Ch 1. The Real and Complex Number Systems

• $\R$, real number : Order set
  1. Field ($+, \cdot$ )
  2. Ordered set ( $<, > )$

Ordered set

• Def.

  Let $S$ be a set.

  An order on $S$ is a relation, denoted by $<$, with the following properties

  (i) If $x, y \in S$, then one and only one of the statements $x<y, x=y, x>y$ is true.

  (ii) If $x, y, z \in S$, $x<y$, $y<z$, then $x<z$

  We write $x\leq y$ if $x<y$ or $x=y$

  An ordered set is a set $S$ in which an order is defined

(Least) Upper/Lower bound

• Def.

  Let $S$ be an ordered set and $E\sube S$ (1) $E$ is called bounded above(위로 유계) if $\exists \beta \in S$ s.t. $\forall x \in E, x\leq \beta$

    In this case, $\beta$ is called on **upper bound**(상계) of $E$.
  

  (2) $E$ is called bounded below(아래로 유계) if $\exists \alpha \in S$ s.t. $\forall x \in E, x\geq \alpha$

    In this case, $\alpha$ is called on **lower bound**(하계) of $E$.
  

• Def. Let $S$ be an ordered set and $E \sube S$

  (1) Supp. $E$ is bounded above.

    $\alpha \in S$ is called the **least upper bound**, or the **supremum** of $E$ if
  
    (i) $\alpha$ is an upper bound of $E$
  
    (ii) $\gamma < \alpha \Rightarrow \gamma$ is not upper bound of $E$
  
    We write $\alpha = \sup E$ in this case
  

  (2) Supp. E is bounded below.

    $\alpha \in S$ is called the **greatest lower bound** or the **infimum** of $E$ if 
    (i) $\alpha$ is a lower bound of $E$
  
    (ii) $\gamma > \alpha \Rightarrow \gamma$  is not a lower bound of $E$.
  

• Ex 1)

  Let $S=Q$.

  $A={p\in Q : p>0, p^2 <2 }$

  $B={p\in Q : p>0, p^2 > 2 }$

  (i) $A$ is bounded above but $A$ has no least upper bound (in $S$)

  (ii) $B$ is bounded below but $B$ has no greatest lower bound (in $S$)

  pf)

  (i) $p\in A$

    Supp. $p>2$ Then $p^2>4$, which contradicts $p^2<2$.
  
    Hence, $p\leq 2$. Therefore, $A$ is bounded above.
  

  • Supp. $A$ has a least upper bound $\alpha \in Q$

    1. Supp. $\alpha^2<2$, Take a positive integer $\displaystyle n>\frac{2\alpha +1 } {2-\alpha^2}$

       Then $\displaystyle (\alpha+{1\over n})^2 =\alpha^2 + {2\over n}\alpha + {1\over n^2} \leq \alpha^2 + {2\over n} \alpha + {1\over n} = \alpha^2 + {2\alpha+1\over n} <2$

       Hence, $\displaystyle \alpha + {1\over n} \in A$ and $\displaystyle \alpha<\alpha+{1\over n}$.

       This contradicts to the fact that $\alpha$ is an upper bound.

       → $\alpha$ 보다 큰 값을 가져왔는데 $A$의 원소이므로 $\alpha$ 는 upper bound가 될 수 없음.

    2. Supp. $\alpha^2>2$, Take a positive integer $\displaystyle n>{2\alpha \over \alpha^2-2}$

       Then $\displaystyle (\alpha - {1\over n} )^2 = \alpha^2 - {2\over n} \alpha + {1\over n^2} > \alpha^2 - {2\over n} \alpha>2$

       For any $p\in A$, $\displaystyle p^2<2<(\alpha-{1\over n} )^2$. Hence, $\displaystyle p<\alpha-{1\over n}$

       Therefore, $\displaystyle\alpha-{1\over n}$ is an upperbound of $A$ and $\displaystyle\alpha-{1\over n} <\alpha$ This contradicts to the minimality of $\alpha$.

    3. Supp. $\alpha^2=2$

       Let $\displaystyle\alpha={n\over m}$, ($m,n \in \Z$, $m\neq 0$, $m, n$ are relatively prime)

       Then $\displaystyle 2 = {n^2\over m^2}$ and thus $2m^2=n^2$

       Since $2$ divides $2m^2=n^2$, $2$ divides $n$.

       Write $n=2k$ ($k\in \Z$). Then $m^2 = 2k^2$

       Now $2$ divides $2k^2 = m^2$, $2$ divides $m$.

       This contradicts to the fact that $ m, n$ are relatively prime

• Ex 2) Suprimum 이 항상 해당 집합의 원소일 필요는 없다.

  $A = {r\in \mathbb{Q} : r<0}, \sup A = 0$

  $B = {r\in \mathbb{Q} : r\leq 0}, \sup B = 0$

  Check : $0 $ is upper bound

  (1) $r\in A \Rightarrow r\leq 0$ : OK

  (2) $s<0 \Rightarrow s<\displaystyle {s \over 2}, \ {s\over 2} \in A$ $\therefore \ s<0$ is not upper bound

• Ex 3)

  Let $E = \displaystyle {{1\over n} : n\in \mathbb{Z}^{\geq 1} }, \sube \mathbb{Q}$

  $\sup E = 1$

  $\inf E = 0\notin E$

  If $E$ has the maximal element $\alpha$, then $\alpha$ is the supremum of $E$ (The uniqueness of supremum : only identical supremum exists.)

  → The converse is NOT true! ( by Ex 2)

• Ex 4)

  (1) $\displaystyle { 1-{(-1)^n \over n} : n\in \mathbb{Z}^{>0} }$, $\sup = 2,\ \inf = {1\over 2}$

  (2) $\displaystyle S = { {1\over n} - {1\over m} : n, m\in \mathbb{Z}^{>0}}$

    $\inf S = -1$
  
    $\sup S = 1$
  
    Precise proof after archimedean property
  

• Def.

  An ordered set $S$ has the least-upper bound(최소상계성질)

  (or is complete, 완비적) if for every nonempty subset $E$ of $S$ that is bound above.

  $\sup E$ exists in $S$

  Rmk.

  1. $\mathbb{Q}$ does not have the least upper bound property.
  2. $\mathbb{R}$ has the least upper bound property. (by definition)

• Thm.

  Let $S$ be an ordered set with the least-upper-bound property.

  Let $B\sub S$ be a nonempty subset that is bounded below.

  Let $L$ be the set of all lower bounds of $B$.

  Then $\sup L$ exists in $S$ and $\inf B = \sup L$

  In particular, $S$ has the greatest-lower-bound property.

(pf)

(Claim) $\sup L$ exists in $S$

(pf) Since $B$ has a lower bound, $L$ is nonempty.

	Since $B$ is nonempty, take $x\in B$.

	Then for any $y\in L$, $y\leq x$.

	Hence, $x$ is an upper bound of $L$. so $L$ is bounded above.

	By least-upper-bound property, $L$ has the supremum.

(Claim) $\inf B = \sup L$

(pf) 

	1. Take  $x\in B$

		For any $y\in L$, $y\leq x$.

		So $x$ is an upper bound of $L$.

		Hence, $\sup L \leq x$ . (Using the definition of supremum)

		→ $\sup L$ is a lower bound of $B$.

	2. Let $\alpha$ be a lower bound of $B$.

		Then $\alpha \in L$. Hence, $\alpha \leq \sup L$

		→ $\sup L$ is the greatest lower bound of $B$