If I 2 S is finite switch the names of I 2 and J 2 Now I 2 S has infinitely

# If i 2 s is finite switch the names of i 2 and j 2

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a contradiction). If I 2 S is finite, switch the names of I 2 and J 2 . Now I 2 S has infinitely many points. Divide I 2 into two equal intervals I 3 and J 3 . Continue in this way, producing a sequence of nested closed intervals I 1 I 2 I 3 . . . such that I n S has infinitely many elements n N . Write I n = [ a n , b n ]. By the nested intervals theorem, intersectiontext n =1 I n negationslash = . Let x intersectiontext n =1 I n . Claim: x is an accumulation point of S . To see this, given any ǫ > 0 , n N s.t. 1 n < ǫ 2 K (using (c) of the Archimedian property). Now n 2 n by mathematical induction, so 1 2 n 1 n < ǫ 2 K . Hence 2 K 2 n < ǫ . Since the length of [ a n , b n ] is 2 K 2 n , we have b n = a n + 2 K 2 n < a n + ǫ x + ǫ, (recall that x [ a n , b n ]). Similarly, x ǫ b n ǫ < b n 2 K 2 n = a n . So [ a n , b n ] ( x ǫ, x + ǫ ). Since [ a n , b n ] contains infinitely many points in S , so does ( x ǫ, x + ǫ ). Since ǫ > 0 was arbitrary, x is an accumulation point of S by Lemma 3.22. square Homework 3/4. (Go by the webpage for the precise numbers and due dates) (1) Show that the interior of a set is open. Also show that Int( S ) is the largest open set contained in S . (2) Show that ¯ S is the smallest closed set containing S . (3) Homework 3.5 Exercise 1 from the textbook 5th Ed (Exercise 14.1 in 4th Edition). (4) Explain why each of the following sets are not compact: (a) [1 , 3), (b) N , (c) { 1 /n : n N } , (d) { x Q : 0 x 2 } . (5) Which of the 9 types of intervals are compact? (6) Show that any isolated point of S is in Bdy(S). (7) Show that the closure of a bounded set is bounded. (8) Read the proofs above in this Section which we skipped in class. END OF WORK FOR TEST 2 Appendix to Chapter 3. Proof of the Heine-Borel theorem. THE PROOF BELOW IS NOT EXAMINABLE We define an open cover of a set S to be a collection C of open sets whose union contains S . A finite subcover of S from C is a finite collection { C 1 , C 2 , · · · , C n } of
MATH 3333–INTERMEDIATE ANALYSIS–BLECHER NOTES 31 the sets in C , whose union contains S . That is, C k ∈ C for all k , and S ⊂ ∪ n k =1 C k . Note that if C 1 C 2 ⊂ · · · ⊂ C n then C n = n k =1 C k , so S C n . Example. Let S = [0 , 1] and let C = { ( 1 50 , 1 50 ) , ( 1 2 , 3 2 ) , ( 1 4 , 3 4 ) , ( 1 8 , 3 8 ) , ( 1 16 , 3 16 ) , ( 1 32 , 3 32 ) , ( 1 64 , 3 64 ) , ( 1 128 , 3 128 ) , · · · } . It is easy to see that in this last example, C is an infinite collection of open sets whose union contains S . That is, C is an open cover of S . But in fact the first 7 sets listed in C have a union which already contains S . These 7 sets comprise a finite subcover. This example also illustrates the condition in (2) of the Heine-Borel Theorem. Theorem 3.30. (The Heine-Borel theorem) If S is a nonempty set of real numbers then the following are equivalent: (1) S is closed and bounded. (2) S has the following property: whenever C is a collection of open sets whose union contains S , then actually S is contained in the union of a finite number of the sets in C . (In English: every open cover of S has a finite subcover.) Proof. (2) (1) Let C = { ( n, n ) : n N } . By (2), this collection has a finite

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