Completeness and Archimedean Property

Construction

Constructing the rational number

Definition. We define the relation

${\sim }_{\mathbb{Q}}$ on the set

$\mathbb{Z}\times {\mathbb{Z}}^{\ast }$ as follows:

$$(a,b){\sim }_{\mathbb{Q}}(c,d)\text{if and only if}ad=bc,$$
where

$a,b,c$ and

$d$ are integers with

$b,d\ne 0$.

Definition. We define the set of rational numbers to be the quotient set

$$\mathbb{Q}=(\mathbb{Z}\times {\mathbb{Z}}^{\ast })/{\sim }_{\mathbb{Q}}$$

Embed

$\mathbb{Z}$ in

$\mathbb{Q}$

Proposition For

$k\in \mathbb{Z}$

$$\varphi (k)=[(k,1)]$$

Arithmetic in

$\mathbb{Q}$

Definition. Given two rational numbers

$[(a,b)]$ and

$[(c,d)]$,

• we define their sum to be
  
  $$[(a,b)]+[(c,d)]=[(ad+bc,bd)]$$
• we define their product to be
  
  $$[(a,b)]\times [(c,d)]=[(ac,bd)].$$

Show above definition is well-defined

Suppose

$(a_2,b_2)\in [((a_1,b_1))]$ and

$(c_2,d_2)\in [((c_1,d_1))]$. Show

• $[(a_1,b_1)]+[(c_1,d_1)]=[(a_2,b_2)]+[(c_2,d_2)]$
• $[(a_1,b_1)]\times [(c_1,d_1)]=[(a_2,b_2)]\times [(c_2,d_2)]$

Constructing the real number

There are two method to construct real number

1. Dedekind: every non-empty subset has a least upper bound (with respect to
   $\le$ ).
2. Cauchy: every Cauchy sequence converges (with respect to ||).

Here, we focus on the second method. I guess Prof. Chen will prove both of them are equivalent.

Definition. Let

$\left\{a_n\right\}$ and

$\left\{b_n\right\}$ be in

${\mathcal{C}}_{\mathbb{Q}}$. Say they are equivalent (i.e. related) if

$a_n-b_n\to 0$; i.e. if the sequence

$(a_n-b_n)$ tends to 0 .

Proposition Above definition yields an equivalence relation on

${\mathcal{C}}_{\mathrm{Q}}$. That is,

$\left\{a_n\right\}\sim \left\{b_n\right\}$ if and only if

$(a_n-b_n)\to 0$.

Definition. We define the set of real numbers to be the quotient set

$$\mathbb{R}={\mathcal{C}}_{\mathbb{Q}}$$

Embed

$\mathbb{Q}$ in

$\mathbb{R}$

Proposition For

$k\in \mathbb{Z}$

$$\varphi (q)=[\{a_n:a_n=q\}]$$

Arithmetic in

$\mathbb{R}$

Definition. Given two rational numbers

$[\{a_n\}]$ and

$[\{b_n\}]$,

• we define their sum to be
  
  $$[\{a_n\}]+[\{b_n\}]=[\{a_n+b_n\}]$$
• we define their product to be
  
  $$[\{a_n\}]\times [\{b_n\}]=[\{a_n{b}_n\}].$$

There are still many issue about being a field.

• $\{a_n+b_n\}$ and
  $\{a_n{b}_n\}$ is Cauchy?
• multilication inverse exist? e.g.
  $\{(0,0,1,0,\dots )\}$

Completeness

Let ordered field

$F$.

• (CSP) A field
  $F$ is said to satisfy Cauchy converge property if every Cauchy sequence is converge in field
  $F$.
• (LUP) A field
  $F$ is said to be Least Upper Bound Complete if every non-empty subset of
  $F$, which has an upper bound in
  $F$, has a least upper bound in
  $F$.
• (BWP) A field
  $F$ is said to satisfy the Bolzano-Welerstrass Sequence Property if every bounded sequence in
  $F$ has a convergent subsequence.
• (BMP) A field
  $F$ is said to satisfy the Bounded Monotone Sequence Property if every bounded monotone sequence in
  $F$ is convergent.
• (NIP) A field
  $F$ is said to satisfy the Nested Interval Property if when
  $\left\{I_k\right\}$ is a sequence of closed and bounded intervals in
  $F$ which is nested, then the
  ${\cap }_k{I}_k\ne \varnothing$.

Please refer to Excercise 4 in [Courant & John] p.116 for more detail. Five of them are equivalent in

$\mathbb{R}$.

(Theorem) Let A = {(LUP), (BWP), (BMP)}.

• In ordered field, elements in
  $A$ is equivalent. Moreover,
  $A\Rightarrow (\mathrm{NIP})\Rightarrow (\mathrm{CSP})$.
• In Archimedean ordered field, five are equivalent.
• If A holds on ordered field, then A is Archimedean ordered field.
• Any complete ordered Archimedean field is order-isomorphic to
  $\mathbb{R}$.

Exist an example which is ordered field with CSP but not Archimedean.

Non-Archimedean Cauchy completeness

Let

$R[x]$ denote a set of all expression

$$g={\sum }_{k=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_k{x}^k\text{, with }{a}_k\in \mathbb{R}\text{ for all }k\text{, }$$
with some

$K\in \mathbb{Z}$ such that

$a_k=0$ if

$k<K$.
if

$g={\sum }_{k=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_k{x}^k$ and

$h={\sum }_{k=-\mathrm{\infty }}^{\mathrm{\infty }}{b}_k{x}^k$, then

$$\begin{array}{rl}& g+h=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}(a_k+b_k)x^k\\ & gh=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{d}_k{x}^k\text{ where }{d}_k=\sum _{i+j=k}{a}_i{b}_j\cdot \end{array}$$
Define

$g>0$ is the first non-zero coefficient is positive.

Theorem.

$R[x]$ is ordered field

Theorem.

$R[x]$ is non-Archimedean

proof. Let

$g=1$ and

$h=x^{-1}$. We have

$1-x^{-1}<0$, i.e.

$1<x^{-1}$. However, no matter how

$n$ large,

$n-x^{-1}<0$ still holds, i.e.

$n<x^{-1}$.

Let a monotonic bounded sequence

$\{a_n=n{x}^{-1}{\}}_{n>0}$, which is bounded by

$M$. However, given a

$ϵ=\overline{ϵ}{x}^0>0$, we cannot find

$n$ such that

$n{x}^{-1}>M-ϵ$ because

$M-ϵ-n{x}^{-1}<0$

Theorem.

$R[x]$ is Cauchy completeness

proof. Let

$ϵ=\overline{ϵ}{x}^K\in R[x]$ with

$\overline{ϵ}>0$, which implies

$ϵ>0$. Let

$\left\{y_n\right\}$ is a Cauchy sequence in

$R[x]$, there exists an

$M_K\in \mathbb{N}$ such that

$n,m\ge M_K$ implies

$|y_n-y_m|<ϵ$. That is, for

$n,m\ge M_K$ and

$j\le k,|a_{n,j}-a_{m,j}|=0$ and

$|a_{n,k}-a_{m,k}|<\overline{ϵ}$.

$$\begin{array}{rrrr}& y_n& =& a_{nj}{x}^j+\cdots +a_{nK}{x}^K+\cdots \\ & y_m& =& a_{mj}{x}^j+\cdots +a_{mK}{x}^K+\cdots \end{array}$$

Since

$ϵ$ is arbitrary, for all

$K$,

$\{a_{nK}{\}}_K$ is a Cauchy sequence in

$\mathbb{R}$. Since

$\mathbb{R}$ is Cauchy Complete, for each

$K$ there exists

$b_K\in \mathbb{R}$, such that

$\underset{n\to \mathrm{\infty }}{lim}{a}_{n,K}=b_K$. Let

$b=\sum _{k=-\mathrm{\infty }}^{\mathrm{\infty }}{b}_K{x}^K$. We show that

$b\in R[x]$ and

$\underset{n\to \mathrm{\infty }}{lim}{y}_n=b$.

Please refer to Thoerem 2.14 in Morgan (1968) for more detail.

Only one way to completion

$\mathbb{Q}$? (The Completion of a Normed Field)

Let's focus on the absolute value

$|\cdot |$. Any nonzero rational number

$x$ can be represen

$$x=\frac{p^{a}r}{s},$$
where

$r$ and

$s$ are integers not divisible by

$p$. Choose prime

$p$. Define

$$|x{|}_p=p^{-a}.$$
As an example, consider the fraction

$$\frac{140}{297}=2^2\cdot 3^{-3}\cdot 5\cdot 7\cdot {11}^{-1}.$$
It has

$p$-adic absolute values given by

$$\begin{array}{rl}{\left|\frac{140}{297}\right|}_2& =\frac{1}{4}\\ {\left|\frac{140}{297}\right|}_3& =27\\ {\left|\frac{140}{297}\right|}_5& =\frac{1}{5}\end{array}$$

Please refer to MIT lecture note for more detail.
https://en.wikipedia.org/wiki/Ostrowski's_theorem