4100fnlnotes - MATH 4100/6100(Azoff Final Exam Notes Fall...

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Unformatted text preview: MATH 4100/6100 (Azoff) Final Exam Notes Fall 2005 Throughout this exam (X , d) is a metric space; R is equipped with the usual metric. 1 (22 points). Statements a) Define what it means for an ordered set (S, g) to have the least upper bound property. b) Define what it means for a function f : X a X to be continuous at a point p E X. 0) Present a statement involving sequences which is equivalent to discontinuity of f : X a X at the point p E X. Answer. a) Every non—empty subset of S which has an upper bound also has a least upper bound. b) For every 6 > 0, there is a 6 > 0 such that d(f(:L'), f(p)) < 6 whenever d(a:,p) < 6. c) There is a sequence in X and a number 6 > 0 such that the sequence converges to p, but d(f(a:n),f(p)) > 6 for each n E J. [I also accepted “there is a sequence converging to p, so that the image sequence (f does not converge to f (p)”.] 2 (24 points). Compute: 2n+3 1' _1 n a) 11nsup( ) <5n_ 1) x2 a: < l b i1$3da a: where (1 a; ={ 7 _ 2 ) 10 ( ) ( ) 2:172, .T>% In 277. c) nil-£20 (77, + ' Solution. a) ~ 1 $27 '77 g 2 b) Write a : a1 + 1012 where 011(LL') 2 2x2 _ i, a: > % . 0, a: : é . . . while : 1 . Since a1 1s continuous, we have 1, :17 > E '1 % 1 1 4 1 63 3 4 4 i-d «-: 2 d 4 d:— ———=— /0L a1(.L) /0 x $+/l a: a: 80 5 40 80, 2 ‘1 3 _ 63 1 1 _ 131 Whencelo '73 d“ — m + (Z) (s) — m. c) The quickest approach is to note that . n 2” 1 1 11m + 1 = . 1 2 = 2. "200 n [hmnaoofl + —)"] 6 One can also take logarithms and apply l’hopital’s rule. 3 (42 points). Give examples of the following: a) a countable set of irrational numbers, b) an open cover of R which does not admit a finite subcover, c) a countable compact subset of R, d) a disconnected subset of R2 whose closure is connected, e) a bounded continuous function f : R a R which does not attain a maximum value, f) a continuous function f : (0,1) % R which is not uniformly continuous, g) a non—Riemann—integrable function f : [0,1] a R such that f 2 is integrable, Examples. a) fiJ b) {(—n,n):rLEJ} c) {ianJ}U{0} m Nfimmomofi) e) f = arctan /__{L xeoomu )_ ~L xemymg' 4 (16 points). Let E be an uncountable set of real numbers. a) Prove that for some n E J, the closed interval [—n, n] contains uncountany many points of E. b) Prove that E must have a limit point in R. Proof. For each n E J, write E z: E 0 [—n,n]. For Part a), note that if each En were at most countable, then their union E would also be countable, contrary to assumption. For Part b), fix n as in Part a). Since the interval [—n, n] is compact, its infinite subset En must have a limit point p; a fortiori, p is also a limit point of E. 5 (16 points). Suppose E and F are compact subsets of a metric space X. Prove that their union E U F is also compact. Proof. Let V be an open cover of E U F. In particular, V covers E, whence by compactness, there is a finite subcollection 5 of V which also covers E. Similarly, there is a finite subcollection .7: of V which covers F. Then 5 U .7: is a finite subcollection of V which covers E U F. 6 (16 points). Suppose f : R a R is non—decreasing and bounded. Prove that limmnoo f exists. Proof. The set S :2 {f : a: E R} is non—empty and bounded so it has a least upper bound L. Let 6 > 0. Since L — 6 is not an upper bound of S, we can find a real number M such that f (M) > L — 6. Now suppose a: Z M. By monotonicity, L—6 < g g L < L+6, whence —L| < 6 as desired. 7 (16 points). Suppose f : R a R is differentiable everywhere and f’ is bounded. Prove that f is uniformly continuous. Proof. Choose M > 0 so that g M for all a: E R. Let 6 > 0. Take 6 = Now suppose [.23 — y| < 6. Applying the Mean Value Theorem, we find a number 0 so that |f($) — f(:u)| = |(-’L'—y)f’(c)| S le—EA < M5 = 6, as desired. 8 (16 points). Suppose f : [0,1] a R is Riemann integrable and g : [0,1] a R satisfies g(a:)—g(y)| g — for all :17, y 6 [0,1]. Prove that g is also Riemann integrable. Lemma. Let I be a subset of [0, 1]. Then supI g — iang g supI f — inf] f. Proof of the Lemma. Let :17, y E I. By definition of upper and lower bounds, we have — f(y) < supI f — inf] f. Reversing the roles of 11,1 , we in fact have — g sup] f — inf] f. Putting this together with the hypothesis, we get — g(y) g sup I f —— inf 1 f. Holding y fixed. we see that g(y) + sup] f — inf] f is an upper bound for the set : a: E I}. By definition of least upper bound, this yields sup I g g g(y) +supI f —ian f. Transposing, freeing y, and applying the definition of greatest lower bound then completes the proof of the lemma. Proof of the Problem. Let 6 > 0. Apply integrability of f to get a partition P of [0,1] satisfying U(f, P) —L(f, P) < 6. But then the Lemma tells us that U(g, P) — L(g, P) g U(f, P) — L(f, P) < e, and g meets the “convenient criterion” for integrability. 9 (16 points). For each n E J, let fn, gn : R a R with 0 3 f7, g 9,, for all a: E R. Prove that if the series 2 gn converges uniformly, then so does the series 2 fn. Proof. (This is a generalization of the Weierstrass-M test, but that result cannot be used to prove this one because the converse of the WM test is not valid.) Write and (tn) for the sequences of partial sums of given series. Uniform convergence of Z 9,, tells us that the sequence (tn) is uniformly Cauchy. But for all a: E R and m, n E J, we have |sm — sn g |tm — tn(:1:)|, so the sequence (37,) is also uniformly Cauchy. It follows that is also uniformly convergent, which is what it means for the series 2 f" to converge uniformly. 10 (16 points). Suppose .7: is a uniformly bounded, equicontinuous family of functions in C [0, 1], and g : R a R is continuous. Prove that the family of composites ’H :: {g o f : f E 7:} is also equicontinuous. Proof. Choose a positive real number M so that g M for all f E .7: and all a: E R. Let 6 > 0. Since 9 is uniformly continuous on the compact interval [—M, M], we can find an 17 > 0 so that |g(z) — g(w)| < 6 whenever z, w E [—M, M] with [Z — w| < 77. Apply equicontinuity of .7: to find 6 > 0 so that — < 17 whenever f E .7: and [.23 — y] < 6. Now suppose h E ’H and < 6. Then h = gof for some f E .73. We then have < 17, whence W13) — h(2/)l = |g(f($)) — 9(f(y))| < 6, as desired- Bonus (10 points). Give an example of a function f : R a R which is differentiable at 0 but discontinuous at each a 7é 0 E R. x2, :1: E Q Solution. Take = { . ' 0, a: E R\Q ...
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4100fnlnotes - MATH 4100/6100(Azoff Final Exam Notes Fall...

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