Download Math 4317, Final Exam: Analysis Name and section: 1. (20 points

Math 4317, Final Exam: Analysis
Name and section:
1. (20 points) Prove or disprove: The intersection of closed sets is closed. (Fully justify
your answer.)
Solution: Proof: Let {Aα }α∈Γ be an arbitrary collection of closed sets. Then
(∩Aα )c = ∪Acα . That is, the complement of an intersection of closed sets is the
union of the complements of those closed sets. Since the complement of a closed set
is open, this is a union of open sets. Since the union of open sets is open, we see
that (∩Aα )c = ∪Acα is open. That is, the intersection ∩Aα is closed.
Name and section:
2. (20 points) (11A) Define the term compact.
Prove directly from the definition (without using the Heine-Borel Theorem) that if A
and B are compact sets, then A ∪ B is compact.
Solution: A set K is compact if any open cover of K has a finite subcover.
Let U = {Uα } be any open cover of A ∪ B. Since A and B are compact there are
finite subcovers finite subcover:
UA = {Uα1 , . . . , Uαk } ⊂ U
and UB = {Uβ1 , . . . , Uβk } ⊂ U,
which are finite subcovers of A and B respectively. Thus, UA ∪UB is a finite collection
of open sets in U which covers A ∪ B. That is, we have found a finite subcover of
A ∪ B in U. Therefore, A ∪ B is compact.
Name and section:
3. (20 points) Let f1 , f2 , . . . be a sequence of real valued functions on the interval [0, 1].
Define what it means for this sequence of functions to converge pointwise to a given
function f : [0, 1] → R.
Define what it means for the sequence of functions above to converge uniformly to f
Prove or disprove: If fj ∈ C 0 [0, 1] for j = 1, 2, . . . is a sequence of continuous functions
which converge pointwise to a continuous function f : [0, 1] → R, then the sequence
converges uniformly to f .
Solution: Given a sequnce of functions fj : [0, 1] → R, we say fj converges pointwise
to a given function f : [0, 1] → R if for each x ∈ [0, 1] and each given > 0, there is
some N > 0 such that
|fj (x) − f (x)| < whenever j > N.
Given a sequnce of functions fj : [0, 1] → R, we say fj converges uniformly to a given
function f : [0, 1] → R if given > 0, there is some N > 0 such that
|fj (x) − f (x)| < whenever j > N
(and x ∈ [0, 1]).
Disroof/counterexample: Let f = f1 ≡ 0 and for j ≥ 2 set

x ≤ 1/j
 jx
−jx + 2 1/j ≤ x ≤ 2/j
fj (x) =

0
x ≥ 2/j.
Each of these functions is continuous and the limit is clearly continuous.
For any x > 2/j or x = 0, we have |fj (x) − f (x)| = 0. This implies pointwise
convergence.
On the other hand, |fj (1/j) − f (1/j)| ≡ 1 for j ≥ 2. Thus, we do not have pointwise
convergence.
Name and section:
4. (20 points) Write down the correct power series for the function f : R → R given by
f (x) =
1
1 + x2
with center of expansion at x = 0.
Determine the radius of convergence of this series, and clearly justify your answer.
Solution: For |x| < 1, this function is the sum of a geometric series with ratio −x2 .
Therefore, when |x| < 1, we have
∞
∞
X
X
1
2 k
=
(−x
)
=
(−1)k x2k ,
2
1+x
k=0
k=0
and the series is absolutely convergent. Since this is a power series, and the coefficients of a power series are unique, this is the power series of the given function. On
the other hand, when |x| > 1, the individual terms in the series do not converge to
zero. Thus, the radius of convergence must be R = 1.
Name and section:
5. (20 points) (Theorem 27.9) Give a correct statement of the mean value theorem.
Use the mean value theorem to prove: If f ∈ C 0 [a, b] and the derivative f 0 : (a, b) →
exists with f 0 ≥ 0, then f (b) ≥ f (a) with equality only if f is constant.
Solution: MVT: If φ ∈ C 0 [a, b] and the derivative φ0 : (a, b) → R exists, then there
is some ξ ∈ (a, b) such that
φ(b) − φ(a)
φ0 (ξ) =
.
b−a
Consider α and β with a ≤ α < β ≤ b. By the mean value theorem there is some
ξ ∈ (α, β) with
f (β) − f (α) = f 0 (ξ)(β − α) ≥ 0.
This says that f is nondecreasing on [a, b]. As a consequence f (a) ≤ f (b), and if
equality holds then for every x ∈ (a, b), we have f (a) ≤ f (x) ≤ f (b) with equality
throughout. That is, f is constant.
R
Name and section:
6. (10 points) (Bonus) Give an example of a closed and bounded set which is not compact.
Hint: What exactly does the Heine-Borel theorem say?
Solution: The Heine-Borel theorem says that compactness is equivalent to “closed
and bounded” in Rn . Thus, for an example we’ll have to look in a topological space
which is not Rn . Probably the easiest one to consider is (0, 1) ⊂ R in the subspace
topology. Since this is not compact in R, there is an open cover with no finite
subcover. That cover will give an open cover with no finite subcover in the subspace
topology too. However, this set is closed (and bounded) in the subspace topology.
Interesting examples can also be found in C 0 (R) where distance is measured in the
sup norm.