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1. (10 pts) True/False. Indicate whether the statement is True or False. (no explanation required)
TRUE (a) Between any two rational numbers, there is an irrational number.
TRUE (b) Let r be a real number. If 3  r < 3 +  for all  > 0, then r = 3.
TRUE (c)  positive real number t,  positive integer N such that  integer n > N, 0 < 8/n < t.
TRUE (d) If sequence (sn) converges to 0, then the sequence (1/sn) diverges to .
TRUE (e) Given any set S of real numbers, every boundary point of S is either an upper bound or a
lower bound of the set S.
2. (6 pts) Let A = [1, 4] and B = [3, 1]. Define function f: A  B by 𝑓(𝑥) = 1 − 2|𝑥 − 3|.
True or False? (No explanation required.)
TRUE (a) f is a surjective function.
FALSE (b) f is an injective function.
TRUE (c)
Let T = {1, 1}. Then pre-image f 1(T) = {4, 3}.
3. (8 pts) Prove by induction:
1
6∙7
+
1
7∙8
+
1
8∙9
+ + ⋯ + (𝑛
1
+ 5)(𝑛 + 6)
=
𝑛
6(𝑛 + 6)
for all n  N.
Page 1 of 10
For n  1,
LHS 


RHS 


1
1  5 1  6 
1
 6  7 
1
42
1
6 1  6 
1
 6  7 
1
42
Since LHS=RHS, so base step is true.
Let it is true for n  k
1
1
1
1
k


 ... 

6.7 7.8 8.9
 k  5 k  6  6  k  6 
We have to show that it is true for n  k  1
For n  k  1 ,
Page 2 of 10
LHS 
1
1
1
1
1


 ... 

6.7 7.8 8.9
 k  5  k  6   k  6  k  7 

k
1

6  k  6   k  6  k  7 

k k  7  6
6  k  6  k  7 
k 2  7k  6

6  k  6  k  7 

 k  1 k  6 
6  k  6  k  7 

k 1
6 k  7
RHS 
k 1
6  k  1  6
k 1
6 k  7
Since LHS=RHS, so result is true for n  k  1
Hence by induction it is proved.
4. (16 pts) True or False? For each part: If true, prove it carefully. You can apply or cite any of our relevant
definitions, theorems, examples, or exercises in a proof. If false, provide a counterexample and explain it.
(a) If {An : n ∈ N} is any collection of open subsets of R, then the intersection ⋂∞
𝑛=1 𝐴𝑛 is open or bounded.
The statement is not true. According to Lemma, the intersection of any finite collection of open
subsets of R is open. In case of infinite, it is not true.
Counterexample:
Suppose A1, A2, . . . , An are open sets and let A = Tn i=1 Ai . Let x ∈ A be arbitrary. Then x ∈ Ai
for every i = 1, . . . , n. For each i, since Ai is open, there exists ri > 0 such that Bri (x) ⊆ Ai . Let r =
min{r1, r2, . . . , rn}. Then Br(x) ⊆ Bri (x) ⊆ Ai for all i = 1, . . . , n, and hence Br(x) ⊆ A. But x ∈
A was chosen arbitrarily, and hence A is an open set.
An infinite intersection of open sets is not necessarily open
(b) If (an) is any Cauchy sequence of real numbers and (bn) is any decreasing sequence of positive
real numbers, then the sequence (an  5bn) converges.
Page 3 of 10
Since an is any Cauchy sequence of real numbers, so by theorem we know that every Cauchy
sequence of real numbers converges to a limit.
Since bn is any decreasing sequence of positive real numbers, so it is bounded. Then the sequence is
convergent.
By a limit theorem, we know the limit of the difference of two convergent sequences is the
difference of their limits. So, an  5bn converges. The statement is true.
Counterexample
5. (15 pts) Complete the following table. For each set, determine the infimum, minimum,
maximum, and supremum (if they exist). (If there is not a finite value or it does not exist, record an
X instead). Indicate whether the set is open, closed, neither, or both. (explanation optional)
Set
Infimum
Minimum
Maximum
Supremum
Open / closed/
neither / both
NO
NO
NO
NO
NEITHER
4
NO
NO
4
OPEN
0
7
NO
7
BOTH
(−1)𝑛 + 1
𝑆 = {6 +
∶ 𝑛𝜖𝐍}
𝑛
𝑆 = ⋂[−
𝑛∈𝑁
𝑆 = ⋃(
𝑛∈𝑁
1
1
,3 +
)
𝑛
2
√𝑛
5𝑛 − 4
1
,7− ]
𝑛
𝑛
Page 4 of 10
6. (12 pts) For each of the following sequences (sn), state whether the sequence converges or diverges. If the
sequence converges, determine the limit. If the sequence diverges to  or to – state the infinite limit. Show
work / explanation. You may cite textbook/notes theorems, examples, and/or homework textbook exercises if
relevant (by source and page numbers).
(a) 𝑠𝑛 = 𝑛3 (0.3)𝑛
Consider the sequence Sn  n3  0.3 :
n
Let S  0.3 , so
Sn  S  n3  0.3  0.3   for   1
n
So, sequence is convergent.
Thus,
lim n3  0.3  0
n
n 
(b)
𝑛!
𝑠𝑛 =
100𝑛
Consider the sequence S n 
S n 1 
n  1!
100n 1
n!
:
100n
So,
lim
n 
S n 1
Sn
n  1!
n 1
 lim 100
n 
n!
100n
n  1! 100n

n  100 n 1
n!
 lim
 lim
n 
n  1 n ! 1

100 n !
n 1
n  100
n 1
 lim

n  100
 lim
Page 5 of 10
Thus, by ratio test sequence is not convergent. So, lim S n  
n 
(c) 𝑠𝑛 = 𝑛[2𝑛 + (−2)𝑛 ]
Consider the sequence sn  n[2n  (2) n ] :
If lim sn  0 , the sequence sn diverges
x 
lim Sn  lim n[2n  (2)n ]
n 
n 
 22
0
Hence, sequence S n is divergent.
7. (3 pts) For each of the sequences in the previous problem, state the subsequential limits (which
could be infinite).
Sequence part in #6
(a)
Subsequential Limits
Subsequential limits =0,
Limit inferior  0
Limit superior  
(b)
Subsequential limits =,1
Limit inferior  
Limit superior  1
(c)
Subsequential limits =-, 0
Limit inferior  0
Limit superior  does not exist
8. (7 pts) Claim: If 𝑠1 = 2 and 𝑠𝑛+1 = −1 + 2𝑠𝑛 for all n in N, then lim 𝑠𝑛 = 1 .
Proof:
Let 𝑠1 = 2 and 𝑠𝑛+1 = −1 + 2𝑠𝑛 for all n in N.
Taking the limit of both sides of the equation, we have
Page 6 of 10
lim 𝑠𝑛+1 = lim(−1 + 2𝑠𝑛 )
lim 𝑠𝑛+1 = −1 + 2 lim 𝑠𝑛 Call the limit s.
We have 𝑠 = −1 + 2𝑠, so, solving for s, we get 1 = 𝑠 .
We conclude that lim 𝑠𝑛 = 𝑠 = 1.
(a) Is the Claim true or false? Explain carefully.
(b) Critique the proof. Is it complete? Does it prove the claim? What are the flaws, if any? Be very
specific.
9. (8 pts) Use the  definition (page 2 of the Week 3 Sequences document or page 44 of Lebl) to prove that
lim
(−1)𝑛 sin(𝑛2 )
√0.2 𝑛
𝑛→∞
= 0.
(1)n sin(n2 )
Consider the limit lim
:
n 
0.2n
Let   0 if there exist a   0 such that for 0  x  x0   and f  x   f  x0    then
lim f  x   f  x0 
x 
Choose n0   2 ,
0  n  2  
So,
(1) n sin(n 2 ) (1) sin( 2 )

0.2n
0.2
2
f  n   f  n0  

(1) (1) n sin(n 2 ) ( 1) sin( 2 )

n

0.2

(1) 1 1
 2
0.2 n 

(1)  2  n
2
0.2 n
2
1
 2  n  2  n 
n 2
(1)  2  n 


 2  n 



 For

  n    
(1)   n  
2

2
Page 7 of 10
Hence,
lim f  n   f  n0 
n 
 f  2 
(1) sin( 2 )

0.2
(1)  0 

0.2
0
2
sin   0
(1) n sin(n 2 )
0
n 
0.2n
Therefore, lim
Page 8 of 10
10. (7 pts) Let S be a subset of R. Claim: If int S  , then S is uncountable.
(Given a set S of real numbers, if its interior is a nonempty set, then the set S is uncountable.)
PART (a): Fill in the blanks below, to prove the Claim. Remark: The notation and terminology refer to my notes and
the textbooks which comprise our course resources.
Proof:
Let S be a subset of R. Suppose uncountable___________________.
Then there exists some real number x  int S.
By definition of interior point, there is a neighborhood W of x such that W  ___ S. (choose = , , or  )
By definition of neighborhood, W is a bounded _uncoutable___________ interval. (choose open or closed
)
Interval W has the same cardinality as the interval (0, 1), by Week 2 Homework, #_Week 2 Homework
_____.
The interval (0, 1) is ______uncountable______________ (choose countable or uncountable),
so W must be _______uncountable_____________ (choose countable or uncountable).
and since W _  __ S (choose = , , or  ), S must be __uncountable_______________ (choose countable
or uncountable).
PART (b): State the contrapositive of the Claim. Is the contrapositive true or false? (no explanation required)
11. (8 pts) Consider the following claim.
𝜋 −𝑛
Claim: The sequence (𝑠𝑛 ) given by 𝑠𝑛 = (− )
4
converges to 0.
Proof:
Let 𝑠𝑛 =
𝜋 −𝑛
(− 4) . The sequence of
ratios
𝑠𝑛+1
𝑠𝑛
=
𝜋 −𝑛+1
4
𝜋 −𝑛
(− )
4
(− )
=−
𝜋
4
𝜋
converges to − 4 < 1 .
so by the Ratio test for sequences (Week 3 sequence notes, page 9, or Lemma 2.2.12(iii) Lebl),
the sequence (𝑠𝑛 ) converges to 0.
(a) Is the Claim true or false? Explain carefully.
From the ratio test, for given sequence S n , if lim
n 
Sn 1
 L  1then lim S n  0 .then sequence S n
n 
Sn
converges to zero.
n
 4
Given S n     , so
 
Page 9 of 10
 4
Sn 1    
 
 n 1
Then
 n 1
lim
n 
S n 1
Sn
 4
 

 lim 
n
n 
 4
 
 
4
 lim 
n 


4

4
3.14
 1.273  1

n
 4
So, sequence Sn     not converges to zero. Hence claim is false.
 
(b) Critique the proof. Is it complete? Does it prove the claim? What are the flaws, if any? Be very specific.
Page 10 of 10