Analysis I

CU Boulder

T1. Understanding Analysis, 2nd ed. by Stephen Abbott
T2. (Optional) Analysis I, 3rd ed. by Terence Tao

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 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

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

Two lowest Homework grades will be dropped.

Homework: 30%         Midterm I: 20%         Midterm II: 20%         Final Exam: 30%

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.