The Real Numbers

Naturals

$N = {1, 2, 3, 4, 5, \dots}$

Integers

$Z = {\dots, - 3, - 2, - 1, 0, 1, 2, 3, \dots}$

Rational number

any number that can be expressed in the form $\frac{p}{q}$ where $p, q \in Z$ and $q \neq 0$

1.1.1 pf $\sqrt{2}$ is not rational

\begin{aligned} {(\frac{p}{q})}^{2} = 2 where gcd(p,q)=1 \\ p^{2} = 2 q^{2} ⟹ p is even by Euclid \\ p = 2 r ⟹ 2 r^{2} = q^{2} \\ so q is even which is a contradiction \end{aligned}

Field

$Q$ makes up a field
any set where addition and multiplication are well-defined operators:

Commutative, associative, distributive, additive identity, additive inverse, multiplicative inverse, multiplicative identity, and closure.

Ordered
$x \leq y$ or $y \leq x$
if $x \leq y$ and $y \leq x$ then $x = y$
if $x \leq y$ and $y \leq z$ then $x \leq z$
if $y \leq z$ then $x + y \leq x + z$
if $0 \leq x$ and $0 \leq y$ then $0 \leq x y$

We assume that $R$ is an ordered field that contains $Q$ .

Functions

Given two sets $A$ and $B$ a function from $A$ to $B$ is a mapping that takes each element $x \in A$ with a single element of $B$ . So $f : A \to B$

Range: ${y \in B : y = f (x) for some x \in A}$

Drichlet's unruly function

g (x) = {\begin{cases} 1 if x \in Q \\ 0 if x \notin Q \end{cases}

Absolute Value Function

| x | = {\begin{cases} x if x \geq 0 \\ - x if x < 0 \end{cases}

Properties

\begin{aligned} | a b | = | a | | b | \\ | a + b | \leq | a | + | b | triangle inequality \end{aligned}

by subbing $a = (a + c - c)$ into the triangle inequality we get:

| a - b | \leq | a - c | + | c - b |

Axiom of Completeness

Every nonempty set of real numbers that is bounded above has a least upper bound

Bounded Above

$A \subseteq R$ is bounded above if there exists a number $b \in R$ such that $a \leq b$ for a $a \in A$

Least Upper bound (supremum)

$s \in R$ is the least upper bound for $A \subseteq R$ if it meets both criteria:

1 $s$ is an upper bound for $A$
2 if $b$ is any upper bound for $A$ , then $s \leq b$

Greatest lower bound is the infimum.

Maximum/Minimum

$a_{0}$ is a maximum of the set $A$ if it is an element of $A$ and $a_{0} \geq a$ for all $a \in A$ .

$a_{1}$ is a minimum if $a_{1} \in A$ and $a_{1} \leq a$ for every $a \in A$ .

When a maximum exists it is a supremum and when a minimum exists it is a infimum.

Example 1.3.7

$A \subseteq R$ is nonempty and bounded above, let $c \in R$ . Now we have the set:

c + A = {c + a : a \in A}

Claim: sup( $c + A$ ) = $c$ + sup $A$

$s = sup A ⟹ a \leq s$ for all $a \in A$

a + c \leq s + c

so $s + c$ is an upper bound.

Let $b$ be another upper bound for $c + A$ .

\begin{matrix} c + a \leq b \\ a \leq b - c \end{matrix}

so $b - c$ is an upper bound for $a$ so $s < b - c$ since it is sup $A$ . So $s + c < b$ . Therefore $s + c$ is sup $(c + A)$ .

Lemma 1.3.8 Supremum

Assuming $s \in R$ is an upper bound for $A \subseteq R$ ,

then $s = sup A$ iff for every choice of $ε > 0$ , there exists an element $a \in A$ such that $s - ε < a$ .

Proof

$\Rightarrow$
Assume $s = sup A$ . $s - ε < s$ so $s - ε$ is not an upper bound for $A$ . Then there must be $a \in A$ so that $s - ε < a$ .

$\Leftarrow$
$s$ is an upper bound. For any $ε > 0$ , $\exists a$ where $s - ε < a$ so $s - ε$ is no longer a upper bound. Let $b < s$ so $s - ε$ for some $ε > 0$ . Therefore $b$ is not an upper bound, therefore $s$ is the lowest upper bound. $s = sup A$ .

Consequences of Completeness

Thm 1.4.1 Nested Interval Property

For each $n \in N$ assume that there is some closed interval $I_{n} = [a_{n}, b_{n}] = {x \in R : a_{n} \leq x \leq b_{n}}$
If $I_{1} \supseteq I_{2} \supseteq I_{3} \supseteq \dots$ then $⋂_{n = 1}^{\infty} I_{n} \neq \emptyset$

Proof

Let $A = {a_{n} : n \in N}$ which is the lower bound of each interval. Since they are nested each $b_{n}$ serves as an upper-bound to $A$ . Let $x = sup A$ which exists by the axiom of completeness.

So for a particular $I_{n}$ we have $a_{n} \leq x$ and $x \leq b_{n}$ which means $x \in I_{n}$ for every choice of $n \in N$ . Therefore the intersection is non-empty.

Thm 1.4.2 Archimedean Property

Statement 1: Given any number $x \in R$ $\exists$ an $n \in N$ where $n > x$

Statement 2: Given any $y > 0$ there exists an $n \in N$ satisfying $\frac{1}{n} < y$

Proof

Proof (statement 1)
States that $N$ is not bounded above. For contradiction assume that $N$ is bounded above. $N \subset R$ so by the axiom of completeness there exists an $α = sup N$ . So, $α - 1$ is no longer an upper bound so there is an $n \in N$ so that $α - 1 < n$ therefore $α < (n + 1) \in N$ which is a contradiction.

( $N$ is closed under addition)

(statement 2) From our previous proof let $x = \frac{1}{y} ⟹ \frac{1}{y} < n$ for some $n \in N$ .

Thm 1.4.3 Density of Q in R

For every two real numbers $a$ and $b$ with $a < b$ $\exists$ $r \in Q$ such that $a < r < b$ .

Proof

Proof want a $m \in Z$ and $n \in Z$ so that

\begin{matrix} (1) & a < \frac{m}{n} < b \end{matrix}

by Archimedean Property $\exists n \in N$ where

\begin{matrix} (2) & \frac{1}{n} < b - a \end{matrix}

Inequality (1) is equivalent to $n a < m < n b$

Pick an $m \in Z$ so that

m - 1 \overset{(3)}{\leq} n a \overset{(4)}{<} m

Inequality (4) gives us $a < \frac{m}{n}$

Inequality (2) is equivalent to $a < b - \frac{1}{n}$ .

\begin{aligned} by (3) & m & \leq n a + 1 \\ < n (b - \frac{1}{n}) + 1 \\ = n b \end{aligned}

so $n < n b ⟹ \frac{m}{n} < b$ and we are done.

Corollary 1.4.4 Irrational Dense in R

Given any two real numbers $a < b$ there exists a $t \in R ∖ Q$ satisfying $a < t < b$

Proof

$r \in Q$ if $x$ is irrational then so is $r + x$ otherwise if $r + x \in Q$ then $r + x - r$ since $Q$ is a field which contradicts $r$ being irrational.

So let for any $a, b \in R$ we have $a - \sqrt{2} < a$ and $b - \sqrt{2} < b$ .

by density in $R$ $\exists r \in Q$ so that

\begin{aligned} a - \sqrt{2} & < r < b - \sqrt{2} \\ a & < r + \sqrt{2} < b \end{aligned}

and $r + \sqrt{2}$ is irrational.

Thm 1.4.5 Existence of Square Roots

There exists an $α \in R$ such that $α^{2} = 2$
Proof

T = {t \in R : t^{2} < 2}

Assume $α = sup T$
Assume for contradiction that
$α^{2} < 2$

\begin{aligned} {(a + \frac{1}{n})}^{2} & = α^{2} + \frac{2 α}{n} + \frac{1}{n^{2}} \\ < α + \frac{2 α}{n} + \frac{1}{n} \\ = α^{2} + \frac{2 α + 1}{n} \end{aligned}

\begin{array}{r} α^{2} + \frac{2 α + 1}{n} < 2 + \frac{2 α + 1}{n} \end{array}

By Archimedes there is a $\frac{1}{n_{0}} < \frac{2 - a^{2}}{2 α + 1}$ so

\begin{aligned} \frac{2 α + 1}{n_{0}} & < 2 - α^{2} \\ {(α + \frac{1}{n_{0}})}^{2} & < α^{2} + (2 - α^{2}) \end{aligned}

so $(α + \frac{1}{n}) \in T$ contradicting that $sup T = α$

Assume now that $α^{2} > 2$

\begin{aligned} {(α - \frac{1}{n})}^{2} & = α^{2} - \frac{2 α}{n} + \frac{1}{n^{2}} \\ > α^{2} - \frac{2 α}{n} \end{aligned}

by Archimedes there is a $n_{0}$ such that

\begin{aligned} \frac{1}{n_{0}} & < \frac{α^{2} + 2}{2 α} so \\ - \frac{2 α}{n_{0}} & > - (a^{2} + 2) \end{aligned}

so therefore

\begin{array}{r} {(α - \frac{1}{n_{0}})}^{2} > a^{2} - (a^{2} + 2) \end{array}

$α - \frac{1}{n_{0}} \notin T$ and $α - \frac{1}{n_{0}} < α$ so $a - \frac{1}{n_{0}}$ is a lower bound.