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Chapter 3 The Real Numbers
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Section 3.5 Compact Sets
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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 (a) The interval S = (0, 2) is not compact. To see this, let An = (1/n, 3) for each n . If 0 x 2, then by the Archimedean property (c), there exists p such that 1/p x. ) ( A4 ) ( A3 ) ( A2 Thus x Ap and F = {An : n } is an open cover for S. ) ( A1 S | 1 2 3 ( ) But if G = is any finite subfamily of F , and if m = max {n1, …, nk}, then It follows that the finite subfamily G is not an open cover of (0, 2) and (0, 2) is not compact.
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Example 3.5.2 (b) Let S = {x1, …, xn} be a finite set and let F = {A: A } be any open cover of S. F x1 x2 x3 xn For each i = 1, …, n, there is a set from F that contains xi , since F is an open cover. It follows that the subfamily also covers S. We conclude that any finite set is compact. In proving a set is compact we must show that every open cover has a finite subcover. It is not sufficient to pick a particular open cover and extract a finite subcover. Because of this, it is often difficult to show directly that a given set satisfies the definition of being compact. Fortunately, the classical Heine-Borel theorem gives us a much easier characterization to use for subsets of .
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The Heine-Borel Theorem
Lemma 3.5.4 If S is a nonempty closed bounded subset of , then S has a maximum and a minimum. Proof: Since S is bounded 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 point of S, then for some 0 we have N *(m; ) S = . But m is the least upper bound of S, so N (m; ) S . (If this intersection were empty, then m – 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 = min S. Theorem 3.5.5 The Heine-Borel Theorem A subset S of is compact iff S is closed and bounded. Proof: First, let us suppose that S is compact. For each n , let In = ( n, n). Then each In is open and so {In : n N} is an open cover of S. Since S is compact, there exist finitely many integers n1, …, nk such that where m = max {n1,…, nk}. It follows that | x | < m for all x S, and S is bounded.
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The Heine-Borel Theorem
A subset S of is compact iff S is closed and bounded. Then there would exist a point p (cl S )\ S. Proof: Next, we assume that S is compact and suppose that S is not closed. For each n , we let Un = (– , p – 1/n ) ( p + 1/n, ). U3 = (– , p – 1/3 ) ( p + 1/3, ). ) ( ) ( U2 = (– , p – 1/2 ) ( p + 1/2, ). ) ( U1 = (– , p – 1 ) ( p + 1, ). ) [ | p S Now each Un is an open set and we have = \{p} S. Thus {Un : n } is an open cover of S. Since S is compact, there exist n1 < n2 < … < nk in such that Furthermore, the Un’s are nested. That is, Um Un if m n. It follows that But then S N ( p; 1/nk) = , contradicting our choice of p (cl S )\S and showing that S must be closed. So far, we have shown that if S is compact, then S is closed and bounded.
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Theorem 3.5.5 The Heine-Borel Theorem
A subset S of is compact iff S is closed and bounded. Proof: For the converse, we suppose that S is closed and bounded. To show that S is compact, let F be an open cover of S. For each x define Sx = {z S : z x} = S (– , x] and let B = {x : Sx is covered by a finite subcover of F }. Examples: [ ] Sz = S [ ] Sy [ ] Sx Sw = w [ ] S x y z d Since S is closed and bounded, Lemma implies that S has a minimum, say d. Then Sd = {d }, and this is certainly covered by a finite subcover of F . Thus d B and B is nonempty. If we can show that B is not bounded above, then it will contain a number z greater than sup S. But then Sz = S, and since z B, we can conclude that S is compact.
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Theorem 3.5.5 The Heine-Borel Theorem
A subset S of is compact iff S is 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 S both lead to contradictions. If m S, then since F is an open cover of S, there exists F0 in F such that m F0. F1, …, Fk ( ) F0 [ ] S x1 m x2 Since F0 is open, there exists an interval [x1, x2] in F0 such that x1 < m < x2. Since x1 < m and m = sup B, there exist F1, …, Fk in F that cover But then F0, F1, …, Fk cover , so that x2 B. This contradicts m = sup B.
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Theorem 3.5.5 The Heine-Borel Theorem
A subset S of is compact iff S is 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 not bounded above and let m = sup B. On the other hand, if m S, then since S is closed there exists an > 0 such that N (m; ) S = . ( ) N (m; ) [ ] S m But then Sm – = Sm + /2. Since m – B, we have m + /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.
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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 contains infinitely many points, then there exists at least one point in that is an accumulation point of S.
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Theorem 3.5.6 The Bolzano-Weierstrass Theorem
If a bounded subset S of contains infinitely many points, then there exists at least one point in 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 (a), so by the Heine-Borel Theorem 3.5.5, S is compact. Since S has no accumulation points, given 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, …, xn in S such that {N (x1), …, N (xn)} covers S. But so S = {x1, …, xn}. This contradicts S having infinitely many points.
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Theorem 3.5.7 Let F = {K : A } be a family of compact subsets of Suppose that the intersection of any finite subfamily of F is nonempty. Then {K : A } . Proof: For each A , let F = \ K . Since each K is compact, it is closed and its complement F is open. Choose a member K of F and suppose that no point of K belongs to every K . That is, the sets F form an open cover of K. Then every point of K belongs to some F . Since K is compact, there exist finitely many indices 1, …, n such that But by Exercise (d), so a contradiction. Thus some point in K belongs to each K , and {K : A } .
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[ ] [ ] [ ] [ ] Corollary 3.5.8 The Nested Intervals Theorem
Let F = {An : n } be a family of closed bounded intervals in such that An +1 An for all n Then [ ] A1 [ ] A2 [ ] A3 [ ] A4 Proof: Given any n1 < n2 < … < nk in , we have Thus Theorem implies that
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