MATH 409 Lecture 1

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Begin Exam 1 content

Lecture Slides

Office appointments: Generally available for TWR afternoons

Topics:

• Axioms
• Point Set Theory
• Compactness, completeness, and connectedness
• Continuity and Uniform Continuity
• Sequencies, series
• Differentiatiability
• Theory of Riemann integration

Challenge 1

50 pts. No deadline

Let $f:\mathbb {R} \to \mathbb {R}$ be an infinitely differentiable function. Suppose that for any point $x\in \mathbb {R}$ there iexists a derivative of $f$ that vanishes at $x$:

$f^{(n)}\left(x\right)=0\,\!$ for some $n\geq 1$

Prove that $f$ is polynomial.

Note: A polynomial can be uniquely characterized as an infinitely differentiable function $f:\mathbb {R} \to \mathbb {R}$ such that $f^{(n)}\left(x\right)\equiv 0$ (identically zero) for some $n\geq 1$.

Smaller Challenge

What is the name of the river on the cover of the book.

Axioms of an Ordered Field

Real Line

Study of calculus of functions begins with study of domain: real numbers $\mathbb {R}$(real line)

The real line is a mathematical object rich with structure:

• algebraic structure (4 operations: add, subtract, multiplication, and division)
• ordering (when choosing any 3 points, one is located between the other two
• metric structure (measurable distance between points)
• continuity (we can get from one point to another in a continuous way)

Axiomatic Model

Provides solid foundation for all subsequent developments

Three postulates, each consisting of one or several axioms

To verify adequacy, prove that axioms are consistent (i.e. there exists an object satisfying them), and categorical (i.e. object is, in a sense, unique)

Axioms chosen among basic properties of real numbers:

1. Formalizes algebraic structure
2. Formalizes ordering
3. Formalizes continuous structure

Note: Metric structure can be formaziled in terms of other structures

Field

Motivated by the real numbers $\mathbb {R}$ and complex numbers $\mathbb {C}$

Informally, a field is a set with 4 arithmetic operations (+, −, ×, ÷) that have roughly the same properties as those of real (or complex) numbers.

Notion of field is important for linear algebra. Members of a field can serve as a set of scalars for a vector space.

Formally, A field is a set $F$ equipped with two closed binary operations:

• addition: $(a,b)\in F\times F~\mapsto ~a+b\in F$
• multiplication: $(a,b)\in F\times F~\mapsto ~a\cdot b\in F$

Which adhere to the following axioms:

F1.	$a+b=b+a$ for all $a,b\in F$	commutativity of addition
F2.	$(a+b)+c=a+(b+c)$ for all $a,b,c\in F$	associativity of addition
F3.	There exists an element of $F$, denoted $0$, such that $a+0=0+a=a$ for all $a\in F$	additive identity
F4.	For any $a\in F$, there exists an element of $F$, denoted $-a$, such that $a+(-a)=(-a)+a=0$	additive inverse
F1'.	$a\cdot b=b\cdot a$ for all $a,b\in F$	commutativity of multiplication
F2'.	$(a\cdot b)\cdot c=a\cdot (b\cdot c)$ for all $a,b,c\in F$	associativity of multiplication
F3'.	There exists an element of $F$ different from $0$, denoted $1$, such that $a\cdot 1=1\cdot a=a$ for all $a\in F$	multiplicative identity
F4'.	For any $a\in F$, $a\neq 0$, there exists an element of $F$, denoted $a^{-1}$, such that $a\cdot a^{-1}=a^{-1}\cdot a=1$	multiplicative inverse
F5.	$a\cdot (b+c)=(a\cdot b)+(a\cdot c)$ for all $a,b,c\in F$	distributive property

Subtraction and division are then defined as compound operations:

${\begin{aligned}a-b&=a+(-b)\\a/b&=a\cdot b^{-1}\end{aligned}}$

Postulate 1: The set of real numbers $\mathbb {R}$ is a field.

Other examples of fields include:

• complex numbers $\mathbb {C}$
• rational numbers $\mathbb {Q}$
• Rational functions $\mathbb {R} (x)$ in variable $x$ with real coefficients (e.g. $f(x)={\frac {a_{n}\,x^{n}+a_{n-1}\,x^{n-1}+\dots +a_{1}\,x+a_{0}}{b_{m}\,x^{m}+b_{m-1}\,x^{m-1}+\dots +b_{1}\,x+b_{0}}}$ for $a_{i},b_{j}\in \mathbb {R}$ and $b_{m}\neq 0$)
• Field of two elements $\mathbb {F} =\left\{{\bar {0}},{\bar {1}}\right\}$

Properties

Expand

Theorem. The zero 0 is unique.

Proof. Suppose $z_{1}$ and $z_{2}$ are both zeroes, so $a+z_{1}=z_{1}+a=a$ and $a+z_{2}=z_{2}+a=a$ for all $a\in F$.

Then $z_{1}+z_{2}=z_{2}$ and $z_{1}+z_{2}=z_{1}$. Hence $z_{1}=z_{2}$.

quod erat demonstrandum

Expand

Theorem. For any $a\in F$, the negative $-a$ is unique.

Proof. Suppose $b_{1}$ and $b_{2}$ are both negatives of $a$. Let's compute the sum of $b_{1}+a+b_{2}$ in two ways:

1. $\left(b_{1}+a\right)+b_{2}=0+b_{2}=b_{2}$
2. $b_{1}+\left(a+b_{2}\right)=b_{1}+0=b_{1}$

By associativity of addition, $b_{1}=b_{2}$.

quod erat demonstrandum

Expand

Theorem. [Cancellation Law]. $a+c=b+c$ implies $a=b$ for any $a,b,c\in F$

Proof. If $a+c=b+c$ then $(a+c)+(-c)=(b+c)+(-c)$. By associativity, $(a+c)+(-c)=a+(c+(-c))=a+0=a$ and $(b+c)+(-c)=b+(c+(-c))=b+0=b$.

Hence $a=b$.

quod erat demonstrandum

Expand

Theorem. $0\cdot a=0$ for any $a\in F$

Proof. Start with $0\cdot a+a=0=0\cdot a+1\cdot a$. Apply the distributive law to get $(0+1)\cdot a=1\cdot a=a$

Let's write this as $0+a$. By the cancellation law, $0\cdot a=0$

quod erat demonstrandum

Expand

Theorem. $(-1)\cdot a=-a$ for any $a\in F$

Proof. There are two ways to add: $a+(-1)\cdot a=(-1)\cdot a+a$. This is equal to $(-1)\cdot a+1\cdot a=(-1+1)\cdot a=0\cdot a=0$ by the previous property.

quod erat demonstrandum

Other properties:

• $-(-a)=a$ for all $a\in F$
• $(-1)\cdot (-1)=1$
• $-(a-b)=b-a$ for all $a,b\in F$
• The unity 1 is unique. (same proof as uniqueness of 0)
• For any $a\neq 0$, the inverse $a^{-1}$ is unique.
• (Cancellation law) $ac=bc$ implies $a=b$ whenever $c\neq 0$.
• For any $a,b\in F$, the equality $a\,b=0$ implies that either $a=0$ or $b=0$.

Relations

Recall that the Cartesian product $X\times Y$ of two sets $X$ and $Y$ is the set of all ordered pairs $(x,y)$ such that $x\in X$ and $y\in Y$.

The Cartesian square $X\times X$ is denoted $X^{2}$

A relation $R$ on a set $X$ is (identified with) a subset of its Cartesian square $R\subseteq X^{2}$.

If $(x,y)\in R$ then we say $x$ is related to $y$ (in the sense of $R$ or by $R$) and write $xRy$.

Examples:

• Equality: $xRy\iff x=y$
• Not equal to: $xRy\iff x\neq y$ (complement of equality)
• Less than: $xRy\iff x<y$
• Less than or equal to: $xRy\iff x\leq y$
• Is contained in: $xRy\iff x\in y$
• Divides: $xRy\iff y/x\in \mathbb {Z}$ for $x\in \mathbb {Z}$

Properties

Let $R$ be a relation on a set $x$

• Reflexive: $xRx$ for all $x\in X$
• Symmetric: $xRy$ implies $yRx$
• Antisymmetric: $xRy$ and $yRx$ cannot hold simultaneously
• Weakly antisymmetric: $xRy$ and $yRx$ imply that $x=y$
• Transitive: for all $x,y,z\in X$, $xRy$, $yRz$ imply $xRz$

Partial Ordering

Relation $R$ on a set $X$ is a partial ordering (or partial order, or simply order) if $R$ is reflexive, weakly antisymmetric, and transitive:

• $xRx$
• $xRy$ and $yRx$ imply $x=y$
• $xRy$ and $yRz$ imply $xRz$

(e.g. less than or equal to)

Strict Partial Ordering

A relation $R$ on a set $X$ is a strict partial order (or strict order) if $R$ is antisymmetric and transitive

• $xRy$ implies $\neg (yRx)$
• $xRy$ and $yRz$ imply $xRz$

(e.g. less than)

Linear / Total Ordering

An order $R$ on a set $X$ is called linear (or total) if for any elements $x,y\in X$ at least one of the following statements hold:

• $xRy$
• $yRx$, or
• $x=y$

Postulate 2: There is a relation on the set of real numbers $\mathbb {R}$, denoted $<$, that is a strict linear order. Moreover, this order and arithmetic operations on $\mathbb {R}$ satisfy the following axioms:

OA.	$a<b$ implies $a+c<b+c$ for all $a,b,c\in \mathbb {R}$
OM1.	$a<b$ and $c>0$ imply $a\,c<b\,c$ for all $a,b,c\in \mathbb {R}$
OM2.	$a<b$ and $c<0$ imply $bc<ac$ for all $a,b,c\in \mathbb {R}$
the axioms OM1 and OM2 can be replaced by a third
OM.	$0<a$ and $0<b$ imply $0<ab$ for all $a,b\in \mathbb {R}$

Auxiliary notation

• $a>b$ means that $b<a$
• by $a\leq b$, we mean that $a<b$ or $a=b$.
• By $a<b<c$, we mean that $a<b$ and $b<c$

Properties of Linearly Ordered Fields

• $a>0\implies -a<0$
• $a<b\implies a-b<0$
• $(a>0)\wedge (b<0)\implies ab<0$
• $(a<0)\wedge (b<0)\implies ab>0$
• $a\neq 0\implies a^{2}>0$ (where $a^{2}=a\cdot a$)
• $0<1$
• $(0<a<b)\wedge (0<c<d)\implies a\,c<b\,d$
• $0<a<b\implies a^{-1}>b^{-1}>0$