Chapter 2 7 Let U be open in a metric space M. Show that U =cl(U) \ bd(U). Is this true for every set in M? Since U is open, M\U is closed and U ⊂ cl(U) but U ∫ cl(U) Then U ⊂ cl(U) \ cl(U) ∩ cl(M\U) = cl(U) \ bd(U) If x ∈ cl(U) \ U then since U is open, ∀x ∈ cl(U) \U is an accumulation point of U that is not in U. ⇒ x ∈ M\U since M\U is closed and M\U = cl(M\U) So cl(U) ∩ cl(M\U) is the set of those accumulation points of U ⇒ cl(U) \ (cl(U) ∩ cl(M\U)) = cl(U) \ bd(U) ⊂ A Hence, the open set U = cl(U) \ bd(U) This is not necessarily true for closed set such as [0,1] 8 Let S ⊂  be nonempty, bounded below, and closed. Show that inf(S) ∈ S. By the definition of inf and of accumulation points we have that: ∀ε > 0 ∃s ∈ S such that |s – inf(S)| < ε So inf(S) is an accumulation point of S. Since the set is closed, it contains its accumulation points by Theorem 2.4.2. Hence, inf(S) ∈ S. 10 Determine which of the following statements are true. a int(cl(A)) =int(A) False. Counterexample: int(cl()) =  but int() = ∅ b cl(A) ∩ A = A True, since cl(A) = A ∪ { accumulation points of A} and A ∪ { accumulation points of A} ∩ A = A c cl(int(A)) = A False, if A is open then cl(int(A)) would be closed. d bd(cl(A)) = bd(A) False. Counterexample: bd(cl()) = bd() = ∅ but bd() =  e If A is open, then bd(A) ⊂ M \ A True, since M\A is closed cl(M\A) = M\A and then bd(A) = cl(A) ∩ cl(M\A) = cl(A) ∩ M\A and thus bd(A) ⊂ M\A 27 Suppose an ≥ 0 and an → 0 as n → ∞. Given any ε > 0, show that there is a subsequence bn of an such that ε<∑∞=1nnb .There exists a subsequence bn such that max{ b1,..,bn} < ε2n ⇒ 01122εεε==<∑∑∞=∞= nnnnb 28 Give examples of a An infinite set in  with no accumulation points  b A nonempty subset of  that is contained in its set of accumulation points (0,1) ⊂ [0,1] c A subset of  that has infinitely many accumulation points but contains none of them A = { n + 1k | n = 1,2,3…., k = 2,3,4….} d A set A such that bd(A) = cl(A) A = {1} 31 Let A’ denote the set of accumulation points of a set A. Prove that A’ is closed. Is (A’)’ = A’ for all A? A is closed ⇔ the accumulation points of A belong to A ⇒ If A’ does not belong to A, A is open M\cl(A) is open and cl(A) = A ∪ { accumulation points of A } ⇒ A = cl(A) ⇔ A is closed If M\cl(A) is open and A\A’ is open then M∪A\A’ is open so A’ is closed. Therefore, (A’)’ = A’. However, this is not true for all sets, take A = { 1n | n ∈  } 39 Let S ⊂  be bounded above and below. Prove that sup(S) – inf(S) = sup{ x-y | x ∈ S and y ∈ S}. sup(S) – inf(S) = sup{ x – y | x ∈ S and y ∈ S } u = sup(S) v = inf(S) “⇒” u ≥ x, ∀x ∈ S v ≤ y, ∀y ∈ S ⇒ u – v ≥ x – y ∀x,y ∈ S ⇒ u – v is an upper bound of { x – y | y ∈ S and x ∈ S } ⇒ u – v ≥ sup(x-y) ⇒ sup(S) – inf(S) ≥ sup{x-y | x ∈ S and y ∈ S }“›” Since sup{x –y | x ∈ S and y ∈ S } ⇒ x is an upper bound of S ⇒ y is a lower bound of S ⇒ x ≥ sup(S), y ≤ inf(S) ⇒ sup(S) –inf(S) = sup{ x – y | x ∈ S and y ∈ S } 43 Let x1 = 3 ,….,xn = 3 + xn-1 . Compute limn→∞ xn. x1 = 3, xn = 3+xn-1 lim xn = ()21321131213*41121+=+=++ 52 Test the following series for convergence a ∑∞=−+01kkke kkeke−−≤−1 ∑∞=−0kke is geometric and converges. Hence, the original series converges as well by the comparison test. b ∑∞=+021kkk ∑∑∞=∞=>+⇒+>+00222212111kkkkkkkkkk By the comparison the series diverges. c ∑∞=+−+02131knnn converges. series original thehence and converges 22 test,series-p By the2 3, n for 013n since 1313133232222∑∞=≤>≥+−+−<+−+knnnnnnnnn d ∑∞=−−+01)/2(tan)log()1log(kkkk ∞=++=++=+∞→∞→∞→kkkkkkkkkkkkk23322)4(8lim8/4111/111lim)/2arctan()/11log(lim Hence, the series is divergent.e ∑∞=−>00 real, ),sin(knαα Using the p-series test we get ∑∑∞=−∞=−>NnNnnn )sin(21αα and the series diverges. f ∑∞=033knn By the ratio test: 13131)1(33)1(33133<→+=++nnnnnn Hence, the series