2.31 Definition By an open cover of a set E in a metric space X we mean a collection {G α } of open subsets of X such that E α G α.

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1 Chapter 2. Basic Topology. 2.3 Compact Sets Definition By an open cover of a set E in a metric space X we mean a collection {G α } of open subsets of X such that E α G α Definition A subset K of a metric space X is said to be compact if every open cover of K contains a finite subcover. More explicitly, the requirement is that if {G α } is an open cover of K, then there are finitely many indices α 1,...,α n such that K G α1... G αn. NOTE: Every finite set is compact. NOTE: We will say that K is compact relative to X if the requirements of Definition 2.32 are met. 1

2 2.33 Theorem Suppose K Y X. Then K is compact relative to X if and only if K is compact relative to Y. Proof Suppose K is compact relative to X, i.e., every open (relative to X) cover of K contains a finite subcover (relative to X). Let {G α } be an open (relative to Y ) cover of K. Then for every α, there exists an open subset, H α,ofx such that G α = H α Y. Since {G α } covers K, then K α G α = α (H α Y ) α H α, which implies that {H α } is an open (relative to X) cover of K. Since K is compact relative to X, there are α 1,...,α n such that Since K Y, K H α1... H αn. K (H α1... H αn ) Y =(H α1 Y )... (H αn Y ) = G α1... G αn. Then {G α1,...,g αn } is a finite subcover (relative to Y )ofk. 2

3 Now suppose that K is compact relative to Y, i.e., every open (relative to Y ) cover of K contains a finite subcover (relative to Y ). Let {V α } be an open (relative to X) cover of K. For every α, let W α = V α Y. Note that, for every α, W α is open relative to Y, and hence {W α } forms an open (relative to Y ) cover of K. Since K is compact relative to Y, then there exist α 1,...,α n such that {W α1,...,w αn } is a finite subcover (relative to Y )of K. Since W α1 V αn for i =1,...,n, then {V α1,...v αn forms a finite subcover (relative to X) of K. 3

4 2.34 Theorem Compact subsets of metric spaces are closed. Proof Suppose K is a compact subset of a metric space X. Let p K c. For every q K, let W q and V q be open neighborhoods of q and p, respectively, such that W q V q (we can do this by choosing the radius of each neighborhood to be less than d(p, q), which is nonzero since p q and X is a metric space. Note that {W q } is an open cover of K. Since K is compact, then there exist a finite number of points q 1,...,q n, such that {W q1,...,w qn } is a finite subcover of K. Since K W q1... W qn, which implies that (W q1... W qn ) c K c, and V αi W c α i for i =1,...,n, then V = V α1... V αn W c α 1... W c α n =(W α1... W αn ) c K c. Since V is a neighborhood of p contained in K c, then p is an interior point of K c. Since p was arbitrarily chosen from K c, then K c is open, which implies that K is closed. 4

5 2.35 Theorem Closed subsets of compact sets are compact. Proof Suppose F K X, F is closed (relative to X), and K is compact. Let {V α } be an open cover of F. Note that Ω = {V α } {F c } is an open cover of K. Since K is compact, there exists a finite subcover Φ of K. Note also that since F K, then Φ covers F. Thus, F is compact. Corollary If F is closed and K is compact, then F K is compact Theorem If {K α } is a collection of compact subsets of a metric space X such that the intersection of every finite subcollection of {K α } is nonempty, then K α is nonempty. Corollary If {K n } is a sequence of nonempty compact sets such that K n+1 K n (n =1, 2, 3,...), then n=1 K n is not empty. 5

6 2.37 Theorem If E is an infinite subset of a compact set K, then E has a limit point in K. Proof Suppose, for the purposes of contradiction, that E has no limit point in K. Then, for every q in K, there exists a neighborhood V q such that V q E {q}. {V q } is an open cover of K, but any finite subcollection of {V q } is then a subset of a finite set, which could not contain the infinite set K (K is infinite since it includes an infinite set, E). This contradicts the compactness of K Theorem If {I n } is a sequence of intervals in R 1, such that I n+1 I n (n =1, 2, 3,...), then n=1 is not empty { } Theorem Let k be a positive integer. If {In is a sequence of k-cells such that I n+1 I n (n =1, 2, 3,...), then n=1 is not empty. 6

7 2.40 Theorem Every k-cell is compact Theorem If a set E in R k has one of the following properties, then it has the other two: (a) E is closed and bounded (b) E is compact (c) Every infinite subset of E has a limit point in E Theorem (Weierstrass) Every bounded infinite subset of R k has a limit point in R k. 7

8 p. 44 #14 Give an example of an open cover of the segment (0, 1) which has no finite subcover. p. 44 #15 Show that Theorem 2.36 and its Corollary become false (in R 1, for example) if the word compact is replaced by closed or by bounded. 8

9 A union of finitely many compact sets is compact. 9

10 Say whether each of the following statements is true or false. (a) If K is a subset of a metric space, and some open covering of K has a finite subcovering, then K is compact. (b) If we regard the set of open segments {(x 1, x+1) x [ 10, 10]} as an open covering of the subset [ 10, 10] of the metric space R, then the set of open segments {(x 1 2,x+1 2 ) x [ 10, 10]} is a subcovering. (c) If we regard the set of open segments {(x 1,x +1) x [ 10, 10]} as an open covering of the subset [ 10, 10] of the metric space R, then the set of open segments {(x 1,x+ 1) x = 10, 9,...,0,...,9, 10} is a subcovering. (d) If we regard the set of open segments {(x 1,x+1) x [ 10, 10]} as an open covering of the subset [ 10, 10] of the metric space R, then the set of open segments {(x 1,x+ 1) x = 10, 8,...,0,...,8, 10} is a subcovering. 10