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Proof of Uniuquness of Complete Archimedean Ordered Field.

How we determine complete Archimedean Ordered field?

Definition of Field

[math]\displaystyle{ (F, +, \cdot ) }[/math] is called a field if the properties hold:

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

Definition of ordered field

[math]\displaystyle{ (F, +, \cdot, \leq) }[/math] is an ordered field if it is a field satisfying the order axioms:

1. [math]\displaystyle{ x \leq x }[/math] (Reflexive)
2. [math]\displaystyle{ x \leq y \land y \leq x \to x = y }[/math] (Antisymmetry)
3. [math]\displaystyle{ x \leq y \land y \leq z \to x \leq z }[/math] (Transivity)
4. [math]\displaystyle{ x \nleq y \to y \leq x }[/math] (Linear order)
5. [math]\displaystyle{ x \leq y \to x+z \leq y+z }[/math]
6. [math]\displaystyle{ x \leq y \land z \geq 0 \to xz \leq yz }[/math]

Definition of Completeness

Definition of Cauchy Sequence

A sequence [math]\displaystyle{ \{ a_n \} : \mathbb{N} \to F }[/math] in an ordered field is a Cauchy sequence if [math]\displaystyle{ \forall \epsilon \gt 0 }[/math], [math]\displaystyle{ \exist N ( N \in \mathbb{N} \land \forall k,l(k,l\gt N \to | a_k - a_l | \lt \epsilon)) }[/math]. (For any positive number -> sufficient large term difference -> close to 0 )

Definition of Convergence

A sequence [math]\displaystyle{ \{ a_n \} }[/math] converges to [math]\displaystyle{ A \in F }[/math] if [math]\displaystyle{ \forall \epsilon \gt 0 }[/math], </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 [math]\displaystyle{ \forall x, y in F, \exist n=(1+ \cdot \cdot \cdot +1) \to nx \gt y }[/math].

Proof of Theorem

Real Number system is a complete Archimedean ordered field

Rational number system is the smallest ordered field.

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

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

By above statement, the ordered field contains [math]\displaystyle{ \mathbb{Q} }[/math]. Now find the set [math]\displaystyle{ \{C(\mathbb{Q})_N \} = \{ \{ a_n \} \in Seq(\mathbb{Q}) | \forall \epsilon\gt 0, \exist N ( \forall k,l\gt N \to |a_k -a_l |\lt \epsilon) \} }[/math] of rational Cauchy sequence, and set the equivalence relation [math]\displaystyle{ \sim_R }[/math] by [math]\displaystyle{ \{ a_n \} \sim_R \{b_n \} }[/math] if for arbitrary ε>0, there is N satisfying n>N implies [math]\displaystyle{ | a_n - b_n |\lt \epsilon }[/math](Converges the same limit - considered as equivalent set).

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

Define the embedding of [math]\displaystyle{ \mathbb{Q} }[/math] into [math]\displaystyle{ \bar{C(\mathbb{Q})}/\sim_R \cong \mathbb{R} }[/math] by [math]\displaystyle{ a_n = q }[/math] for all [math]\displaystyle{ q \in \mathbb{Q} }[/math]. Then for any [math]\displaystyle{ \{ a_n \} \in \bar{C(\mathbb{Q})} }[/math] there is [math]\displaystyle{ s, N \in \mathbb{N} }[/math] such that [math]\displaystyle{ \forall n\gt N , a_n \lt s }[/math]. Therefore, [math]\displaystyle{ [ \{ a_n \} ] \lt [ \{ s+1 \}] }[/math] ,and the set [math]\displaystyle{ \bar{C(\mathbb{Q})}/\sim_R \cong \mathbb{R} }[/math] satisfies Archimedean property.

Uniqueness of complete ordered field with Archimedean property.