Advanced Calculus
Spring 2025
HOMEWORK 2
Due Date: Thursday, February 13, uploaded to Gradescope
Instructions: Collaboration with classmates is encouraged, but the final submission should
be your own work. Explain your steps or otherwise show all work for full credit.
Reference: Chapter 2.1-2.4 of Advanced Calculus
1. Prove or give a counterexample.
(a) If the sequence {an + bn } converges, then the sequences {an } and {bn } also
converge.
(b) If the sequences {an + bn } and {an } converge, then the sequence {bn } also converges.
(c) If the sequence {|an |} converges, then the sequence {an } also converges.
2. Prove that the set [1, ∞) is closed.
3. Prove that Z is closed.
4. Prove that the set of irrational numbers is not closed.
5. Prove or give a counterexample.
(a) A subsequence of a bounded sequence is bounded.
(b) Every sequence from the set Q ∩ (0, 1) has a convergent subsequence.
(c) If {ank } is a subsequence of a convergent sequence {an }, then {ank } converges.
(d) If a sequence {an } contains a convergent subsequence, then the full sequence
{an } converges.
(e) A subsequence of a monotone sequence is monotone.
6. Suppose the sequence {an } is monotone and that it has a convergent subsequence.
Use the Monotone Convergence Theorem to prove that {an } converges.
7. The following question was posed to ChatGPT, along with the response. If ChatGPT’s response is correct, you don’t have to do anything. If it is not correct, identify
what mistake(s) it made and provide a correct proof.
Question: Suppose a sequence {xn } contains a convergent subsequence. Is {xn }
bounded?
Yes, a sequence {xn } that contains a convergent subsequence is bounded. Let’s
prove this.
Proof:
We are given that the sequence {xn } contains a convergent subsequence. Suppose
that the convergent subsequence is {xnk }, where nk is a strictly increasing sequence
of indices. Let the limit of this subsequence be L, i.e.,
lim xnk = L.
k→∞
We want to show that the sequence {xn } is bounded.
Step 1: Boundedness of the convergent subsequence
Since the subsequence {xnk } converges to L, for any ϵ > 0, there exists an integer
N such that for all k ≥ N ,
|xnk − L| < ϵ.
This means that for all k ≥ N , the terms of the subsequence {xnk } lie within the
interval (L − ϵ, L + ϵ), which is a bounded interval.
Hence, the subsequence {xnk } is bounded.
Step 2: Boundedness of the entire sequence {xn } Now consider the entire sequence
{xn }. We know that the subsequence {xnk } is bounded, so there exists some constant M1 > 0 such that for all k,
|xnk | ≤ M1 .
Next, consider the terms of {xn } that are not part of the subsequence {xnk }.
We can focus on a finite number of terms of the sequence between the indices
n1 , n2 , . . . , nN −1 , and since this is a finite set of terms, there is a maximum absolute
value for these terms. Let M2 be the maximum of the absolute values of the first
N − 1 terms of {xn }:
M2 = max{|x1 |, |x2 |, . . . , |xnN −1 |}.
Thus, the entire sequence {xn } is bounded by M = max(M1 , M2 ). Specifically, for
every n,
|xn | ≤ M.
Conclusion:
Since we have shown that the entire sequence {xn } is bounded, we conclude that
if a sequence contains a convergent subsequence, it must be bounded.
Thus, the sequence {xn } is bounded.
(Optional, ungraded) If you would like additional problems to solve, consider the following from Fitzpatrick’s textbook. Difficult problems are marked with stars.
Section 2.1: 18∗∗
Section 2.2: 1,4
Section 2.3: 2,3,4
Section 2.4: 2,3,10