Lecture 1: The real number system, $\mathbb{R}$

Welcome to Real Analysis! This is a proof-based class that is concerned with proving properties of the real number system. Over this course, we will study the following sets of numbers, which comprise the real numbers:

Theorem (Existance of Reals): There exists a totally ordered field satisfying the completeness axiom, which contains the rationals $\mathbb{Q}$ as a subfield.

We aren't going to prove this yet. Let's break down some of the terminology. What is:

A field
Totally ordered
The completeness axiom

Let's also take the chance to get on the same page regarding some symbols commonly used in proofs:

$\forall$ means "for all."
Example: $\forall x > y$ means "for all values of the variable x that are greater than y."
$\in$ means "is an element of."
Example: $\forall x \in \mathbb{R}$ means "for all values of the variable x that are real" or "for all elements of the reals (which we weill represent by the abstraction $x$)."
s.t. or $|$ means "such that."
Example: $\forall x \in \mathbb{R} | x < 7$ means "for all values of the variable x that are real and such that $x < 7$"
$\exists$ means "there exists."
Example: $\exists x \in \mathbb{R} | x < 7$ means "there exists a value of the variable x that is real and such that $x < 7$"

Definition: The real number line with addition and multiplication is a field. For every $a,b \in \mathbb{R}$, $a+b$ and $ab$ are real.

The field axioms of addition and multiplication satisfy the following properties:

Addition is commutative: $ a + b = b + a$, $ \forall a,b \in \mathbb{R}$
Addition is associative: $a + b = b + a$, $ \forall a,b \in \mathbb{R}$
Addition has a neutral (identity) element. There exists something (let's call it zero.) such that $0 + a = a + 0 = a, \forall a \in \mathbb{R} $
Addition is symmetric: $\forall a \in \mathbb{R}$, $\exists$ some $-a$ s.t. $a + (-a) = (-a) + a = 0$

Just considering the real number line and addition, we can say that $(\mathbb{R}, +)$ is a group. Now throw in multiplication.

Multiplication is associative: $a \cdot b = b \cdot a$ $\forall a,b \in \mathbb{R}$
Multiplication is distributive: $(a \cdot b) \cdot c = a \cdot (b \cdot c)$ $\forall a,b,c \in \mathbb{R}$
Multiplciation has a neutral element: There exists something (let's call it one.) such that $1 \cdot a = a \cdot 1 = a, \forall a \in \mathbb{R} \setminus \{0\}$
Multiplcation is symmetric: $\forall a \in \mathbb{R} \setminus \{0\}$, $\exists$ $a^{-1}$ s.t. $a \cdot a^{-1} = a^{-1} \cdot a = 1$
Multiplication is commutative: $ a \cdot b = b \cdot a$ $\forall a,b \in \mathbb{R}$

It's just a matter of notation to extend the above statements to subtraction and division. Subtraction: $a-b = a + (-b)$, $ \forall a,b \in \mathbb{R}$. Division: $\frac{a}{b} = a \cdot b^{-1}$, $\forall a \in \mathbb{R}$, $\forall b \in \mathbb{R} \setminus \{0\}$

Let's use fancy terms to describe sets of numbers and operators that we already know!

$(\mathbb{Q}, +, \cdot)$ is a field. It satisfies all of the things listed above.
$(\mathbb{Z}, +, \cdot)$ is a ring. It satisfies all of the things listed above except the symmetric property. $2^{-1} = \frac{1}{2} \notin \mathbb{Z}$.
$(\mathbb{N}, +, \cdot)$ satisfies the associative properties, the commutative properties, the multiplicative neutral element, and the distributive property. But doesn't have enough structure to be a field or a ring.

There are consequences of the field properties:

0 is unique.
$ \forall a \in \mathbb{R}$, $0 \cdot a = a \cdot 0 = 0 $
1 is unique.
$\forall a \in \mathbb{R}$, if $a \cdot b = 0$, then either $a = 0$ or $b = 0$
0 is unique.

Hey! We're just one-third the way through that first theorem. Go look at some cats in sinks.

TOTALLY ORDERED: $\forall a,b \in \mathbb{R}$, it must be that $a < b$ or $a > b$ or $a = b$

Here are some consequences of the real line being totally ordered:

$\forall a \in \mathbb{R}$, $a \neq 0$, $a^{2} > 0$. In particular, $1 = 1 \cdot 1 = 1^{2} > 0$
$\forall a \in \mathbb{R} \setminus \{0\}$, $a > 0 \implies a^{-1} > 0$
if $a < b$, then $a \cdot c < b \cdot c$ if $c > 0$ and $a \cdot c > b \cdot c$ if $c < 0$
if $a>b$, then $\forall c \in \mathbb{R}$, $a+c>b+c$

Proof: Consequence 4:

Suppose $a > b$. Then $a-b>0$.

Notice $a-b = a -b +c -c = (a+c) - (b+c)$

So $(a+c) - (b+c) >0 \iff a + c > b + c$

And we're done!

We'll save the completeness axiom for Lecture 2, and end with some definitions.

Definition: $\forall a,b \in \mathbb{R}$, the "maximum" of $a$ and $b$ is: \[ max(a,b) = \begin{cases} a & \quad \text{if } a \geq b \\ b & \quad \text{if } a < b\\ \end{cases} \]

Definition: $\forall a,b \in \mathbb{R}$, the "minimum" of $a$ and $b$ is: \[ min(a,b) = \begin{cases} b & \quad \text{if } a \geq b \\ a & \quad \text{if } a < b\\ \end{cases} \]

Definition: A set $S \subset \mathbb{R}$ is an aggregate of numbers.

Examples: $S = \{1, 2, 3\}$ or $S = \mathbb{N}$