Math 131AReal AnalysisQuiz 2Instructions: You have 50 minutes to complete the quiz. There are three problems, worth atotal of 40 points. You may not use any books or notes. Partial credit will be given for progresstoward correct proofs. Write your solutions in the space below the questions. If you need morespace use the back of the page. Do not forget to write your name and UID in the space below.Name:Student ID number:Question Points Score1 102 123 18Total: 40Problem 1. 10pts.Suppose S is a nonempty bounded subset of R. Then the set −S defined by−S := {−s : s ∈ S}is also a nonempty bounded subset of R.Prove that sup(S) ≤ −inf(−S).[In fact, they are equal but proving this is more work.]You can have 3 points for writing nothing.On the other hand, any severe logical fallacies which demonstrate writing down randomnonsense in the hope of scavenging points will result in an immediate 0. Demonstratingthat you understand relevant definitions, what you need to verify to answer the question,and clearly identifying any points at which you become stuck is a far better strategy!Solution: We will show that −inf(−S) is an upper bound for sup(S).Let s ∈ S. By definition of −S, we have −s ∈ −S.Since inf(−S) is a lower bound for −S, we have −s ≥ inf(−S). Thus, s ≤ −inf(−S).We have now demonstrated that −inf(−S) is an upper bound for S.Since sup S is the least upper bound we have sup(S) ≤ −inf(−S).Problem 2. 12pts.Suppose that (sn)∞n=1and (tn)∞n=1are sequences of real numbers, that∀n ∈ N, tn≥12,and that (sn) and (tn) both converge to 1.Prove that the sequencesntn∞n=1converges to 1.[For full credit, you CANNOT use the algebra of limits without proof. It is probablyquicker to prove the result directly.]You can have 3 points for writing nothing.On the other hand, any severe logical fallacies which demonstrate writing down randomnonsense in the hope of scavenging points will result in an immediate 0. Demonstratingthat you understand how to write proofs which verify the relevant definition, and clearlyidentifying any points at which you become stuck is a far better strategy!Solution:Let  > 0.Since (sn) converges to 1, there exists an N1so that n > N1implies |sn− 1| <4.Since (tn) converges to 1, there exists an N2so that n > N2implies |tn− 1| <4.Let N = max{N1, N2} and n > N. Thensntn− 1=sn− tntn≤ 2|sn− tn| ≤ 2|sn− 1| + |1 − tn|< 24+4= .Problem 3.Give counter-examples or find problems with the following statements.[Answers should be short and to the point; waffling will not be rewarded.](a) [3pts.] If S = {0}, then sup S = S.(b) [3pts.] If S is a nonempty bounded subset of R then sup S = max S.(c) [4pts.] If S and T are nonempty bounded subsets of R andST := {st : s ∈ S, t ∈ T },then ST is bounded andsup(ST ) = max{sup S sup T, inf S inf T}.(d) [4pts.] Let T be a nonempty bounded subset of R, andT0= {x ∈ R : there exists an S ⊆ T such that x = sup S}.Then T0⊆ T ∪ {sup T }.(e) [4pts.] If (sn) and (tn) are sequences of positive numbersand (√sn−√tn)∞n=1converges to 0, then (sn− tn)∞n=1converges to 0.Solution:(a) sup S is the number 0. S is the set containing 0.(b) Some sets S do not have maximum elements. S = [0, 1) is one such example.(c) Let S = {1, 2} and T = −S = {−2, −1}. ThenST = {−4, −2, −1}.So sup(ST ) = −1.But sup(S) sup(T ) = 2 ·(−1) = −2 and inf(S) inf(T ) = 1 ·(−2) = −2.(d) Let T = [0, 1) ∪ [2, 3) andT0= {x ∈ R : there exists an S ⊆ T such that x = sup S}.Let S = [0, 1). Then S ⊂ T and sup S = 1. So 1 ∈ T0, but 1 /∈ T ∪ {sup T }.(e) Let sn= n + 1 and tn= n.Then (sn− tn)∞n=1converges to 1 and 1 6= 0.Since√n + 1 −√n =1√n+1+√n, (√sn−√tn)∞n=1converges to