Chapter The Real Numbers Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.5, Slide Section 3.5 Compact Sets Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.5, Slide Definition 3.5.1 A set S is said to be compact if whenever it is contained in the union of a family F of open sets, then it is contained in the union of some finite number of the sets in F If F is a family of open sets whose union contains S, then F is called an open cover of S If G ⊆ F and G is also an open cover of S, then G is called a subcover of S Thus S is compact iff every open cover of S contains a finite subcover Example 3.5.2 To see this, let An = (1/n, 3) for each n ∈ (a) The interval S = (0, 2) is not compact • • • • • • ( ( ( S ( ) A4 ) A3 ) A2 ) A1 |( | )| | { ABut G, A=nk } n1 , if If 0  we have N *(m; ε) ∩ S = ∅ (If this intersection were empty, But m is the least upper bound of S, so N (m; ε) ∩ S ≠ ∅ then m – ε would be an upper bound of S.) Together these imply m ∈ S, so again we have m = max S Similarly, inf S ∈ S, so inf S = S ♦ Theorem 3.5.5 The Heine-Borel Theorem A subset S of is compact iff S is closed and bounded For each n ∈ Proof: First, let us suppose that S is compact S ⊆ Then each I is open and n ∞ Un =1 I n , , let In = (− n, n) so {I  : n ∈ N} is an open cover of S n Since S is compact, there exist finitely many integers n1, …, nk such that S ⊆ ( I n1 ∪ L ∪ I nk ) = I m , where m = max {n ,…, nk} Copyright © 2013, 2005, 2001 Pearson Education, Inc It follows that | x | < m for all x ∈ S, and S is bounded Section 3.5, Slide Theorem 3.5.5 The Heine-Borel Theorem A subset S of is compact iff S is closed and bounded Proof: Next, we assume that S is compact and suppose that S is not closed For each n ∈ would exist a point p ∈ (cl S )\ S ) Then there , we let U = (– ∞, p – 1/n ) ∪ ( p + 1/n, ∞) n U = (– ∞, p – 1/3 ) ∪ ( p + 1/3, ∞) ( ) U = (– ∞, p – 1/2 ) ∪ ( p + 1/2, ∞) ( U = (– ∞, p – ) ∪ ( p + 1, ∞) ( ) )| [ S p Now each Un is an open set and we have ∞ Un =1U n = \{p} ⊇ S Thus {Un : n ∈ } is an open cover of S Since S is compact, there exist n1 < n2 such that N  (m;  ε ) ∩ S = ∅ N (m; ε ) ) ( [ S ] • [ S ] m But then Sm – ε = Sm + ε/2 Since m – ε ∈ B, we have m + ε /2 ∈ B, which again contradicts m = sup B Since the possibility that B is bounded above leads to a contradiction, we must conclude that B is not bounded above, and hence S is compact ♦ Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.5, Slide In Example 3.4.15 we showed that a finite set will have no accumulation points We also saw that some unbounded sets (such as ) have no accumulation points As an application of the Heine-Borel theorem, we now derive the classical Bolzano-Weierstrass theorem, which states that these are the only conditions that can allow a set to have no accumulation points Theorem 3.5.6 The Bolzano-Weierstrass Theorem If a bounded subset S of at least one point in Copyright © 2013, 2005, 2001 Pearson Education, Inc contains infinitely many points, then there exists that is an accumulation point of S Section 3.5, Slide 10 Theorem 3.5.6 The Bolzano-Weierstrass Theorem If a bounded subset S of one point in contains infinitely many points, then there exists at least that is an accumulation point of S Proof: Let S be a bounded subset of containing infinitely many points and suppose that S has no accumulation points Then S is closed by Theorem 3.4.17(a), so by the Since S has no accumulation points, given Heine-Borel Theorem 3.5.5, S is compact any x ∈ S, there exists a neighborhood N  (x) of x such that S ∩ N (x) = {x} N(x) ( S ) ( • ) •x ( ) ( • Now the family {N ) •  (x) : x ( ) • ( ) • ••• ••• ∈ S} is an open cover of S, and since S is compact there exist x1, …, x n in S such that {N (x1), …, N (xn)} covers S But S ∩ [ N ( x1 ) ∪ ×××∪ N ( xn )] = {x1, K , xn }, so S = {x1, …, xn} This contradicts S having infinitely many points ♦ Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 3.5, Slide 11 Theorem 3.5.7 Let F = {Kα : α ∈  A } be a family of compact subsets of Suppose that the intersection of any finite subfamily of F Proof: For each α ∈ A , let Fα = its complement Fα is open is nonempty Then ∩ {Kα : α ∈ A } ≠ ∅ \ Kα Since each Kα is compact, it is closed and Choose a member K of F and suppose that no point of K belongs to every Kα Then every point of K belongs to some Fα That is, the sets Fα form an open cover ofSince K K is compact, there exist finitely α1,  …,  α K ⊆ ( Fα1 ∪ ×××∪ Fα n ) n such that But Fα1 ∪ ×××∪ Fα n = ( = by \ Kα1 ) ∪ ×××∪ ( \ Kα n ) \ ( Kα1 ∩ ×××∩ Kα n ), K ∩ ( Kα1 ∩ ×××∩ Kα n ) = ∅, Exercise 2.1.26(d), so Thus some point in K belongs to each Kα , and Copyright © 2013, 2005, 2001 Pearson Education, Inc many indices a contradiction   {Kα : α ∈ A } ≠ ∅ ♦ Section 3.5, Slide 12 Corollary 3.5.8 The Nested Intervals Theorem Let F = {An : n ∈  } be a family of closed bounded intervals in An +1 ⊆ An for all n ∈ ∞ I A n =1 n Then ≠ ∅ [ A1 ] [ A2 such that ] [ A3 ] [ A4 ] • • • Proof: Given any n1 < n2 < … < nk in Thus Theorem 3.5.7 implies that Copyright © 2013, 2005, 2001 Pearson Education, Inc I , we have ∞ A n =1♦ n I k A i =1 ni = Ank ≠ ∅ ≠ ∅ Section 3.5, Slide 13 ... F = {An : n ∈ is any finite subfamily of F , and if An1 ∪ L ∪ Ank = Am = It follows that the finite subfamily G Copyright © 2013, 2005, 2001 Pearson Education, Inc Thus x ∈ Ap and } is an open... above and nonempty, m = sup S exists by the completeness axiom We want to show that m ∈ S If m is an accumulation point of S, then since S is closed, we have m ∈ S and m = max S If m is not an accumulation... closed and bounded Proof: We have Sx = {z ∈ S : z ≤ x} and B = {x : Sx is covered by a finite subcover of F } Suppose that B is bounded above and let m = sup B We shall show that m ∈ S and m ∉