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
Problems:
1) Prove that if
|an | is converges, then
|an |2 also converges.
2) Prove that
n=1
1
1
= .
n(n + 1)(n + 2)
4
Hint: Use a telescope trick, i. e. an argument of the form
n=1 an
n=1 an
=
n=1

18.100B Problem Set 5
Due Friday October 20, 2006 by 3 PM
Problems:
1) Let M be a complete metric space, and let X M. Show that X is complete if and only if X
is closed.
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
Problems:
1) Give an example of an open cover of the set E = cfw_(x1 , x2 ) R2 : x2 + x2 < 1 R2 which
1
2
has no nite subcover. (As usual, R2 is equipped with standard Euclidean metric.)
2
2) Consid

18.100B Problem Set 1 Solutions
Sawyer Tabony
1) The proof is by contradiction. Assume r Q such that r2 = 12. Then we may write r as a
b
with a, b Z and we can assume that a and b have no common factors. Then
a 2 a2
12 = r2 =
= 2 ,
b
b
2 = a2 .
so 12b
No

18.100B Problem Set 3 Solutions
Sawyer Tabony
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
Problems.
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
1
u = sup S, then for eve

18.100B Problem Set 2
Due Friday September 22, 2006 by 3 PM
Problems:
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
Xiaoguang Ma
Problem 1.
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.
Problem 2.
Proof. Notice that
|x| |y| |x y| |x| |y| |x y| and |y| |x| |x

18.100B Problem Set 5 Solutions
Sawyer Tabony
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
Xiaoguang Ma
Solution/Proof of Problem 1. Since
|an | converges, lim |an | = 0. So N
n
N such that for n N , |an | < 1. Thus for n N we have |a2 | |an | and
n
2
by the comparison theorem and the convergence of
|an | we conclude that
|an

18.100B Problem Set 9
Due Friday December 1, 2006 by 3 PM
Problems:
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
Problems:
1) Let (fn ) be the sequence of functions on R dened as follows.
2
f0 (t) = sin t and fn+1 (t) = fn (t) + 1 for n N.
3
Show that fn 3 uniformly on R. What can you say if we choose f0 (t)

SOLUTIONS TO PS 8
Xiaoguang Ma
Solution/Proof of Problem 1. From f (x) = f (x2 ), we have
1
1
1
f (x) = f (x 2 ) = f (x 4 ) = = f (x 2n ).
1
Now let yn = x 2n , and assume x = 0, so lim yn = 1. Since f is continuous, we
n
have if x = 0
f (x) = lim f (x) =

18.100B Problem Set 9 Solutions
Sawyer Tabony
1) First we need to show that
1
nx + 1
converges pointwise but not uniformly on (0, 1). If we x some x (0, 1), we have that
1
lim fn (x) = lim
= 0.
n
n nx + 1
Thus the fn converge pointwise. However, consider

18.100B Problem Set 7 Solutions
Sawyer Tabony
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
i
convergence of
(1)i ai = a0 a1 + a2 .
i=0
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
Problems:
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
Problems:
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
Xiaoguang Ma
Solution/Proof of Problem 1. Consider the open set
1
2
2
2
Bn = (x1 , x2 ) R : x1 + x2 < 1
.
n
Then we can see that E Bn because for any point (x, y) E, x2 + y 2 < 1, we
1
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
Problems:
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

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18.100B Analysis I
Fall 2010
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18.100B/C Practice Final Exam
Closed book, no calculators.
YOUR NAME:
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.
YOUR NAME:
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.
GRADING
1.
/15
2.
/

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
2
quantities is positive. Thus x + 0 < x + x (because y < z implies x + y < x + z for all x), i.e.,
2
2

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,
we write
= 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
n=1
|an |p < ;
n=1
(an )
n=1
whereas consists of all those sequences a =
such that supnN |an | < . We
p
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
3
3
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 < .
j=1
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
A any index set), one can nd a nite subcover E
U , with A A a nite subset.
A
(b) E = cfw_e1 , . . . , eN
Given an open cover E
U , f