16 Questions on Analysis I with Solution - Final Exam | MATH 554, Exams of Mathematics

Download 16 Questions on Analysis I with Solution - Final Exam | MATH 554 and more Exams Mathematics in PDF only on Docsity! Math 554, Final Exam Summer 2004 Write your answers as legibly as you can on the blank sheets of paper provided. Use only one side of each sheet. Take enough space for each problem. Turn in your solutions in the order: problem 1, problem 2, . . . ; although, by using enough paper, you can do the problems in any order that suits you. There are 16 problems. Problems 1, 2, 3, and 4 are worth 7 points each. Problems 5 through 16 are worth 6 points each. The exam is worth a total of 100 points. If I know your e-mail address, I will e-mail your grade to you. If I don’t already know your e-mail address and you want me to know it, then send me an e-mail. Otherwise, get your course grade from VIP. I will post the solutions on my website shortly after the class is finished. 1. Define Cauchy sequence. Use complete sentences. The sequence {an} is a a Cauchy sequence if for all ε > 0 , there exists n0 such that whenever n,m > n0 , then |an − am| < ε . 2. Let f : E → R be a function which is defined on a subset E of R . Define lim x→p f(x) = L . Use complete sentences. (Be sure to tell me what kind of a thing p is, and what kind of a thing L is.) Let f : E → R be a function which is defined on a subset E of R . Assume that p is a limit point of E and L is a real number. We say that lim x→p f(x) = L if for all ε > 0 , there exists δ > 0 such that whenever |x− p| < δ , x 6= p , and x ∈ E , then |f(x) − L| ≤ ε . 3. Define continuous. Use complete sentences. Let E be a subset of R . The function f : E → R is continuous at the point p of E , if, for all ε > 0 , there exists δ > 0 , such that whenever |x − p| < δ and x ∈ E , then |f(x) − f(p)| < ε . 4. STATE either version of the Bolzano-Weierstrass Theorem. (version 1.) Every bounded infinite set of real numbers has a limit point in R . (version 2.) Every bounded sequence of real numbers has a convergent subsequence. 5. PROVE either version of the Bolzano-Weierstrass Theorem. Proof of version 1. Let S be a bounded infinite subset of R , and let I be a finite closed interval which contains S . Cut I in half. At least one of the resulting two closed subintervals of I contains infinitely many elements of S . Call this interval I1 . Continue in this manner to build the closed interval In , for each natural number n , with the length of In equal to 1/2n times the length of I and In contains infinitely many elements of S . The nested interval property of R tells us that the intersection 2 ∞⋂ n=1 In is non-empty. Let p be an element of ∞⋂ n=1 In . We will show that p is a limit point of S . Given ε > 0 , there exists n large enough that the length of In is less than ε . We know that p ∈ In . It follows that In ⊆ Nε(p) . Furthermore, there is at least one element q of S with q 6= p and q ∈ Nε(p) ; since In ∩ S is infinite. Proof of version 2. Let {an} be a bounded sequence of real numbers. There are two cases to consider depending upon the cardinality of the set {an | n ∈ N} . Case 1: {an | n ∈ N} is finite. In this case, it is clear that some subsequence of {an} is constant. Case 2: {an | n ∈ N} is infinite. We apply version 1 of the Bolzano-Weierstrass Theorem. The set {an | n ∈ N} has a limit point p . Pick n1 with |an1 − p| < 1 . Pick n2 > n1 with |an1−p| < 12 . Every open neighborhood of p contains infinitely many elements of {an | n ∈ N} . We continue in this manner to pick nk > nk−1 with |ank − p| < 1k . We see that the subsequence {ank} of the sequence {an} converges to p . 6. PROVE that every Cauchy sequence converges. Let {an} be a Cauchy sequence. It is easy to see that {an} is bounded. Indeed, there exists n0 with |an − an0 | < 1 for all n > n0 . In this case, |an| ≤ M = max{|an0 | + 1, |a1|, . . . , |an0−1|} for all n . Version 2 of the Bolzano- Weierstrass Theorem guarantees the existence of a convergent subsequence {ank} of {an} . Let a be the limit of the subsequence {ank} . We will prove that the entire sequence {an} converges to a . Let ε > 0 be fixed, but arbitrary. The subsequence {ank} converges to a ; so, there exists k0 such that, whenever k ≥ k0 , then |ank − a| < ε2 . The sequence {an} is a Cauchy sequence; so, there exists n1 such that, whenever n,m ≥ n1 , then |an − am| ≤ ε2 . Pick n0 ≥ max{n1, nk0} . If n > n0 , then we may choose k with k > k0 and nk > n0 . We now have: |an − a| ≤ |an − ank | + |ank − a| ≤ ε 2 + ε 2 = ε. 7. Let E be a set which is not closed. PROVE that E is not compact by constructing an open cover of E which does not admit a finite subcover. The set E is not closed; so, some limit point p of E is not contained in E . We construct a sequence {pn} in E which converges to p and which has no other limit points. Pick p1 ∈ E with |p1 − p| < 1 . Once pn has been found, pick pn+1 in E with |pn+1−p| < 12 |pn−p| . The fact that p is a limit point of E guarantees the existence of pn+1 . Let S be the set {pn | n ∈ N} . We have constructed pn in a manner which guarantees that S is infinite and the only limit point of S is p . For each x ∈ E , x is not a limit point of S , so we may pick εx so that Nεx(x) contains at most one element of S . Let U = {Nεx(x) | x ∈ E} . It is clear that U is an open cover of E . It is also clear that no finite subset of U can cover E ; since a finite subset of U can cover only a finite subset of the infinite set S .