + 1)}∞j=1
is an open covering of the set S = [5, 9]. The collection D = {(j − 1, j + 1)}∞j=5
is a subcovering.
However, the collection
E = {(4, 6), (5, 7), (6, 8), (7, 9), (8, 10)}
is a finite subcovering.
Theorem 4.32 A set S ⊆ R is compact if and only if every open covering C = {Oα}α∈A of S has a finite subcovering.
Proof: Assume that S is a compact set and let C = {Oα}α∈A be an open covering of S.
By Theorem 4.27, S is closed and bounded. Therefore it holds that a = inf S is a finite real number, and an element of S. Likewise, b = sup S is a finite real number and an element of S. Write I = [a, b]. The case a = b is trivial so we assume that a < b.
Set
A = {x ∈ I : C contains a finite subcover that covers S ∩ [a, x]} . ThenA is nonempty since a ∈ A. Let t = sup A. Then some element O0of C contains t. Let s be an element of O0to the left of t. Then, by the definition
4.3. COMPACT SETS 75 of t, s is an element of A. So there is a finite subcovering C′ of C that covers S∩ [a, s]. But then D = C′∪ {O0} covers S ∩ [a, t], showing that t = sup A lies in A. But in fact D even covers points to the right of t. Thus t cannot be the supremum ofA unless t = b.
We have learned that t must be the point b itself and that therefore b∈ A.
But that says that S∩[a, b] = S can be covered by finitely many of the elements ofC. That is what we wished to prove.
For the converse, assume that every open covering of S has a finite subcov-ering. Let {aj} be a sequence in S. Assume, seeking a contradiction, that the sequence has no subsequence that converges to an element of S. This must mean that for every s∈ S there is an ǫs> 0 such that no element of the sequence sat-isfies 0 <|aj− s| < ǫs. Let Is= (s− ǫs, s + ǫs). The collectionC = {Is} is then an open covering of the set S. By hypothesis, there exists a finite subcovering Is1, . . . Isk of open intervals that cover S. But each Isℓ could only contain at most one element of the sequence{aj}—namely, sℓitself. We conclude that the sequence has only finitely many distinct elements, a clear contradiction. Thus the sequence does have a convergent subsequence whose limit is in S.
Example 4.33 If A ⊆ B and both sets are nonempty then A ∩ B = A 6= ∅.
A similar assertion holds when intersecting finitely many nonempty sets A1 ⊇ A2⊇ · · · ⊇ Ak; it holds in this circumstance that∩kj=1Aj = Ak.
However, it is possible to have infinitely many nonempty nested sets with null intersection. An example is the sets Ij = (0, 1/j). Certainly Ij ⊇ Ij+1 for all j yet
\∞ j=1
Ij=∅ .
By contrast, if we take Kj= [0, 1/j] then
\∞ j=1
Kj ={0} .
The next proposition shows that compact sets have the intuitively appealing property of the Kjs rather than the unsettling property of the Ijs.
Proposition 4.34 Let
K1⊇ K2⊇ · · · ⊇ Kj ⊇ . . . be nonempty compact sets of real numbers. Set
K =
\∞ j=1
Kj.
ThenK is compact and K 6= ∅.
76 CHAPTER 4. BASIC TOPOLOGY Proof: Each Kj is closed and bounded henceK is closed and bounded. Thus K is compact. Let xj∈ Kj, each j. Then{xj} ⊆ K1. By compactness, there is a convergent subsequence{xjk} with limit x0∈ K1. However,{xjk}∞k=2⊆ K2. Thus x0∈ K2. Similar reasoning shows that x0 ∈ Kmfor all m = 1, 2, . . . . In conclusion, x0∈ ∩jKj=K.
Exercises
1. Let K be a compact set and let U be an open set that contains K. Prove that there is an ǫ > 0 such that, if k∈ K, then the interval (k − ǫ, k + ǫ) is contained in U .
2. Let K be compact and L closed, and assume that the two sets are disjoint.
Show that there is a positive distance between the two sets.
3. Let K be a compact set. Let δ > 0. Prove that there is a finite collection of intervals of radius δ that covers K.
4. Let K be a compact set. LetU = {Uj}kj=1be a finite covering of K. Show that there is a δ > 0 so that, if x is any point of K, then the disc or interval of center x and radius δ lies entirely in one of the Uj.
5. Prove that the intersection of a compact set and a closed set is compact.
6. Assume that we have intervals [a1, b1]⊇ [a2, b2]⊇ · · · and that limj→∞|aj− bj| = 0. Prove that there is a point x such that x ∈ [aj, bj] for every j.
7. If K in R is compact then show thatcK is not compact.
8. Prove that the intersection of any number of compact sets is compact.
The analogous statement for unions is false.
9. Let U ⊂ R be any open set. Show that there exist compact sets K1 ⊂ K2⊂ · · · so that ∪jKj = U .
10. Produce an open set U in the real line so that U may not be written as the decreasing intersection of compact sets.