COMPACTNESS VS. SEQUENTIAL COMPACTNESS
The aim of this handout is to provide a detailed proof of the equivalence between the two
denitions of compactness: existence of a nite subcover of any open cover, and existence of a limit
point of any innite subset.
18.100B Problem Set 6
Due Friday October 27, 2006 by 3 PM
1) Prove that if
|an | is converges, then
|an |2 also converges.
2) Prove that
n(n + 1)(n + 2)
Hint: Use a telescope trick, i. e. an argument of the form
18.100B Problem Set 5
Due Friday October 20, 2006 by 3 PM
1) Let M be a complete metric space, and let X M. Show that X is complete if and only if X
2) a) Show that a sequence in an arbitrary metric space (xn ) converges if and only i
18.100B Problem Set 4
Due Friday October 6, 2006 by 3 PM
1) Give an example of an open cover of the set E = cfw_(x1 , x2 ) R2 : x2 + x2 < 1 R2 which
has no nite subcover. (As usual, R2 is equipped with standard Euclidean metric.)
18.100B Problem Set 1 Solutions
1) The proof is by contradiction. Assume r Q such that r2 = 12. Then we may write r as a
with a, b Z and we can assume that a and b have no common factors. Then
a 2 a2
12 = r2 =
= 2 ,
2 = a2 .
18.100B Problem Set 3 Solutions
1) We begin by dening d : V V R such that d(x, y) = x y. Now to show that this
function satises the denition of a metric. d(x, y) = x y 0 and
d(x, y) = 0 x y = 0 x y = 0 x = y
So the function is positive denit
18.100B Problem Set 1
Due Friday September 15, 2006 by 3 PM
1) (10 pts) Prove that there is no rational number whose square is 12.
2) (10 pts) Let S be a non-empty subset of the real numbers, bounded above. Show that if
u = sup S, then for eve
18.100B Problem Set 2
Due Friday September 22, 2006 by 3 PM
1) (10 pts) Prove that the empty set is a subset of every set.
2) (10 pts) If x, y are complex, prove that
|x| |y| |x y|.
(Hint: This is equivalent to proving the following two inequali
SOLUTIONS TO PS2
Proof. It is true that for any two sets A, B, the intersection A B is a subset of
A. Now consider = A Ac . So is a subset of A for any set A.
Proof. Notice that
|x| |y| |x y| |x| |y| |x y| and |y| |x| |x
18.100B Problem Set 5 Solutions
1) We have X M, with M complete. X is complete if and only if every Cauchy sequence of X
converges to some x X. Let (xi ) be Cauchy, with xi X. M being complete implies that
xi y M. Therefore y is a limit poin
SOLUTIONS TO PS6
Solution/Proof of Problem 1. Since
|an | converges, lim |an | = 0. So N
N such that for n N , |an | < 1. Thus for n N we have |a2 | |an | and
by the comparison theorem and the convergence of
|an | we conclude that
18.100B Problem Set 9
Due Friday December 1, 2006 by 3 PM
1) Let fn (x) = 1/(nx+1) and gn (x) = x/(nx+1) for x (0, 1) and n N. Prove that fn converges
pointwise but not uniformly on (0, 1), and that gn converges uniformly on (0, 1).
2) Let fn (x
18.100B Problem Set 10
Due Friday December 8, 2006 by 3 PM
1) Let (fn ) be the sequence of functions on R dened as follows.
f0 (t) = sin t and fn+1 (t) = fn (t) + 1 for n N.
Show that fn 3 uniformly on R. What can you say if we choose f0 (t)
SOLUTIONS TO PS 8
Solution/Proof of Problem 1. From f (x) = f (x2 ), we have
f (x) = f (x 2 ) = f (x 4 ) = = f (x 2n ).
Now let yn = x 2n , and assume x = 0, so lim yn = 1. Since f is continuous, we
have if x = 0
f (x) = lim f (x) =
18.100B Problem Set 9 Solutions
1) First we need to show that
nx + 1
converges pointwise but not uniformly on (0, 1). If we x some x (0, 1), we have that
lim fn (x) = lim
n nx + 1
Thus the fn converge pointwise. However, consider
18.100B Problem Set 7 Solutions
1) We have ai > 0 and ai+1 ai for all i = 0, 1, 2, ., and lim ai = 0, and we want to show the
(1)i ai = a0 a1 + a2 .
So we dene sn to be the partial sums of the rst n + 1 terms of the sum:
18.100B Problem Set 8
Due Thursday November 9, 2006 by 3 PM
1) Let f : [0, ) R be continuous, and suppose
f (x2 ) = f (x)
holds for every x 0. Prove that f has to be a constant function.
Hint: Show that f (0) = f (x) if 0 x < 1, and f (1) = f (x
18.100B Problem Set 7
Due Friday November 3, 2006 by 3 PM
1) Consider an innite series with alternating signs
a0 a1 + a2 a3 + a4 . . . =
(1)n an .
Prove the Leibnitz criteria for convergence:
If ai > 0 for all i, ai+1 ai and i ai = 0, then the s
SOLUTIONS TO PS4
Solution/Proof of Problem 1. Consider the open set
Bn = (x1 , x2 ) R : x1 + x2 < 1
Then we can see that E Bn because for any point (x, y) E, x2 + y 2 < 1, we
can nd an n big enough such that x2 + y 2 < 1 n , i.
18.100B Problem Set 3
Due Friday September 29, 2006 by 3 PM
1) (10 pts) In vector spaces, metrics are usually dened in terms of norms which measure the length
of a vector. If V is a vector space dened over R, then a norm is a function from vecto
18.100B/C Practice Final Exam
Closed book, no calculators.
This is a 180-minute exam. No notes, books, or calculators are permitted. Point values
(out of 100) are indicated for each problem. There is a (hard) bonus question, Problem 9,
at the e
Practice Quiz 2
18.100B R2 Fall 2010
Closed book, no calculators.
This is a 30 minute in-class exam. No notes, books, or calculators are permitted. Point
values are indicated for each problem. Do all the work on these pages.
18.100B/C: Fall 2010
Solutions to Practice Final Exam
1. Suppose for sake of contradiction that x > 0. Then 1 x > 0 because the product of two positive
quantities is positive. Thus x + 0 < x + x (because y < z implies x + y < x + z for all x), i.e.,
THE WEIERSTRASS PATHOLOGICAL FUNCTION
Until Weierstrass published his shocking paper in 1872, most of the mathematical world
(including luminaries like Gauss) believed that a continuous function could only fail to
be differentiable at some collection of i
CONSTRUCTION OF R
1. M OTIVATION
We are used to thinking of real numbers as successive approximations. For example,
= 3.14159 . . .
to mean that is a real number which, accurate to 5 decimal places, equals the above
string. To be precise, this m
p IS COMPLETE
Let 1 p , and recall the denition of the metric space p :
For 1 p < , p =
sequences a = (an ) in R such that
|an |p < ;
whereas consists of all those sequences a =
such that supnN |an | < . We
dened the p-norm as the func
The Devils Staircase
Recall the usual construction of the Cantor set: C0 = [0, 1], C1 = [0, 1 ] [ 2 , 1], and in
general Cn is a disjoint union of 2n closed intervals, each of length 3n , constructed from
Cn1 by deleting the open-middle-third of each
CONTINUOUS ALMOST EVERYWHERE
Denition 1. Let be a subset of R. We say that has measure 0 if, for each > 0, there
is a sequence of balls (Bj = Brj (cj )jN with radii rj > 0 (and centres cj R), such that
jN Bj and rj < .
The balls here are open interva
18.100B Fall 2010
Practice Quiz 1 Solutions
1.(a) E X is compact if, given any open cover E
U by open sets U X (with
A any index set), one can nd a nite subcover E
U , with A A a nite subset.
(b) E = cfw_e1 , . . . , eN
Given an open cover E
U , f