Midterm Exam with Problem Solution - Real Analysis | MATH 4100, Exams of Quantitative Techniques

Download Midterm Exam with Problem Solution - Real Analysis | MATH 4100 and more Exams Quantitative Techniques in PDF only on Docsity! MATH 4100/6100 MIDTERM EXAM Throughout, we let X be a metric space with metric d. Problem 1. Let p ∈ X and r > 0. Show that {x : d(x, p) > r} is open. Solution Put U := {x : d(x, p) > r}, and suppose x ∈ U . Put  := r − d(x, p) > 0. We show that N(x) ⊂ U ; since this will hold for every x ∈ U it will follow that U is open. Given y ∈ N(x), the triangle inequality yields d(y, p) ≤ d(x, p) + d(x, y) < d(x, p) +  = d(x, p) + (r − d(x, p) = r. Thus y ∈ U , as claimed. Problem 2. Let (an), (bn), (cn) be Cauchy sequences in X . Prove that a) limn→∞ d(an, bn) exists b) limn→∞ d(an, cn) ≤ limn→∞ d(an, bn) + limn→∞ d(bn, cn). Solution a) We will show that d(an, bn) is itself a Cauchy sequence of real numbers, and there- fore converges. Let  > 0, and let N ∈ N be large enough that m,n ≥ N =⇒ d(am, an) <  2 and d(bm, bn) <  2 . Then for m,n ≥ N , the triangle inequality yields d(am, bm) ≤ d(am, an) + d(an, bn) + d(bn, bm) <  2 + d(an, bn) +  2 = d(an, bn) + , and similarly d(an, bn) < d(am, bm) + . Therefore |d(am, bm)− d(an, bn)| < . b) By the triangle inequality d(an, cn) ≤ d(an, bn) + d(bn, cn) for every n ∈ N. Since all of the limits exist by part a), lim n→∞ d(an, cn) ≤ lim n→∞ (d(an, bn) + d(bn, cn)) = lim n→∞ d(an, bn) + lim n→∞ d(bn, cn) as claimed. Problem 3. Let (xn) be a sequence in R, and suppose that lim supxn < c. Prove that there is N ∈ N such that n ≥ N =⇒ xn < c. 1