University of London The London School of Economics and Political Science
ABSTRACT MATHEMATICS
MATHEMATIC MA203

Fall 2014
Solutions
116
Solution to Exercise 3.29.
(1) True.
Since the series
X
n=1
an  converges, we have necessarily that
lim an  = 0.
n
Thus there is an N N such that for all n > N , an  < 1. But now for n > N ,
a2n  = an an  an ,
and so it follows
University of London The London School of Economics and Political Science
ABSTRACT MATHEMATICS
MATHEMATIC MA203

Fall 2014
Solutions to the exercises from Chapter 4
129
Solution to Exercise 4.33. Let X = (0, 1], Y = R with the usual Euclidean metrics. Define f : (0, 1]
R by
1
(x (0, 1]).
f (x) =
x
Then f is continuous. Take (xn )nN to be the Cauchy sequence n1 nN . Then f (x
University of London The London School of Economics and Political Science
ABSTRACT MATHEMATICS
MATHEMATIC MA203

Fall 2014
Solutions to the exercises from Chapter 4
121
Solutions to the exercises from Chapter 4
Solution to Exercise 4.3. That (1) implies (2 ) is immediate since (1) implies (2), and (2) implies (2 ).
Now suppose that (2 ) holds. Let (xn )nN be a sequence contai
University of London The London School of Economics and Political Science
ABSTRACT MATHEMATICS
MATHEMATIC MA203

Fall 2014
Solutions
134
Solutions to the exercises from Chapter 5
Solution to Exercise 5.9. We have
f (a)
(a) = det f (a)
f (b)
g(a)
g(a)
g(b)
1
1 =0
1
since the first and second rows of the matrix are linearly dependent. Similarly,
f (b) g(b) 1
(b) = det f (a) g(a
University of London The London School of Economics and Political Science
ABSTRACT MATHEMATICS
MATHEMATIC MA203

Fall 2014
Solutions
Solutions to the exercises from Chapter 1
Solution to Exercise 1.7.
(D1) As d(x, y) is either 0 or 1 for all x, y X, clearly d(x, y) 0.
Also, for all x X, d(x, x) = 0 by definition.
If x, y X and x 6= y, then d(x, y) = 1 6= 0. So if d(x, y) = 0,
University of London The London School of Economics and Political Science
ABSTRACT MATHEMATICS
MATHEMATIC MA203

Fall 2014
Solutions to the exercises from Chapter 3
109
Solutions to the exercises from Chapter 3
Solution to Exercise 3.3. We have
(2n + 1) (2n 1)
1
1
1
= tan1
tan1
,
tan1 2 = tan1
2n
1 + (2n + 1)(2n 1)
2n 1
2n + 1
and so the partial sums telescope to give
X
tan1
University of London The London School of Economics and Political Science
ABSTRACT MATHEMATICS
MATHEMATIC MA203

Fall 2014
Solutions
104
Solution to Exercise 2.34. (If part) Suppose that (an )nN converges to 0. Let > 0. Then there
exists an N N such that whenever n > N , an 0 < . As an 0, we have that an < for all n > N .
But by the definition of an , this means that for an
University of London The London School of Economics and Political Science
ABSTRACT MATHEMATICS
MATHEMATIC MA203

Fall 2014
Solutions
90
Solution to Exercise 1.25. Consider the open ball B(x, r) = cfw_y X : d(x, y) < r in X. If y B(x, r),
then d(x, y) < r. Define r = r d(x, y) > 0. We claim that B(y, r ) B(x, r). Let z B(y, r ). Then
d(z, y) < r = r d(x, y) and so d(x, z) d(x,
University of London The London School of Economics and Political Science
ABSTRACT MATHEMATICS
MATHEMATIC MA203

Fall 2014
Solutions to the exercises from Chapter 2
97
Solutions to the exercises from Chapter 2
Solution to Exercise 2.5. Let > 0. Since (an )nN is a Cauchy sequence, there exists an N N
such that for all n, m > N , an am  < . In particular for n > N , we have m
University of London The London School of Economics and Political Science
ABSTRACT MATHEMATICS
MATHEMATIC MA203

Fall 2014
140
Solutions
Solution to Exercise 5.29. (From the calculation done in the given hint, we guess that the derivative
of g at c is the map Rn h 7 2f (c)f (c)h, and we prove our claim below.)
We have for x Rn that
(f (x)2 (f (c)2 2f (c)f (c)(x c)
(f (x) + f