实数构造

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有理数

实数

$Definition$ . (Sequences). Let $m$ be an integer. A sequence $(a_{n})_{n = m}$ of rational numbers is any function from the set ${n \in Z : n \geq m}$ to $Q$ , i.e., a mapping which assigns to each integer $n$ greater than or equal to $m$ , a rational number $a_{n}$ . More informally, a sequence $(a_{n})_{n = m}$ of rational numbers is a collection of rationals $a_{m}, a_{m + 1}, a_{m + 2}, \dots$ .

$Definition$ . (Cauchy sequences). A sequence $(a_{n})_{n = m}$ of rational numbers is said to be a Cauchy sequence iff for every $ϵ > 0$ , there exists an $N \geq 0$ such that $d (a_{j}, a_{k}) \leq ϵ$ for all $j, k \geq N$ .

$Definition$ . (Bounded sequences). Let $M \geq 0$ be rational. A finite sequence $a_{1}, a_{2}, \dots, a_{n}$ is bounded by $M$ iff $∣ a_{i} ∣ \leq M$ for all $1 \leq i \leq n$ . An infinite sequence $(a_{n})_{n = m}$ is bounded by $M$ iff $∣ a_{i} ∣ \leq M$ for all $i \geq m$ .

$Lemma$ . Finite sequences are bounded

$Lemma$ . $∣ x - y ∣ <= M \Rightarrow ∣ x ∣ <= ∣ y ∣ + M$

$Lemma$ . (Cauchy sequences are bounded). Every Cauchy sequence $(a_{n})_{n = m}$ is bounded.

$Lemma$ . $n \in N \land x \in Q \Rightarrow n x = \sum_{i = 1}^{n} x$ , 提示：自然数到整数的同构映射、整数到有理数的同构映射、数学归纳法

令 $C = {x \in \prod_{i \in N} Q ∣ x_{i} is a Cauchy sequence}$ ，即由 $N$ 索引的 $Q$ 的笛卡尔积。

定义 $C$ 上的一种关系 $\sim_{c}$ ， $x \sim_{c} y$ 当且仅当对于任意 $ϵ \in Q \land ϵ > 0$ ，存在 $N \in N$ ，使得对于任意 $n \geq N$ ，有 $∣ x_{n} - y_{n} ∣ \leq ϵ$

$Lemma$ . $\sim_{c}$ 是等价关系

$\sim_{c}$ 在 $C$ 上的所有等价类所构成的集合为 $C$ ， $C$ 中的元素记作 $[(a_{n})_{n = 0}]$ ，简记为 $[a_{n}]$

实数加法

$Lemma$ . if $(a_{n})_{n = 0}$ and $(b_{n})_{n = 0}$ are Cauchy sequences, then $(a_{n} + b_{n})_{n = 0}$ is also a Cauchy sequence

在 $C$ 上定义加法

$[(a_{n})_{n = 0}] + [(b_{n})_{n = 0}] = [(a_{n} + b_{n})_{n = 0}]$

简记为 $[a_{n}] + [b_{n}] = [a_{n} + b_{n}]$

容易验证上述定义的加法是良定义的

上述定义的加法具有以下性质

加法交换律
加法结合律
加法单位元 $[0_{n}]$ ， $0_{n}$ 为 $Q$ 中的零序列
加法逆元 $[(a_{n})_{n = 0}] + [(- a_{n})_{n = 0}] = [(a_{n} - a_{n})_{n = 0}] = [0_{n}]$

因此 $< C, + >$ 构成一个交换群

note: 需要验证逆元的良定义性，即若两个数表示的等价类相等，则它们的逆元表示的等价类也相等

实数乘法

$Lemma$ . if $(a_{n})_{n = 0}$ and $(b_{n})_{n = 0}$ are Cauchy sequences, then $(a_{n} b_{n})_{n = 0}$ is also a Cauchy sequence

证明：by definition

任取一个 $ϵ \in Q \land ϵ > 0$

因为柯西列都是有界，所以 $∣ a_{n} ∣ \leq M_{1}$ ， $∣ b_{n} ∣ \leq M_{2}$ ，因此它们有共同上界 $M = ma x (M_{1}, M_{2})$

对于 $ϵ_{a} = ϵ / (2 M)$ ，存在 $N_{a}$ ，使得对于任意 $n, m \geq N_{a}$ ，有 $∣ a_{n} - a_{m} ∣ \leq ϵ_{a}$

对于 $ϵ_{b} = ϵ / (2 M)$ ，存在 $N_{b}$ ，使得对于任意 $n, m \geq N_{b}$ ，有 $∣ b_{n} - b_{m} ∣ \leq ϵ_{b}$

取 $N = ma x (N_{a}, N_{b})$ ，对于任意 $n, m \geq N$ ，有

$∣ a_{n} b_{n} - a_{m} b_{m} ∣ = ∣ a_{n} b_{n} - a_{n} b_{m} + a_{n} b_{m} - a_{m} b_{m} ∣ = ∣ a_{n} ∣∣ b_{n} - b_{m} ∣ + ∣ b_{m} ∣∣ a_{n} - a_{m} ∣ \leq M ∣ b_{n} - b_{m} ∣ + M ∣ a_{n} - a_{m} ∣ \leq ϵ_{a} + ϵ_{b} = ϵ$

在 $C$ 上定义乘法

$[(a_{n})_{n = 0}] \times [(b_{n})_{n = 0}] = [(a_{n} b_{n})_{n = 0}]$

简记为 $[a_{n}] \times [b_{n}] = [a_{n} b_{n}]$

容易验证上述定义的乘法是良定义的

$Definition$ . (Sequences bounded away from zero). A sequence $(a_{n})_{n = 0}$ of rational numbers is said to be bounded away from zero iff there exists a rational number $c > 0$ such that $∣ a_{n} ∣ \geq c$ for all $n \geq 0$ .

$Lemma$ . Let $x$ be a non-zero real number. Then $x = [(a_{n})_{n = 0}]$ for some Cauchy sequence $(a_{n})_{n = 0}$ which is bounded away from zero.

证明：
由 $x = [b_{n}] \neq = [0_{n}]$ ,存在 $ϵ \in Q \land ϵ > 0$ ，使得对于任意 $N$ , 都存在 $n_{0} > N$ ,使得 $∣ b_{n_{0}} ∣ \geq ϵ$

因为 $x$ 是柯西列，所以存在 $N_{1}$ ，使得对于任意 $n, m \geq N_{1}$ ，有 $∣ b_{n} - b_{m} ∣ \leq ϵ /2$

取 $N = ma x (N_{1}, n_{0})$ ，对于任意 $n \geq N$ ，有 $∣ b_{n_{0}} - b_{n} ∣ \leq ϵ /2$ ，因此 $∣ b_{n} ∣ \geq ∣ b_{n_{0}} ∣ - ∣ b_{n_{0}} - b_{n} ∣ \geq ϵ - ϵ /2 = ϵ /2$

构造一个新的数列 $a_{i} = b_{n_{0}}, i < n_{0}$ ， $a_{i} = b_{i}, i \geq n_{0}$

易得 $[a_{n}] = [b_{n}] = x$ ,且 $∣ a_{n} ∣ \geq ϵ /2$

即 $(a_{n})_{n = 0}$ 是 bounded away from zero 的 cauchy sequence

$Lemma$ . 若 $a_{n}$ 是 bounded away from zero 的 cauchy sequence，则 $1/ a_{n}$ 也是 cauchy sequence. 按柯西列的定义证明

$Lemma$ . 若 $a_{n} \sim_{c} b_{n}$ ，且 $a_{n}, b_{n}$ 都是 bounded away from zero 的 cauchy sequence，则 $1/ a_{n} \sim_{c} 1/ b_{n}$ ，即逆元是良定义的。

上述定义的乘法具有以下性质

乘法交换律
乘法结合律
乘法单位元 $[1_{n}]$ ， $1_{n}$ 为 $Q$ 中的单位序列
乘法逆元，若 $[a_{n}] \neq = [0_{n}]$ ，且 $(a_{n})_{n = 0}$ 是 bounded away from zero 的 cauchy sequence，则 $[a_{n}]$ 有乘法逆元 $[b_{n}]$ ，其中 $b_{n} = \frac{1}{a _{n}}$
乘法对加法的分配律

因此 $< C, +, \times >$ 构成一个交换除环，即域

$Definition$ . $i (n) : Q \to C$ ， $i (q) = [q_{n}]$ ， $\forall q_{i} = q$

$i (q_{1} + q_{2}) = [(q_{1} + q_{2})_{n}] = [(q_{1})_{n}] + [(q_{2})_{n}]$

$i (q_{1} q_{2})$ = $[(q_{1} q_{2})_{n}] = [(q_{1})_{n}] [(q_{2})_{n}]$

若 $i (q_{1}) = i (q_{2})$ ， $[(q_{1})_{n}] = [(q_{2})_{n}] \Rightarrow q_{1} = q_{2}$ ，因此 $i$ 是单射

因此 $i$ 是 $Q$ 到 $C$ 的同构映射

有序域

$Definition$ . Let $(a_{n})_{n = 0}$ be a sequence of rationals. We say that this sequence is positively bounded away from zero iff we have a positive rational $c > 0$ such that $a_{n} \geq c$ for all $n \geq 0$ (in particular, the sequence is entirely positive). The sequence is negatively bounded away from zero iff we have a positive rational $c > 0$ such that $a_{n} \leq - c$ for all $n \geq 0$ (in particular, the sequence is entirely negative).

$Definition$ . A real number $x$ is said to be positive iff it can be written as $x = [(a_{n})_{n = 0}]$ for some Cauchy sequence $(a_{n})_{n = 0}$ which is positively bounded away from zero. （ $x$ is said to be negative iff it can be written as $x = [(a_{n})_{n = 0}]$ for some Cauchy sequence $(a_{n})_{n = 0}$ which is negatively bounded away from zero.）

$P = {x \in C ∣ x is positive}$

可以验证 $P$ 满足如下性质

$\forall x, y \in P, x + y \in P, x y \in P$
$\forall x \in C, x \in P, x = 0, - x \in P$ 三者有且只有一个成立

因此 $< C, +, \times >$ 构成一个有序域，即 $< R, +, \times >$

可以定义 $<$ 为 $x < y ⟺ y - x \in P$

(抽象代数)有序域(环)上的性质对 $< C, +, \times >$ 都成立

Trichotomy: $x < y, x = y, x > y$ 三者有且只有一个成立
Transitivity: $x < y, y < z \Rightarrow x < z$
Addition preservation: $x < y \Rightarrow x + z < y + z$
Multiplication preservation: $x < y, z > 0 \Rightarrow x z < yz$
Negation reverses order: $a > b \Rightarrow - a < - b$
$a > 0 ⟺ a^{- 1} > 0$
$0 < a, 0 < b \to 0 < a / b$
$0 < a < 1 \to 1 < 1/ a$
$- 1 < a < 0 \to 1/ a < - 1$

$lemma$ . $x, y \in Q \land x < y \Rightarrow i (x) < i (y)$ ,即 $i (q) = [(q_{n})_{n}]$ 是从 $Q$ 到 $C$ 的一个子集的序同构

至此将 $< C, +, \times >$ 记作 $< R, +, \times >$ ,简记为 $R$

$Proposition$ . Let $a_{0}, a_{1}, a_{2}, ...$ be a Cauchy sequence of non-negative rational numbers. Then $[(a_{n})_{n = 0}]$ is a non-negative real number.

$Corollary$ . Let $(a_{n})_{n = 0}$ and $(b_{n})_{n = 0}$ be Cauchy sequences of rationals such that $a_{n} \geq b_{n}$ for all $n \geq 0$ . Then $[(a_{n})_{n = 0}] \geq [(b_{n})_{n = 0}]$ .

$Proposition$ . Let $x$ be a positive real number. Then there exists a positive rational number $q$ such that $q \leq x$ , and there exists a positive integer $N$ such that $x \leq N$ .

$Corollary$ . Let $x$ be a real number, and let $ϵ$ be a positive real number. Then there exists a positive integer $M$ such that $M ϵ > x$ .

$Proposition$ . Given any two real numbers $x < y$ , we can find a rational number $q$ such that $x < q < y$ .

$Definition$ . $∣ x ∣ = x$ if $x \geq 0$ , else $∣ x ∣ = - x$

(抽象代数)有序环上的绝对值有以下性质

$∣ x ∣ \geq 0$
$∣ - x ∣ = ∣ x ∣$
$∣ x ∣ = 0 ⟺ x = 0$
$x < ∣ x ∣$
$∣ x + y ∣ \leq ∣ x ∣ + ∣ y ∣$
$∣ x y ∣ = ∣ x ∣∣ y ∣$

用绝对值定义距离 $d (x, y) = ∣ x - y ∣$ ，则有以下性质

$d (x, y) \geq 0$ . Also, $d (x, y) = 0$ if and only if $x = y .$
$d (x, y) = d (y, x)$ .
$d (x, z) \geq d (x, y) + d (y, z)$ .

$Lemma$ . Let $(a_{n})_{n = 0}$ be a Cauchy sequence of rationals, and let $x$ be a real number. Show that if $a_{n} \leq x$ for all $n \geq 0$ , then $[a_{n}] \leq x$ . Similarly, show that if $a_{n} \geq x$ for all $n \geq 0$ , then $[a_{n}] \geq x$ .

$Definition$ . (Upper bound). Let $E$ be a subset of $R$ , and let $M$ be a real number. We say that $M$ is an upper bound for $E$ , iff we have $x \leq M$ for every element $x$ in $E$ .

$Definition$ . (Least upper bound). Let $E$ be a subset of $R$ , and $M$ be a real number. We say that $M$ is a least upper bound for $E$ iff (a) $M$ is an upper bound for $E$ , and also (b) any other upper bound $M$ for $E$ must be larger than or equal to $M$ .

$Proposition$ . Let $E$ be a subset of $R$ . Then $E$ can have at most one least upper bound.

$Theorem$ . (Existence of least upper bound). Let $E$ be a non-empty subset of $R$ . If $E$ has an upper bound, (i.e., $E$ has some upper bound $M$ ), then it must have exactly one least upper bound.

目标：构造一个实数(有理数的 cauchy 列 $(a_{n})_{n = 0}$ )，使得其为最小上界

从非空集 $E$ 中取一个元素 $x_{0}$

对于每一个 $n \in N$

从有理数的 archimedean 性，可以找到一个整数 $U_{n}$ ，使得 $U_{n} (1/ (1 + n)) > M$
同理可以找到一个整数 $L_{n}$ ，使得 $L_{n} (1/ (1 + n)) < x_{0}$

对于每个 $n$ , $S_{n} = {i \in Z ∣ i / (1 + n) is an upper bound of E}$ ,有最小元，记作 $K_{n}$

$a_{n} = K_{n} / (1 + n)$ 的每一项都是 $E$ 的上界
$b_{n} = (K_{n} - 1) / (1 + n)$ 的每一项都不是 $E$ 的上界

接下来要证明 $a_{n}$ 和 $b_{n}$ 是 cauchy 列，且它们是 cauchy 等价的，即 $[a_{n}] = [b_{n}]$

$a_{n} - a_{n^{'}} = K_{n} / (1 + n) - K_{n^{'}} / (1 + n^{'}) \geq K_{n} / (1 + n) - (K_{n} - 1) / (1 + n) \geq 1/ (1 + n)$

$a_{n} - b_{n^{'}} = K_{n} / (1 + n) - K_{n^{'}} / (1 + n^{'}) \leq (K_{n^{'}} - 1) / (1 + n^{'}) - K_{n^{'}} / (1 + n^{'}) \leq - 1/ (1 + n^{'})$

因此对于任意的 $ϵ \in Q \land ϵ > 0$ ，存在 $N \in N \land N > 1/ ϵ$ ，使得对于任意 $n, n^{'} \geq N$ ，有 $∣ a_{n} - a_{n^{'}} ∣ \leq 1/ N \leq ϵ$ ，即 $a_{n}$ 是 cauchy 列

容易证明 $a_{n} \sim_{c} b_{n}$ ； $b_{n}$ 也是 cauchy 列;因此 $[a_{n}] = [b_{n}]$

因为 $a_{n}$ 的每一项都是 $E$ 的上界，所以 $[(a_{n})_{n = 0}]$ 是 $E$ 的上界

因为 $b_{n}$ 的每一项都不是 $E$ 的上界，所以 $[(b_{n})_{n = 0}]$ 小于 $E$ 的每一个上界

$s = [a_{n}] = [b_{n}]$ 满足最小上界性质，即 $s$ 是 $E$ 的最小上界

拓展实数域

$\overline{R} = R \cup {- \infty, + \infty}$
其中 $- \infty$ 和 $+ \infty$ 是两个新的元素，且满足 $- \infty < x < + \infty$ 对所有 $x \in R$ 成立

$Definition$ . (上确界) if E has a upper bound, then the least upper bound of $E$ is the supremum of $E$ ,denoted by $s u p (E)$ , if E not has a upper bound, then the supremum of $E$ is $+ \infty$

因为有最小上界属性的全序集都有最大下界属性，因此可以定义下确界

$Definition$ . (下确界) if E has a lower bound, then the greatest lower bound of $E$ is the infimum of $E$ ,denoted by $in f (E)$ , if E not has a lower bound, then the infimum of $E$ is $- \infty$

$Lemma$ . $in f (E) = - s u p (- E)$

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