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程序代写案例-HW 2
时间：2022-07-10
20 2. HW 2: Due via Gradescope Sun July 10 at 11:59 pm. Sections 2.1-2.4.
2.4. Exercises: 2.4.1[YL], 2.4.2[YL], 2.4.4[YL], 2.4.5[CL], 2.4.6[CL],
2.4.7[CL], 2.4.8[CL], 2.4.10[BC].
Section 2.4 Learning Objectives:
See the Typed notes on Section 2.4 in the Files section of Canvas for a summary
of this section.
(1) Definition of Subsequence.
(2) Proposition: If a sequence converges to a limit, then every subsequence con-
verges to the same limit.
(3) Every sequence has a monotone subsequence.
(4) Every bounded sequence has a convergent subsequence.
(5) Sequential Compactness Theorem: Every sequence of numbers a ≤ xn ≤ b
has a subsequence that converges to some number a ≤ x∞ ≤ b.
Exercise 2.4.1. [Sol. By YL]
(a). True. Notice that the bound of a sequence applies to every term of the
sequence ai < M , for some number M . Since this property applies to every
term in any subsequence, the subsequence is also bounded (by the same M).
(b). True. Notice that the monotonicity of a sequence applies to every term of
the sequence, that (in the case of monotone increasing) ai ≥ aj if i < j, so
it also applies to any subsequence, since the order is preserved (i < j implies
ni < nj).
(c). True. For any ε > 0, the existence of N such that |an − a| < ε for n ≥ N
implies that |ank − a| < ε for k ≥ N , since nk ≥ k for all k ∈ N for any
subsequence.
(d). False. Consider an = (−1)n. The subsequence consisting of the terms with
even indices converges to 1. But {an} does not converge.
Exercise 2.4.2. [Sol. By YL]
(a). True. Notice that any sequence defined in (0, 1) will be bounded by 0 below
and 1 above. Then by Theorem 2.33, we know that there exist a convergent
subsequence. Note, however, the subsequence may converge to a number
outside of the set.
(b). False. Consider an =
1
n , where n ≥ 2, which converges to 0.
(c). False. Consider an = n (Of course, integers are rational numbers too). To
show there is no subsequence converging, suppose for contradiction that a
subsequence ank converges to a. By definition of convergence, there exist N
such that |ank−a| < 1 for k > N.We also notice that, nk ≥ k, in other words,
ank ≥ ak. Now picking m = max(N +1, ⌊a⌋+2), we see that anm ≥ am = m,
so anm − a ≥ m− a > 1, a contradiction
(d). True. Otherwise say it converges to some a < 0, then for ε = |a|2 there exist
N , such that when n > N, |an − a| < |a|2 . However, this implies an < 0, and
a contradiction occurs.
2.4. Exercises: 2.4.1[YL], 2.4.2[YL], 2.4.4[YL], 2.4.5[CL], 2.4.6[CL], 2.4.7[CL], 2.4.8[CL], 2.4.10[BC].21
(e). False. Consider an = n, and we just shown that there is no subsequence
converges.
Exercise 2.4.4. [Sol. By YL]
(a). We notice that the sequence is monotonically decreasing, so the all indices are
peak.
(b). We notice that for every even index n, an = 1 ≥ ai = 1 for any even i ∈ N,
and an = 1 ≥ aj = −1 for any odd j ∈ N. So each even index is a peak index.
Clearly, each odd integer fails to be a peak index. Indeed, if n is odd, then
an = −1 < 1 = an+1. Hence, the peak indices are exactly the even indices.
(c). We notice that the sequence is unbounded in positive direction, so there could
not exist any n such that an ≥ ai for all i > n. (In our case, we may consider
an+1 and an+2 for example; one of them is larger than an.)
(d). Similarly to (b), the peak indices are exactly the even indices.
Exercise 2.4.5. [Sol. by CL] Let {an} be a strictly increasing sequence. By
definition, for any n, we have an < an+1 and therefore n cannot be a peak index.2
Exercise 2.4.6. [Sol. by CL] Let {an} be monotonically decreasing, this
means for any n ∈ N, an ≥ an+1. And it follows from induction that for any n ≤ m
we have an ≥ am. So each n ∈ N is clearly a peak index. 2
Exercise 2.4.7. [Sol. by CL] (⇒) This is trivial since: If a monotone sequence
is bounded, then the sequence itself is a bounded subsequence.
(⇐) It suffices to prove this for monotonically increasing sequences. (The case
of monotonically decreasing sequences is exactly analogous.)
Suppose {an} is a monotonically increasing sequence that has a bounded sub-
sequence {ank}. Then there is a number M such that
ank ≤M for any k ∈ N.
For any k ∈ N we have k ≤ nk, and hence
ak ≤ ank ≤M,
which means the sequence {an} is also bounded by M from above. But any mono-
tonically increasing sequence is also bounded by its first term from below, so {an}
is bounded both below and above and hence it is bounded. 2
Exercise 2.4.8. [Sol. by CL] We know that any convergent sequence is
bounded, so {an} has a bounded subsequence. Use Exercise 2.4.7 we can see
that {an} itself is bounded. So by The Monotone Convergence Theorem
{an} converges. 2
Exercise 2.4.10. [Sol. by BC] Statement:
{an} does not converge to a if and only if there exists ε > 0 and there exists a
subsequence {ank} such that
(2.13) |ank − a| ≥ ε for every index k ∈ N.
22 2. HW 2: Due via Gradescope Sun July 10 at 11:59 pm. Sections 2.1-2.4.
Proof of the forward implication ⇒. Suppose that {an} does not converge
to a. Then the negation of the following definition is true: For every ε > 0 there
exists N ∈ N such that |an − a| < ε for all n ≥ N .
That is, we have:
There exists ε > 0 such that for any N ∈ N there exists n ≥ N such that
|an − a| ≥ ε.
We need to deduce from this statement the statement about subsequences. In
fact, this is easy:
By the statement with N = 1, there exists n1 such that |an1 − a| ≥ ε.
Next, by the statement with N = n1 + 1, there exists n2 > n1 such that
|an2 − a| ≥ ε.
By the same (similar) argument, there exists n3 > n2 such that |an3 − a| ≥ ε.
By continuing in this way we obtain the desired subsequence. As usual, we
omit the rigorous proof, which is by induction.
Proof of the backward implication ⇐. Suppose that there exists ε > 0
and there exists a subsequence {ank} such that
(2.14) |ank − a| ≥ ε for every index k ∈ N.
We want to show that {an} does not converge to a. Suppose for a contradiction
that {an} converges to a. Then, for ε as in the statement above, by the definition
of an → a we have that there exists N such that |ak − a| < ε for all k ≥ N . Since
{ank} is a subsequence of {an}, we have that nk ≥ k. Therefore, nN ≥ N , which
implies that |anN − a| < ε, which is a contradiction. This completes the proof.
2.4. Exercises: 2.4.1[YL], 2.4.2[YL], 2.4.4[YL], 2.4.5[CL], 2.4.6[CL],
2.4.7[CL], 2.4.8[CL], 2.4.10[BC].
Section 2.4 Learning Objectives:
See the Typed notes on Section 2.4 in the Files section of Canvas for a summary
of this section.
(1) Definition of Subsequence.
(2) Proposition: If a sequence converges to a limit, then every subsequence con-
verges to the same limit.
(3) Every sequence has a monotone subsequence.
(4) Every bounded sequence has a convergent subsequence.
(5) Sequential Compactness Theorem: Every sequence of numbers a ≤ xn ≤ b
has a subsequence that converges to some number a ≤ x∞ ≤ b.
Exercise 2.4.1. [Sol. By YL]
(a). True. Notice that the bound of a sequence applies to every term of the
sequence ai < M , for some number M . Since this property applies to every
term in any subsequence, the subsequence is also bounded (by the same M).
(b). True. Notice that the monotonicity of a sequence applies to every term of
the sequence, that (in the case of monotone increasing) ai ≥ aj if i < j, so
it also applies to any subsequence, since the order is preserved (i < j implies
ni < nj).
(c). True. For any ε > 0, the existence of N such that |an − a| < ε for n ≥ N
implies that |ank − a| < ε for k ≥ N , since nk ≥ k for all k ∈ N for any
subsequence.
(d). False. Consider an = (−1)n. The subsequence consisting of the terms with
even indices converges to 1. But {an} does not converge.
Exercise 2.4.2. [Sol. By YL]
(a). True. Notice that any sequence defined in (0, 1) will be bounded by 0 below
and 1 above. Then by Theorem 2.33, we know that there exist a convergent
subsequence. Note, however, the subsequence may converge to a number
outside of the set.
(b). False. Consider an =
1
n , where n ≥ 2, which converges to 0.
(c). False. Consider an = n (Of course, integers are rational numbers too). To
show there is no subsequence converging, suppose for contradiction that a
subsequence ank converges to a. By definition of convergence, there exist N
such that |ank−a| < 1 for k > N.We also notice that, nk ≥ k, in other words,
ank ≥ ak. Now picking m = max(N +1, ⌊a⌋+2), we see that anm ≥ am = m,
so anm − a ≥ m− a > 1, a contradiction
(d). True. Otherwise say it converges to some a < 0, then for ε = |a|2 there exist
N , such that when n > N, |an − a| < |a|2 . However, this implies an < 0, and
a contradiction occurs.
2.4. Exercises: 2.4.1[YL], 2.4.2[YL], 2.4.4[YL], 2.4.5[CL], 2.4.6[CL], 2.4.7[CL], 2.4.8[CL], 2.4.10[BC].21
(e). False. Consider an = n, and we just shown that there is no subsequence
converges.
Exercise 2.4.4. [Sol. By YL]
(a). We notice that the sequence is monotonically decreasing, so the all indices are
peak.
(b). We notice that for every even index n, an = 1 ≥ ai = 1 for any even i ∈ N,
and an = 1 ≥ aj = −1 for any odd j ∈ N. So each even index is a peak index.
Clearly, each odd integer fails to be a peak index. Indeed, if n is odd, then
an = −1 < 1 = an+1. Hence, the peak indices are exactly the even indices.
(c). We notice that the sequence is unbounded in positive direction, so there could
not exist any n such that an ≥ ai for all i > n. (In our case, we may consider
an+1 and an+2 for example; one of them is larger than an.)
(d). Similarly to (b), the peak indices are exactly the even indices.
Exercise 2.4.5. [Sol. by CL] Let {an} be a strictly increasing sequence. By
definition, for any n, we have an < an+1 and therefore n cannot be a peak index.2
Exercise 2.4.6. [Sol. by CL] Let {an} be monotonically decreasing, this
means for any n ∈ N, an ≥ an+1. And it follows from induction that for any n ≤ m
we have an ≥ am. So each n ∈ N is clearly a peak index. 2
Exercise 2.4.7. [Sol. by CL] (⇒) This is trivial since: If a monotone sequence
is bounded, then the sequence itself is a bounded subsequence.
(⇐) It suffices to prove this for monotonically increasing sequences. (The case
of monotonically decreasing sequences is exactly analogous.)
Suppose {an} is a monotonically increasing sequence that has a bounded sub-
sequence {ank}. Then there is a number M such that
ank ≤M for any k ∈ N.
For any k ∈ N we have k ≤ nk, and hence
ak ≤ ank ≤M,
which means the sequence {an} is also bounded by M from above. But any mono-
tonically increasing sequence is also bounded by its first term from below, so {an}
is bounded both below and above and hence it is bounded. 2
Exercise 2.4.8. [Sol. by CL] We know that any convergent sequence is
bounded, so {an} has a bounded subsequence. Use Exercise 2.4.7 we can see
that {an} itself is bounded. So by The Monotone Convergence Theorem
{an} converges. 2
Exercise 2.4.10. [Sol. by BC] Statement:
{an} does not converge to a if and only if there exists ε > 0 and there exists a
subsequence {ank} such that
(2.13) |ank − a| ≥ ε for every index k ∈ N.
22 2. HW 2: Due via Gradescope Sun July 10 at 11:59 pm. Sections 2.1-2.4.
Proof of the forward implication ⇒. Suppose that {an} does not converge
to a. Then the negation of the following definition is true: For every ε > 0 there
exists N ∈ N such that |an − a| < ε for all n ≥ N .
That is, we have:
There exists ε > 0 such that for any N ∈ N there exists n ≥ N such that
|an − a| ≥ ε.
We need to deduce from this statement the statement about subsequences. In
fact, this is easy:
By the statement with N = 1, there exists n1 such that |an1 − a| ≥ ε.
Next, by the statement with N = n1 + 1, there exists n2 > n1 such that
|an2 − a| ≥ ε.
By the same (similar) argument, there exists n3 > n2 such that |an3 − a| ≥ ε.
By continuing in this way we obtain the desired subsequence. As usual, we
omit the rigorous proof, which is by induction.
Proof of the backward implication ⇐. Suppose that there exists ε > 0
and there exists a subsequence {ank} such that
(2.14) |ank − a| ≥ ε for every index k ∈ N.
We want to show that {an} does not converge to a. Suppose for a contradiction
that {an} converges to a. Then, for ε as in the statement above, by the definition
of an → a we have that there exists N such that |ak − a| < ε for all k ≥ N . Since
{ank} is a subsequence of {an}, we have that nk ≥ k. Therefore, nN ≥ N , which
implies that |anN − a| < ε, which is a contradiction. This completes the proof.
