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Homework assignments

Homework 1

acd

Let $F=\{0, 1\}$, i.e. $F$ is a set with only two elements, called “$0$” and “$1$”. Define addition and multiplication in $F$ by \[ 0+0=0,\quad 0+1=1+0=1,\quad 1+1=0, \] and \[ 0\cdot0 = 0\cdot 1 = 1\cdot 0 = 0, \quad 1\cdot 1=1. \] Is $F$ a field? Is $F$ an ordered field?

If $A\subset\R$ is a nonempty subset that is bounded from below, then consider $B=\{-x \mid x\in A\}$, and show that $\inf A + \sup B = 0$.

This is problem 5 from chapter 1 in Rudin. See Roger Cooke’s solutions

If $A$ and $B$ are bounded and non empty subsets of $\R$, then show that $A\subset B \implies \sup A \leq \sup B$.

Let $A, B\subset \R$ be bounded and nonempty sets of real numbers. Let $C = \{a+b \mid a\in A, b\in B\}$ and $D = \{ab \mid a\in A, b\in B\}$.

Prove or disprove: $\sup C = \sup A + \sup B$.
True.
First show that $\sup A + \sup B$ is an upper bound for $C$. This is true because every number in $C$ is of the form $a+b$ with $a\in A$ and $b\in B$. Since $\sup A$ is an upper bound for $A$ we have $a\leq \sup A$; similarly, we have $b\leq \sup B$. Therefore $a+b\leq \sup A + \sup B$, and we have shown that $\sup A+\sup B$ is an upper bound for $C$.

Next show that there is no smaller upper bound, i.e. suppose $m\lt \sup A+\sup B$ and show that $m$ is not an upper bound. To do this we define $r=(\sup A + \sup B)-m$. Since $m\lt \sup A + \sup B$, we know that $r\gt0$. The set $A$ must contain a number $a$ with $a\gt \sup A - \frac12 r$ because $\sup A$ is the least upper bound of $A$. Similarly, there is a $b\in B$ with $b\gt \sup B - \frac12 r$. Then $a+b\in C$ and $a+b\gt (\sup A + \sup B) - r = m$, which implies that $m$ is not an upper bound for $C$. The conclusion is that $\sup A+ \sup B$ is the lowest upper bound for $C$.
Prove or disprove: $\sup D = (\sup A)(\sup B)$.
Not true. If $A=[-2, -1]$ and $B=[-2,-1]$ then $D=[1,4]$ so that $\sup A = \sup B = -1$, while $\sup D = 4$, which is not the same as $(\sup A)(\sup B)$.

(Variations on the Archimedean property)

Show that the set $A=\{2^n: n\in \N\}$ is unbounded, i.e. show that for every real number $x$ there is an integer $n$ with $2^n\gt x$.
Suppose $A$ were bounded. Then $m=\sup A$ exists, by the least upper bound axiom. The set $A$ must contain a number in the interval $(\frac12 m, m]$, for otherwise $\frac12m$ would also be an upper bound for $A$ and $m$ would not be the least upper bound. Suppose then that $2^n\in A$ is the number in the interval $(\frac12 m, m]$. Then $2^{n+1}$ also belongs to $A$, while $2^{n+1}=2\cdot 2^n \gt 2\cdot \frac12m = m$. This contradicts the fact that $m$ is an upper bound for $A$.
Show that the set $A=\{n^2 \mid n\in\N\}$ is unbounded.
For every $n\in\N$ we have $n^2\geq n$, which follows by multiplying both sides of the inequality $n\geq 1$ with $n$ and concluding $n^2\geq n\geq 1$. Given any $x\in \R$ choose a natural number $n\gt x$. Then $n^2\in A$ and $n^2\geq n\gt x$. Thus $A$ is unbounded.
Show that the set $A=\{n! \mid n\in\N\}$ is unbounded.
Same argument. First show that $n!\geq n$.
Show that the set $A=\{\sqrt{n} \mid n\in\N\}$ is unbounded.
The set $A$ contains $\N$ as a subset: $A\subset \N$. Let $x\in\R$ be any number. Then choose a natural number $n\in\N$ with $x\lt n$. Since $n=\sqrt{n^2}$ we have $n\in A$, and therefore $x$ cannot be an upper bound for $A$. Thus $A$ is unbounded.
Let $A=\{a_1, a_2, a_3, \ldots\}$ be a set of real numbers where $a_{n+1}\ge a_n+1$ holds for all $n\in\N$. Show that $A$ is unbounded.

Use the Archimedean property to show that for any pair of real numbers $x\lt y$ there is a rational number $r\in \Q$ with $x\lt r\lt y$.

Let \[ a_n = 0.\overbrace{11\dots11}^{n\;{\sf digits}} = \frac{1}{10} + \frac{1}{10^2} + \cdots + \frac{1}{10^n}. \] and let $A= \{a_1, a_2, a_3, \dots\}$. Show that $x=\sup A$ exists, and show that $9x=1$. (Rudin: see section 1.22 on page 11.)

Let $x_0$, $x_1$, $x_2$,…be a sequence of positive real numbers that satisfy $x_{n}\leq \frac23 x_{n-1}$ for all $n\geq 1$. Show that there is an integer $N\in\N$ such that $x_N\lt 1$.

The idea is that $x_n \leq (\frac 23)^n x_0$, and that $(\frac23)^n$ goes to zero as $n\to\infty$. Here we will prove this without using the words “limit” and “converge” because we have not defined them yet.

Let $m=\inf A$ where $A=\{x_1, x_2, x_3, \dots\}$. Since $0$ is a lower bound for $A$ the greatest lower bound $m$ for $A$ must exist. We will show that $m=0$. If $m$ were positive then, since $m$ is the greatest lower bound of $A$, the set $A$ must contain a number $x_n \in [m, \frac32m)$. But then $A$ also contains $x_{m+1}\leq \frac 23x_n \lt \frac23\cdot\frac32m = m$. Since $A$ apparently contains the number $x_{n+1}$ which is less than $m$, $m$ cannot be a lower bound for $A$, which contradicts the assumption $m=\inf A$. Hence we have shown that $\inf A=0$.

Since $0$ is the greatest lower bound for $A$, the number $1$ is not a lower bound, and therefore there is an $x_n$ with $x_n\lt 1$.

Let $A = \{x_0, x_1, x_2, \dots\}$ where $x_0$, $x_1$, $x_2$, … are positive real numbers that satisfy $x_{n+1} = \frac{x_n}{x_n+1}$ for all $n\geq 0$. Show that $\inf A=0$.

Homework 2

Chapter 2: Problems 2, 3, 4 about countable and uncountable sets.

Chapter 2: Problem 11 about “metric spaces”

Consider the metric $d_5$ on $\R$ from problem 11, chapter 2. Find $B_{1/2}(1)$, i.e. find the “ball with radius $1/2$, centered at $x=1$” for this metric.

Is $d(x, y) = |\arctan(x) - \arctan(y)|$ a metric on the set $\R$?

Prove that the set $\{(x, y)\in\R^2 : 2x+3y \gt 0\}$ is an open subset of $\R^2$.

Homework 3

Consider the sequence of real numbers given by $a_n=n^{-1}$.

Show that $a_n\to 0$, directly from the definition.
Given $\varepsilon\gt0$ choose an integer $N$ such that $N\gt 1/\varepsilon$. Then for all $n\ge N$ one has $0\lt a_n \le 1/N \lt \varepsilon$.
Show that if a sequence $x_n$ in a metric space $X$ converges, then any subsequence of $x_n$ also converges and has the same limit.
Let $y_k = x_{n_k}$ be a subsequence of $x_n$, and let $L$ be the limit of $x_n$. In particular this implies that $n_k\ge k$ for all $k$. Given $\varepsilon\gt0$ we choose $N\in\N$ such that $|x_n-L|\lt \varepsilon$ for all $n\ge N$. Then, if $k\ge N$ we have $n_k\ge k\ge N$ and thus $|y_k-L| = |x_{n_k}-L| \lt \varepsilon$.
Show that $b_n=n^{-2}\to 0$ without using the definition of convergence.
$b_n = a_{n^2}$, so $b_n$ is a subsequence of the sequence $a_n=1/n$. Therefore it converges, and it has the same limit: $\lim_{n\to\infty} 1/n^2 = 0$.

Let $x_n$ be a convergent sequence of points in a metric space $(X,d)$. Show that the sequence is bounded (i.e. there is a point $a\in X$ and there is a number $R\gt0$ such that $x_n\in B_R(a)$ for all $n\in\N$).

Let $x_n\to p$, i.e. let $p$ be the limit of the sequence $x_n$. Choose $\varepsilon=1$. There is an $N\in\N$ such that for all $n\ge N$ on has $d(x_n, p)\lt \varepsilon=1$. Thus, if \[ R=\max\left\{1, d(x_1, p), \dots, d(x_{N-1}, p)\right\}, \] then all $x_k$ satisfy $d(x_k, p)\le R$.

Let $x_n$ be a sequence of points in a metric space $(X,d)$. Show that $x_n\to x$ holds if and only if $d(x_n, x)\to0$.

Consider the sequence of real numbers $a_n = d(x_n, p)$. We are asked to show that $a_n\to0 \iff x_n\to p$.

Assume $a_n\to 0$. To show $x_n\to p$, let $\varepsilon\gt 0$ be given. By assumption there is an $N\in\N$ such that $|a_n|\lt \varepsilon$ for all $n\ge N$. This implies $d(x_n, p)\lt \varepsilon$ for all $n\ge N$. Therefore $x_n\to p$.

Assume $x_n\to p$. To show $a_n\to 0$, let $\varepsilon\gt 0$ be given. By assumption there is an $N\in\N$ such that $d(x_n, p)\lt \varepsilon$ for all $n\ge N$. This implies $|a_n|\lt \varepsilon$ for all $n\ge N$. Therefore $a_n\to 0$.

Let $(X,d)$ be a metric space, $C\subset X$ a closed subset, and $x_n\in C$ a sequence of points with $x_n\to x$. Show that $x\in C$.

In order to reach a contradiction, assume that $x\not\in C$. The complement $C^c$ of $C$ is open in $X$, so there is an $\varepsilon\gt0$ such that $B_\varepsilon(x)\subset C^c$. Since $x_n\to x$, there is an $N_\varepsilon\in\N$ for such that $x_n\in B_\varepsilon(x)$ for all $n\ge N$. But then we would have $x_n\not\in C$ for $n\ge N$, which is a contradiction. We conclude that $x\in C$ after all.

Let $x_n\in\R^k$ and $y_n\in \R^k$ be two sequences of vectors for which $x_n\to x$ and $y_n\to y$. Show that $(x_n, y_n) \to (x,y)$. Here $(x,y)$ is the dot product, or inner product of the two vectors. In Rudin’s notation (page 16): the assignment is to show that $x_n\cdot y_n \to x\cdot y$.

The sequences $x_n\in\R^k$ and $y_n\in\R^k$ converge, so they are bounded, i.e. there is an $R\gt0$ such that $\|x_n\|\le R$, and $\|y_n\|\le R$ for all$n\in\N$. We also have \begin{align*} \left|x_n\cdot y_n - x\cdot y\right| &=\left| x_n\cdot y_n - x_n\cdot y + x_n\cdot y - x\cdot y\right| \\ &\le \left| x_n\cdot y_n - x_n\cdot y\right| + \left|x_n\cdot y - x\cdot y\right| & \text{triangle }\le\\ &\le \left| x_n\cdot (y_n - y)\right| + \left|(x_n - x)\cdot y\right| \\ &\le \| x_n\|\,\| y_n - y\| + \|x_n - x\|\,\| y\| & \text{Cauchy }\le \\ &\le R\| y_n - y\| + R \|x_n - x\| \end{align*} Since $x_n$ and $y_n$ converge we have \[ \lim_{n\to\infty} R\| y_n - y\| + R \|x_n - x\| = 0 \] and hence \[ \lim_{n\to\infty} \left|x_n\cdot y_n - x\cdot y\right| =0. \]

Let $A\subset\R$, and let $\{x_n\}_{n\in\N}$ be a sequence of upper bounds for $A$, i.e. assume that every $x_n$ is an upper bound for $A$. Show that if $x_n\to x$, then $x$ is also an upper bound for $A$.

Suppose $x$ is not an upper bound for $A$. Then there is a number $a\in A$ with $a\gt X$. Let $\varepsilon= a-x$. Since $x_n\to X$, there is an $N$ such that for all $n\ge N$ one has $|x_n-x|\lt \varepsilon$. This implies, for $n\ge N$, that $x_n \lt x+\varepsilon = a$. Hence $x_n$ is not an upper bound for $A$ if $n\ge N$, contradicting what was given.

[Honors only] For any two points $x,y\in\R^n$ we define \[ d_1(x, y) = |x_1-y_1| + \cdots + |x_n-y_n|. \]

Show that $d_1$ defines a metric on $\R^n$.
Let $d_2$ be the usual Euclidean metric. Prove that for any two points $x,y\in\R^n$ one has \[ d_2(x, y) \leq d_1(x,y) \leq \sqrt{n}\;d_2(x, y). \]
Prove that a sequence of points $\{x_i\in\R^n\}_{i\in\N}$ converges in the $d_1$ distance if and only if it converges in the Euclidean metric. (Notation: denote the components of $x_i\in\R^n$ by $(x_{i1}, \ldots, x_{in})$.)

Homework 4

Let $E = \{x\in\mathbb{Q} : 0\le x\le 1\}$, i.e. $E$ consists of all rational numbers in the interval $[0,1]$. Find a sequence of points in $E$ that has no subsequence that converges to some point in $E$.

For each of the following sets say if it is compact or not. Give a short reason. You may quote theorems from the textbook, or from class (if they are from lecture then you must include the statement). All these sets are subsets of the real line or the euclidean plane $\R^2$. In all cases the intended metric is the usual Euclidean distance $d(x,y) = |x-y|$.

$E = (-\infty, 0]$.
$E = \{1\} \cup \{ \frac{n}{n+2} : n\in\N\}$

Consider the parabolic segment $E = \{(x, x^2) \in\R^2 : |x|\le 1\}$. Show directly from the definition of a compact subset (summary here) that $E$ is a compact subset of $\R^2$. (Hint: show that if $t_i\in\R$ is a sequence of real numbers with $t_i\to a$, then the points $(t_i, t_i^2)\in\R^2$ converge to the point $(a, a^2)\in\R^2$.)

Let $\R\supset I_1\supset I_2 \supset I_3 \supset \cdots$ be a decreasing sequence of intervals, i.e. each $I_k\subset \R$ is an interval and for every $k\in\N$ we have $I_{k+1}\subset I_k$. Suppose each $I_k$ is open and nonempty. Is it true that the intersection $\cap_{k\in\N} I_k$ is not empty?

(Honors only) There is a metric space $X$ which contains a countable set of points $E = \{x_n : n\in\N\}$ for which the distance between any two points is $d(x_n, x_m) = 1$ (for all $n\neq m$).

Show that for any $p\in X$ the ball $B_{1/2}(p)$ cannot contain more than one point of the set $E$.
Can $X$ be $\R^2$ or $\R^3$? (Hint: try to draw a set with three points in the plane so that the distance between any two of them is exactly 1. Then try to draw such a set with four points. Can you find four points in three dimensional Euclidean space that all have unit distance to each other?)
Show that $E$ is a closed and bounded subset of $X$.
Show that $E$ is not a compact subset of $X$.
Show that $E$ is complete.

Homework 5

Let $x_n$ be a bounded sequence of real numbers.

Suppose that $x_{n_k}$ is a convergent subsequence. Show that \[ \liminf x_n \leq \lim x_{n_k} \leq \limsup x_n. \]
Show that there is a subsequence $x_{n_k}$ that converges to $\limsup x_n$.
Suppose $A\gt \limsup x_n$. Show that $\{n\in\N \mid x_n\geq A\}$ is finite.
Suppose $B\lt \limsup x_n$. Show that $\{n\in\N \mid x_n\geq B\}$ is infinite.

In this problem assume that square roots of non–negative real numbers have been shown to exist and have been shown to satisfy the properties familiar from algebra. Let $x_n = \sqrt{n}$.

Show that $x_{n+1}-x_n\to0$.
Is $x_n$ a Cauchy sequence?

a. $0\leq \sqrt{n+1}-\sqrt{n} = \frac{1}{\sqrt{n+1}+\sqrt{n}} \lt \frac1{\sqrt n}$ so $x_n\to0$.

b. Cauchy sequences in $\R$ always converge, so $x_n=\sqrt n$ cannot be a Cauchy sequence. Alternatively, Cauchy sequences in any metric space are bounded, so $x_n=\sqrt n$ cannot be a Cauchy sequence.

Show that for any two bounded sequences of real numbers $x_n$ and $y_n$ one has \[ \limsup \bigl(x_n+y_n\bigr) \leq \limsup x_n + \limsup y_n. \]
By definition \[ \limsup_{n\to\infty}(x_n+y_n) = \lim_{n\to\infty} \Bigl(\sup\{x_k+y_k : k\ge n\}\Bigr). \] For every $k\ge m$ we have \[ x_k \le \sup\{x_l : l\ge m\}, \qquad y_k \le \sup\{y_l : l\ge m\}. \] Hence, for all $k\ge m$, \[ x_k+y_k \le \sup\{x_l : l\ge m\} + \sup\{y_l : l\ge m\}. \] The quantity on the right does not depend on $k$ and thus it is an upper bound for $\{x_k+y_k : k\ge m\}$. It is therefore not smaller than the least upper bound: \[ \sup\{x_k+y_k : k\ge m\} \le \sup\{x_l : l\ge m\} + \sup\{y_l : l\ge m\}. \] If we now take the limit for $m\to\infty $ on both sides we get \[ \limsup_{k\to\infty} x_k+y_k \le \limsup_{k\to\infty} x_k + \limsup_{k\to\infty} y_k. \]
Find two sequences $x_n$ and $y_n$ with $\limsup x_n = \limsup y_n =1$ and $\limsup (x_n+y_n) =0$.
There are many choices. One possible choice is $x_n=(-1)^n$, and $y_n=-x_n$. Then $\limsup x_n =\limsup y_n =1$ while $x_n+y_n = 0$ for all $n\in\N$, so that $\limsup (x_n+y_n) = 0$.
If $A, B\subset \R$ are bounded and nonempty sets of real numbers,, and if $C = \{a+b \mid a\in A, b\in B\}$, is it then possible that $\sup C \lt \sup A + \sup B$? (This part of the problem addresses the difference between sets of real numbers and sequences of real numbers.)
This was a problem in Homework 1.

Let $x_n$ be the sequence defined by $x_1=1$, and $x_{n+1} = \frac{3x_n}{1+x_n}$.

Prove by induction that $x_n\leq 2$ for all $n\in\N$.
Prove by induction that $x_n$ is an increasing sequence.
Compute $\lim x_n$.

(Hint: consider the function $f(x) = 3x/(1+x) = 3 - 3/(1+x)$, and show that $x\lt x'$ implies $f(x)\lt f(x')$ for all $x, x'\geq0$. Also consider $f(1)$ and $f(2)$. )

Homework 6

Suppose $X$ is a metric space, and that $x_n\in X$ is a sequence in $X$ for which $\lim_{n\to\infty} d(x_n, x_{n+1}) = 0$. Is such a sequence always a Cauchy sequence?

No. See problem 1 from the previous homework assignment.

Which of the following series converge? In this problem assume that the standard functions from calculus (sine, cosine, natural logarithm, exponential function) have been defined and that their familiar properties have been proved.

$\sum_{n=1}^\infty \frac{n}{n^3+1}$
$\sum_{n=1}^\infty \left(\frac{n}{2n^2-1}\right)^2$
$\sum_{n=1}^\infty \frac{1}{n (\ln n)^2}$
$\sum_{n=1}^\infty \frac{\ln n}{n^2}$
$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{\ln (n+1)}$
$\sum_{n=1}^\infty \frac{\sin(an)}{\ln (n+1)}$ where $a\in\R$ is some constant.

For which values of $x\in\R$ does the series \[ f(x) \stackrel {\sf def}= \sum_{n=1}^\infty \frac{x^n}{1+x^{2n}} \] converge?

Let $\sum a_n$ and $\sum b_n$ be two convergent series.

Show that $\sum (a_n+b_n)$ also converges.
If the two series $\sum a_n$ and $\sum b_n$ converge absolutely, does it follow that $\sum (a_n+b_n)$ also converges absolutely?
Suppose the series $\sum a_n$ converges. Does it follow that $\sum a_{2n}$ converges?
No. Consider the series $a_n = (-1)^{n+1}/n$.
Suppose the series $\sum a_n$ converges absolutely. Does it follow that $\sum a_{2n}$ converges absolutely?

(Honors only)

Suppose $x_n\in\R$ is a sequence that converges to $\lim x_n=L$ and consider the sequence $a_n = (x_1+\cdots+x_n)/n$ (i.e. $a_n$ is the average of the first $n$ numbers in the sequence $x_n$). Show that $a_n\to L$.
If $x_n\in\R$ is any sequence and $a_n = (x_1+\cdots+x_n)/n$ converges, does it then follow that the sequence $x_n$ also converges?
Suppose $a_n$ is a bounded sequence of real numbers, and define for any $r\in (0, 1)$ the quantity \[ A(r) = (1-r) \sum_{n=1}^\infty a_n r^n. \] Show that the series defining $A(r)$ converges if $0\leq r\lt 1$. Then show that if the sequence $a_n$ converges, then \[ \lim_{r\nearrow1} A(r) = \lim_{n\to\infty} a_n \]