MATH 210 Exam 1 Solution, Fall 2016

Math 210, Introduction to Analysis- Fall 2016-2017 Quiz1 Solutions Question 1. (i) A ⊂ X is said to be compact if for every open cover {Oi}i of A, (A ⊂ � i Oi), there exists a finite subcover Oi1, Oi2, . . . , Oik s.t A ⊂ Oi1 ∪ Oi2 ∪ · · · ∪ Oik. (ii) We need to show that ∀ϵ > 0, ∃n0 ∈ N s.t n ≥ n0 =⇒ | n√ 5 − 1| < ϵ . However | n√ 5 − 1| = 5 − 2 n√ 5n−1 + n√ 5n−2 + · · · + n√ 5 + 1 ≤ 4 n. Then ∀ϵ > 0, ∃ n0 = 8 ϵs.tn ≥ n0 =⇒ | n√ 5 − 1| < ϵ. (iii) Let {Oi}i be an arbitrary open cover of A. Since 1 ∈ A ⊂ � i Oi then ∃i0 s.t 1 ∈ Oi0. Now Oi0 is open, then ∃r > 0 s.t B(1, r) = (1 − r, 1 + 1) ⊂ Oi0. Since n√ 5 −→ 1 by part (ii), then for ϵ = 1, ∃n0 s.t n > n0 =⇒ n√ 5 ∈ (1 − r, 1 + r) ⊂ Oi0. We have finitely many remaining terms : for every n = 1, . . . n0 ∃ in s.t n√ 5 ∈ Oin. Also, ∃iN s.t 5 ∈ OiN. We conclude A ⊂ Oi0 ∪ Oi1 ∪ Oi2 · · · ∪ Oin0 ∪ OiN. Thus there exists a finite subcover of A. Question 2. For n = 2k even, x2k = 1 + 2k√ 3 −→ 1 + 1 = 2 as k −→ ∞. For n = 2k + 1 odd, x2k = −1 + 2k√ 3 −→ −1 + 1 = 0 as k −→ ∞. We need to find the set of all sub sequential limits of {xn}. Let {xnk} be a subsequence of {xn} s.t xnk −→ l. But according to whether nk is even or odd, xnk will converge to 0 or 2, and by uniqueness of limit, l = 0 or l = 2. Then the set of all subsequential limits of {xn} is A = {0, 2}. so lim sup xn = sup A = 2, and lim inf xn = inf A = 0. Another method. Let yn = sup{xn, xn+1, . . . } = sup{xn, xn+1, . . . , x2n, x2n+1, . . . } = sup{(−1)n + n√ 3, (−1)n+1 n+1√ 3, . . . , 1 + 2n√ 3, −1 + 2n+1√ 3, . . . }. lim sup xn = lim n−→∞ yn = lim k−→∞ sup{1 + 2k√ 3, −1 + 2k+1√ 3, . . . } = 1 + 1 = 2. and Let zn = inf{xn, xn+1, . . . } = sup{xn, xn+1, . . . , x2n, x2n+1, . . . } = inf{(−1)n+ n√ 3, (−1)n+1 n+1√ 3, . . . 2n√ 3, −1 + 2n+1√ 3, . . . }. lim inf xn = lim n−→∞ yn = lim k−→∞ inf{1 + 2k√ 3, −1 + 2k+1√ 3, . . . } = −1 + 1 = 0. Question 3. a) S has the least upper bound property iff for every subset E ⊂ S that is bounded above, sup E exists and sup E ∈ S. b)A = [0, √z]. Let’s show first that √ 7 ̸ inQ: Suppose not, i.e √ 7 = p q s.t (p, q) = 1 then p2 = 7q2 thus 7|p2 =⇒ 7|p =⇒ p = 7k. Thus p2 = 49k2 = 7q2 =⇒ 7k2 = q2 =⇒ 7|q2 =⇒ 7|q because 7 is prime. Thus (p, q) ≥ 7, this is a contradiction. Hence √ 7 /∈ Q. Suppose α = sup A exists in Q, then α ̸= √ 7. If α < √ 7, and since Q is dense in R, then ( √ 7, α) ∩ Q ̸= φ thus ∃β ∈ (α, √ 7) ∩ Q thus β ∈ A and β > α = sup A. This is a contradiction. If α > √ 7 Since Q is dense in R then ( √ 7, α) ∩ Q ̸= φ then ∃γ ∈ ( √ 7, α) ∩ Q. Thus γ ∈ Q and γ > √ 7, then γ is an upper bound of A in Q and γ > α = sup A. This is again a contradiction. Hence sup A doesn’t exists in Q. In R. √ 7 is an upper bound of A in R then A is bounded above in R. We know that R has the least upper bound property thus α = sup A exists in R. Also sup A ≤ √ 7. As before, If 1 2 α < √ 7, and since Q is dense in R, then ( √ 7, α) ∩ Q ̸= φ thus ∃β ∈ (α, √ 7) ∩ Q thus β ∈ A and β > α = sup A. This is a contradiction. Then α = sup A = √ 7 in R. c) Q doesn’t have the least upper bound property since by b), the set A ⊂ Q is bounded above in Q (for example 3 is an upper bound for A in Q) and sup A doesn’t exist in Q. Question 4. We have lim n−→∞ sin n2 + (−1)nn n3 + 1 = 0and lim n−→∞ 2√n − 3 √n + 4 −→ 2 Proof. ∀ϵ > 0, find n0 s.t n ≥ n0 =⇒ d∞ � (sin n2 + (−1)nn n3 + 1 , 2√n − 3 √n + 4 ), (0, 2) � < ϵ i.e max ����� sin n2 + (−1)nn n3 + 1 − 0 ����, ���� 2√n − 3 √n + 4 − 2 ���� � < ϵ. But | sin n2+(−1)nn n3+1 − 0 ���� ≤ 1+n n3+1 ≤ 2n n3 = 2 n2. Pick n1 s.t 2 n2 1 < ϵ i.e n1 = �� 2 ϵ � + 1. Also ���� 2√n − 3 √n + 4 − 2 ���� = ���� 2√n − 3 − 2√n − 8 √n + 4 ���� ≤ 2|√n − 3 − √n| √n + 4 + 8 √n + 4 = 2. 6 (√n + 4)(√n − 3 + √n) + 8 √n + 4 ≤ 6 n + 8 √n = 14 √n. We want this to < ϵ, so take take n2 = � (14)2 ϵ2 � + 1, then ∀n ≥ n2, ���� 2√n−3 √n+4 − 2 ���� < ϵ. Put n0 = max(n1, n2) then ∀n ≥ n0, we have d∞ � (sin n2 + (−1)nn n3 + 1 , 2√n − 3 √n + 4 ), (0, 2) � < ϵ □ Question 5. a) i. x is a limit point of E ⊂ X iff ∀r > 0, B(x, r) � E\{x} ̸= φ. ii. E = E ∪ E′ where E′ is the set of all limit points of E. b) Claim: E = {(x, y) ∈ R2; x ≤ 1}: Let M(1, b) ∈ R2; Let’s show M ∈ E : ∀r > 0 consider the point N(x, b), we need N to be in Bd1 ∩ E : so take x < 1 and d1(M, N) < r i.e |x − 1| + |b − b| = |x − 1| < r so take 1 − r < x < 1 + r and x < 1 i.e 1 − r < x < 1, pick for example x = 1 − r 2 then N(1 − r 2, b) ∈ Bd1(M, r) ∩ E, hence the line (x = 1) ⊂ E. We’ll now show that any point M(a, b′) with a > 1 will not be in E: for r = a−1 2 , Bd1(M, r) ∩ E = φ since any point in B(M, r) will have its x-coordinate ≥ 1 + a−1 2 = 1 2 + a 2 > 1 s such M(a, b′); a > 1 is not in E. Thus E = {(x, y) ∈ R2; x ≤ 1}