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:
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. T
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 doe
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
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
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
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
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 ), doe
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) i
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
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, the
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
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 (
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 functi
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)
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 in