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

공책 문서입니다. 독자연구성 서술이나 난해한 서술이 있을 수도 있음에 유의해주세요.

이 작은숲 문서의 출처는 [[Korean Study Note]]입니다.

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:

$x + y = y + x$ (commutativity of addition)
$(x + y) + z = x + (y + z)$ (associativity of addition)
$\exists 0 (0 \in F \land x + 0 = x)$ (existence of addictive identity)
$\forall x (x \in F \to (\exists - x (x \in F \land x + (- x) = 0))$ (existence of addictive inverse)
$x \cdot y = y \cdot x$ (Commutativity of Multiplication)
$(x \cdot y) \cdot z = x \cdot (y \cdot z)$ (Associativity of Multiplication)
$\exists 1 (1 (\neq 0) \in F \land x \cdot 1 = x)$ (Existence of multiplicative identity)
$\forall x (x (\neq 0) \in F \to (\exists x^{- 1} (x^{- 1} \in F \land x \cdot x^{- 1} = 1))$ (Existence of multiplicative inverse)
$x \cdot (y + z) = x \cdot y + x \cdot z$ (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:

$x \leq x$ (Reflexive)
$x \leq y \land y \leq x \to x = y$ (Antisymmetry)
$x \leq y \land y \leq z \to x \leq z$ (Transivity)
$x ≰ y \to y \leq x$ (Linear order)
$x \leq y \to x + z \leq y + z$
$x \leq y \land z \geq 0 \to x z \leq y z$

Definition of Completeness

Definition of Cauchy Sequence: A sequence ${a_{n}} : ℕ \to F$ in an ordered field is a Cauchy sequence if $\forall ϵ > 0$ , $\exists N (N \in ℕ \land \forall k, l (k, l > N \to | a_{k} - a_{l} | < ϵ))$ . (For any positive number -> sufficient large term difference -> close to 0 )
Definition of Convergence: A sequence ${a_{n}}$ converges to $A \in F$ if $\forall ϵ > 0$ , </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 $\forall x, y i n F, \exists n = (1 + \cdot \cdot \cdot + 1) \to n x > y$ .

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 $ℕ$ 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 $ℚ$ contains natural numbers. Also, we will prove that it is the smallest one containing $ℕ$ . Suppose F contains all of natural numbers, then F contains -n, the addctive inverse of $n \in m a t h b b N$ . 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, $ℚ \subset F$ .

Real number system is the smallest complete ordered field with Archimedean property.: By above statement, the ordered field contains $ℚ$ . Now find the set ${C (ℚ)_{N}} = {{a_{n}} \in S e q (ℚ) | \forall ϵ > 0, \exists N (\forall k, l > N \to | a_{k} - a_{l} | < ϵ)}$ 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 $| a_{n} - b_{n} | < ϵ$ (Converges the same limit - considered as equivalent set).; Now consider the set $\bar{C (ℚ)} = {C (ℚ)_{N}} / \sim_{R}$ Then it is equivalent to the set $ℝ$ . Take a strict order < on ${C (ℚ)_{N}}$ a ${a_{n}} < {b_{n}}$ if $\exists ϵ > 0, N (\forall n > N \to a_{n} \leq b_{n} - ϵ)$ . Then < satisfies the linearity in $\bar{C ℚ}$ . That's because $a_{n} ≮ b_{n}$ and $b_{n} ≮ a_{n}$ implies $\forall ϵ > 0, N, \exists n > N | a_{n} - b_{n} | < ϵ$ . Since {a_n} and {b_n} are all Cauchy sequence, we can assume $| a_{p} - b_{p} | \leq | a_{p} - a_{n} | + | a_{n} - b_{n} | + | b_{n} - b_{p} | < 3 ϵ$ for any p>N. Therefore, we can assume ${a_{n}} \sim_{R} {b_{n}}$ and < becomes a linear order in $\bar{C (ℚ)} / \sim_{R} ≅ ℝ$ .; Define the embedding of $ℚ$ into $\bar{C (ℚ)} / \sim_{R} ≅ ℝ$ by $a_{n} = q$ for all $q \in ℚ$ . Then for any ${a_{n}} \in \bar{C (ℚ)}$ there is $s, N \in ℕ$ such that $\forall n > N, a_{n} < s$ . Therefore, $[{a_{n}}] < [{s + 1}]$ ,and the set $\bar{C (ℚ)} / \sim_{R} ≅ ℝ$ satisfies Archimedean property.

Uniqueness of complete ordered field with Archimedean property.