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Definition: Maximum

Let $S $ be a nonempty subset of $\R $. If $\exists s_0 \in S$ such that $\forall s \in S, s \leq s_0$, then we call $s_0 $ the maximum of $S$ and write $s_0 = \max{ S }$.

Definition: Minimum

Let $S $ be a nonempty subset of $\R $. If $\exists s_0 \in S$ such that $\forall s \in S, s \geq s_0$, then we call $s_0 $ the minimum of $S$ and write $s_0 = \min{ S }$.

Definition: Upper Bound

$S$ is a subset of a ordered field $F$. If exists an element $m \in F$ such that $\forall s \in S, m \geq S$, then we call $m $ an upper bound for $S$.

If $x$ is an upper bound for $S $ and $x \in S $, then $x$ is maximum of $S$.

Definition: Lower Bound

$S$ is a subset of a ordered field $F$. If exists an element $m \in F$ such that $\forall s \in S, m \leq S$, then we call $m$ an lower bound for $S$.

If $x$ is an lower bound for $S $ and $x \in S $, then $x$ is minimum of $S$.

Definition: Least Upper Bound (Supremum)

The supremum of $S$ is the minimum of the set of $S$’s upper bounds, and is denoted by $\supr{ S } $.

Definition: Greatest Lower Bound (Infimum)

The infimum of $S$ is the maximum of the set of $S$’s lower bounds, and is denoted by $\infi{ S } $.

Note

By definition,

Maxima are suprema.

Minima are infima.

Note

$\N$ and $\Q$ have the same size, but $\R$ is bigger.

Note

By the definition of supremum and infimum,

$inf(\emptyset) = + \infty$,

$sup(\emptyset) = - \infty$.

Axioms: Completeness Axiom

If $S \subseteq \R, s \neq \emptyset$, $S$ is bounded above, then $S$ has a supremum.

Corollary

If $S \subseteq \R, s \neq \emptyset$, $S$ is bounded below, then $S$ has a infinum.

Proof 
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Let $S$ have lower bound $x$. Then $(-S) = \set{ -y \mid y \in S }$.

$x \leq y, \forall y \in S, -x \geq -y$.

$-x$ is an upper bound to $-S$.

$\therefore$ by completeness, there is a $u = \supr{ -s }$.

$u \geq -y, \forall y \in S$.

$\forall z \geq -y, \forall y \in S, z \geq u$.

Then $y \geq -u, \forall y \in S $(lower bound)

$\exists -z \leq y \forall y $

$Then -z \leq -u $

Note: Completeness Axiom Does Not Hold for $\Q $

The set $A = \set{ r \in \Q \mid 0 \leq r \leq \sqrt[ ]{ 2 }} $ is a set of rational number, but it DOES NOT have a supremum that is a rational number.

Note

Showing a number isn’t algebraic isn’t easy.

e.g. How to show $\pi, e $ are not algebraic?

Example

$\set{ \frac{ 1 }{ n }}_{n \geq 1} $

$0 $ is a lower bound of $S $.

$1 = \supr{ S }$,

$0 = \infi{ S }$.

Could $\infi{ S }$ be some very small $r > 0$?

The completeness axiom is called completeness is because it fills in the “gap”.

Properties: Archimedean Property

$a,b > 0.\ \exists n, na > b $.

Proof 
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Assume on the contrary, $\forall n, na \leq b$.

Then, $b$ is an upper bound to $S = \set{ na \mid n \in \N } $.

By completeness axiom, $\exists s_0, s_0 = sup(S)$.

Since $a > 0, s_0 < s_0 + a$.

$\therefore s_0 - a < s_0 $.

$s_0 - a $ is not upper bound to $S$.

$\exists n, s_0 -a < na $.

$s_0 < \pare{n+1} a $.

$s_0 < (n + 1)a \leq s_0 $.

$(n+1) a \in S$, so $s_0 $ is not an upper bound for $S $. CONTRADICTION.

$\therefore a, b > 0.\ na > b. $

Example

$a \leq b \leq c \leq d \implies a \leq d$, but why?

Theorem : Denseness of $\Q$

$$\forall a, b \in \R , a < b, \exists r \in \Q, a < r < b $$

Proof 
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