Supremum, Infimum

1. Definition : Supremum

$E \subset \R$ and $E \ne \empty$ 이라고 가정하자.
(1) $E$ is bounded above $\iff$ $\forall a \in E, \exist\ M\in \R$ s.t. $a\leq M$.
$M$ is called upper-bound of $E$.
(2) $s\in\R$ is a supremum of $E$ $\iff$ $s$ is an upper-bound of $E$ and $s\leq M$ for all upper-bounds of $E$.
(다시 말해, upperbound들 중 가장 작은 개체가 supremum이다.)

1.1. Theorem : Approximation Property for Suprema

If $E$ has a (finite) supremum, then $\forall\epsilon\gt0$, $\exist\ a\in E$ s.t. $supE-\epsilon<a\le supE$.
pf)
Theorem이 False라고 가정하면, $\exist\epsilon_0, \forall a \in E$ s.t. $a\notin (supE-\epsilon_0, supE]$이다. 그렇다면, $s:=supE-\epsilon_0$ 는 $E$의 upper-bound 라고 할 수 있다. 따라서 $supE\leq s=supE-\epsilon_0<supE$ 라는 contradiction이 생기므로, Theorem은 True이다.

1.2. Completenes Axiom

If $E\subset \R$ is non-empty and bounded above, then $E$ has a (finite) supremum.

1.3. Theorem

If $E\subset\Z$ has a supremum, then $sup E\in E$.
pf)
$s:=supE$라고 정의하자. Approx. Property for Suprema에 의해, 우리는 $x_0\in E$ s.t $s-1<x_0\le s_0$ 를 하나 선택할 수 있다.
Claim : $s=x_0$
If $s\ne x_0$, choose $x_1\in E$ s.t. $x_0<x_1\le s$ (by Approx. Property for Suprema). Then, $0<x_1-x_0\le s-x_0<1$ and hence $x_1-x_0\in (0,1) \cap\Z$ which is contradiction!
( Remark: There is no $n\in\Z$ s.t $0<n<1$. )

1.4. Theorem : Archimedes Principle

If $a,b\in\R, a>0$, then $\exist n \in\N$ s.t $b<n*a$.
pf)
Case-1 : $b<a$. Set n=1. $b<n*a=1*a=a$
Case-2 : $b\ge a$. Consider $E=\{k\in\N:k*a\le b\}$. Then $1\in E\subseteq\N$ and $E$ is bounded above by $\frac{a}{b}$. By the Completenes Axiom and Thm 1.3, $s:=supE$ exist and $s\in E\subseteq\N$. Set $n:=s+1$, then $n\in\N$ and $n\notin E$. Thus $b<n*a$.

2. Definition : Infimum

$\empty\ne E\subseteq\R$ 이라고 하자.
(1) $E$ is bounded below $\iff$ $\exist m\in\R, \forall a\in E, m\le a.$ $m$ is lower-bound of $E$.
(2) $t\in\R$ is an infimum of $E$ $\iff$ $t$ is a lower-bound of $E$ and $\forall m\in\{x:x\ is\ lower\ bound\ of\ E\}, m\le t$. ( lower-bound 중 가장 큰 개체가 infimum이다. )
(3) $E$ is bounded $\iff$ $E$ is bounded both above and below.

2.1. Reflection

Reflection of $E$란, $E\subseteq\R$에 대하여,
$-E = \{x\in\R :x=-a\ for\ some\ a\in E\}$를 말한다.

2.2 Theorem : Reflection Principle

For $\empty\ne E\subseteq\R$, $E$ has an infimum $\iff$ $-E$ has a supremum.
In this case, $inf(E)=-sup(-E)$
pf)
(1) "$\Leftarrow$" : Suppose that $-E$ has a $sup(-E)=:s.$ Then $-s$ is a lower-bound of $E.$ Let $m$ be an any lower-bound of $E$. Since $s\le -m$, $m\le-s.$ Thus, $-s$ is the infimum of $E.$
(2) "$\Rightarrow$": By reflection, $-E$ is bounded above. By Completenes Axiom, $-E$ has a supremum.

3. Theroem : Well-ordering Principle

If $\empty\ne E\subseteq\N$, then $E$ has a least element $x$ i.e. $x\in E.$
pf)
0 is a lower-bound of $E.$ By reflection, Completenes Axiom and Theorem 1.3, $-E$ has a $sup(-E)=:s\in -E.$ By Reflection Principle, $E$ has an $inf(E)=-s\in E$ which is the least element.