Due to COVID-19, we have changed this course to an online format using Canvas,
so I've stopped updating this page since 03/16/2020.
# Analysis I

[Lecture 1]

[Lecture 2]

[Lecture 3]

[Lecture 4]

[Lecture 5]

[Lecture 6]

[Lecture 7]

you should redo the exercises that the grader commented on.

This will be necessary for your receiving at least 55% on the this 3rd homework.

[Lecture 8]

[Lecture 9]

[Lecture 10]

[Lecture 11]

[Lecture 12]

[Lecture 13]

[Lecture 14]

[Lecture 15]

[Lecture 16]

[Lecture 17]

[Lecture 18]

[Lecture 19]

[Lecture 20]

[Lecture 21]

Topics covered: Lectures 1-12

(Exam is closed-book.)

Sample Exam // Solutions

Exam // Solutions

If $\sup B < \inf A$, then any number in the open interval $(\sup B, \inf A)$ is a lower bound of $A$, hence must belong to $B$; but an element of $B$ cannot be strictly greater than $\sup B$.

If $\inf A< \sup B$, let $c$ be any number that belongs to the interval $(\inf A, \sup B)$. Therefore, by the meaning of $\inf$ and $\sup$, there exists $a\in A$ and $a< c$; there also exists $b\in B$ and $b>c$. This leads to $a < b$. However, this cannot happen, since $b$ is a lower bound of $A$.

For $f$, note that each number $x\in (0,1)$ admits a binary representation. For example, $$ 3/4 = 1/2+1/4 = (.110000\cdots)_2. $$ Now suppose that some $x\in (0,1)$ has a binary representation $(.0100101000\cdots)_2$, in which precisely the 2nd, 5th and 7th digits are 1, we want to define $f(x)$ to be the subset $\{2,5,7\}$ of $\mathbb{N}$. Using this idea, you'll find a construction of $f$. Is this $f$ 1-1?

Now suppose that you have a subset $S\subset \mathbb{N}$. You may associate a binary number to $S$ in the following way. You start by checking whether $1\in S$. If $1\in S$, you write down .01, if $1\notin S$, you write .00. Next, if $2\in S$, you continue as .**01; if $2\notin S$, you continue as .**00, etc. In this manner, for example, the subset $\{1,3,4,7\}\subset \mathbb{N}$ corresponds to the binary number $$ (.01000101000001)_2. $$ Doing this avoids generating a binary number that has all digits being 1 starting from some point. Does this construction give a map $g$ that's 1-1?

If $x_1 = 0$, then $2\in S$ and $1\notin S$;

If $x_1 = 1$, then $1\in S$ and $2\notin S$;

If $x_2 = 0$, then $4\in S$ and $3\notin S$;

If $x_2 = 1$, then $3\in S$ and $4\notin S$;

etc.

This guarantees that as long as $(x_1,x_2,\dots)$ and $(y_1,y_2,\dots)$ are not identical, the corresponding subsets constructed admit no inclusive relations between them.

(b) One can define $z_n = \inf\{a_k: k\ge n\}$. If $(a_n)$ is bounded, then so is $z_n$. Moreover, $(z_n)$ satisfies $z_{n+1}\ge z_n$ for all $n$, since as $n$ increases you're taking infimum for a smaller set. It follows that $(z_n)$ converges. We just call the limit $\liminf a_n$.

(c) Indeed, for each $n$, $z_n\ge y_n$. By the order limit properties, we have $\lim z_n\le \lim y_n$, that is, $\liminf a_n\le \limsup a_n$.

(d) For each $n$ we have $$z_n\le a_n\le y_n.$$ Therefore, if $\liminf a_n = \limsup a_n =a$, then the squeeze theorem (Exercise 2.3.3) implies that $\lim a_n = a$. Conversely, if $\lim a_n = a$, then for any $\epsilon >0$, there exists $N$ such that $|a_n - a|<\epsilon$ for all $n\ge N$. This implies that, for such $n$, both $\liminf a_n$ and $\liminf a_n$ belong to the interval $[a-\epsilon, a+\epsilon]$. Since $\epsilon$ is arbitrary, we must have that $\liminf a_n = \liminf a_n = a$.

For part (d), you want to show that if $(a_n)$ satisfies: there exists a subsequence converging to each $1/n$, then there is a subsequence converging to $0$. To see this, suppose that $(a_{n^{(k)}_\ell})_{\ell = 1}^\infty$ is a subsequence of $(a_n)$ that converges to $1/k$. We can construct a new sequence in the following manner.

First, choose $\ell_1$ such that $0 < a_{n^{(1)}_{\ell_1}}<2$;

next, choose $\ell_2$ such that $0 < a_{n^{(2)}_{\ell_2}}<1$ and $n^{(2)}_{\ell_2}>n^{(1)}_{\ell_1}$,

$\dots$

In the $k$-th step, choose $\ell_k$ such that $0< a_{n^{(k)}_{\ell_k}}<2/k$ and $n^{(k)}_{\ell_k}>n^{(k-1)}_{\ell_{k-1}}$.

The subsequence $$\left( a_{n^{(k)}_{\ell_k}}\right)_{k = 1}^\infty$$ is a subsequence of $(a_n)$ that converges to $0$.

For part (d), we can assume that the original sequence is increasing. Suppose that $(a_{n_k})$ is a subsequence whose limit is $a$. It is necessary that $a_{n_k}\le a$ for all $k\in \mathbb{N}$.

Now for any $\epsilon>0$, there exists a $K$ such that for all $k\ge K$, $|a_{n_k} - a|<\epsilon$. For any $m\ge n_k$, we can always find an $\ell$ such that $\ell>k$ and $m\le \ell$. Thus, by the assumption, $$ a_{n_k}\le a_m\le a_{n_\ell}\le a $$ for all $m\ge n_k$. In particular, $|a_m - a|<\epsilon$. This proves that $(a_n)\rightarrow a$.

In the infinite case, you can find a subsequence consisting of peak terms. This subsequence must be decreasing.

(ii) You may find the following inequality useful $$|(x_{n+1} + y_{n+1}) - (x_n + y_n) |\le |x_{n+1} - x_n|+|y_{n+1} - y_n|.$$

For part (b), by assumption, $\sum b_n$ is convergent, so the sequence of its partial sums is bounded. It follows that the sequence of partial sums of $\sum a_n$ is bounded and increasing. The conclusion follows by the Monotone Convergence Theorem.

(c) Impossible by an algebraic limit property. Consider $\sum (x_n+y_n) - \sum x_n$.

(d) Consider $x_n$ that satisfies $x_n = \frac{1}{n}$ when $n$ is even, and $x_n = 0$ when $n$ is odd.

(b) Since $\lim n^2a_n$ exists, the sequence $(n^2a_n)$ is bounded, say $$|n^2a_n| = n^2a_n \le M $$ for all $n\in \mathbb{N}.$ Therefore $a_n<\frac{M}{n^2}$ for all $n$. Since $\sum\frac{1}{n^2}$ converges, by comparison test, $\sum a_n$ also converges.

(b) False. Example: $a_n = \frac{(-1)^n}{\sqrt{n}}$, $b_n = \frac{(-1)^n}{\sqrt{n}}$.

(c) True. If $\sum n^2a_n$ converges, then $\lim n^2 |a_n| = 0$. By Exercise 2.7.7(b), which also works when $a_n\ge 0$, $\sum|a_n|$ converges. In other words, $\sum a_n$ converges absolutely.

If we define: $$ S_m:= \{a\in S: \epsilon_a\ge 1/m\}, $$ all points in $S$ can thus be organized in the following manner: $$ S = \bigcup_{m = 1}^\infty S_m. $$ Since $S$ is uncountable, there exists an $M\in \mathbb{N}$ such that $S_M$ is uncountable. All points in $S_M$ belong to $$ \mathbb{R} = \bigcup_{n\in \mathbb{Z}} \left[\frac{n}{2M}, \frac{n+1}{2M}\right], $$ which is a countable union. So one of the intervals $\left[\frac{n}{2M}, \frac{n+1}{2M}\right]$ must contain at least two numbers, say, $a,b$, in $S_M$, which implies that $a,b$ are at most $1/2M$ apart. On the other hand, since $a,b$ both belong to $S_M$ and are distinct, they need to be at least $1/M$ apart. This is a contradiction.

# Analysis I

CU Boulder

Instructor: Yuhao Hu

Email: yuhao.hu@colorado.edu

Office: Math 225

Office Hours: WF 4:00-5:45pm

Lectures: MWF 9:00-9:50am at HLMS 255

## Overview

In some sense, mathematical analysis was introduced as a new language of mathematics in order
to bring rigor to the subject of calculus (e.g., to understand the meaning of "limit").
The development of this 'language', like the invention of a telescope,
allows one to make more accurate observations,
put forward precise questions and propositions, and construct useful theories and techniques.
These new questions, theorems, etc., in turn,
enrich not only the language but also the body of mathematics that's tied to it.
Almost 200 years after Cauchy's 1821 book

This course is a beginning step towards understanding analysis. We start with a bit of set theory, followed by an introduction to the structure of real numbers. We develop language/tools that allow us to make sense of limit, which further allows us to take a more precise view of functions and their continuity. This course ends with basic theories of differentiation and integration.

*Cours d'analyse*, analysis is now a 'universe' in itself.This course is a beginning step towards understanding analysis. We start with a bit of set theory, followed by an introduction to the structure of real numbers. We develop language/tools that allow us to make sense of limit, which further allows us to take a more precise view of functions and their continuity. This course ends with basic theories of differentiation and integration.

## Textbook

**T1.**

*Understanding Analysis, 2nd ed.*by Stephen Abbott

**T2. (Optional)**

*Analysis I, 3rd ed.*by Terence Tao

## Schedule

(weekly, only the first meeting of the week is dated; "S" stands for "Section" in**T1**)**01/13**Number systems, sets (S1.1); functions, logic (S1.2)

**01/22**Characterization of $\mathbb{R}$, axiom of completeness (S1.3); Consequences of completeness (S1.4)

**01/27**Cardinality (S1.5); Cardinality (cont.); Cantor's theorem (S1.6) and infinite sum puzzles (S2.1)

**02/03**Limit of a sequence (S2.2); limit properties (S2.3); Monotone convergence theorem (S2.4)

**02/10**Bolzano-Weierstraß theorem (S2.5); the Cauchy criterion (S2.6); Infinite Series (S2.7)

**02/17 (Midterm I on Friday)**Open and closed sets (S3.2)

**02/24**Compact sets (S3.3); Cantor set (S3.1)

**03/02**Functional limits (S4.2); continuous functions (S4.3); uniform continuity (S4.4)

**03/09**

**03/16**

**03/23-27 (Spring break; no class)**

**03/30**

**04/06 (Midterm II on Friday)**

**04/13**

**04/20**

**04/27**

## Homework

(Due on each**Friday**, unless specified otherwise.)**Homework 1**(Lectures 1-2) due Jan. 24

[Lecture 1]

**Reading:**pp. 1-7 (note: The default textbook is

*'Understanding Analysis'*.)

**Exercises:**[p. 11, Sec. 1.2: 1, 2, 3(a,c)]

[Lecture 2]

**Reading:**pp. 7-11

**Exercises:**[p. 12, Sec. 1.2: 6(a,b), 8, 10, 11, 12]

**Homework 2**(Lectures 3-4) due Jan. 31

[Lecture 3]

**Reading:**pp. 14-18

**Exercises:**[pp. 18-19, Sec. 1.3: 1(a), 2(a,c), 3, 5, 7, 8(c,d), 9]

[Lecture 4]

**Reading:**pp. 20-23

**Exercises:**[p. 24, Sec. 1.4: 2, 3, 4, 5, 6 ,8]

**Homework 3**(Lectures 5-7) due Feb. 7

[Lecture 5]

**Reading:**pp. 25-29

**Exercises:**[pp. 29-31, Sec. 1.5: 4, 5, 6, 7(b), 8, 10]

[Lecture 6]

**Reading:**(see previous reading)

**Exercises:**[p. 31, Sec. 1.5: 9]

[Lecture 7]

**Reading:**pp. 32-35 and 39-42

**Exercises:**[pp. 34-35, Sec. 1.6: 6, 7, 9, 10]

**Homework correction:**If you received a 45 or lower on the 1st homework,

you should redo the exercises that the grader commented on.

This will be necessary for your receiving at least 55% on the this 3rd homework.

**Homework 4**(Lectures 8-10) due Feb. 14

[Lecture 8]

**Reading:**pp. 42-47

**Exercises:**[pp. 47-48, Sec. 2.2: 1, 2, 4, 7]

[Lecture 9]

**Reading:**pp. 49-54

**Exercises:**[pp. 54-55, Sec. 2.3: 1, 2, 3, 4, 6, 7, 10]

[Lecture 10]

**Reading:**pp. 56-58 (may skip Theorem 2.4.6); Basel problem (link)

**Exercises:**[pp. 59-61, Sec. 2.4: 1, 2, 3, 7]

**Homework 5**(Lectures 11-13) due Feb. 21

[Lecture 11]

**Reading:**pp. 62-64

**Exercises:**[pp. 65-66, Sec. 2.5: 1, 2(c,d), 5, 8]

[Lecture 12]

**Reading:**pp. 66-68

**Exercises:**[p. 70, Sec. 2.6: 2, 4(a,b), 5]

[Lecture 13]

**Reading:**pp. 71-75 (may skip "rearrangements")

**Exercises:**[p. 77, Sec. 2.7: 2, 3, 4, 7, 8]

**Homework 6**(Lectures 14-15) due Feb. 28

[Lecture 14]

**Reading:**no new readings

**Exercises:**[p. 78, Sec. 2.7: 9, 11]

[Lecture 15]

**Reading:**pp. 88-93

**Exercises:**[pp. 93-94 , Sec. 3.2: 2, 3, 4, 6(a,c), 10]

**Homework 7**(Lectures 16-18) due Mar. 06

[Lecture 16]

**Reading:**no new readings

**Exercises:**[p. 94, Sec. 3.2: 7, 8, 11, 13]

[Lecture 17]

**Reading:**pp. 96-97

**Exercises:**[pp. 99-100, Sec. 3.3: 2(a,b,d,e), 5, 6, 8]

[Lecture 18]

**Reading:**pp. 85-88, p. 98

**Exercises:**[p. 100, Sec. 3.3: 4, 12, 13] (Note: For 3.3.12, you may use the Heine-Borel Theorem.)

**Homework 8**(Lectures 19-21) due Mar. 13

[Lecture 19]

**Reading:**pp. 115-119

**Exercises:**[pp. 120-122, Sec. 4.2: 2(a,b), 5(b,d), 6, 7, 8(a,b), 10]

[Lecture 20]

**Reading:**pp. 122-126

**Exercises:**[pp. 126-129, Sec. 4.3: 1, 2, 4, 6, 7(a), 8, 9]

[Lecture 21]

**Reading:**no new readings

**Exercises:**no new exercises

## Exams

**Midterm I:**2/21 Friday

Topics covered: Lectures 1-12

(Exam is closed-book.)

Sample Exam // Solutions

Exam // Solutions

**Midterm II:**4/10 Friday

**Final:**TBD

## Grading

Two lowest Homework grades will be dropped.

**Homework:**30%**Midterm I:**20%**Midterm II:**20%**Final Exam:**30%## Frequently asked questions

(You may find in this section hints/comments for selected homework exercises.)**1.2.6(b)**The inequality $(a+b)^2\le (|a|+|b|)^2$ reduces to $ab\le |ab|$.

**1.3.3(a)**To see that $\sup B = \inf A$, you may argue by showing that both $\sup B < \inf A$ and $\sup > \inf A$ are impossible.

If $\sup B < \inf A$, then any number in the open interval $(\sup B, \inf A)$ is a lower bound of $A$, hence must belong to $B$; but an element of $B$ cannot be strictly greater than $\sup B$.

If $\inf A< \sup B$, let $c$ be any number that belongs to the interval $(\inf A, \sup B)$. Therefore, by the meaning of $\inf$ and $\sup$, there exists $a\in A$ and $a< c$; there also exists $b\in B$ and $b>c$. This leads to $a < b$. However, this cannot happen, since $b$ is a lower bound of $A$.

**1.4.8(d)**The answer is "impossible". Note that each $$ J_n := \bigcap_{k = 1}^n I_k $$ is a closed interval. They are also nonempty by assumption. Moreover, the sequence $J_n$ is nested. So $\cap_{n = 1}^\infty J_n \ne \emptyset$. Finally, you need to show $$ \bigcap_{n = 1}^\infty J_n = \bigcap_{n = 1}^\infty I_n $$ to complete the proof.

**1.5.4(c)**You may construct a map $f: [0,1)\rightarrow(0,1)$ by requiring: $$ f(0) = 1/2, $$ $$ f(1/2) = 1/3, $$ $$ f(1/3) = 1/4, $$ $$\cdots$$ and $f(x) = x$ if $x$ is not zero or of the form $1/n$ ($n\in \mathbb{N}$). You need to argue that this $f$ is 1-1 and onto.

**1.5.7(b)**Each element in $S$ is of the form $(x,y)$, where $x,y\in (0,1)$. So you can express them uniquely as $$ x = 0.x_1x_2x_3\cdots, \quad y = 0.y_1y_2y_3\cdots $$ in such a way that however far you look, you always see digits that are not $9$. This is to avoid a representation such as $0.4999\cdots$. Then you may define a map $f: S\rightarrow (0,1)$ by demanding $$ f(x,y) = 0.x_1y_1x_2y_2\cdots. $$ You need to show that this map is 1-1 but not onto. For the "onto" part, you may ask, by the construction above, which pair $(x,y)$ is mapped to the number $0.1219191919\cdots$?

**1.5.8**Write $B$ as $$ B = \bigcup_{n = 1}^\infty \left(B\cap \left[\frac{1}{n+1}, \frac{1}{n}\right]\right) \cup (B\cap [1,\infty)) $$ This allows you to see $B$ as a countable union of sets. Now suppose that $B$ is uncountable, one of the sets that occur in the union above must be uncountable (because a countable union of countable sets remains countable). Suppose that, for some $N$, $$ B\cap \left[\frac{1}{N+1}, \frac{1}{N}\right]$$ is uncountable. Thus, in $B$, there are uncountably many elements that are at least $1/(N+1)$. Choosing any $3N+3$ among them would make a sum greater than $2$. Contradiction.

**1.5.10(a)**The set $C$ can be seen as a countable union: $$ C = \bigcup_{n=1}^\infty \left(C\cap \left[\frac{1}{n}, 1\right]\right) \cup (C\cap\{0\}). $$ Since $C$ is uncountable, one of the sets that appear in the union above must be uncountable. In other words, there exists an $n$ such that $C \cap [1/n, 1]$ is uncountable.

**1.5.10(b)**The answer is "no". Suppose that $C\cap [\alpha, 1]$ is uncountable. Then by part (a), essentially, there exists some $\beta\in (\alpha, 1)$ such that $C\cap (\beta,1)$ is uncountable. Let $\gamma$ be any number that belongs to $(\alpha,\beta)$. It follows that $C\cap[\gamma,1]$ is uncountable. Since $\gamma>\alpha$, this violates the meaning of $\alpha$.

**1.6.9**From working on Ex 1.5.7, reading the results in Ex 1.5.11, and knowing that $\mathbb{R}\sim (0,1)$, you may realize that, to show that $P(\mathbb{N})\sim \mathbb{R}$, it suffices to construct two maps: $$ f: (0,1)\rightarrow P(\mathbb{N}) $$ and $$ g: P(\mathbb{N})\rightarrow (0,1), $$ such that both $f$ and $g$ are 1-1.

For $f$, note that each number $x\in (0,1)$ admits a binary representation. For example, $$ 3/4 = 1/2+1/4 = (.110000\cdots)_2. $$ Now suppose that some $x\in (0,1)$ has a binary representation $(.0100101000\cdots)_2$, in which precisely the 2nd, 5th and 7th digits are 1, we want to define $f(x)$ to be the subset $\{2,5,7\}$ of $\mathbb{N}$. Using this idea, you'll find a construction of $f$. Is this $f$ 1-1?

Now suppose that you have a subset $S\subset \mathbb{N}$. You may associate a binary number to $S$ in the following way. You start by checking whether $1\in S$. If $1\in S$, you write down .01, if $1\notin S$, you write .00. Next, if $2\in S$, you continue as .**01; if $2\notin S$, you continue as .**00, etc. In this manner, for example, the subset $\{1,3,4,7\}\subset \mathbb{N}$ corresponds to the binary number $$ (.01000101000001)_2. $$ Doing this avoids generating a binary number that has all digits being 1 starting from some point. Does this construction give a map $g$ that's 1-1?

**1.6.10(b)**You may consider representing a function $f: \mathbb{N}\rightarrow\{0,1\}$ as a binary number in $[0,1]$.

**1.6.10(c)**By part (b), all functions from $N$ to $\{0,1\}$ form a uncountable set. We can associate to each such function, represented as a sequence of 0 and 1: $(x_1,x_2,\dots)$, a subset $S$ in $\mathbb{N}$ in the following way.

If $x_1 = 0$, then $2\in S$ and $1\notin S$;

If $x_1 = 1$, then $1\in S$ and $2\notin S$;

If $x_2 = 0$, then $4\in S$ and $3\notin S$;

If $x_2 = 1$, then $3\in S$ and $4\notin S$;

etc.

This guarantees that as long as $(x_1,x_2,\dots)$ and $(y_1,y_2,\dots)$ are not identical, the corresponding subsets constructed admit no inclusive relations between them.

**2.3.2(b)**Since $x_n\rightarrow 2$, for any $\epsilon >0$, there exists $N$ such that for any $n\ge N$, both $|x_n - 2|<\epsilon$ and $x_n>1$ hold. Therefore, for such $n$, we have $$ \left|\frac{1}{x_n} - \frac{1}{2}\right| = \left|\frac{x_n-2}{2x_n}\right| \le \frac{\epsilon}{2}. $$ To conclude, $1/x_n\rightarrow 1/2$.

**2.4.7**(a) Since $(a_n)$ is bounded, so is $(y_n)$. Furthermore, $y_{n+1}\le y_{n}$ for all $n$ since as $n$ increases you're taking supremum for a smaller set. By the Monotone Convergence Theorem, $(y_n)$ converges.

(b) One can define $z_n = \inf\{a_k: k\ge n\}$. If $(a_n)$ is bounded, then so is $z_n$. Moreover, $(z_n)$ satisfies $z_{n+1}\ge z_n$ for all $n$, since as $n$ increases you're taking infimum for a smaller set. It follows that $(z_n)$ converges. We just call the limit $\liminf a_n$.

(c) Indeed, for each $n$, $z_n\ge y_n$. By the order limit properties, we have $\lim z_n\le \lim y_n$, that is, $\liminf a_n\le \limsup a_n$.

(d) For each $n$ we have $$z_n\le a_n\le y_n.$$ Therefore, if $\liminf a_n = \limsup a_n =a$, then the squeeze theorem (Exercise 2.3.3) implies that $\lim a_n = a$. Conversely, if $\lim a_n = a$, then for any $\epsilon >0$, there exists $N$ such that $|a_n - a|<\epsilon$ for all $n\ge N$. This implies that, for such $n$, both $\liminf a_n$ and $\liminf a_n$ belong to the interval $[a-\epsilon, a+\epsilon]$. Since $\epsilon$ is arbitrary, we must have that $\liminf a_n = \liminf a_n = a$.

**2.5.1(c,d)**For part (c), try the following sequence $$(1,1,1/2,1,1/2,1/3,1, 1/2,1/3,1/4,\dots)$$ Note that each $1/n$ occurs infinitely many times in this sequence.

For part (d), you want to show that if $(a_n)$ satisfies: there exists a subsequence converging to each $1/n$, then there is a subsequence converging to $0$. To see this, suppose that $(a_{n^{(k)}_\ell})_{\ell = 1}^\infty$ is a subsequence of $(a_n)$ that converges to $1/k$. We can construct a new sequence in the following manner.

First, choose $\ell_1$ such that $0 < a_{n^{(1)}_{\ell_1}}<2$;

next, choose $\ell_2$ such that $0 < a_{n^{(2)}_{\ell_2}}<1$ and $n^{(2)}_{\ell_2}>n^{(1)}_{\ell_1}$,

$\dots$

In the $k$-th step, choose $\ell_k$ such that $0< a_{n^{(k)}_{\ell_k}}<2/k$ and $n^{(k)}_{\ell_k}>n^{(k-1)}_{\ell_{k-1}}$.

The subsequence $$\left( a_{n^{(k)}_{\ell_k}}\right)_{k = 1}^\infty$$ is a subsequence of $(a_n)$ that converges to $0$.

**2.5.2(c,d)**For part (c), suppose that $|a_n|< M $ for all $n\in \mathbb{N}$. Bolzano-Weierstrass gives a subsequence $a_{n_k}$ which converges to $b\in [-M,M]$. Now the original sequence diverges, so there exists an $\epsilon>0$ such that outside $V_\epsilon(b)$, there are infinitely many terms in the sequence. These infinitely many terms must occur in the union $$[-M,b-\epsilon]\cup [b+\epsilon, M]$$, which is again a bounded set. Bolzano-Weierstrass thus give another subsequence of $(a_n)$ whose limit $a$ is within $[-M,b-\epsilon]\cup [b+\epsilon, M]$. In fact, $|a-b|\ge\epsilon$, we have found two subsequences of $(a_n)$ that converge to distinct limits.

For part (d), we can assume that the original sequence is increasing. Suppose that $(a_{n_k})$ is a subsequence whose limit is $a$. It is necessary that $a_{n_k}\le a$ for all $k\in \mathbb{N}$.

Now for any $\epsilon>0$, there exists a $K$ such that for all $k\ge K$, $|a_{n_k} - a|<\epsilon$. For any $m\ge n_k$, we can always find an $\ell$ such that $\ell>k$ and $m\le \ell$. Thus, by the assumption, $$ a_{n_k}\le a_m\le a_{n_\ell}\le a $$ for all $m\ge n_k$. In particular, $|a_m - a|<\epsilon$. This proves that $(a_n)\rightarrow a$.

**2.5.8(b)**The set of peak terms is either finite (including empty) or infinite. In the finite case, as long as $n_1$ is large enough, $a_{n_1}$ is not a peak term; this allows you to find $n_2>n_1$ such that $a_{n_2}>a_{n_1}$. Continuing this gives you a subsequence that's increasing.

In the infinite case, you can find a subsequence consisting of peak terms. This subsequence must be decreasing.

**2.6.5**(i) Consider the sequence $(s_n)$ where $$s_n = 1+\frac{1}{2}+\frac{1}{3} +\cdots+\frac{1}{n}.$$ This sequence is pseudo-Cauchy.

(ii) You may find the following inequality useful $$|(x_{n+1} + y_{n+1}) - (x_n + y_n) |\le |x_{n+1} - x_n|+|y_{n+1} - y_n|.$$

**2.7.3**For part (a), note that for $m< n$, $$|a_m+\cdots + a_n|\le |b_m+\cdots+b_n|,$$ since $a_k, b_k$ are all non-negative. Thus, if $\sum b_n$ is convergent, then for any given $\epsilon>0$ there exits an $N$ such that for all $m,n\ge N$, $$|a_m+\cdots + a_n|\le |b_m+\cdots+b_n| < \epsilon.$$ It follows that $\sum a_n$ is convergent, by Cauchy's criterion.

For part (b), by assumption, $\sum b_n$ is convergent, so the sequence of its partial sums is bounded. It follows that the sequence of partial sums of $\sum a_n$ is bounded and increasing. The conclusion follows by the Monotone Convergence Theorem.

**2.7.4(b,c,d)**(b) Example: $x_n = \frac{(-1)^n}{n}$, $y_n = (-1)^n$.

(c) Impossible by an algebraic limit property. Consider $\sum (x_n+y_n) - \sum x_n$.

(d) Consider $x_n$ that satisfies $x_n = \frac{1}{n}$ when $n$ is even, and $x_n = 0$ when $n$ is odd.

**2.7.7**(a) To start with, $\ell>0$ by an order limit property. Now, by the assumption, there exists an $N$ such that for all $n\ge N$, $$na_n>\frac{\ell}{2}.$$ In other words, $a_n>\frac{\ell}{2n}$ for all $n\ge N$. Since $\sum\frac{1}{n}$ diverges, $\sum a_n$ also diverges.

(b) Since $\lim n^2a_n$ exists, the sequence $(n^2a_n)$ is bounded, say $$|n^2a_n| = n^2a_n \le M $$ for all $n\in \mathbb{N}.$ Therefore $a_n<\frac{M}{n^2}$ for all $n$. Since $\sum\frac{1}{n^2}$ converges, by comparison test, $\sum a_n$ also converges.

**2.7.8(a,b,c)**(a) True. $\sum|a_n|$ converges implies that there exists an $N$ such that for all $n\ge N$, $|a_n|<1$. Therefore, for all $n\ge N$, $|a_n^2|\le |a_n|$. It follows that $\sum a_n^2$ also converges, by comparison test (essentially, as we really start comparing from the $N$-th term).

(b) False. Example: $a_n = \frac{(-1)^n}{\sqrt{n}}$, $b_n = \frac{(-1)^n}{\sqrt{n}}$.

(c) True. If $\sum n^2a_n$ converges, then $\lim n^2 |a_n| = 0$. By Exercise 2.7.7(b), which also works when $a_n\ge 0$, $\sum|a_n|$ converges. In other words, $\sum a_n$ converges absolutely.

**2.7.9(a)**By the assumption, $\lim \left|\frac{a_{n+1}}{a_n}\right| = r$. Let $\epsilon = r' - r>0$. There exists an $N$ such that for all $n\ge N$, $$\left|\frac{a_{n+1}}{a_n}\right|< \epsilon + r = r'.$$ This is equivalent to $|a_{n+1}|< |a_n|r' $.

**2.7.11(a,b,c)**An idea is to create two series, each one has partial sums that are unbounded, but the terms $\min\{a_n,b_n\}$ are actually $1/n^2$. $$ \begin{array}{ccccccccccccc} 1& \color{orange}{1/2^2} & {1/2^2} &\cdots & 1/2^2 & \color{orange}{1/7^2} & \color{orange}\cdots & \color{orange}{1/42^2} & {1/42^2}&\cdots&{1/42^2}&\cdots\\ \color{orange} 1 & 1 & \color{orange}{1/3^2} & \color{orange}\cdots & \color{orange}{1/6^2} & 1/6^2 &\cdots & 1/6^2 & \color{orange}{1/43^2}&\cdots &\color{orange}{1/1806^2}&\cdots \end{array} $$ Here a pattern is that each orange segment ends with $$ 1, \quad 1\times 2 = 2,\quad 2\times 3 = 6, \quad 6\times 7 = 42, \quad 42\times 43 = 1806, \quad etc. $$ This ensures that each black segment adds up to $1$.

**3.2.10(iii)**Let $S$ be the set of isolated points in the original set $A$. Now for each point $a\in S$, since it is isolated, there exists an $\epsilon_a>0$ such that $$ V_{\epsilon_a}(a)\cap A = \{a\}. $$ In other words, any other point in $A$ must be at least $\epsilon_a$-apart from the point $a$. We fix these $\epsilon_a$'s for all $a\in S$.

If we define: $$ S_m:= \{a\in S: \epsilon_a\ge 1/m\}, $$ all points in $S$ can thus be organized in the following manner: $$ S = \bigcup_{m = 1}^\infty S_m. $$ Since $S$ is uncountable, there exists an $M\in \mathbb{N}$ such that $S_M$ is uncountable. All points in $S_M$ belong to $$ \mathbb{R} = \bigcup_{n\in \mathbb{Z}} \left[\frac{n}{2M}, \frac{n+1}{2M}\right], $$ which is a countable union. So one of the intervals $\left[\frac{n}{2M}, \frac{n+1}{2M}\right]$ must contain at least two numbers, say, $a,b$, in $S_M$, which implies that $a,b$ are at most $1/2M$ apart. On the other hand, since $a,b$ both belong to $S_M$ and are distinct, they need to be at least $1/M$ apart. This is a contradiction.