Math 131 A, Lecture 3 Real Analysis Sample Midterm 2 Instructions: You have 50 minutes to complete the exam. There are five problems, worth a total of fifty points. Write your solutions in the space below the questions. If a question is in multiple parts, clearly label each part. If you need more space use the back of the page. Do not forget to write your name in the space below. Name: Question Points Score 1 10 2 10 3 10 4 10 5 10 Total: 50 Problem 1. (a) [5pts.] Define a dominant term of a sequence (sn ). Solution: We say that sm is a dominant term of a sequence if sm > sn for all n > m. (b) [5pts.] Suppose that (sn ) has finitely many dominant terms. Prove that (sn ) contains an increasing subsequence (snk ). Solution: Let {sm1 , · · · sm` } be the finitely many dominant terms of (sn ). Let sn1 = sm` +1 . Then sn1 is not dominant, so there is some n2 > n1 such that sn1 ≤ sn2 . Similarly, sn2 is not dominant, so there is some n3 > n2 such that sn2 ≤ sn3 . We may use the same procedure inductively to choose the remainder of the subsequence. Problem 2. (a) [5pts.] Define lim sup sn . Solution: The limit supremum of (tn ) is lim sup tn = limN →∞ sup{tn : n > N }. (b) [5pts.] Suppose (sn ) is a bounded sequence such that sn ≥ 0 for all n. Prove that lim sup sn < +∞. Solution: Since (sn ) is bounded, there exists an M such that sn < M for all n. In particular, for any n ∈ N, we have sup{sn : n > N } ≤ M . Therefore lim sup sn = limN →∞ sup{sn : n > N } ≤ M . Problem 3. (a) [5pts.] State the Cauchy criterion for series. P Solution: The Cauchy criterion states that a series an convergesP if and only if for every > 0, there exists N such that n > m > N implies that | nk=m ak | < . P (b) [5pts.] Let (an ) be a sequence such that all an ≥ 0 and the series an converges. Suppose there is a finite subset F = {n1 , . . . , nr } ⊆ N such that the sequence (bn ) has the properties that |bnk | > ank for nk ∈ F and |bn | ≤ an for n 6∈ F . Prove that P bn converges. Solution: Let N1 = max F . Let > 0. Since (an ) converges, it satisfies the Cauchy criterion, so there is some N2 such that n > m > N2 implies P | Pnk=m ak | < P . Let N =Pmax{N1 , N2P }. Then n > m > P N implies that | nk=m bk | ≤ nk=m |bk | < nk=m ak = | nk=m ak | < . So bn also satisfies the Cauchy condition, hence converges. Problem 4. 10pts. Let f (x) = 0 when x is rational and f (x) = 1 when x is rational. Prove that f is discontinuous at every point on the reals. Solution: Let x be irrational. Then since every interval contains a rational number, for n ∈ N, we may choose xn ∈ (x − n1 , x) such that xn is rational. Then lim xn = x, but lim f (xn ) = lim 1 = 1 6= 0 = f (x). So f is discontinuous at x. A similar argument picking a sequence of irrational numbers shows that f is discontinuous at every rational point as well. Problem 5. (a) [5pts.] State the Bolzano-Weierstrauss Theorem. Solution: Every bounded sequence of real numbers has a convergent subsequence. (b) [5pts.] Given that (sn ) ⊆ [a, b] and lim sn = s implies s ∈ [a, b], prove that if f (x) is uniformly continuous on a bounded set S, then f is bounded on S. Solution: Suppose for the sake of contradiction that f is unbounded on S. Then for every n ∈ N, since f is unbounded on S, we may choose xn ∈ S such that f (xn ) > n. Now (xn ) is a bounded sequence since S is bounded, so by Bolzano-Weierstrauss, (xn ) has a convergent subsequence (xnk ). Since (xnk ) is convergent, it is Cauchy, so since f is uniformly continuous on S and uniformly convergent functions map Cauchy sequences to Cauchy sequences, (f (xnk )) is Cauchy. But f (xnk ) → ∞, a contradiction. So f must be bounded on S.