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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
id «: 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 WeierstrassM 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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