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

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 $a,b\in P,a+b,ab\in P$ (2) For any $a\in \mathbb{R}$, either $a\in P,a\notin P,a=0$ (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, $N\subseteq S$ by definition.

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 $Q$

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. Equivalently, $\mathbb{R}\backslash \mathbb{Q}\ne \varnothing$.

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 for any upper bound $c^{\prime }$ of $S$, we have $c\ge c^{\prime }$. We denote such $c$ by $l.u.bS$ or $\boxed{\sup S}$.

Completeness Axiom:

For any $S\subseteq \mathbb{R}$ 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 $S\in \mathbb{R}$, 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 $\sup S=2$). Also note that $\sup S$ is unique. We should prove this.

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 $c\ge c^{\prime }$, 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. Since $\sup T$ exists, the lower bounds of $S$ are negatives of upper bounds of $T$, then $-c=\inf S$.

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 $\epsilon =\frac{1}{c}$, wee 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 $k>b-\frac{1}{2},k\in \mathbb{N}$.

By definition, $k+1$ exists, and $k+1>b-\frac{1}{2}+1>b$. Thus, we have a contradiction.

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 eixsts $k\in \mathbb{Z}$ 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 $c+1\ge 0$, 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 $k^{\prime }<c+1\Rightarrow k\ge c$. Then $k^{\prime }-k<c+1-c=1$, 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 $m\in [nb-1,nb)\equiv \frac{m}{n}\in [b-\frac{1}{n},b)\subseteq (a,b)$.

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

Corollary:

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

Proof:

Let $a,b\in \mathbb{R},a<b$. By the Theorem, $q\in \mathbb{Q}$ such that $q\in (\frac{a}{\sqrt{2}},\frac{b}{\sqrt{2}})$. Thus, $\sqrt{2}q\in (a,b)$, where $\sqrt{2}q$ is irrational.

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 $$-|a|\le a\le |a|-|b|\le b\le |b|$$

Prop.:

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

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

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

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

Lecture 5

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

Sequences & Convergence

Definition:

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

Notation:

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

Example:

$\{\frac{1}{n}\},\{1+(-1{)}^n\},\{a_n\}\text{ where an is the largest integer that is less than n3}$, 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 $n\in \mathbb{N},n\le N$, $|a_n-a|<\epsilon$.

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