HOMEWORK 8 FOR MA108A
DUE DATE: 4PM, THURSDAY DECEMBER 2, 2010
(1) Let f : 2 R be given by f (y ) = n (yn + yn /n). f is well dened (i.e. the series converges), as
we know that elements of 2 are also elements of 4 , and the functions is bo
COMPLETING METRIC SPACES USING CAUCHY-SEQUENCES
FOKKO VAN DE BULT
Suppose (M, d) is a metric space. We want to show it has a completion (we
do not concern ourselves with uniqueness in this note). The main idea is to add
points which can serve as limits fo
MA108A FALL 2013
(1) Problem 40 of Chapter 5
A := cfw_(x, y ) : x 0; y = ex and B := cfw_(x, y ) : x 0; y = 0.
Then both A and B are nonempty closed sets, and A B = .
Moreover, for any > 0, choose x0 = log + 1. Then
d(A, B )
MA108A FALL 2013
(1) Problem 26 of Chapter 2
Proof Let x = 1 23xnn and y = 1 23yn . Suppose f (x) >
P 1 xn
f (y ). Then n=1 2n > n=1 2n . Let kP mincfw_i : xi > yi .
P 1 yn
Then k < 1 since otherwise we have
MA108 HOMEWORK 3 SOLUTIONS
1. Problem 1
Let y Ac , then d( x, y) > r and thus d( x, y) > r + for some
that B (y) Ac and so Ac is open and thus A is closed as desired.
> 0. It is then follows
Let M be a set with more than one point. Dene a function d on M
MIDTERM FOR MA108A DUE DATE: 4PM, TUESDAY NOVEMBER 2, 2010
Read the instructions on the separate sheet before you begin! (1) Let C 1 [0, 1] be the vectorspace of all continuously differentiable functions on the interval [0, 1]. We know that f = supx[0,1]
FINAL FOR MA108A
DUE DATE: 4PM, FRIDAY DECEMBER 10, 2010
Read the instructions on the separate sheet before you begin!
(1) Suppose f : X Y is a continuous map from a compact metric space X to an arbitrary metric
space Y .
(a) Show that for any A X we have
HOMEWORK 3 FOR MA108A
DUE DATE: 4PM, TUESDAY OCTOBER 19, 2010
(1) Let G be open and D dense in M . Show that G D is dense in G. Show that this is not necessarily
true if G is not open.
Proof. To show that G D is dense in G, it suces to show th
HOMEWORK 5 FOR MA108A
DUE DATE: 4PM, TUESDAY NOVEMBER 9, 2010
(1) Let 1 p < . Let c R . Show that Sc = cfw_x
only if c p .
What goes wrong if we take p = ?
| |xn | cn is a compact subset of
Proof. First we show that if 1 p < and c
HOMEWORK 7 FOR MA108A
DUE DATE: 4PM, TUESDAY NOVEMBER 23, 2010
(1) (a) Let fn f , where fn : X Y , with X a compact metric space. Suppose fn forms an
equicontinuous sequence (i.e. cfw_fn is equicontinuous). Show that fn f .
(b) X compac
HOMEWORK 4 FOR MA108A
DUE DATE: 4PM, TUESDAY OCTOBER 26, 2010
(1) (Combination of 5.46 and 5.51) Let H be the Hilbert cube, dened as the set of sequences x
with x 1, with metric
2n |xn yn |
dH (x, y ) =
(This is a metric by Set 2 problem
HOMEWORK 6 FOR MA108A
DUE DATE: 4PM, TUESDAY NOVEMBER 16, 2010
(1) Suppose fn : M N is continuous for all n (with M and N metric spaces). Suppose fn f .
Suppose xn M such that xn x. Show that fn (xn ) f (x).
Proof. Let us rst remark that f
MA 108A, HOMEWORK 3 SOLUTIONS
11. dim Mn,m (R) = mn. The standard basis for M is the collection of
mn matrices Eij which have 1 in the position (i, j ) and 0 everywhere
else. Let us show that A is a norm. Clearly A 0 for all A
and O = 0 (O is the zero mat
MA 108A, HOMEWORK 4 SOLUTIONS
16. Assume f has no xed points. Then g (x) = |f (x) x| > 0 on K
and so attains a strictly positive minimum value at some x0 K . But
in this case f (x0 ) = x0 and so g (f (x0 ) < g (x0 ), i.e. g (x0 ) is not a
minimum. Thus a
MA 108A, HOMEWORK 6 SOLUTIONS
26. Dh = (hu , hv ), where hu =
g (f ), fu and hv =
g (f ), fv .
27. f (x) is dierentiable at every point. For x = 0 we have fx (x, y ) =
4 1/ 3
x sin(y/x) x2/3 y cos(y/x) and fy (x, y ) = x1/3 cos(y/x). Thus the
HOMEWORK 2 FOR MA108A
DUE DATE: 4PM, TUESDAY OCTOBER 12, 2010
(1) Let f : R0 R0 be a concave strictly increasing function with f (0) = 0. Moreover let an denote
a sequence of positive real numbers. Dene a function d : R R R cfw_ as
ak f (|xk y
HOMEWORK 1 FOR MA108A
DUE DATE: 4PM, TUESDAY OCTOBER 6, 2009
(1) (a) Consider the sequence dened recursively by x1 = 1 and xn+1 = x2 + x1 . Show that xn
converges and determine its limit.
(b) Given two positive numbers x and y , such that
MA 108A, HOMEWORK 7 SOLUTIONS
31. A 2 = sup x =1 Ax, Ax = sup x =1 At Ax, x = max (At A).
If At = A then max (At A) = max (A2 ) = 2 (A) 0
and so A = |max (A)|. (Note that A2 has the same eigenvectors as
A and the eigenvalues of A2 are the squares of e