---
title: Completeness and Archimedean Property
tags: Course
description:
---
# Completeness and Archimedean Property
## Construction
### Constructing the rational number
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**Definition.** We define the relation $\sim_{\mathbb{Q}}$ on the set $\mathbb{Z} \times \mathbb{Z}^*$ as follows:
\begin{equation}
(a, b) \sim_{\mathbb{Q}}(c, d) ~\text{if and only if}~ a d=b c,
\end{equation}
where $a, b, c$ and $d$ are integers with $b, d \neq 0$.
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**Definition.** We define the set of rational numbers to be the quotient set
\begin{equation*}
\mathbb{Q}=\left(\mathbb{Z} \times \mathbb{Z}^*\right) / \sim_{\mathbb{Q}}
\end{equation*}
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Embed $\mathbb{Z}$ in $\mathbb{Q}$
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**Proposition** For $k\in\mathbb{Z}$
\begin{equation}
\phi(k)=[(k, 1)]
\end{equation}
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Arithmetic in $\mathbb{Q}$
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**Definition.** Given two rational numbers $[(a, b)]$ and $[(c, d)]$,
- we define their sum to be
\begin{equation*}
[(a, b)]+[(c, d)]=[(a d+b c, b d)]
\end{equation*}
- we define their product to be
\begin{equation*}
[(a, b)] \times[(c, d)]=[(a c, b d)] .
\end{equation*}
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:::spoiler 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
- $[(a_1,b_1)]+[(c_1,d_1)]=[(a_2,b_2)]+[(c_2,d_2)]$
- $[(a_1,b_1)]\times [(c_1,d_1)]=[(a_2,b_2)]\times[(c_2,d_2)]$
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### 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 $\leq$ ).
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.

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**Definition.** Let $\left\{a_n\right\}$ and $\left\{b_n\right\}$ be in $\mathscr{C}_{\mathbb{Q}}$. Say they are equivalent (i.e. related) if $a_n-b_n \rightarrow 0$; i.e. if the sequence $\left(a_n-b_n\right)$ tends to 0 .
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**Proposition** Above definition yields an equivalence relation on $\mathscr{C}_{\mathrm{Q}}$. That is, $\left\{a_n\right\} \sim\left\{b_n\right\}$ if and only if $\left(a_n-b_n\right)\to 0$.
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**Definition.** We define the set of real numbers to be the quotient set
\begin{equation*}
\mathbb{R}=\mathscr{C}_{\mathbb{Q}}
\end{equation*}
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Embed $\mathbb{Q}$ in $\mathbb{R}$
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**Proposition** For $k\in\mathbb{Z}$
\begin{equation}
\phi(q)=[\{a_n:a_n=q\}]
\end{equation}
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Arithmetic in $\mathbb{R}$
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**Definition.** Given two rational numbers $[\{a_n\}]$ and $[\{b_n\}]$,
- we define their sum to be
\begin{equation*}
[\{a_n\}]+[\{b_n\}]=[\{a_n+b_n\}]
\end{equation*}
- we define their product to be
\begin{equation*}
[\{a_n\}] \times[\{b_n\}]=[\{a_nb_n\}] .
\end{equation*}
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There are still many issue about being a field.
- $\{a_n+b_n\}$ and $\{a_nb_n\}$ is Cauchy?
- multilication inverse exist? *e.g.* $\{(0,0,1,0,\dots)\}$
## Completeness
Let ordered field $F$.
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- **(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 $\cap_{k}I_k\neq \varnothing$.
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Please refer to Excercise 4 in [Courant & John] p.116 for more detail. Five of them are equivalent in $\mathbb{R}$.
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**(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 $\mathbb{R}$.
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:::warning
Exist an example which is ordered field with CSP but not Archimedean.
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## Non-Archimedean Cauchy completeness
Let $R[x]$ denote a set of all expression
\begin{equation*}
g={\sum}_{k={-\infty}}^{\infty} a_k x^k \text {, with } a_k \in \mathbb{R} \text { for all } k \text {, }
\end{equation*}
with some $K\in\mathbb{Z}$ such that $a_k=0$ if $k<K$.
if $g={\sum}_{k={-\infty}}^{\infty} a_k x^k$ and $h={\sum}_{k={-\infty}}^{\infty} b_k x^k$, then
\begin{equation*}
\begin{aligned}
&g+h=\sum_{k=-\infty}^{\infty}\left(a_k+b_k\right) x^k \\
&gh=\sum_{k=-\infty}^{\infty} d_k x^k \text { where } d_k=\sum_{i+j=k} a_i b_j \cdot
\end{aligned}
\end{equation*}
Define $g>0$ is the first non-zero coefficient is positive.
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**Theorem.** $R[x]$ is ordered field
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**Theorem.** $R[x]$ is non-Archimedean
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*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.* $n<x^{-1}$.
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Let a monotonic bounded sequence $\{a_n=nx^{-1}\}_{n>0}$, which is bounded by $M$. However, given a $\epsilon=\bar{\epsilon}x^{0}>0$, we cannot find $n$ such that $nx^{-1}>M-\epsilon$ because $M-\epsilon-nx^{-1}<0$
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**Theorem.** $R[x]$ is Cauchy completeness
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*proof.* Let $\epsilon=\bar{\epsilon}x^K \in R[x]$ with $\bar{\epsilon}>0$, which implies $\epsilon>0$. Let $\left\{y_n\right\}$ is a Cauchy sequence in $R[x]$, there exists an $M_K\in\mathbb{N}$ such that $n, m \geq M_K$ implies $\left|y_n-y_m\right|<\epsilon$. That is, for $n, m \geq M_K$ and $j \leq k,\left|a_{n, j}-a_{m, j}\right|=0$ and $\left|a_{n, k}-a_{m, k}\right|<\bar{\epsilon}$.

\begin{equation}
\begin{array}{rrr}
&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}
\end{equation}

Since $\epsilon$ 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 $\lim _{n \rightarrow \infty} a_{n, K}=b_K$. Let $b=\sum_{k=-\infty}^{\infty} b_K x^K$. We show that $b \in R[x]$ and $\lim _{n \rightarrow \infty} y_n=b$.

Please refer to Thoerem 2.14 in [Morgan (1968)](https://scholarworks.umt.edu/cgi/viewcontent.cgi?article=9237&context=etd) 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
\begin{equation*}
x=\frac{p^a r}{s},
\end{equation*}
where $r$ and $s$ are integers not divisible by $p$. Choose prime $p$. Define
\begin{equation*}
|x|_p=p^{-a} .
\end{equation*}
As an example, consider the fraction
\begin{equation*}
\frac{140}{297}=2^2 \cdot 3^{-3} \cdot 5 \cdot 7 \cdot 11^{-1} .
\end{equation*}
It has $p$-adic absolute values given by
\begin{equation*}
\begin{aligned}
\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{aligned}
\end{equation*}

Please refer to [MIT lecture note](https://math.mit.edu/classes/18.095/2015IAP/lecture1/padic.pdf) for more detail.
https://en.wikipedia.org/wiki/Ostrowski%27s_theorem