Math 2001: Real analysis
Class test, Oct 25, 2016
Name:
Student ID:
• You have 75 minutes to solve six problems.
• Show your complete work.
• The maximal score for this test is 25. Scores for individual problems add up to 27,
so you can get a perfect score even if you loose two points.
• No aids permitted: this is a closed book, closed notes test.
Problem Your score Max. score
1
4
2
4
3
4
4
5
5
5
6
5
Total
25
(1) Use mathematical induction to prove that
n2 (n + 1)2
1 + 2 + 3 + . . . + (n − 1) + n =
4
for all natural n.
3
3
3
3
3
For n = 1 the left-hand side is 13 = 1, while the right-hand side is 12 · 22 /4 = 1, so
the result holds true. Suppose it holds true for some n = k and check n = k + 1:
Solution:
13 + 23 + . . . + k 3 + (k + 1)3 = (13 + 23 + . . . + k 3 ) + (k + 1)3
k 2 (k + 1)2
+ (k + 1)3
4
(k + 2)2 (k + 1)2
(k 2 + 4(k + 1))(k + 1)2
=
,
=
4
4
=
as desired.
(2) (a) Give the denition of the supremum of a set S .
(b) Use this denition and the Archimedean property of theset or real numbers to
n
: n ∈ N is equal to 1.
prove that the supremum of the set S =
n+1
(a) A number m is the supremum of S if and only if m ≥ s for all s ∈ S , but for
any ε > 0 there is s ∈ S such that s > m − ε.
(b) Since n ≤ n + 1 and n + 1 > 0, we have n/(n + 1) ≤ 1 for all n ∈ N, so 1 is an upper
bound of S . Let ε > 0. We need to nd n ∈ N such that n/(n + 1) > 1 − ε. Equivalently:
1 − 1/(n + 1) > 1 − ε, or 1/(n + 1) < ε, or n + 1 > 1/ε, or n > 1/ε − 1. Since N is not
bounded above (this is the Archimedean property), such a number n ∈ N indeed exists, and
thus sup S = 1.
Solution:
(3) (a)
(b)
(c)
(d)
Give an example
Give an example
Give an example
Give an example
of a set
of a set
of a set
of a set
A ⊆ R such that int(cl A) 6⊆ A.
B ⊆ R such that int(cl B) 6⊇ B .
C ⊆ R such that cl(int C) 6⊆ C .
D ⊆ R such that cl(int D) 6⊇ D.
Note: In this problem you are only required to give expressions for the sets
int(cl A)
A, cl A,
etc. Do not prove these expressions!
One set to rule them all is A = (0, 1) ∪ (1, 2) ∪ {3}. Then cl A = [0, 2] ∪ {3} and
int(cl A) = (0, 2), which neither contains nor is contained in A. Similarly, int A = (0, 1)∪(1, 2)
and cl(int A) = [0, 2], which again neither contains nor is contained in A.
Solution:
Fun problem: What is the largest possible number of distinct sets one can get by repeatedly applying
interior and closure operations to a given set
complement problem.
S?
This is a variant of the Kuratowski's closure-
(4) (a) Give the denition of the limit of a sequence.
1
x
(b) Use this denition to prove that if lim an = , then lim (xan ) = 1.
n→∞
n→∞
Do not use any of the limit theorems here, refer directly to the denition.
(a) A number x is the limit of a sequence (an ) if for any ε > 0 there is N ∈ N such
that the inequality |an − x| < ε is satised for all n ≥ N .
(b) Clearly, x 6= 0 (otherwise the assumption does not make sense). Apply the denition for
ε/|x| instead of ε: there is N ∈ N such that |an − 1/x| < ε/|x| for all n ≥ N . Since |x| > 0,
we get |x| · |an − 1/x| < ε, that is, |an x − 1| < ε for all n ≥ N . This proves that (an x)
converges to 1.
Solution:
(5) Prove that
n2
= 1.
n→∞ n2 − (−1)n
lim
State briey each limit theorem that you use.
We have n2 /(n2 + 1) ≤ n2 /(n2 − (−1)n ) ≤ n2 /(n2 − 1) for n ≥ 2. Furthermore,
limn→∞ (n2 /(n2 +1) = limn→∞ (1/(1+1/n2 )) = 1/(1+0) = 1, and similarly limn→∞ (n2 /(n2 −
1) = 1. By the squeeze theorem, limn→∞ (n2 /(n2 − (−1)n ) = 1.
Squeeze theorem: if an ≤ bn ≤ cn for n large enough and limn→∞ an = limn→∞ cn = x, then
limn→∞ bn exists and it is equal to x.
Arithmetic rules for limits: limn→∞ (an /bn ) = (limn→∞ an )/(limn→∞ bn ), limn→∞ (an bn ) =
(limn→∞ an ) · (limn→∞ bn ), limn→∞ (an + bn ) = limn→∞ an + limn→∞ bn .
Simple limits: limn→∞ (1/n) = 0.
Solution:
(6) Let s1 = 0 and sn+1 =
and nd its limit.
s3n + 6
for n ∈ N. Prove that (sn ) is a convergent sequence
7
Note: the three solutions of the equation
x3 − 7x + 6 = 0
are
−3, 1
and
2.
Observe that s2 = (03 + 6)/7 = 6/7 > s1 . We claim that (sn ) is increasing, that
is, sn > sn−1 for n ≥ 2. We already noticed that for n = 2; furthermore, if sn > sn−1 , then
sn+1 = (s3n + 6)/7 > (s3n−1 + 6)/7 = sn . Our claim follows by induction.
If (sn ) is bounded, then it is convergent, and its limit x satises x = limn→∞ sn+1 =
limn→∞ (s3n + 6)/7 = (x3 + 6)/7, that is, x3 − 7x + 6 = 0. It follows that x = −3, x = 1 or
x = 2. Since −3 < s0 < 1 and (sn ) is increasing, we may suppose that the limit is 1; so let
us try prove that (sn ) is bounded above by 1.
Clearly, s1 < 1. Furthermore, if sn < 1, then sn+1 = (s3n + 6)/7 < (13 + 6)/7 = 1. Thus,
by induction, sn < 1 for all n, as desired.
We conclude that (sn ) is increasing, and bounded above by 1, and therefore convergent.
Furthermore, the limit is greater than s1 = 0, but not greater than the upper bound 1. The
only possibility is limn→∞ sn = 1.
Solution:
(Extra space for solutions)