Solution to Warmup Exercise 3  Math 1530  RUBIN  August 25, 2003
Negate the following statements.
a) For all x, S (x) holds.
NEGATION: There exists x for which S (x) does not hold. (Symbollically: x
S (x)
does not hold.)
b) For some x, S (x) holds.
NE
Solutions to Midterm Exam  Math 1530  Fall, 2001  Rubin
1. a) A number b is the supremum of a sequence xn if xk b for all k N and b is
less than or equal to every other number with this property. The lim sup of xn
is the supremum of the set of cluster
Solutions to Final Exam  Math 1530  Fall, 2001  Rubin
1. a) FALSE: If an is the sequence cfw_1, 1, . . . and bn is the sequence cfw_1, 1, 1, 1, . . .,
then an 1 and bn is bounded but an bn = bn does not converge.
b) FALSE: If an = n1/4 0 but bn = n1/2
FINAL EXAM
12/13/2001
MATH 1530
Dr. Jonathan Rubin
Name:
You have 120 minutes to complete this exam. Be sure to justify your reasoning and to complete
all parts of all problems.
Score: 1.
/20
3.
/30
4.
/40
5.
/30
6.
/20
7.
/35
Total:
/175
FINAL
Math1530 
Solutions to Homework for Week 1  Math 1530  Fall, 2003  Rubin
1.2 I will simply consider b), since a) is a special case of b). First, note that if x0 = 0,
then xn = 0 n, so the sequence converges to 0, and if x0 = 1, then xn = 1 n, so
the sequence con
Solutions to Homework for Week 2  Math 1530  Fall, 2003  Rubin
1.4.1 There are many ways to do this. For example, for m = n, we have

1
1
(m n)(m + n)
(2 maxcfw_m, n)2
4
2 = 
<
=
2
2 n2
2 n2
n
m
m
m
(mincfw_m, n)2
Thus, for xed > 0, if we choose
Solutions to Homework for Week 3  Math 1530  Fall, 2003  Rubin
1.7.1  pg. 98. # 10. The bounded metric is dened by
(x, y ) = d(x, y )/(1 + d(x, y )
for d(x, y ) itself a metric. We need to check 4 properties to conclude that is a
metric; let x, y, z M
Solutions to Homework for Week 4  Math 1530  Fall, 2003  Rubin
2.2.1 Let A = [0, 1) (1, 2] and B = cfw_1. Then int(A B ) = (0, 2), while int(A) int(B ) =
(0, 1) (1, 2), which, unlike int(A B ), does not contain the point cfw_1.
2.2.2, pg. 145 # 12c. We
Solutions to Homework for Week 5  Math 1530  Fall, 2003  Rubin
2.5.1  pg. 144, #9b. We need to show that cl(A) = M \ int(M \ A). First, we show that
cl(A) M \ int(M \ A). By denition, int(M \ A) is open. Thus, M \ int(M \ A) is
closed. If x A, then x
Solutions to Homework for Week 6  Math 1530  Fall, 2003  Rubin
2.9.1 (pg. 148, # 47). We can form a sequence of rectangles Rn with areas 1/n for each
n N as follows: for each n, let Rn have width 1 and height 1/n. If we line these
rectangles up on the
Solutions to Homework for Week 9  Math 1530  Fall, 2003  Rubin
4.1.1 a) Fix
> 0 and x0 ]0, 1[. For any x ]0, 1[, we have
x3 x3  = x x0 (x2 + xx0 + x2 ) 3x x0 
0
0
Thus, if we choose < /3, then x x0  < x3 x3  < . Therefore x3 is continuous
0
o
Solutions to Homework for Week 10  Math 1530  Fall, 2003  Rubin
4.4.1. The objective is to give an example of a discontinuous function f (x) dened on a
compact domain K I such that for all x K , f (x) > inf cfw_f (x) : x K . Let
R
K = [0, 2]. Let
x, x
Solutions to Homework for Week 11  Math 1530  Fall, 2003  Rubin
pg. 234, #25. Suppose that m M are constants such that m f (x) M on
(0, 1). Consider an arbitrary sequence xn 0 from above (i.e. xn (0, 1) for
all n) and note that for each n, the tangent
Solutions to Homework for Week 14  Math 1530  Fall, 2003  Rubin
Section 5.4, question 1. a) We have f (x) = 1x (1/t) dt for x > 0. For arbitrary xed
a > 0, let h(b) = f (ab) f (a) f (b) be a function dened for b > 0. Then h(1) =
f (a) f (a) f (1) = f (
Solutions to Homework for Weeks 1213  Math 1530  Fall, 2003  Rubin
pg. 316, # 2. a) On all of I  sin x 
R, k
to 0.
1
k
0 as k , so sin x/k converges uniformly
b) On (0, 1), for > 0, 1/(kx + 1) < for kx + 1 > 1/ , which occurs for k >
(1/x)(1/ 1).
MIDTERM EXAM I
10/22/2001
MATH 1530
Dr. Jonathan Rubin
1. a) State the denitions of sup xn (the supremum of xn ) and of lim sup xn for a sequence
xn in I .
R
b) Prove that if a sequence xn in I has innitely many terms greater than a xed
R
number M , then