작은숲:공책/실해석/Uniqueness of Complete Archimedean Ordered Field

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Proof of Uniuquness of Complete Archimedean Ordered Field.

How we determine complete Archimedean Ordered field?

Definition of Field

$(F,+,\cdot )$ is called a field if the properties hold:

1. $x+y=y+x$ (commutativity of addition)
2. $(x+y)+z=x+(y+z)$ (associativity of addition)
3. $\exists 0(0\in F\land x+0=x)$ (existence of addictive identity)
4. $\forall x(x\in F\to (\exists -x(x\in F\land x+(-x)=0))$ (existence of addictive inverse)
5. $x\cdot y=y\cdot x$ (Commutativity of Multiplication)
6. $(x\cdot y)\cdot z=x\cdot (y\cdot z)$ (Associativity of Multiplication)
7. $\exists 1(1(\neq 0)\in F\land x\cdot 1=x)$ (Existence of multiplicative identity)
8. $\forall x(x(\neq 0)\in F\to (\exists {x}^{-1}({x}^{-1}\in F\land x\cdot {x}^{-1}=1))$ (Existence of multiplicative inverse)
9. $x\cdot (y+z)=x\cdot y+x\cdot z$ (Distribution of multiplication associated with addition)

Definition of ordered field

$(F,+,\cdot ,\leq )$ is an ordered field if it is a field satisfying the order axioms:

1. $x\leq x$ (Reflexive)
2. $x\leq y\land y\leq x\to x=y$ (Antisymmetry)
3. $x\leq y\land y\leq z\to x\leq z$ (Transivity)
4. $x\nleq y\to y\leq x$ (Linear order)
5. $x\leq y\to x+z\leq y+z$
6. $x\leq y\land z\geq 0\to xz\leq yz$

Definition of Completeness

Definition of Cauchy Sequence

A sequence $\{a_{n}\}:\mathbb {N} \to F$ in an ordered field is a Cauchy sequence if $\forall \epsilon >0$, $\exists N(N\in \mathbb {N} \land \forall k,l(k,l>N\to |a_{k}-a_{l}|<\epsilon ))$. (For any positive number -> sufficient large term difference -> close to 0 )

Definition of Convergence

A sequence $\{a_{n}\}$ converges to $A\in F$ if $\forall \epsilon >0$, </math> \exist N (N \in \mathbb{N} land \forall n (n>N \to |a_n -A| < \epsilon )) </math> (For all sufficiently large N -> a_N is arbitrary close to A )

Definition of Completeness

An ordered field(or set) F is complete if every Cauchy sequence is convergent to some value in F. (Sequence that converges to certain point -> converges to an element in F)

Archimedean Property

An ordered field has Archimedean property if $\forall x,yinF,\exists n=(1+\cdot \cdot \cdot +1)\to nx>y$.

Proof of Theorem

Real Number system is a complete Archimedean ordered field

Rational number system is the smallest ordered field.

We know $1>0$ because $1\neq 0$ but $1\cdot 1=1\geq 0$. Using the logic, we should find that $N=1+\cdot \cdot \cdot +1>0$ for any $N\in F$. Thus, we can deduce that the set of natural numbers $\mathbb {N}$ by producing the union of 0 and $\sum _{i=1}^{N}1$ for 1 ∈ F is contained in the field F. Also, the set of rational number $\mathbb {Q}$ contains natural numbers. Also, we will prove that it is the smallest one containing $\mathbb {N}$. Suppose F contains all of natural numbers, then F contains -n, the addctive inverse of $n\in mathbb{N}$. Also, for n≠0, F contains $1/n\in F$. Since F is closed under addition and multiplication, 구문 분석 실패 (알 수 없는 함수 "\F"): {\displaystyle \forall n, m(\neq 0) \in mathbb{N}, ~ \pm n\cdot m^-1 in \F } . Therefore, $\mathbb {Q} \subset F$.

Real number system is the smallest complete ordered field with Archimedean property.

By above statement, the ordered field contains $\mathbb {Q}$. Now find the set $\{C(\mathbb {Q} )_{N}\}=\{\{a_{n}\}\in Seq(\mathbb {Q} )|\forall \epsilon >0,\exists N(\forall k,l>N\to |a_{k}-a_{l}|<\epsilon )\}$ of rational Cauchy sequence, and set the equivalence relation $\sim _{R}$ by $\{a_{n}\}\sim _{R}\{b_{n}\}$ if for arbitrary ε>0, there is N satisfying n>N implies $|a_{n}-b_{n}|<\epsilon$(Converges the same limit - considered as equivalent set).

Now consider the set ${\bar {C(\mathbb {Q} )}}=\{C(\mathbb {Q} )_{N}\}/\sim _{R}$ Then it is equivalent to the set $\mathbb {R}$. Take a strict order < on $\{C(\mathbb {Q} )_{N}\}$ a $\{a_{n}\}<\{b_{n}\}$ if $\exists \epsilon >0,N(\forall n>N\to a_{n}\leq b_{n}-\epsilon )$. Then < satisfies the linearity in ${\bar {C\mathbb {Q} }}$. That's because $a_{n}\nless b_{n}$ and $b_{n}\nless a_{n}$ implies $\forall \epsilon >0,N,\exists n>N|a_{n}-b_{n}|<\epsilon$. Since {a_n} and {b_n} are all Cauchy sequence, we can assume $|a_{p}-b_{p}|\leq |a_{p}-a_{n}|+|a_{n}-b_{n}|+|b_{n}-b_{p}|<3\epsilon$ for any p>N. Therefore, we can assume $\{a_{n}\}\sim _{R}\{b_{n}\}$ and < becomes a linear order in ${\bar {C(\mathbb {Q} )}}/\sim _{R}\cong \mathbb {R}$.

Define the embedding of $\mathbb {Q}$ into ${\bar {C(\mathbb {Q} )}}/\sim _{R}\cong \mathbb {R}$ by $a_{n}=q$ for all $q\in \mathbb {Q}$. Then for any $\{a_{n}\}\in {\bar {C(\mathbb {Q} )}}$ there is $s,N\in \mathbb {N}$ such that $\forall n>N,a_{n}<s$. Therefore, $[\{a_{n}\}]<[\{s+1\}]$ ,and the set ${\bar {C(\mathbb {Q} )}}/\sim _{R}\cong \mathbb {R}$ satisfies Archimedean property.

Uniqueness of complete ordered field with Archimedean property.