Lecture 3

Def: We say that $S$ is bounded above by $M$ (ie $M$ is an upper bound for $S$) if $\forall x \in S$, $x \leq M$

A set bounded above admits many upper bounds. We are usually interested in the least upper bound.

Assume $S$ is bounded above. Then by the completeness axiom, if $S$ is not empty, then $S$ has a least upper bound (= supremum).

The defining properties of the supremum:

1. $L = sup S \iff L$ is an upper bound for $S$
2. $L = sup S \iff \forall M \in \mathbb{R}$, if $M$ is an upper bound for $S$, then $L \leq M$.

In Lecture 2, we proved the following characterization: \[ L = sup S \iff \begin{cases} \forall x \in S, x \leq L \\ \forall \epsilon > 0, \exists x \in S such that L - \epsilon < x \\ \end{cases} \]

Def: an open interval $(a, b)$, for $a, b \in \mathbb{R}$, $a < b$ is that the set $(a, b) = \{x \in \mathbb{R}, a < x < b \}$

An open interval does not include it's end points.

Def: a closed interval $[a, b]$, for $a, b \in \mathbb{R}$, $a < b$ is that the set $(a, b) = \{x \in \mathbb{R}, a \leq x \leq b \}$

We can also have half-closed, half open intervals:

• $[a, b) = \{x \in \mathbb{R}$, $a \leq x < b$ \}
• $(a, b] = \{x \in \mathbb{R}$, $a < x \leq b$ \}

Example: $S = (0, 1)$, $1 \notin S$ so we can show $sup S = 1$.

Definition: $S$ is bounded below by $m \in \mathbb{R}$ if $\forall x \in S$, $m \leq x$.

Definition: $S$, bounded below, admits a greatest lower bound (aka GLB or infinum), if there exists $l$ such that:

1. $l$ is a lower bound for $S$
2. $\forall m \in \mathbb{R}$ that are lower bounds of $S$, $m \leq l$

Remember, by the completeness axiom, every $S \neq \emptyset$ bounded above admits a supremum.

Theorem: every $S \subset \mathbb{R}$, $S \neq \emptyset$ bounded below, admits a greatest lower bound or infinum.

Let's prove this using a reflection argument.

Proof:

Let $S \subset \mathbb{R}$, $S \neq \emptyset$, bounded below.

Define $S' = \{-x, x\in S\}$

Claim: $S'$ is bounded above.

Let $m$ be a lower bound for $S$. Then $\forall x \in S$, $m \leq x \iff -x \leq -m$

This means: $\forall x' \in S'$, $x' \leq -m \implies -m$ is an upper bound for $S'$.

By the completeness axiom, $S'$ has an infinum, called $L$.

Claim: $-L = inf S$ since $L = sup S'$ then:

1. $\forall x \in S'$, $x \leq L$
2. $\forall M$, U.B of $S'$, $L \leq M$

Let $x \in S$. Then $-x \in S'$, so by (1) $-x \leq L \iff -L \leq x$.

So $\forall x \in S$, $-L \leq x$. $-L$ is a lower bound of $S$.

Let $m$ be a lower bound for $S$. Claim: $m \leq -L$

$m$ is a lower bound for $S \iff \forall x \in S$, $m \leq X \iff -m \geq -x \iff \forall x' \in S'$, $x' \leq -m$.

So $-m$ is an upper bound for $S'$, since $L = sup S'$

so $L \leq -m \iff m \leq -L$. $\blacksquare$

We can derive everything about lower bounds based on stuff about upper bounds.

NEW THING!!

The distribution of $\mathbb{Q}$ and $\mathbb{Z}$

Definition (the distribution of $\mathbb{Z}$): $\forall c \in \mathbb{R}$, $\exists ! n \in \mathbb{Z}$, $c \leq n < c + 1 \iff n \in [c, c+1)$

Definition (the 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)

$\mathbb{Q}$ populated \mathbb{R} in a dense way.

Thm (Archimedian property): The following equivalent statements hold true:

1. $\forall c > 0$, $\exists n \in \mathbb{N}$ such that $n > c$
2. $\forall \epsilon > 0$, $\exists n \in \mathbb{N}$ such that $0 < \frac{1}{n} < \epsilon$

Proof:

WTS: (1) $\iff$ (2) given $c>0$ and $\epsilon = \frac{1}{c} > 0$.

If $n$ is such that $n > c$, then $\frac{1}{n} < \epsilon = \frac{1}{c}$ and vice versa. so (1) $\iff$ (2), as we said. We only need to prove (1)

By contradiction, suppose (1) does not hold. $\exists c > 0$, $\forall n \in \mathbb{N}$, $n \leq C \implies \mathbb{N}$ is bounded above $\implies$ by the completeness axiom, $\exists$ sup $\mathbb{N} = L$.

Since $L$ is the supremum, $L - \frac{1}{2}$ is not an upper bound for $\mathbb{N}$.

$\exists n \in \mathbb{N}$ such that $L - \frac{1}{2} < n$

Then $L < L + \frac{1}{2} = L - \frac{1}{2} + 1 < n + 1$

But $n + 1 \in \mathbb{N}$ and $L$ is an upper bound for $\mathbb{N}$ which is a contradiction! $\blacksquare$

Proposition 1: $\forall n \in \mathbb{Z}$, there is no integer inside $(n, n +1)$

Recall: For $S \subset \mathbb{R}$ bounded above and not empty, sup $S$ always exists, but max $S$ may not always exist.

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

Proof:

By the completeness axiom, sup $S$ = $a$ always exists. $a$ is the least upper bound, so $a - 1$ is not an upper bound for $S$. So there exists $m \in S$ such that $a - 1 < m \iff a < m + 1$. So $\forall n \in S$, $n \leq a < m+1$ so $S \subset (-\infty, m + 1) = \{x \in \mathbb{R}$ such that $x < m + 1\}$

But by Prop 1, $(m, m+1)$ has no integer in it. So $S \cap (m, m +1) = \emptyset$. $S \subset (-\infty, m+1)$ with $S \cap (m, m +1) = \emptyset$ so $S \subset (-\infty, m]$ since $m \in S$. $m$ is an upper bound for $S$ and since $m \in S$, $m = max S$. $\blacksquare$