Lecture 4

The supremum of a set $S$ is the least upper bound of a non-empty set, bounded above. It has the following characterization: \[ L = sup S \iff \begin{cases} \forall x \in S, & \quad x \leq L \\ \forall M \in \mathbb{R}, & \quad \text{if } M \text{ upper bound of } S \implies L \leq M \\ \end{cases} \]

By the completeness axiom, if $S$ bounded above, sup $S$ always exists.

Definition: Max $S$ is the maximal element of $S$ (if it exists) that satisfies (a) $max S \in S$ and (b) $x \leq max S$.

Claim: If $max S$ exists, then $max S = sup S$.

Suppose $max S$ exists by (b), $\forall x \in S$, $x \leq max S \implies max S$ is an upper bound of $S$, then $sup S \leq max S$.

Now $max S \in S$ and $sup S$ is an upper bound for $S \implies max S \leq sup S$. $\implies max S = sup S$. $\blacksquare$

Back to density theorems!

• Distribution of $\mathbb{Z}$: $\forall c \in \mathbb{R}$, $\exists! n \in \mathbb{Z}$, $c \leq n < c + 1 \iff n \in [c, c+1)$
• Distribution of $\mathbb{Q}$: $\forall a,b$ such that $a < b$ and $a,b \in \mathbb{R}$, $\exists x \in \mathbb{Q}, x \in (a,b)

Prop 2 from last time: If $S \subset \mathbb{Z}$, $S \neq \emptyset$, bounded above, then max $S$ exists.

Proof of Distribution of $\mathbb{Z}$:

Let $c \in \mathbb{R}$ and define $s \{n | n \in \mathbb{Z}, n << +1 \}$

$s \neq \emptyset$: if $c \geq 0$: then $0 \in S$ so $S \neq \emptyset$

$s \neq \emptyset$: if $c < 0$: by Arch law, there exists $n \in \mathbb{N}$ such that $n > -(c +1) \implies -n < c+1$ i.e. $-n \in S$

$S$ is bounded above by construction by $c +1$. $S \subset \mathbb{Z}$ by construction. By Prop 2, $max S = k$ exists. $k = max S \in S$ so by the defining property of $S$, $k < c + 1$.

Now will show$k \geq c$. By contradiction, assume $k < c$, then $k +1 < c +1$, $k + 1 \in \mathbb{Z}$. So $k +1 \in S$. Since $k = Max S$, $k + 1 \leq k$. Here's the contradiction. $k \geq c$.

This proves the existence of $k$, but not its uniqueness. Suppose $k,k'$ saisfy $k,k' \in \mathbb{Z}$. $c \leq k < c + 1$ and $c \leq k' < c + 1 \iff -c-1 < -k' \leq -c$.

$c-c-1 < k -k' < 1$

$-1 < k -k' < 1$

Suppose without loss of generality, $k > k'$. then $0 < k - k' < 1 \implies k - k' \in \mathbb{Z}$ which is a contradiction! This proves uniqueness, and we are done. $\blacksquare$

Definition: Let $U \subset \mathbb{R}$, we say that $U$ is dense in $\mathbb{R} \iff \forall a < b$, $a,b \in \mathbb{R}$ exists $x \in U$, $a < x < b$.

With this definition, the distribution of $\mathbb{Q}$ is equivalent to $\mathbb{Q}$ is dense in $\mathbb{R}$.

Proof of Distribution of $\mathbb{Q}$:

Let $a,b \in \mathbb{R}$, $a < b \implies b-a > 0$. By the Archimedian Law, there exists $q \in \mathbb{N}$ such that $\frac{1}{q} < b - a$. $\iff 1 < q(b-a) = qb-qa$.

by the distribution of $\mathbb{Z}$, there exists a unique $p \in \mathbb{Z}$ such that $qb -1 \leq p < qb$. Since $qb - 1 > qa$, we have $qa < p< qb \iff a < \frac{p}{q} < b$. So $\mathbb{Q}$ is dense in $\mathbb{R}$. $\blacksquare$

Corollary: The irrationals, $\mathbb{R} \setminus \mathbb{Q}$ are dense in $\mathbb{R}$.

Fun fact, there are more irrationals than rationals. Think about transcendental darts!

Proof:

WTS: $\forall a,b \in \mathbb{R}$, $\exists x \in \mathbb{R} \setminus \mathbb{Q}$, $a < x< b$, and use that $\sqrt{2}$ is irrational.

Let $a < b$. Then $\frac{a}{\sqrt{2}} < \frac{b}{\sqrt{2}}$ by the density of $\mathbb{Q}$, there exists a rational $\frac{p}{q}$ such that $\frac{a}{\sqrt{2}} < \frac{p}{q} < \frac{b}{\sqrt{2}} \iff a < \sqrt{2} \frac{p}{q} < b$

We need to show that $\sqrt{2} \frac{p}{q}$ is not a rational number. By contradiction, let $\sqrt{2} \frac{p}{q} \in \mathbb{Q}$. $\frac{p}{q} \in \mathbb{Q}$ Since $\mathbb{Q}$ is stable by multiplication, then $\sqrt{2} \cdot \frac{p}{q} \cdot \frac{q}{p} = \sqrt{2} \in \mathbb{Q}$. And here is the contradiction. $\blacksquare$

New Thing!! Inequalities and Identities

Def: The absolute value of $x$ means that for $x \in \mathbb{R}$, $|x| = x$ if $x \geq 0$ and $|x| = -x$ if $x < 0$, ie = max$(x, -x)$.

Example: |2| = 2, |-3| = 3

Property: $\forall x \in \mathbb{R}$, $|x| \geq 0$.

Claim: If $c \in \mathbb{R}$, $d \geq 0$, "$|c| \leq d$" $\iff -d \leq c \leq d$

"$|c| \leq d$" $\implies$ max$(c,-c) \leq d \iff c \leq d, -c \leq d \iff c \leq d, -d \leq c$. In particular, $-|x| \leq x \leq |x|$

Triangle inequality: $\forall a,b \in \mathbb{R}$, $|a+b| \leq |a| + |b|$

Proof:

$-|a| \leq a \leq |a|$

$-|b| \leq b \leq |b|$

$-(|a|+|b|) \leq a + b \leq |a|+|b| \iff |a + b| \leq |a| + |b|$

Another way to write the triange inequality...swap out the b with -b

$|a - b| = |a + (-b)| \leq |a| + |-b|= |a| + |b|$

$\forall a,b \in \mathbb{R}, |a - b| \leq |a| + |b|$