Math 431.02: Exam #1
Instructor: Dr. Herzog
February 19th, 2014
1. Let {an } and {bn } be bounded sequences and define sets A, B, and C by
A = {an }, B = {bn }, and C = {an + bn }. Prove that sup C ≤ sup A + sup B.
Give an example to show that strict inequality may hold.
Proof. By the definition of supremum, we have the bound
an + bn ≤ sup A + sup B
for each n. Therefore, sup A + sup B is an upper bound of the set C. Since
sup C is the least upper bound, it follows that
sup C ≤ sup A + sup B.
As an example that strict inequality may hold, let an = (−1)n and
bn = (−1)n+1 . Then an + bn = 0 for all n but sup A = sup B = 1, so
sup C = 0 < 2 = 1 + 1 = sup A + sup B.
2. Let {dn } be a sequence of limit points of the sequence {an }. Suppose
that dn → d as n → ∞. Prove that d is a limit point of {an }.
Proof. See the Quiz # 4 Solutions.
3. Suppose that a function f is continuous at a point c and that f (c) > 0.
Prove that there exists a δ > 0 so that for all x ∈ Dom(f ),
|x − c| ≤ δ inplies f (x) ≥
1
f (c)
.
2
> 0 as well. Since f is continuous
Proof. Since f (c) > 0, we have that f (c)
2
at c, there exists δ > 0 such that for all x ∈ Dom(f )
f (c)
.
2
But note that this implies that for all x ∈ Dom(f ) with |x − c| ≤ δ
|x − c| ≤ δ
|f (x) − f (c)| ≤
implies
f (c)
2
which implies that for all x ∈ Dom(f ) with |x − c| ≤ δ
f (x) − f (c) ≥ −
f (x) ≥
f (c)
,
2
as required.
4. Show that the function f (x) = x1 is NOT uniformly continuous on (0, ∞)
but IS uniformly continuous on any interval [µ, ∞) if µ > 0.
Proof. We first show that f is NOT uniformly continuous on (0, ∞). Let
= 21 . We will show that we cannot choose δ > 0 such that
1
2
for all x, y ∈ (0, ∞). We can accomplish this by exhibiting two sequences
xn , yn ∈ (0, ∞) with the property that |xn − yn | → 0 as n → ∞ but
1
. Since
|f (xn ) − f (yn )| ≥ 1 > 21 for all n. Take xn = n1 and yn = n+1
|xn − yn | ≤ |xn | + |yn | ≤ n2 → 0 as n → ∞, we have by the squeeze theorem
that |xn − yn | → 0 as n → ∞. Note, however, that
|x − y| ≤ δ
|f (x) − f (y)| ≤ =
implies
1
=
2
for all n. Hence, f is NOT uniformly continuous on (0, ∞) (the problem is
at x = 0 where the function blows up!).
|f (xn ) − f (yn )| = |n − (n + 1)| = 1 >
We now show that f IS uniformly continuous on [µ, ∞) where µ > 0 is
fixed. Let > 0 and pick δ = µ2 . Then for all x, y ∈ [µ, ∞) with
|x − y| ≤ δ we have
1 1 x − y ≤ δ ≤ δ = ,
|f (x) − f (y)| = − = x y
xy |x||y|
µ2
finishing the proof.
2
5. Let f be a continuous function on [a, b] and suppose that for every
a1 ∈ [a, b] and b1 ∈ [a, b] that
Z b1
f (x) dx = 0.
a1
Show that f (x) = 0 for all x ∈ [a, b].
Proof. Suppose, to the contrary, that f (c) 6= 0 for some c ∈ [a, b]. Without
loss of generality, suppose that f (c) > 0. By Problem 3 of this exam, since
f is continuous on c, there exists δ > 0 such that
|x − c| ≤ δ implies f (x) ≥
f (c)
;
2
> 0. By picking δ > 0
that is, for all x ∈ (c − δ, c + δ) ∩ [a, b], f (x) ≥ f (c)
2
small enough, either (c, c + δ) ⊂ [a, b] or (c − δ, c) ⊂ [a, b]. If
(c, c + δ) ⊂ [a, b], then
Z c+δ
f (c)
δ>0
f (x) dx ≥
2
c
or if (c − δ, c) ⊂ [a, b], then
Z c
f (x) dx ≥
c−δ
f (c)
δ > 0,
2
a contradiction.
6. Suppose that {xn } is a monotone increasing sequence of points in R and
suppose that xn has a subsequence that converges to a finite limit. Prove
that {xn } converges.
Proof. Let xnk → a ∈ R as k → ∞. We first claim that xn ≤ a for all n.
Suppose, to the contrary, that there exists K such that xK > a. Since xn is
monotone increasing, xn ≥ xK > a for all n ≥ K. In particular,
xnk ≥ xK > a for all nk ≥ K. Hence xnk cannot converge to a as k → ∞
since |xnk − a| ≥ xK − a > 0 for all nk ≥ K, a contradiction. To finish the
proof, let > 0 and pick, by convergence of this subsequence, a N ∈ N such
that k ≥ N implies
a − xnk = |xnk − a| ≤ .
3
Fix k ≥ N . Then for all n ≥ nk , by the above and monotonicity we have
|xn − a| = a − xn ≤ a − xnk = |xnk − a| ≤ ,
as required.
7. Suppose that the sequence {an } has the property that |an+1 − an | ≤ 2−n
for all n. Prove that {an } is a Cauchy sequence.
Proof. Let > 0. We need to show that there exists N ∈ N such that
m, n ≥ N imply
|am − an | ≤ .
Choose N ∈ N such that 2−N +1 ≤ . Then we have for m ≥ n ≥ N
m−1
X
|am − an | = aj+1 − aj j=n
≤
m−1
X
|aj+1 − aj |
j=n
≤
m−1
X
2−j
j=n
−n
=2
≤ 2−N
m−1
X
j=n
∞
X
j=0
4
2n−j
2−j = 2−N +1 ≤ .