↑ index§3 Notes

MATH410

Revised 2026-09-17

Lecture 1

Analysis of Real Numbers $\mathbb{R}$$$\boxed{\text{axioms}}\Rightarrow \boxed{\text{definitions}}\Rightarrow \boxed{\text{Theorems}}$$

Field Axioms

Two operations Addition ($+$), Multiplication ($\times$), $a,b,c\in \mathbb{R}$

Prop. 1: Both $\vec{0}$ and $\vec{1}$ are unique.

Prop. 2: The additive inverses and multiplicative inverses are unique.

Definition: Let $a,b\in \mathbb{R}:b\ne \vec{0}$, $-a:=$ additive inverse of $a$, $b^{-1}:=$ multiplicative inverse of $b$. $$\Rightarrow a+(-a)=\vec{0},b{b}^{-1}=\vec{1}$$

Definition: Difference and Quotient. $$a-b:=a+(-b)$$$$\frac{a}{b}:=a{b}^{-1}\{b\ne 0\}$$

Prop. 3: $$\forall a\in \mathbb{R},a\cdot \vec{0}=\vec{0}$$$$\forall a,b\in \mathbb{R}:ab=\vec{0}\Rightarrow a=\vec{0}\vee b=\vec{0}$$$$\forall a,b\in \mathbb{R},b\ne 0,-(-a)=a,(b^{-1}{)}^{-1}=b,-(b^{-1})=(-b{)}^{-1}$$

Proof:

We are given $0+0=0$. $$a\cdot 0=a\cdot (0+0)=a\cdot 0+a\cdot 0$$

Add the additive inverse of $a\cdot 0$ to both sides$$-(a\cdot 0)+a\cdot 0=-(a\cdot 0)+(a\cdot 0)+(a\cdot 0)$$$$\Rightarrow \boxed{0=a\cdot 0}$$

Positivity Axioms

Let $P\subseteq \mathbb{R}$ be the set of “positive numbers”.

The axiom says:

(1) If $a,b\in P$, then $a+b,ab\in P$

(2) For any $a\in \mathbb{R}$, either $a\in P,-a\in P,a=0$ (all xor)

Definition: Comparison operators

We say $a>b$ if $a-b\in P$, $a\ge b$ if either $a>b$ or $a=b$

Properties:

(1) For each $a\in \mathbb{R},a\ne 0$, then $a^2>0(a^2=a\cdot a)$. In particular, $1>0$.

(2) For each $a\in P,a^{-1}\in P$

(3) If $a>b,c>0$, then $ac>bc$. If $a>b,c<0$, then $ac<bc$.

Proof (of (3)):

$a>b⟺a-b\in P$ by definition. Then, $c(a-b)\in P$ by statement (1) of Positivity Axiom. By distributive property, we have $ac-bc\in P$. By definition, $\boxed{ac>bc}$ as desired.

Definition: Interval Notation

For $a,b\in \mathbb{R},a<b$, $$(a,b):=\{x\in \mathbb{R}:a<x<b\}$$$$[a,b]:=\{x\in \mathbb{R}:a\le x\le b\}$$$$(a,\infty ):=\{x\in \mathbb{R}:a<x\}$$$$(-\infty ,\infty )\equiv \mathbb{R}$$

Other intervals can be derived following established convention.

Lecture 2

Natural Numbers & Mathematical Induction

Natural Numbers

Definition:

A set $S\subseteq \mathbb{R}$ is inductive if

1. $1\in S$

2. $x\in S\Rightarrow x+1\in S$

Definition:

The set of Natural Numbers ($\mathbb{N}$) is the intersection of all inductive subsets of $\mathbb{R}$ (use 1 exists, $x+1$ is the min needed, etc.)

Prop. 1:

$\mathbb{N}$ is the smallest inductive subset of $\mathbb{R}$. That is, for an inductive subset $A$, $A\subseteq \mathbb{N}\Rightarrow A=\mathbb{N}$.

Proof:

First, we show that $\mathbb{N}$ is inductive.

Let $S\subseteq \mathbb{R}$ be any inductive subset. $1\in S$. Then $1\in \mathbb{N}$.

Suppose $x\in \mathbb{N}$. Then for any inductive subset $S\subseteq \mathbb{R}$, we have $x\in S$. By definition, $x+1\in S$ for any $S\subseteq \mathbb{R}$ inductive, following $x+1\in \mathbb{N}$ so $\mathbb{N}$ is inductive.

Then, prove that $\mathbb{N}$ is the smallest. For any $S\subseteq \mathbb{R}$ inductive, $\mathbb{N}\subseteq S$ by definition. In particular, if $A\subseteq \mathbb{N}$ is inductive, then $\mathbb{N}\subseteq A$, so $A=\mathbb{N}$.

Principle of Mathematical Induction

Statement:

For each $n\in \mathbb{N}$, let $S(n)$ be some assertion. Suppose (a) $S(1)$ is true, and (b) $S(k)$ is true $\Rightarrow S(k+1)$ is also true. Then $S(n)$ is true for all $n\in \mathbb{N}$.

Proof:

Define $A:=\{k\in \mathbb{N}:S(k)$ is true$\}$. By assumptions, we know that $A$ is inductive. Also $A\subseteq \mathbb{N}$. By (Prop. 1), $A\equiv \mathbb{N}$.

Example:

Prove for all $n\in \mathbb{N}$$$\sum _{j=1}^{n}j=\frac{n(n+1)}{2}$$

Proof:

We prove by induction.

Base case: $$\sum _{j=1}^{1}j=1=\frac{1(1+1)}{2}=1$$

Inductive step:

Suppose, for some $k\in \mathbb{N}$, ${\sum }_{j=1}^{k}j=\frac{k(k+1)}{2}$. Then$$\sum _{j=1}^{k+1}j=\sum _{j=1}^{k}j+(k+1)=\frac{k(k+1)}{2}+\frac{2(k+1)}{2}=\frac{(k+1)(k+2)}{2}$$

Exercise:

Show that $\mathbb{N}\equiv \{1,2,3,4,\dots \}$.

rough outline: show that $\{1,2,3,4,\dots \}$ is in all inductive subsets, therefore it’s an inductive subset of $\mathbb{N}$ and must therefore be $\mathbb{N}$ by minimality.

Exercise:

Use induction to show that if $m,n\in \mathbb{N}$, then $m+n,mn\in \mathbb{N}$.

Definition:

1. The set of integers ($\mathbb{Z}$) is the union of all natural numbers, their negatives, and 0.

2. The set of rational numbers ($\mathbb{Q}$) is the set $\{x\in \mathbb{R}:x=\frac{m}{n},m,n\in \mathbb{Z},n\ne 0\}$.

3. A number $x$ is irrational if $x$ is not rational.

Properties of $\mathbb{Q}$

1. If $x\in \mathbb{Q}$, then we can write $x=\frac{m}{n},m,n\in \mathbb{Z},n\ne 0$ where either $m$ or $n$ is odd.

2. An integer $n$ is even if $n^2$ is even.

Prop. 2:

There is NO rational number whose square is 2. Combined with the existence of $\sqrt{2}$ (Lecture 3), this gives $\mathbb{R}\backslash \mathbb{Q}\ne \varnothing$.

Proof

We prove by contradiction.

Assume that there exists $x\in \mathbb{Q},x^2=2$. Then by [property 1], $x=\frac{m}{n},m,n\in \mathbb{Z},n\ne 0$, either $m$ or $n$ odd, $(\frac{m}{n}{)}^2=2$. Then $m^2=2n^2$. Hence $m^2$ is even. By [property 2], then $m$ is even. Therefore, there exists an integer $k$ such that $m=2k$. Also, since $m$ is even, $n$ is odd. Next, $$(2k{)}^2=2n^2\Rightarrow 2k^2=n^2\Rightarrow 2|n\text{by property 2}$$

We have a contradiction, thus the assumption must be false. Hence, there is no $x\in \mathbb{Q}:x^2=2$.

Completeness Axiom

Motivation: It’s a good idea to include irrational numbers in $\mathbb{R}$

Definition:

A nonempty subset $S\subseteq \mathbb{R}$ is bounded above if there exists $c\in \mathbb{R}$ (denoting $c$ as the upper bound of $S$) such that$$x\le c\text{ for all }x\in S$$

We say that $c$ is the least upper bound/supremum of $S$ if $c$ is an upper bound of $S$ and for any upper bound $c^{\prime }$ of $S$, we have $c\le c^{\prime }$. We denote such $c$ by $l.u.bS$ or $\boxed{\sup S}$.

Completeness Axiom:

For any nonempty $S\subseteq \mathbb{R}$ bounded above, $\sup S$ exists.

Lecture 3

Completeness Axiom (continued)

Example:

Consider a set $S=\{x\in \mathbb{R}|x\ge 0,x^2<2\}$.

Observe that $S\subseteq \mathbb{R}$, and it is bounded above (let’s say it’s bounded by perhaps 2 or 3). In addition, $S$ is non-empty, since $1\in S$.

Thus, by the completeness axiom, $\sup S$ exists. (One can then show that $(\sup S{)}^2=2$; see HW 1.1 #17.)

Prop.:

Let $c>0$. Then the equation $x^2=c$, $x>0$, has a unique solution.

Proof:

Existence: apply the same argument to $S=\{x\in \mathbb{R}\mid x\ge 0,x^2<c\}$ (HW 1.1 #17).

Uniqueness: Suppose $\exists a,b>0$ such that $a^2=b^2=c$. Then we know $a^2-b^2=0=(a+b)(a-b)$. By positivity, we know $(a+b)>0$ since $a,b>0$. Then $a-b=0\Rightarrow a=b$.

Definition:

For $c>0$, if $x^2=c$ with $x>0$, define $\sqrt{c}:=x$. Also define $\sqrt{0}:=0$.

Definition:

Let $S\subseteq \mathbb{R}$ be nonempty, and $S$ is bounded below if there exists a number $c$ such that $c\le x\forall x\in S$.

$S$ is bounded if it is bounded above and below.

We say that $c$ is the greatest lower bound or “infimum” of $S$, denoted $\inf S$, if $c$ is a lower bound of $S$ and $c\ge c^{\prime }$, where $c^{\prime }$ is any lower bound of $S$.

Theorem: (Existence of infimum)

Suppose $S\subseteq \mathbb{R}$ is nonempty and bounded below. Then $\inf S$ exists.

Proof:

Let $T=\{x\in \mathbb{R}|-x\in S\}$. Recall that for any $x,b\in \mathbb{R},x\le b$ iff $-x\ge -b$. Then $b$ is a lower bound of $S$ iff $-b$ is an upper bound of $T$. Hence, $S$ is bounded below iff $T$ bounded above. By the Completeness Axiom, $c:=\sup T$ exists. Since the lower bounds of $S$ are negatives of upper bounds of $T$, $-c=\inf S$.

Distribution of Integers ($\mathbb{Z}$) and Rationals ($\mathbb{Q}$)

Theorem: (Archimedean Property)

The following two equivalent statements hold.

(1) For any $c$, there exists $n\in \mathbb{N}$ such that $n>c$.

(2) For any $\epsilon >0$, there exists $n\in \mathbb{N}$ such that $\frac{1}{n}<\epsilon$.

Proof:

First, show equivalence of the two statements.

For $c,\epsilon >0$ such that $\epsilon =\frac{1}{c}$, we see that for any $n\in \mathbb{N}$, $n>c⟺\frac{1}{n}<\frac{1}{c}=\epsilon$.

Then, show that statement (1) is true. Proof by contradiction.

Assume that $\exists c>0$ such that for any $n\in \mathbb{N},n\le c$. This means that $\mathbb{N}$ is bounded above. By the completeness axiom, define $b:=\sup \mathbb{N}$ exists.

Then, $b-\frac{1}{2}$ is NOT an upper bound of $\mathbb{N}$. Thus, there exists a number $k>b-\frac{1}{2},k\in \mathbb{N}$.

Since $\mathbb{N}$ is inductive, $k+1\in \mathbb{N}$, and $k+1>b-\frac{1}{2}+1>b$. This contradicts $b$ being an upper bound of $\mathbb{N}$.

Prop.:

For any $n\in \mathbb{Z}$, there is no integer $k$ such that $k\in (n,n+1)$.

Proof:

We start with $n=0$. That is, we show that there are no integers in $(0,1)$. Define the set $S=\{k\in \mathbb{N}|k\ge 1\}$. Then $S$ is inductive as well as $S\equiv \mathbb{N}$. We know $k\ge 1\forall k\in \mathbb{N}$. Hence, $-k\le -1\forall k\in \mathbb{N}$. Finally, $0\notin (0,1)$. (proved the base case)

For any $n\in \mathbb{Z}$, prove by contradiction. Suppose there exists $k\in \mathbb{Z}$ such that $k\in (n,n+1)$. Then $k-n\in (0,1)$. But we know $k-n\in \mathbb{Z}$. This contradicts the $n=0$ case.

Prop.:

Suppose $S\subseteq \mathbb{Z}$ is nonempty and bounded above. Then $S$ has a maximum.

Proof:

By Completeness Axiom, let $a:=\sup S$. Then $a-1$ is NOT an upper bound of $S$. This means $\exists m\in S$ such that $m>a-1\Rightarrow a<m+1$. $$\Rightarrow S\subseteq (-\infty ,m+1)$$

Since $S\subseteq \mathbb{Z}$, there can be no element in $S$ that lies in $(m,m+1)$ (by the previous proposition)$$\Rightarrow S\subseteq (-\infty ,m],\text{m is the maximum}$$

Lecture 4

Distributions of Integers and Rationals.

Remark:

In general, for any $S\subseteq \mathbb{R}$, it is possible that $\sup S\ne \max S$. In fact, the maximum of $S$ might not exist. Take, for example, $S=(0,1),\sup S=1$. But there is no maximum.

Theorem:

For any $c\in \mathbb{R}$, there is a unique $k\in \mathbb{Z}$ such that $k\in [c,c+1)$.

Proof:

Existence: Define the set $S=\{n\in \mathbb{Z}\mid n<c+1\}$. S is nonempty. If $c+1>0,0\in S$. If $c+1\le 0$, then by the Archimedean Property, there exists $n\in \mathbb{N}$ such that $n>-(c+1)$. Then $-n<c+1\Rightarrow -n\in S$.

Also, show that $S$ is bounded above. Directly state that $c+1$ is an upper bound. By Prop (maximum if nonempty and bounded above in Integers), let $k$ be the maximum of $S$.

In particular, $k<c+1$. Also, note that $k\ge c$– if not, then $k<c$, but then $k+1<c+1$, contradicting the maximality of $k$. Therefore, $k\in [c,c+1)$.

Uniqueness: Suppose that both $k,k^{\prime }\in [c,c+1)$. We may also assume $k<k^{\prime }$ WLOG. Then $0<k^{\prime }-k$. Also, since $k^{\prime }<c+1$ and $k\ge c$, we have $k^{\prime }-k<c+1-c=1$, so $k^{\prime }-k\in (0,1)$. By one of the propositions above, this is impossible since both are integers (closed under addition). So we must have $k=k^{\prime }$.

Definition:

A set $S\subseteq \mathbb{R}$ is said to be dense in $\mathbb{R}$ if for any $I=(a,b),a<b$, there exists $s\in S$ such that $s\in I$.

Theorem:

$\mathbb{Q}$ is dense in $\mathbb{R}$.

Proof:

Let $a,b\in \mathbb{R},a<b$. By the Archimedean Property, there exists $n\in \mathbb{N}$ such that $\frac{1}{n}<b-a$. Now, by the previous Theorem, there exists $m\in \mathbb{Z}$ such that $m\in [nb-1,nb)\Rightarrow \frac{m}{n}\in [b-\frac{1}{n},b)\subseteq (a,b)$.

This is because $\frac{1}{n}<b-a\iff b-\frac{1}{n}>a$.

Corollary:

The irrationals are dense in $\mathbb{R}$.

Proof:

Let $a,b\in \mathbb{R},a<b$. By the Theorem, there exists $q\in \mathbb{Q}$, $q\ne 0$, such that $q\in (\frac{a}{\sqrt{2}},\frac{b}{\sqrt{2}})$ (if $0$ lies in this interval, apply the Theorem to $(0,\frac{b}{\sqrt{2}})$ instead). Thus, $\sqrt{2}q\in (a,b)$, where $\sqrt{2}q$ is irrational: if $\sqrt{2}q=r\in \mathbb{Q}$, then $\sqrt{2}=\frac{r}{q}\in \mathbb{Q}$, contradicting Prop. 2.

Inequalities & Identities

Definition:

Let $x\in \mathbb{R}$, we define the absolute value of $x,|x|$:$$|x|=\{\begin{array}{ll}x,& x\ge 0\\ -x,& x<0\end{array}$$

Note:

If $c,d\in \mathbb{R},d\ge 0$, then $|c|\le d\iff -d\le c\le d$.

If $x\in \mathbb{R},-|x|\le x\le |x|$.

Theorem: (Triangle Inequality)

Given $a,b\in \mathbb{R}$, $$|a+b|\le |a|+|b|.$$

Proof:

By the Note, the statement is equivalent to $$-|a|-|b|\le a+b\le |a|+|b|$$

By the Note, we know that $$-|a|\le a\le |a|-|b|\le b\le |b|$$

Adding these two inequalities gives the result.

Prop.:

Given $a,r\in \mathbb{R},r>0$, the following are equivalent:

(a) $|x-a|<r$.

(b) $a-r<x<a+r$.

(c) $x\in (a-r,a+r)$.

Proof:

(b)$\iff$(c) by definition of interval notation. (a)$\iff$(b): by the Note (with strict inequalities), $|x-a|<r⟺-r<x-a<r⟺a-r<x<a+r$.

Lecture 5

Useful Formulae: Read pg. 18, 19 from Textbook. Learn the formulae, finish inequalities.

Sequences & Convergence

Definition:

A sequence of real numbers is a real-valued function whose domain is $\mathbb{N}$.

Notation:

$\{a_n\}$, where $n$ is the index of the sequence and $a_n$ is the n-th term of it.

Example:

$\{\frac{1}{n}\},\{1+(-1{)}^n\},\{a_n\}\text{ where an is the largest integer that is less than or equal to n3}$, define $a_1=1$, given $a_n$ defined let $a_{n+1}=\{\begin{array}{ll}{a}_n+\frac{1}{n}& a_n^2\le 2\\ a_n-\frac{1}{n}& a_n^2>2\end{array}$, Given $r\in \mathbb{R}$, define $\{s_n\}$ as $s_n={\sum }_{j=1}^n{r}^j$ ($r=1\Rightarrow s_n=n,r\ne 1\Rightarrow s_n=\frac{r-r^{n+1}}{1-r}$).

More generally, given $\{c_n\}$, let $s_n={\sum }_{j=1}^n{c}_j$. Then $\{s_n\}$ formed in this way is called a series. $s_n$ is called the $n$-th partial sum.

Definition:

We say that a sequence $\{a_n\}$ converges to a number $a\in \mathbb{R}$, if $\forall \epsilon >0$, there exists $N\in \mathbb{N}$ such that for any $n\in \mathbb{N},n\ge N$, $|a_n-a|<\epsilon$.

Prop.:

If $\{a_n\}$ converges, then it must converge to a unique number.

Proof:

Suppose $\{a_n\}$ converges to both $a,a^{\prime }\in \mathbb{R},a<a^{\prime }$. Choose $\epsilon <\frac{a^{\prime }-a}{2}$. Then, the intervals $(a-\epsilon ,a+\epsilon ),(a^{\prime }-\epsilon ,a^{\prime }+\epsilon )$ are disjoint. By convergence, there exist $N_1,N_2\in \mathbb{N}$ such that $$a_n\in (a-\epsilon ,a+\epsilon ),n\ge N_1{a}_m\in (a^{\prime }-\epsilon ,a^{\prime }+\epsilon ),m\ge N_2$$

Let us choose $N=\max \{N_1,N_2\}$. Then for any $n\ge N$, $a_n\in (a-\epsilon ,a+\epsilon )\cap (a^{\prime }-\epsilon ,a^{\prime }+\epsilon )$. This is a contradiction, since the intervals are disjoint. Then $a=a^{\prime }$.

Notation:

If $\{a_n\}$ converges to $a$, we say $a$ is the limit. $$\underset{n\to \infty }{\lim }{a}_n=a.$$

Example:

Show that ${\lim }_{n\to \infty }\frac{1}{n}=0$.

Proof:

Let $\epsilon >0$. We want to show there exists $N\in \mathbb{N}$ such that for all $n\ge N$, $|\frac{1}{n}-0|<\epsilon \Rightarrow \frac{1}{n}<\epsilon$.

By the Archimedean Principle, there exists $N\in \mathbb{N}$ such that $\frac{1}{N}<\epsilon$. Then for all $n\ge N$, $$\frac{1}{n}\le \frac{1}{N}<\epsilon .$$

Example:

$\{(-1{)}^n\}$ does not converge.

Proof:

Suppose $\{(-1{)}^n\}$ converges to some $a\in \mathbb{R}$. Choose $\epsilon <1$. By convergence, there exists $N\in \mathbb{N}$ such that $(-1{)}^n\in (a-\epsilon ,a+\epsilon )\subseteq (a-1,a+1)\forall n\ge N$. This is a contradiction, since the interval has length strictly less than 2, which cannot contain both $1$ and $-1$.

Example:

Show that ${\lim }_{n\to \infty }(\frac{2}{n^2}+\frac{3}{n}+3)=3$.

Proof:

Let $\epsilon >0$. We want some $N\in \mathbb{N}$ such that $\forall n\ge N$$$|\frac{2}{n^2}+\frac{3}{n}+3-3|<\epsilon \Rightarrow |\frac{2}{n^2}+\frac{3}{n}|<\epsilon$$

By the positivity axiom, we have$$|\frac{2}{n^2}+\frac{3}{n}|=\frac{2}{n^2}+\frac{3}{n}\le \frac{2}{n}+\frac{3}{n}=\frac{5}{n}$$

Then it suffices to choose $N\in \mathbb{N}$ such that for any $n\ge N$, $\frac{5}{n}<\epsilon$. Such $n$ can be found by Archimedean Principle: choose $N$ such that $\frac{1}{N}<\frac{\epsilon }{5}$.

Lemma: (Comparison Lemma)

Let $\{a_n\}$ converge to $a$, then $\{b_n\}$ converges to $b$ if there exists $c\ge 0$ and $N_1\in \mathbb{N}$ such that for all $n\ge N_1$$$|b_n-b|\le c|a_n-a|.$$

Proof:

Let $\epsilon >0$. We need $\exists N\in \mathbb{N}:\forall n\ge N,|b_n-b|<\epsilon$. If $c=0$, then we are done. If $c>0$, by convergence of $\{a_n\}$, choose $N_2$ such that $\forall n\ge N_2,|a_n-a|<\frac{\epsilon }{c}$.

Further, choose $N=\max \{N_1,N_2\}$, for all $n\ge N$ we have $|b_n-b|\le c|a_n-a|<c\cdot \frac{\epsilon }{c}=\epsilon .$

Lecture 6

Archimedean Property:

For any $a\in \mathbb{R}$, $\exists k\in \mathbb{N}$ such that $k>a$.

Maximum Existence:

A maximum of a subset $S$ exists if there is an element $m\in S$ such that $x\le m$ for all $x\in S$.

Definition (dense):

A set $S\subseteq \mathbb{R}$ is dense in $\mathbb{R}$ if for any $I=(a,b),a<b$, there exists $s\in S$ such that $s\in I$.

Convergence (of a sequence):

$\{a_n\}$ converges to $a\in \mathbb{R}$ if $\forall \epsilon >0$, there exists $N\in \mathbb{N}$ such that $|a_n-a|<\epsilon$ for all $n\ge N$.

Comparison Lemma:

If $\{a_n\}$ converges to $a\in \mathbb{R}$, then $\{b_n\}$ converges to some $b\in \mathbb{R}$ if there exists some $C\ge 0$ and $N\in \mathbb{N}$ such that $$|b_n-b|\le C|a_n-a|\text{for all }n\ge N.$$

Limit Laws

Theorem: (Sum Property)

Suppose $\{a_n\}$ converges to $a$ and $\{b_n\}$ converges to $b$. Then $\{a_n+b_n\}$ converges to $a+b$. Notationally, $$\underset{n\to \infty }{\lim }(a_n+b_n)=\underset{n\to \infty }{\lim }{a}_n+\underset{n\to \infty }{\lim }{b}_n.$$

Proof:

Fix $\epsilon >0$. By convergence of both $\{a_n\}$ and $\{b_n\}$ there exist $N_1,N_2\in \mathbb{N}$ such that $$|a_n-a|<\frac{\epsilon }{2}\forall n\ge N_1,|b_n-b|<\frac{\epsilon }{2}\forall n\ge N_2.$$

Choose $N=\max \{N_1,N_2\}$. Then, for all $n\ge N$, both inequalities hold. Then, by the Triangle Inequality, for all $n\ge N$, $$|(a_n+b_n)-(a+b)|=|(a_n-a)+(b_n-b)|\le |a_n-a|+|b_n-b|<\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon .$$

Lemma A:

Suppose a sequence $\{a_n\}$ converges to $a$. Then, for any $\alpha \in \mathbb{R}$, $$\underset{n\to \infty }{\lim }(\alpha \cdot a_n)=\alpha \cdot a.$$

Proof:

Note that $|\alpha a_n-\alpha a|=|\alpha ||a_n-a|$ for all $n\ge 1$. Then apply the Comparison Lemma with $b_n=\alpha \cdot a_n$, $b=\alpha \cdot a$, and $C=|\alpha |$.

Lemma B:

If $\{a_n\},\{b_n\}$ converge to 0, then $\{a_n\cdot b_n\}$ also converges to 0.

Proof:

Fix $\epsilon >0$. By convergence, $\exists N_1,N_2\in \mathbb{N}$ such that $$|a_n|<\sqrt{\epsilon },n\ge N_1,|b_n|<\sqrt{\epsilon },n\ge N_2.$$

Now set $N=\max \{N_1,N_2\}$; for all $n\ge N$ both inequalities hold. wwts: $|a_n{b}_n-0|<\epsilon$. So, $$|a_n{b}_n-0|=|a_n{b}_n|=|a_n||b_n|<\sqrt{\epsilon }\sqrt{\epsilon }=\epsilon .$$

Note: $$\underset{n\to \infty }{\lim }{a}_n=a⟺\underset{n\to \infty }{\lim }(a_n-a)=0.$$

Theorem: (Product Property)

We have that $\lim (a_n{b}_n)=\lim (a_n)\lim (b_n)$, with $n\to \infty$ everywhere.

Proof: (denote $\lim (a_n)=a$ and $\lim (b_n)=b$)

Define ${\alpha }_n=a_n-a$ and ${\beta }_n=b_n-b$. Then $\lim ({\alpha }_n)=0=\lim ({\beta }_n)$, so the product is also 0 by Lemma B: $\lim ({\alpha }_n\cdot {\beta }_n)=0$. Expanding, $${\alpha }_n{\beta }_n=a_n{b}_n-a{b}_n-b{a}_n+ab\Rightarrow a_n{b}_n={\alpha }_n{\beta }_n+a{b}_n+b{a}_n-ab.$$

Every term on the right converges (the first by Lemma B, the next two by Lemma A), so by the Sum Property, $$\lim (a_n{b}_n)=0+ab+ba-ab=ab.$$

Prop.:

Suppose a sequence $\{b_n\}$ of nonzero numbers converges to $b\ne 0$. Then, $$\underset{n\to \infty }{\lim }\frac{1}{b_n}=\frac{1}{b}.$$

Pf Strategy: (use Comparison Lemma)

W.w.t.s. there’s some $C\ge 0$, $N_1\in \mathbb{N}$ such that $|\frac{1}{b_n}-\frac{1}{b}|\le C|b_n-b|$ for all $n\ge N_1$.

Proof:

Observe, $$|\frac{1}{b_n}-\frac{1}{b}|=|\frac{b-b_n}{b\cdot b_n}|=\frac{1}{|b\cdot b_n|}\cdot |b-b_n|(\star )$$

Suppose we can show there exists some $N_1\in \mathbb{N}$ such that $$|b_n|>\frac{|b|}{2}\text{ for all }n\ge N_1.(\star \star )$$

Then by $(\star )$, for all $n\ge N_1$, $$|\frac{1}{b_n}-\frac{1}{b}|=\frac{1}{|b|}\cdot \frac{1}{|b_n|}\cdot |b_n-b|\le \frac{2}{|b{|}^2}|b_n-b|.$$

Choose $C=\frac{2}{|b{|}^2}$. Now, it suffices to prove $(\star \star )$:

Since ${\lim }_{n\to \infty }{b}_n=b$, there exists $N_1\in \mathbb{N}$ such that $|b_n-b|<\frac{|b|}{2}$ for all $n\ge N_1$. $(\star \star \star )$

Recall $||x|-|y||\le |x-y|$. Now, for all $n\ge N_1$, $$|b_n|=|b-(b-b_n)|\ge ||b|-|b-b_n||\ge |b|-|b-b_n|>|b|-\frac{|b|}{2}=\frac{|b|}{2}$$

where the strict inequality is by $(\star \star \star )$.

Theorem: (Full Quotient Property)

Suppose $\lim (a_n)=a$ and $\lim (b_n)=b$, $b_n\ne 0\forall n\in \mathbb{N}$, and $b\ne 0$. Then, $$\lim (\frac{a_n}{b_n})=\frac{\lim (a_n)}{\lim (b_n)}=\frac{a}{b}.$$

Proof:

By the previous Prop. and the Product Property, $$\lim (\frac{a_n}{b_n})=\lim (a_n\cdot \frac{1}{b_n})=\lim (a_n)\cdot \lim (\frac{1}{b_n})=\frac{a}{b},$$

since we know $\{\frac{1}{b_n}\}$ converges to $\frac{1}{b}$.

Prop.: (Linearity Property)

For any $\alpha ,\beta \in \mathbb{R}$, $$\lim (\alpha a_n+\beta b_n)=\alpha \lim (a_n)+\beta \lim (b_n).$$

Proof:

Lemma A + Sum Property.

Polynomials

Definition:

For $k\in \mathbb{Z}$ nonnegative and $c_0,\dots ,c_k\in \mathbb{R}$, the function $$p(x)=\sum _{i=0}^k{c}_i{x}^i$$

is called a polynomial. In particular, if $c_k\ne 0$, we say $p(x)$ has degree $k$.

Example:

$p(x)=3x^2+6x+1$, with $(c_0,c_1,c_2)=(1,6,3)$ and $k=2$, so degree 2.

Prop.: (Polynomial Property)

If $\lim (a_n)=a$, then for any polynomial $p(x)$, $\{p(a_n)\}$ converges to $p(a)$.

Proof (hint): Induction (on degree $k$).

Lecture 7

Sequences & Sets

Recall:

A set $S\subseteq \mathbb{R}$ is bounded if there exists some $M\in \mathbb{R},M\ge 0$ s.t. $|x|\le M\forall x\in S$.

Definition:

A sequence $\{a_n\}$ is bounded if there exists some $M\in \mathbb{R},M\ge 0$ such that $|a_n|\le M$ for all indices $n$.

Theorem:

Convergent sequences are bounded.

Proof:

Assume ${\lim }_{n\to \infty }{a}_n=a$. Let $\epsilon =1$, then there exists $N\in \mathbb{N}$ s.t. for all $n\ge N$, $|a_n-a|<1$. Then by the Triangle Inequality, $|a_n|=|a_n+a-a|=|a+(a_n-a)|\le |a|+|a_n-a|<1+|a|$ for all $n\ge N$.

Choose $M=\max \{|a_1|,|a_2|,\dots ,|a_{N-1}|,1+|a|\}$. Then $|a_n|\le M$ for all $n\in \mathbb{N}$.

Note:

Bounded sequences might not converge. (Example just take $\{(-1{)}^n\}$).

Convergent Sequences & Density

Recall:

A set $S\subseteq \mathbb{R}$ is dense if every $I=(a,b)$ contains a member of $S$.

Definition:

A sequence $\{a_n\}$ is in a set $S$ if $a_n\in S$ for all $n\in \mathbb{N}$.

Prop.:

$S\subseteq \mathbb{R}$ is dense iff every $x\in \mathbb{R}$ is a limit of some sequence in $S$.

Proof:

($\Rightarrow$) Assume $S$ is dense. Fix $x\in \mathbb{R}$. For any $n\in \mathbb{N}$, consider $(x,x+\frac{1}{n})\subseteq \mathbb{R}$. By density, there exists some $s_n\in (x,x+\frac{1}{n})$ such that $s_n\in S$. Then consider $\{s_n\}$. We know for all $n\in \mathbb{N}$, $|s_n-x|<\frac{1}{n}$. Then by Comparison Lemma, ${\lim }_{n\to \infty }{s}_n=x$.

($\Leftarrow$) Fix an arbitrary interval $I=(a,b)\subseteq \mathbb{R}$. Set $x=\frac{a+b}{2}$. By assumption, there is $\{s_n\}$ in $S$ such that ${\lim }_{n\to \infty }{s}_n=x$. Choose $\epsilon =\frac{b-a}{2}$. Then there exists $N\in \mathbb{N}$ such that $|s_N-\frac{b+a}{2}|<\epsilon$$$\Rightarrow a=\frac{b+a}{2}-\frac{b-a}{2}<s_N<\frac{b+a}{2}+\frac{b-a}{2}=b$$

Thus, $s_N\in (a,b)$, and we are done.

Theorem:

Every real number is a limit of a rational sequence.

Proof:

We know $\mathbb{Q}$ is dense in $\mathbb{R}$ and the previous proposition.

Lemma:

Suppose $\{d_n\}$ converges to $d$ and $d_n\ge 0$ for all $n\in \mathbb{N}$. Then, $d\ge 0$.

Proof:

Suppose $d<0$. Choose $\epsilon =-\frac{d}{2}>0$. There exists a $N\in \mathbb{N}$ such that $|d_N-d|<\epsilon$. In particular, $d_N<d+\epsilon =\frac{d}{2}$. This is a contradiction of $d_n\ge 0$.

Theorem:

Let $\{c_n\}$ be in $[a,b]$. If ${\lim }_{n\to \infty }{c}_n=c$, then $c\in [a,b]$.

Proof:

By the Linearity Property, $$\underset{n\to \infty }{\lim }(b-c_n)=b-\underset{n\to \infty }{\lim }{c}_n=b-c\ge 0$$

(using the Lemma from before)

$$\underset{n\to \infty }{\lim }(c_n-a)=c-a\ge 0$$

Hence $a\le c\le b$.

Definition:

$S\subseteq \mathbb{R}$ is closed if whenever a sequence $\{a_n\}$ in $S$ converges to some number $a$, then $a\in S$.

Example:

$\{x\ge 0\}=[0,\infty )$ is closed. We know that all closed intervals $[a,b]$ are closed, but $(0,1]$ is not closed: $\{\frac{1}{n}\}$ is in $(0,1]$ but converges to $0\notin (0,1]$.

Example:

$\mathbb{Q}$ is not closed.

We claim that we can find a sequence $\{q_n\}\subseteq \mathbb{Q}$ such that ${\lim }_{n\to \infty }{q}_n=\sqrt{2}$.

Indeed, every real number is a limit of a rational sequence (Theorem above), so such $\{q_n\}$ exists. But $\sqrt{2}\notin \mathbb{Q}$ by Prop. 2 (Lecture 2), so $\mathbb{Q}$ is not closed.

Lecture 8

Monotone Convergence Theorem

Definition:

A sequence $\{a_n\}$ is said to be monotonically increasing if $a_{n+1}\ge a_n\forall n\in \mathbb{N}$, and monotonically decreasing if $a_{n+1}\le a_n\forall n\in \mathbb{N}$.

We say $\{a_n\}$ is monotone if it is either monotonically increasing or monotonically decreasing.

Theorem: (Monotone Convergence Theorem)

A monotone sequence is bounded iff it converges. In particular, $\{a_n\}$ converges to $\sup \{a_n\mid n\in \mathbb{N}\}$ if monotonically increasing and $\inf \{a_n\mid n\in \mathbb{N}\}$ if monotonically decreasing.

Proof:

It suffices to show that a bounded monotone sequence converges. We show the increasing case.

Let $S=\{a_n\mid n\in \mathbb{N}\}$. By assumption, $S$ is bounded above, therefore by the Completeness Axiom the supremum $L=\sup S$ exists. Our goal is to show ${\lim }_{n\to \infty }{a}_n=L$.

Fix $\epsilon >0$. We want to show $\exists N\in \mathbb{N}\mid \forall n\ge N,L-\epsilon <a_n<L+\epsilon$.

Since $L$ is the supremum and $\epsilon >0$, we have $a_n\le L<L+\epsilon$. In addition, $L-\epsilon$ is NOT an upper bound, therefore $\exists N\in \mathbb{N}$ such that $a_N>L-\epsilon$. By monotonicity, $\forall n\ge N,a_n\ge a_N>L-\epsilon$.

Example:

Harmonic series: $s_n={\sum }_{k=1}^n\frac{1}{k}$. Show that $\{s_n\}$ does NOT converge.

Solution:

First, observe that $s_n$ is monotonically increasing. By MCT, it suffices to show that $\{s_n\}$ is not bounded above.

Observe: $s_2=1+\frac{1}{2}$, $s_4=s_2+\frac{1}{3}+\frac{1}{4}\ge s_2+\frac{1}{2}$, $s_8=s_4+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\ge s_4+\frac{1}{2}$. So we claim that $s_{2^n}\ge 1+\frac{n}{2},\forall n\in \mathbb{N}$.

For each $k\in \mathbb{N}$, there are $2^{k-1}$ indices $i$ such that $2^{k-1}<i\le 2^k$. For each $i$, $\frac{1}{i}\ge \frac{1}{2^k}$:$$\sum _{2^{k-1}<i\le 2^k}\frac{1}{i}\ge \sum _{2^{k-1}<i\le 2^k}\frac{1}{2^k}=\frac{1}{2}$$

Therefore, for $n\in \mathbb{N}$, $s_{2^n}=1+{\sum }_{2\le i\le 2^n}\frac{1}{i}$$$=1+\sum _{k=1}^n\sum _{2^{k-1}<i\le 2^k}\frac{1}{i}\ge 1+\sum _{k=1}^n\frac{1}{2}=1+\frac{n}{2}.$$

By the Archimedean Property, $1+\frac{n}{2}$ exceeds any given real number for $n$ large enough, so the sequence is not bounded above.

Example:

Consider $\{a_n\}$ defined recursively:$$a_1=3,a_{n+1}=\frac{1}{4-a_n}$$

Show that $\{a_n\}$ converges and compute the limit.

Solution:

Prove by induction that $\{a_n\}$ monotonically decreases.

Base case: $a_2=\frac{1}{4-a_1}=1<a_1$.

Inductive step: suppose $a_n\le a_{n-1}$. Since $a_n\le a_{n-1}\le a_1=3<4$, both denominators are positive, so $a_{n+1}=\frac{1}{4-a_n}\le \frac{1}{4-a_{n-1}}=a_n$.

Now, we show that $\{a_n\}$ is bounded below. Once again, use induction (show that $a_n\ge \frac{1}{4}\forall n\in \mathbb{N}$)

Base case: $a_1=3\ge \frac{1}{4}$.

Inductive step: $a_{n+1}=\frac{1}{4-a_n}\ge \frac{1}{4-\frac{1}{4}}\ge \frac{1}{4}$.

By the Monotone Convergence Theorem, ${\lim }_{n\to \infty }{a}_n=a$ ($\{a_n\}$ converges). Now, compute the limit (the Quotient Property applies since $a\le a_1=3$, so $4-a\ne 0$). $$a_{n+1}=\frac{1}{4-a_n}\Rightarrow \underset{n\to \infty }{\lim }{a}_{n+1}=\underset{n\to \infty }{\lim }\frac{1}{4-a_n}\Rightarrow a=\frac{1}{4-a}$$$$\Rightarrow a^2-4a+1=0\Rightarrow a=\frac{4\pm \sqrt{16-4}}{2}=2\pm \sqrt{3}$$

We know the sequence is monotonically decreasing, with $a_1=3$, so $2+\sqrt{3}>3$ cannot be the limit. Thus, the limit is $a=2-\sqrt{3}$.

Proposition:

Let $c\in \mathbb{R}$ such that $|c|<1$. Then ${\lim }_{n\to \infty }{c}^n=0$.

Proof:

Suffices to consider $c>0$ (the case $c=0$ is trivial, and for $c<0$, $|c^n-0|=|c{|}^n$). Since $c\in (0,1),c^n$ decreases. Also, it is bounded below by 0 (because $c$ is a positive number). By MCT, the limit ${\lim }_{n\to \infty }{c}^n=l$ exists. $l=\inf \{c^n\mid n\in \mathbb{N}\}$.