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Compactness
Handout #6 – 3/11/
Defn. Suppose that K ⊆ IR. A collection G of open subsets such that
K ⊆
⋃ O∈G
O.
is called an open cover of K. K has a finite subcover from G if there exist O 1 , O 2 ,... , On in G for which
K ⊆
⋃n j=
Oj.
Defn. K is called compact, if each open cover G of K has a finite subcover.
Theorem. The continuous image of a compact set is compact. Proof. Suppose f : K → IR is continuous and K is compact. Each open cover C of f [K] can be drawn back to an open cover C˜ of K, by considering the sets
O˜ := f −^1 (O), O ∈ C.
K compact implies that we may draw a finite subcover from C˜. Each of these members is the inverse image (under f ) from a member of C. These form the desired subcover of f [K]. 2
Theorem. (Heine-Borel) Suppose that a ≤ b, then the interval [a, b] is compact. Proof. Let C be an open cover for [a, b] and consider the set
A := {x| [a, x] has an open cover from C}.
Note that A 6 = ∅ since a ∈ A. Let γ := lub(A). It is enough to show that γ > b, since if x 1 ∈ A and a ≤ x ≤ x 1 , then x ∈ A. Suppose instead that γ ≤ b, then there must be some O 0 ∈ C such that γ ∈ O 0. But O 0 is open, so there exists δ > 0 so that Nδ(γ) ⊆ O 0. Since γ is the least upper bound for A, then there is an x ∈ A such that γ − δ < x ≤ γ. But x ∈ A implies there are members O 1 ,... , On of C whose union covers [a, x]. The collection O 0 , O 1 ,... , On covers [a, γ + δ/2]. Contradiction, since γ is the least upper bound for the set A. 2
Theorem. Each closed subset C of a compact set K is compact. Proof. Let G˜ be an open cover for C. Let O 0 be the complement of C, then O 0 is open and G := G ∪ {O˜ 0 } is an open cover for K. There is a finite subcover of G which covers K and hence C. This subcover (dropping O 0 if it appears) is the desired finite subcover for C. 2
Defn. Suppose {an} is a sequence. A sequence {bk} is called a subsequence of {an} if there exists a strictly increasing sequence of natural numbers
n 1 < n 2 <... < nk <...
such that bk = ank , k = 1, 2 ,...
Theorem. Suppose that K ⊆ IR, then TFAE:
a.) K is compact, b.) K is closed and bounded, c.) each sequence in K has a subsequence which converges to a member of K, d.) (Bolzanno-Weierstrass) each infinite subset of K has a limit point in K.
Proof. (a) ⇒ (b) : To show that K is bounded, consider the open cover of K con- sisting of the collection of nested open intervals On := (−n, n), n ∈ IN. To show that K is closed, let x 0 be a limit point of K. Assume to the contrary that x 0 6 ∈ K. Consider the open cover of K consisting of the collection of nested open sets On := {x ∈ IR||x − x 0 | > 1 /n}, n ∈ IN. Any finite subcollection which would cover K would have union whose complement would be a neighborhood of x 0 not intersecting K. This shows that x 0 could not be a limit point of K. (b) ⇒ (d) : We use the ‘divide and conquer’ method, better known as the ‘bisection’ method. Let A be an infinite subset of K. Since K is bounded, there is an interval [a, b] such that K ⊆ [a, b]. Inductively define the closed subintervals as follows. Let [a 0 , b 0 ] := [a, b]. Either the left or right half of [a 0 , b 0 ] contains an infinite number of members of K. In the case that it is the right half, set [a 1 , b 1 ] := [(b 0 + a 0 )/ 2 , b 0 ]. Set [a 1 , b 1 ] equal to the left half of [a 0 , b 0 ] otherwise. Inductively, let [an+1, bn+1] be the half of [an, bn] which contains an infinite number of members of A. Notice that the length of this interval is (b − a)/ 2 n+1, that the an’s satisfy an ≤ an+1 ≤... < b and so must converge to some real number a ≤ x 0 ≤ b. Each neigborhood of x 0 will contain one of the intervals [an, bn] and hence will contain an infinite number of members of A, i.e. x 0 is a limit point of A. This also shows that x 0 is a limit point of the closed set K and must therefore belong to K. (d) ⇒ (c) : Let {xn}∞ n=1 be a sequence in K. If the sequence’s image is finite, then we may construct a constant subsequence which has the value which we may choose as any of the values of {xn}∞ n=1 which is repeated infinitely often. Otherwise, let A be the range of the sequence. Then A is an infinite subset of K. By the Bolzanno- Weierstrass property, A must have a limit point (x 0 say) which belongs to K. For each k ∈ IN , we may find an integer nk larger than those previously picked (i.e., n 1 ,... , nk− 1 ), so that |xnk − x 0 | < 1 /k. This is the desired subsequence.
