MA 355
Homework 1
For each subset of R, give its maximum, minimum, supremum, and inmum, if they exist:
cfw_1, 3: min and inf=1, max and sup=3
n
1
n+1 : n N : min and inf= 2 , max=none, sup=1
(, 4): min and inf=none, max=none, sup=4
p.21: #1. (a) Suppose
MA 355
Homework 10 solutions
#1 Use mean value theorem to establish:
the
1
a) 1 < 51 7 < 7
8
Look at f (x) = 49 + x on [0, 2]. Then by MVT there is c (0, 2) such that f (c) = f (2)f (0) =
2
1
1
1
which implies f (2) f (0) = 49+c . Thus 51 7 = 49+c . Since
MA 355
Homework 9 solutions
# 1 Prove that f (x) = x is uniformly continuous on [0, ).
Step I: Notice that f (x) is uniformly continuous on [0, 2] because [0, 2] is compact. So given, > 0, > 0
such that |f (x) f (y)| < for |x y| < .
x+y
|xy|
Step II: Not
MA 355
Homework 10
#1 Use mean value theorem to establish:
the
1
a) 1 < 51 7 < 7
8
b) | cos(x) cos(y)| |x y| for x, y R
#2 Suppose i) f is continuous for x 0, ii) f (x) exists for x > 0, iii) f (0) = 0, iv) f is monotonically increasing. Dene g(x) = f (x)
MA 355
Homework 8 solutions
#1 Dene f : R R by f (x) = x2 3x + 5. Use the denition (of continuity) to prove that f is
continuous at 2.
WTS: limx2 f (x) = 3 = f (2).
Notice: If < 1, then |x1| = |x2+1| |x2|+|1| 2. So |x2 3x+2| = |x1|x2| < 2|x2|.
i) So give
MA 355
Homework 7 solutions
2 3x + 1 = 11. Notice if |x 5| < and 1, then
#1 Use the denition to prove limx5 x
|x+2| = |x5+7| |x5|+7 8. So consider two cases: A) Given > 0 (and 8), choose = 8
2 3x+111| = |x5|x+2| < 8|x5| < 8 = 8 for |x5| < = .
(thus < 1) a
MA 355
Homework 6 solutions
1
#1 Prove the sequence s1 = 1, sn+1 = 4 (sn + 5) where n N is monotone and bounded. Then
nd the limit.
Claim: Bounded above by 2 and below by 0. The lower bound is clear. Prove the upper bound by
1
induction: We know s1 = 1. A
MA 355
Homework 5 solutions
#1 Use the denition of convergence to show limn 3n+1 = 3.
n+2
5
5
Let > 0. Choose N = 5 then for n > N , | 3n+1 3| < n < N < .
n+2
n
n+1 = 0.
= 1 . The
2
# 2 Show limn
Let > 0. Choose N
n
for n > N , | n+1 | <
n
n
>
1
n
<
1
N
=
MA 355
Homework 4 solutions
#1 Prove: If S is a nonempty closed, bounded subset of R, then S has a maximum and a
minimum.
Pf: Since S is bounded above, m = supS exists by the completeness of R. Since m is the least
upper bound for S, given any > 0, m is n
MA 355
Homework 3 solutions
#1Find the interior of the following sets:
1
:nN :
n
(ii) [0, 3] (3, 5) : (0, 5)
(i)
(iii) [0, 2] [2, 4] :
#2 Classify the following sets as open, closed or neither:
1
: n N : neither
n
(ii) N : closed
(i)
(iii) cfw_x : x2 > 0
MA 355
Homework 2 solutions
#1 Let A, B, C be sets and let f : A B, g : B C be functions. Prove: If f is onto B and g is
onto C, then g f : A C is onto C.
Suppose f is onto B and g is onto C. Let c C. Then there is a b B such that g(b) = c, since g
is ont
21-355
Real Analysis
First Exam
Fall, 2011
1. Give a precise mathematical denition of:
A set E X is closed
E is closed if every limit point of E is a point in E.
An open cover of a set E in a metric space X
An open cover of a set E in a metric space X i
21-355
Real Analysis
Second Exam
Fall, 2011
1. Give a precise mathematical denition or statement of:
A function f : D R where D R and c D is continuous.
f is continuous at c if > 0 there exists a > 0 such that |f (x) f (c)| < whenever
|x c| < whenever x
MA 355
Homework 11 solutions
b
2
#1 Suppose that f (x) = x for all x [0, b]. Show that f is integrable and that 0 f (x)dx = b2 .
b
b
Consider the partition P = cfw_0, n , 2b , (n1)b , b. So xi = ib and xi = n . Since f is increasing we see
n ,
n
n
(i1)b
i