Chapter 3 1 Which of the following sets are compact? Which are connected? a. {(x1,x2) ∈ 2 | |x1| ≤ 1} Not compact (x2 is not bdd) Connected (since path-connected) b. {x ∈ n | ||x|| ≤ 10 } Compact (since bounded and closed in n) Connected (since path-connected) c. {x ∈ n | 1 ≤ ||x|| ≤ 2 } Compact (since bounded and closed in n) Connected d.  = {integers in } Not compact (since not bounded) Not connected (since there exist open sets U,V that separate , U = (-∞,1/2), V = (1/2,∞)) e. A finite set in  Compact (since bounded and closed in n) Not connected if more than one point f. {x ∈ n | ||x|| = 1} n = 1: Compact (since bounded and closed in n) Not connected (since there exist open sets U,V that separate the set) n ≥ 2: Compact (since bounded and closed in n) Connected (since path-connected) g. Perimeter of the unit square in 2 Compact (since bounded and closed in n) Connected (since path-connected) h. The boundary of a bounded set in  Compact (since bounded and closed in ) Not necessarily connected (depends on the set) i. The rationals in [0,1] Not compact (since not bounded) Not connected (since there exist open sets U,V that separate the set, U = [0,π/4), V = (π/4,1] ) j. A closed set in [0,1] Compact (since bounded and closed in n)Not necessarily connected (depends on the set) 2 Prove that a set A ⊂ n is not connected iff we can write A ⊂ F1 ∪ F2, where F1, F2 are closed, A ∩ F1 ∩ F2 = ∅, F1 ∩ A ∫ ∅, F2 ∩ A ∫ ∅. “⇒” A is not connected U1 = n\F1 U2 = n\F2 Then A ∩ U1 ∩ U2 = ∅, U1 ∩ A ∫ ∅, U2 ∩ A ∫ ∅. And U1, U2 are open. Hence, F1,F2 are closed. “›” Suppose A is connected F1 = n \ U1 F2 = n\U2 Then A ∩ F1 ∩ F2 = ∅, F1 ∩ A ∫ ∅, F2 ∩ A ∫ ∅. And U1, U2 are open. Hence, F1,F2 are closed. 6 Suppose that Fk is a sequence of compact nonempty sets satisfying the nested set property such that diameter(Fk) → 0 as k → ∞. Show that there is exactly one point in ∩{Fk}. (By definition, diameter(Fk) = sup{ d(x,y) | x,y ∈ Fk } ) If diameter(Fk) = 0 (as when k → ∞) then sup{ d(x,y) | x,y ∈ Fk } = 0 ⇒ d(x,y) = 0 ⇒ x = y ⇒ Fk contains only one point since the maximum distance from this point is 0 Since all Fk contain this point and Fk as k → 0 contains only this point, the intersection of all these sequences consists of only this points. ⇒ ∩{Fk} = one point 9 Determine the truth or falsity of the following statements: a. (A is compact in n) ⇒ (n\A is connected) False, A = [0,1] is compact \[0,1] is not connected b. (A is connected in n) ⇒ (n\A is connected) False, A = [0,1] is connected \[0,1] is not connected c. (A is connected in n) ⇒ (A is open or closed) False, A = [0,1) is connected A is neither open nor closed d. (A = {x ∈ n | ||x|| < 1 }) ⇒ (n\A is connected) For : False, A = [-1,1]n \ A is not connected For n: True, A = {x ∈ n | ||x|| < 1 }) Suppose n\A is not connected and therefore that there exist U,V that separate n\A. Then there exists x ∈ bd(U ∩ n\A) with ||x|| ≥ 1. Since ||x|| ≥ 1 we get that x ∈ n\A. Then x – U ∩ n\A and since x ∈ n\A we have that x – U. But since x ∈ n\A we also have that x ∈ V. Since V is open there exists ε > 0 such that D(x,ε) ⊂ V. Since x ∈ bd(U ∩ n\A), D(x,∈) ∩ (U ∩ n\A) ∫ ∅, so there exists y ∈ (U ∩ n\A) ∩ (V ∩ n\A) which is a contradiction. ⇒ A is connected. 20 Prove that a compact subset of a metric space must be closed as follows: Let x be in the complement of A. For each y ∈ A, choose disjoint neighborhoods Uy of y and Vy of x. Consider the open cover {Uy}y∈A of A to show the complement of A is open Let Uy = D(y,d(x,y)2 ), Vy = D(x, d(x,y)2 ). Since A is compact, { Uy | y ∈ A } is an open cover. Since Uy has a finite subcover, let r = inf{ { d(Vy,Uy) | ∀y ∈ A } } Then D(x,r2 ) ⊂ M\A which implies that M\A is open and therefore A is closed. 23 Let  denote the rationals in . Show that both  and the irrationals \ are not connected.  ⊂ (-∞,π) ∪ (π,∞) 1.  ∩ (-∞,π) ∪ (π,∞) = ∅ 2.  ∩ (-∞,π) ∫ ∅ 3.  ∩ (π,∞)∫ ∅ 4.  ⊂ (-∞,π) ∪ (π,∞) Therefore (-∞,π) and (π,∞) are separated and  is not connected. \ ⊂ (-∞,1) ∪ (1,∞) 1. \ ∩ (-∞,1) ∪ (1,∞) = ∅ 2. \ ∩ (-∞,1) ∫ ∅ 3. \ ∩ (1,∞)∫ ∅ 4. \ ⊂ (-∞,1) ∪ (1,∞) Therefore (-∞,1) and (π,1) are separated and \ is not connected. 25 Prove that there is a sequence of distinct integers n1,n2,… → ∞ such that limk→∞ sin nk exists.sin nk is contained in the compact set [-1,1]. Since [-1,1] is bounded and closed, it is compact. Since it is compact, it is sequentially compact by the Bolzano-Weierstrass theorem. Since it is sequentially compact it has a convergent subsequence such that limk→∞ sin nk converges. 28 Let A ⊂ M be connected and contain more than one point. Show that every point of A is an accumulation point of A. Assume x ∈ A is not an accumulation point. Then there exists an open set U containing no other point of A than x. Then A would be separated since for another open set V = A\U it would satisfy the conditions that A ∩ V ∩ U = ∅, A ∩ V ∫ ∅, A ∩ U ∫ ∅ and A ⊂ U ∪ V. This is a contradiction since we know that the set A ⊂ M is connected. Hence, every point of A is an accumulation point of