一些通用的数学概念和记号

最近想着有空的时候把数学捡起来
这篇博文写了分析学中的集合论
先从分析学开始吧

集合论一些结论的证明

$\textbf{solution1}$
$X \times Y = \emptyset \iff (X=\emptyset) \vee (Y =\emptyset)$

$\textbf{proof}$

$$\begin{gathered} \emptyset = (x \in X) \wedge (x \in C_MX) \\ \begin{cases} (x \in X) \wedge (y \in Y) = (x \in X) \wedge (x \in C_MX) \\ (x \in X) \wedge (y \in Y) = (y \in Y) \wedge (y \in C_MY) \end{cases} \end{gathered}$$

$$\begin{gathered} \Rightarrow (x \in X) \wedge (y \in Y) = (x \in C_MX) \wedge (y \in C_MY) \\ \begin{cases} X = C_MX \\ Y=C_MY\end{cases} \Rightarrow (x=\emptyset) \vee (y = \emptyset) \end{gathered}$$

另一方面

$$(X = \emptyset) \vee (Y=\emptyset) \Rightarrow \forall(x,y), \ (x\in C_MX) \vee (y \in C_MY)$$

$$\begin{gathered} \Rightarrow \forall (x,y), \ x \in C_M(X\times Y) \\ \Rightarrow \forall (x,y), \ x \notin X\times Y \end{gathered}$$

$X, Y$ 任意, 所以 $X \times Y = \emptyset$

$\textbf{solution2}$
$(A\times B)\subset (X \times Y) \iff (A \subset X) \wedge (B \subset Y)$

$\textbf{proof}$

$$\begin{gathered} \textbf{Let} \ \textbf{P}=(x \in A) \wedge (y \in B), \ \textbf{Q}=(x\in X) \wedge (y \in Y) \\ P \Rightarrow Q \iff (\lnot p \vee Q) \end{gathered}$$

$1)$

$$\begin{gathered} (\lnot p) \vee ((x \in X) \wedge (y \in Y)) \iff (\lnot p \vee (x\in X)) \wedge (\lnot p \vee (y \in Y)) \\ \iff (p \Rightarrow (x \in X)) \wedge (p \Rightarrow (y \in Y)) \end{gathered}$$

$$\begin{gathered} x \in A \Rightarrow x\in X \iff (A \subset X) \\ y \in B \Rightarrow y\in Y \iff (B \subset Y) \end{gathered}$$

$2)$
根据命题等价性, 右边推左边也同理

$\textbf{solution3}$
$(X \times Y) \cup (Z \times Y) = (X \cup Z) \times Y$

$$\begin{gathered} \forall x \in X, \forall y \in Y, \forall z\in Z, \\ \ \\ (x \in X) \vee (y \in Y) \vee (z \in Z) \vee (y \in Y) \\ \ \\ \iff \{(x, z) | (x \in X) \vee (z \in Z)\} \cup Y \iff (X \cup Z) \times Y \end{gathered}$$

$\textbf{solution4}$
$(X \times Y) \cap (X' \times Y') = (X \cap X') \times (Y \cap Y')$

不妨假设
$u \in (X \cap X'), \ v \in (Y \cap Y')$

$$\begin{gathered} \{(u, v) | (u \in (X \cap X')) \wedge (v \in (Y \cap Y'))\} \\ \ \\ \iff \{(u, v) | (u \in X) \wedge (u \in X') \wedge (v \in Y) \wedge (v \in Y')\} \\ \ \\ \iff \{(u, v) | [(u \in x) \wedge (v \in Y)] \ \wedge \ [(u \in X') \wedge (v \in Y')]\} \end{gathered}$$

令 $A = X \times Y, \ B = X' \times Y'$

$$\begin{gathered} \{(u, v) | (u \in X) \wedge (v \in Y)\} \Rightarrow \{p | p \in A\} \\ \ \\ \{(u, v) | (u \in X') \wedge (v \in Y')\} \Rightarrow \{p | p \in B\} \\ \ \\ \iff \{p | (p \in A) \wedge (p \in B)\} \iff A \cap B \iff (X \times Y) \cap (X' \times Y') \end{gathered}$$

$\textbf{solution5} \ \text{反对称性}$

$$\begin{gathered} S \Delta T=(S \backslash T) \cup(T \backslash S) \\ \ \\ S \backslash T=S \cup(\neg T), \quad T \backslash S=T \cup(\neg S) \\ \ \\ (S \cup T) \cup\left(\neg T \cup \neg S\right)=(S \cup T) \cup(\neg(S \cap T)) \\ \ \\ =(S \cup T) \backslash(S \cap T) \end{gathered}$$

函数的表示

$\textbf{solution1}$
$(A \subset B) \Rightarrow (f(A) \subset f(B))$

$1) (A \subset B) \Rightarrow \quad f(A) \subset f(B)$

$$\begin{gathered} f(A) := \{y \in Y | \exists x, (x \in A) \wedge (y \in f(x))\} \\ \ \\ A \subset B \iff (x \in A \Rightarrow x \in B) \\ \ \\ \Rightarrow \{y \in Y | \exists x, (x \in B) \wedge (y \in f(x))\} = f(B) \end{gathered}$$

$2) \ (f(A) \subset f(B)) \not \Rightarrow (A \subset B)$

$$\begin{gathered} \{y \in Y | \exists x, (x \in A)\} \Rightarrow \{y \in Y | \exists x, x \in B\} \\ \ \\ \exists x, (x \in A) \Longrightarrow \exists x, (x \in B) \\ \ \\ \iff \exists x, \ (x \in A) \wedge (x \not \in B) \end{gathered}$$

$\textbf{solution2}$
$A \neq \emptyset \Rightarrow f(A) \neq \emptyset$

$$\begin{gathered} \exists x, x \in A \\ \ \\ Let \ x_1 \in A \Rightarrow f(x_1) = y_1 \Leftrightarrow \exists y_1 \in Y \\ \ \\ \iff \{\exists x | x \in A \} \Rightarrow \{\exists y | f(x) = y\} \end{gathered}$$

$\textbf{solution3}$
$f(A \cap B) \subset (f(A) \cap f(B))$

$$\begin{gathered} f(A):=\{y \in Y | \exists x \ (x \in (A \cap B)) \wedge (y \in f(x))\} \\ \ \\ \{\exists x, x \in (A \cap B) \wedge y \in f(x)\} \Rightarrow \{\exists x | (x \in A) \wedge (y \in f(x)) \wedge (x \in B) \wedge (y \in f(x)\} \\ \ \\ \Rightarrow f(A) \cap f(B) \end{gathered}$$

$\textbf{solution4}$
$f(A \cap B) = f(A) \cap f(B)$

$$\begin{gathered} f(A \cap B):= \{y \in Y | \exists x, (x \in (A \cap B) \wedge (y \in f(x)))\} \\ \ \\ \{\exists x,y | ((x \in A) \vee (x \in B)) \wedge (y \in f(x))\} \\ \ \\ \iff \{\exists x,y | (x \in A \wedge y \in f(x)) \vee (x \in B \wedge y \in f(x))\} \\ \ \\ \iff f(A) \cup f(B) \end{gathered}$$

$\textbf{solution5}$
$A^{\prime} \subset B^{\prime} \Rightarrow f^{-1}\left(A^{\prime}\right) \subset f^{-1}\left(B^{\prime}\right)$
需要用到定义
$f^{-1}(B)=\{x \in X | f(x) \in B\}$

$$\begin{gathered} A^{\prime} \subset B^{\prime} \iff f(x) \in A^{\prime} \Rightarrow f(x) \in B^{\prime} \iff \left\{x \in X | f(x) \in A^{\prime}\right\} \Rightarrow\left\{x \in X | f(x) \in B^{\prime}\right\} \\ \ \\ f^{-1}(A) \subset f^{-1}(B) \end{gathered}$$

$\textbf{solution6}$
$f^{-1}\left(A^{\prime} \cap B^{\prime}\right)=f^{-1}\left(A^{\prime}\right) \cap f^{-1}\left(B^{\prime}\right)$

$$\begin{gathered} \left\{x \in X |\left(f(x) \in A^{\prime}\right) \wedge\left(f(x) \in B^{\prime}\right)\right\} \\ \ \\ \Leftrightarrow\left\{x \in X\left|f(x) \in A^{\prime}\right\} \cap\left\{x \in X | f(x) \in B^{\prime}\right\}\right.\\ \ \\ \iff f^{-1}\left(A^{\prime}\right) \cap f^{-1}\left(B^{\prime}\right) \end{gathered}$$

$\textbf{solution7}$
$f^{-1}\left(A^{\prime} \cup B^{\prime}\right)=f^{-1}\left(A^{\prime}\right) \cup f^{-1}\left(B^{\prime}\right)$
同理可证

$\textbf{solution8}$
$f^{-1}(A / B)=f^{-1}(A) / f^{-1}(B)$

$$\begin{gathered} \{x \in X |(f(x) \in A) \wedge(f(x) \notin B)\} \iff \left\{x \in X\left|f(x) \in A\} \cap\left\{x \in X | f^{\prime}(x) \notin B\right\}\right.\right. \\ \ \\ \Leftrightarrow f^{-1}(A) / f^{-1}(B) \end{gathered}$$

$\textbf{solution9}$
$f^{-1}\left(C_{Y} A\right)=C_{X} f^{-1}(A)$

$$\begin{gathered} C_{X} f^{-1}(A)=X-\{x \in X |f(x) \in A\} \\ \ \\ =\{x \in X | f(x) \notin A\}=\left\{x \in X | f(x) \in C_{Y} A\right\} = f^{-1}\left(C_{Y} A\right) \end{gathered}$$

$\textbf{solution10}$
$\begin{gathered} \forall A, \ A \subset X\\ \forall B , \ B \subset Y \end{gathered}$
有 $f^{-1}(f(A)) \supset A$

需要用到的很重要的结论是

$$f(A)=\{y \in Y | \exists x, \quad(x \in A) \wedge(y \in f(x))\}$$

$$\begin{gathered} \text { i) } \quad x \in A \Rightarrow f(x) \in f(A) \\ \ \\ \text {ii)} \quad A \subset X \Rightarrow x \in A, x \in X \\ \Rightarrow \{x \in X | f(x) \in f(A)\}=f^{-1}(f(x)) \\ let \ \textbf{M} = \{x \in X | f(x) \in f(A)\} \end{gathered}$$

$$\iff \{x \in A \Rightarrow x \in \textbf{M} \} \Rightarrow A \subset M \iff f^{-1}(f(A)) \supset A$$

$\textbf{solution11}$
$f\left(f^{-1}(B)\right) \subset B$

$$\begin{gathered} M=f^{-1}(B)=\{x \in X | f(x) \in B\} \\ \ \\ f(M)=\{y \in Y, \exists x|(x \in M) \wedge(y \in f(x))\} \\ \ \\ \Rightarrow f(M)=\{y \in Y, \exists x \in X \ | \ (f(x) \in B) \wedge(y \in f(x))\} \end{gathered}$$

所以存在这样的一个 $x$

$$\begin{gathered} \exists x_i, f(x_i) \in B \\ \ \\ \Leftrightarrow\{y \in Y |(f(x) \in B) \wedge(y \in f(x))\} \subset B \\ \ \\ \iff f(f^{-1}(B)) \subset B \end{gathered}$$

函数的其他命题（一）

$\textbf{solution1}$
$\mathcal{R}_1 \subset X \times Y, \quad \mathcal{R}_2 \subset Y \times X$ 满足
$\left(\mathcal{R}_{1} \circ \mathcal{R}_{2}=\Delta_{x}\right) \wedge\left(\mathcal{R}_{2} \circ \mathcal{R}_{1}=\Delta_{Y}\right)$
它们都是函数关系

$$\left\{\begin{array}{l} \mathcal{R}_{1} \subset X \times Y \\ \\ \mathcal{R}_{2} \subset Y \times X \end{array}\right. \Rightarrow \mathcal{R}_{2} \circ \mathcal{R}_{1}= \left\{\exists y,(x \mathcal{R}_1 y) \wedge\left(y \mathcal{R}_{2} x\right)\right\} = \Delta_y$$

不妨设 $\mathcal{R} = \mathcal{R}_2 \circ \mathcal{R}_1$

$$\begin{gathered} \left\{\begin{array}{l} \left.x \mathcal{R} y_{1}=\{\exists y_{1},\quad\left(x \mathcal{R}_1 y_{1}\right) \wedge\left(y_{1}\mathcal{R}_{2} x\right)\right\}=\Delta_{Y} \\ \ \\ x \mathcal{R} y_{2}=\left\{\exists y_{2}, \quad\left(x \mathcal{R}_1y_{2}\right) \wedge\left(y_{2} \mathcal{R}_{2} x\right)\right\}=\Delta_{Y} \end{array}\right. \iff y_1=y_2 \end{gathered}$$

同理可证, $\mathcal{R}_2 \circ \mathcal{R}_1$为函数关系

$\textbf{solution2}$
$\mathcal{R} \subset X^{2}$, 传递性等价于$\mathcal{R} \circ \mathcal{R} \subset \mathcal{R}$

$\text{i)}$ 根据传递性 $(a \mathcal{R} b) \wedge (b \mathcal{R} c) \Rightarrow (a \mathcal{R} c)$
$\exists a \in X, \exists b \in X$, 因为 $\mathcal{R}$ 是建立在 $X \times X$ 上的关系
所以有 $c \in X$

$\text{ii)}$ 有

$$\begin{gathered} \mathcal{R} \circ \mathcal{R}=\{\exists x,(x \mathcal{R} x) \wedge(x \mathcal{R} x)\} \subset\{\forall x | x \in X \times X\} \subset \mathcal{R} \\ \ \\ \iff \mathcal{R} \circ \mathcal{R} \subset \mathcal{R} \end{gathered}$$

$\textbf{solution3}$
$\mathcal{R}' \subset Y \times X \ \text{和} \ \mathcal{R} \subset X \times Y \ \text{满足} \ (y \mathcal{R}' x) \Leftrightarrow (x \mathcal{R} y), \text{则} \ \mathcal{R}' \text{是} \ \mathcal{R} \text{的转置关系}$
$\mathcal{R} \subset X^2 \text{的反对称性等价于} \ \mathcal{R} \cap \mathcal{R}' \subset \Delta_X$

$$\exists a \in X, \exists b \in X, \quad(a \mathcal{R} b) \wedge(b \mathcal{R} a) \Rightarrow a=b$$

$$\begin{gathered} \text{因为} (b \mathcal{R} a) \Leftrightarrow (a \mathcal{R}' b) \iff (a \mathcal{R} b) \wedge (a \mathcal{R}' b) \Rightarrow a = b \iff (\mathcal{R} \cap \mathcal{R}') X = X \\ \ \\ \text{所以有} \mathcal{R} \cap \mathcal{R}' \subset \Delta_X \end{gathered}$$

$\textbf{solution4}$
$\text{当且仅当} \mathcal{R} \cup \mathcal{R}' = X^2 \text{时, 集合} X \text{中的任意两个元素由} \mathcal{R} \subset X^2 \text{相联系}$

$$\begin{gathered} \text{因为} \mathcal{R} \subset X^2, \quad \forall a, b \in X, \quad a \mathcal{R} b \in X \Leftrightarrow b \mathcal{R}' a \in X \\ \ \\ \text{所以} (a \mathcal{R} b) \cup (b \mathcal{R}' a) \in X^2 \\ \ \\ \iff \mathcal{R} \cup \mathcal{R}' = X^2 \end{gathered}$$

$$\text{另一方面}, \mathcal{R} \cup \mathcal{R}', \text{显然} \mathcal{R} \subset X^2, \ \mathcal{R}' \subset X^2$$

函数的其他命题（二）

$f:X \to Y\text{是映射}, y \in Y \text{的原像} f^{-1}(y) \subset X \text{称为 } y \text{上的层}$

$\textbf{solution1}$
$x_1 \in X, x_2 \in X, \text{如果} f(x_1)=f(x_2), \text{认为} x_1, x_2 \text{由关系} \mathcal{R} \subset X^2 \text{相联系, 并记作} x_1 \mathcal{R} x_2, \mathcal{R}\text{是等价关系}$

$$\begin{gathered} \text{i), 很显然} x_1 \mathcal{R} x_2 = x_2 \mathcal{R} x_1 \\ \ \\ \text{ii)} \ x_1 \mathcal{R} a := f(x_1)=f(a), \quad a \mathcal{R} x_2:=f(a)=f(x_2) \\ \ \\ (x_1 \mathcal{R} a) \wedge (a \mathcal{R} x_2) \Rightarrow f(x_1)=f(a)=f(x_2) \Rightarrow f(x_1)=f(x_2) \\ \ \\ \text{是等价关系} \end{gathered}$$

$\textbf{solution2}$
$\text{i)} \quad f: X\to Y \text{的层互不相交, 而所有层的并集是整个集合 } X$

$$\begin{gathered} \text{如果} X \to Y \text{的层相交}, f^{-1}(y_1) = f^{-1}(y_2) = x_1 \\ \ \\ f(x_1) = y_1 \quad f(x_1) = y_2, \text{与映射定义矛盾} \end{gathered}$$

$\text{ii)} \quad f^{-1}(y_1) \cup f^{-1}(y_2) \cup \cdots f^{-1}(y_n) = x_1 \cup x_2 \cup \cdots \cup x_n = X$

映射的性质

$\textbf{solution1}$
$\textbf{映射} \ f:X \to Y \text{是满射, 当且仅当 } \forall B \subset Y, f(f^{-1}(B))= B$

$$\begin{matrix} \text{a)} \quad \forall x \in f^{-1}(B), f(x) \in B \subset Y \\ \ \\ \forall x \in X, f(x) = \bigcup_{i=1} B_{i} = Y \iff f(X) = Y \end{matrix}$$

$$\begin{gathered} \text{b)} \quad \forall B \subset Y, f(f^{-1}(B)) = B, \text{保证了 } f \text{ 是满射} \text{, 下面证明 } f \text{ 为单射} \\ \ \\ \iff f(x_1)=f(x_2) \Rightarrow x_1=x_2 \\ \ \\ \text{不妨设 } f(x_1) = f(x_2) = p \in Y \\ \ \\ f^{-1}(f(x_1)) = f^{-1}(p) = \{x\in A | f(x)=p\} = M_1 \subset A \\ \ \\ f^{-1}(f(x_2)) = f^{-1}(p) = M_2 \subset A \\ \ \\ f^{-1}(p) = M_1 = M_2, \text{ 即它的层是唯一的} \\ \ \\ f(x_1)=f(x_2) \Rightarrow x_1=x_2 \end{gathered}$$

$\textbf{solution2}$
$\text{如果 }\ f:X \to Y \text{ 与 }\ g: Y\to X \text{ 满足} \ g\circ f=e_X, e_X \text{ 为中性元, }$
$\text{则 }\ g \ \text{称为} \ f \ \text{的左逆映射, }\ f \ \text{称为} \ g \ \text{的右逆映射, 可能存在多个单侧逆映射}$

$$\begin{gathered} \left(x_{1}, x_{2}, \ldots \ldots, x_{n} \ldots\right) \stackrel{f_{a}}{\rightarrow}\left(a, x_{1}, x_{n} \ldots x_{n}\right) \\ \ \\ \text{注意到 }a\text{ 的取值可以是任意的} \end{gathered}$$

$\textbf{solution3}$

$f: X \to Y \quad g: Y\to Z\text{ 是双射, 映射 }\ g \circ f : X \to Z\text{ 是双射, 并且 }(g\circ f)^{-1} = f^{-1} \circ g^{-1}$

$$\begin{gathered} \text{i)} \quad \text{先证满射性, }X \stackrel{f}{\rightarrow} Y\stackrel{g} \rightarrow Z \\ \ \\ (g \circ f) X = Z \\ \ \\ g \circ f \text{为满射} \end{gathered}$$

$$\begin{gathered} \text{ii)} \quad \text{再证单射性, }(g \circ f)(x_1)=(g\circ f)(x_2) \Rightarrow g(f(x_1))=g(f(x_2)) \\ \ \\ g \text{为单射, }f(x_1)=f(x_2) \\ \ \\ f \text{为单射, }x_1=x_2 \end{gathered}$$

$$\begin{gathered} \text{iii)} \quad (f^{-1} \circ g^{-1}) \circ (g \circ f) = f^{-1} \circ (g^{-1}\circ g) \circ f= f^{-1} \circ e_Y \circ f \\ \ \\ = f^{-1} \circ f = e_X \end{gathered}$$

$\textbf{solution4}$
$\text{对于任何映射 }\ f : X\to Y, \text{由} x \stackrel {F} \longmapsto (x, f(x))\text{ 定义的映射}$
$F: X \to X \times Y\text{ 是单射}$

$$\begin{gathered} f(x_1) = f(x_2) \Rightarrow x_1 = x_2 \\ \ \\ \forall x_1 \neq x_2, f(x_1) \neq f(x_2) \Rightarrow (x_1, f(x_1)) \neq (x_2, f(x_2)) \\ \ \\ F(x_1)=F(x_2) \Rightarrow x_1 = x_2 \end{gathered}$$

极限理论基础

极限为 $A$, $A$ 的邻域之外, 有有限个点, 在数学上要注意表示形式
比如, 证明极限的唯一性的时候, 要想证明邻域重合

$$\begin{gathered} \forall n > N_1, x_1 \in V(A_1) \\ \ \\ \forall n > N_2, x_2 \in V(A_2) \\ \ \\ \text{不妨设 }\ A_1 < A_2, N_1 < N_2 \\ \ \\ \text{此时, 使得 }\ A_1 + \epsilon = A_2 - \epsilon \\ \ \\ \epsilon = \frac{A_1-A_2}{2}, \text{更一般地, 让 }\ \epsilon = \frac{|A_1 - A_2|}{2} \end{gathered}$$

$\textbf{solution1}$

$$\begin{gathered} \left|A \cdot B-x_{n} y_{n}\right| \leqslant \left|x_{n}\right|\left(\Delta y_{n}\right)+\left|y_{n}\right|\left(\Delta x_{n}\right)+\Delta x_{n} \Delta y_{n} \\ \ \\ \text{构造 }\frac{\epsilon}{3} \end{gathered}$$

$$\begin{gathered} &\left\{\begin{gathered} \Delta x_{n}<\min \left(1, \frac{\varepsilon}{3}\right) \leqslant 1 \\ \Delta y_{n}< \min \left(1, \frac{\varepsilon}{3}\right) \leqslant 1 \end{gathered}\right.\\ &\Rightarrow \quad \Delta x_{n} \Delta y_{n}<\frac{\varepsilon}{3} \end{gathered}$$

$$\begin{gathered} \left\{\begin{gathered} \left|x_{n}\right|<A+\Delta x_{n} \leqslant A+1 \\ \left|y_{n}\right|<B+\Delta y_{n} \leqslant B+1 \end{gathered}\right. \stackrel{|x_n|\Delta y_n \leqslant \frac{\epsilon}{3}}\Longrightarrow \Delta y_{n} \leqslant \min \left(1, \frac{\varepsilon}{3(A+1)}\right) \\ \text{这样所需要的 }\epsilon \text{ 已经构造完毕了} \end{gathered}$$

$\textbf{solution2}$

$$\begin{gathered} \left| \frac{A}{B} - \frac{x_n}{y_n} \right| \leqslant \frac{\left|x_{n}\right|+\Delta x_{n}}{\left|y_{n}\right|-\Delta y_{n}}-\frac{x_{n}}{y_{n}}=\frac{\left|x_{n}\right|\left|y_{n}\right| +\left|y_{n}\right| \Delta x_{n}-\left|x_{n}\right|\left|y_{n}\right|+\left|x_{n}\right| \Delta y_{n}}{\left|y_{n}\right|^{2}-|y _{n} | \Delta y_{n}} \\ \ \\ =\frac{\left|x_{n}\right| \Delta y_{n}+\left|y_{n}\right| \Delta x_{n}}{y_{n}^{2}} \frac{1}{1-\delta\left(y_{n}\right)}, \delta\left(y_{n}\right)=\frac{\Delta y_{n}}{\left|y_{n}\right|} \end{gathered}$$

$\text{构造证明的第一部分, 已知条件 }\left|x_{n}\right|<|A|+\Delta x_{n}, \quad\left|y_{n}\right|>|B|-\Delta y_{n}, \text{ 先凑出}$

$$\frac{1}{1-\delta\left(y_{n}\right)} < ?$$

$$\begin{gathered} \Delta y_{n}<\frac{|B|}{4} \Longrightarrow\left|y_{n}\right|>|B|-\Delta y_{n}>\frac{|B|}{2} \\ \Longrightarrow \delta\left(y_{n}\right)<\frac{1}{2} \Longrightarrow \frac{1}{1-\delta\left(y_{n}\right)}<2 \end{gathered}$$

$$\begin{gathered} \text{接下来只要证明 }\frac{\left|x_{n}\right| \Delta y_{n}}{y_{n}^{2}}<\frac{\varepsilon}{4}, \quad \frac{\left|y_{n}\right| \Delta x_{n}}{y_{n}^{2}}<\frac{\varepsilon}{4}\text{ 即可} \end{gathered}$$

$\textbf{i)}$

$$\begin{gathered} \left|x_{n}\right|<|A|+1, \quad \frac{1}{y_{n}^{2}}<\left(\frac{2}{|B|}\right)^{2}=\frac{4}{B^{2}} \\ \ \\ \text{此时构造出 }\Delta y_{n}<\frac{1}{16} \frac{\varepsilon B^{2}}{(| A |+1)}, \Delta y_{n}<\min \left\{\frac{1}{16} \frac{\varepsilon B^{2}}{(| A |+1)}, \frac{|B|}{4}\right\} \end{gathered}$$

$\textbf{ii)}$

$$\begin{gathered} \frac{1}{y_{n}}<\frac{2}{|B|}, \quad \Delta x_{n} \leqslant \frac{\varepsilon|B|}{8} \\ \Delta x_{n} \leqslant \min \left\{\frac{\varepsilon|B|}{8}, 1\right\} \end{gathered}$$

$\textbf{solution3} \quad \text{对于}\ \forall \epsilon>0, \text{可以求出}\{n, m\} > N, \text{使得 }|x_m-x_n| < \epsilon$

分析中很重要的一点, 集合变小时, 下界不减小, 上界不增大
柯西准则的证明如下

$\textbf{i)} \quad \text{基本数列是收敛数列}$

$$\begin{gathered} \left|x_{m}-x_{n}\right|<\left|x_{m}-A\right|+\left|x_{n}-A\right| \\ \\ \text{if} \quad \forall k>N \quad,\left|x_{k}-A\right|<\frac{\varepsilon}{2} \Longrightarrow\left|x_{m}-x_{n}\right|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon \end{gathered}$$

$\textbf{ii)} \quad \text{收敛数列是基本数列}$

$$\begin{gathered} a_{n}:=\inf _{k \geqslant n} x_{k} \quad b_{n}:=\sup _{k \geqslant n} x_{k} \\ \ \\ a_{n+1}:=\inf _{k \geqslant n+1} x_{k} \quad b_{n+1}:=\sup _{k \geqslant n+1} x_{k} \\ \ \\ \Rightarrow a_{n} \leqslant a_{n+1} \leqslant b_{n+1} \leqslant b_{n} \\ \ \\ \text{根据闭区间套, }[a_n, b_n] \text{有公共点 }A \end{gathered}$$

$$\begin{gathered} \left[a_{n}, b_{n}\right], \quad\left\{\begin{gathered} \left|A-x_{k}\right| \leqslant b_{n}-a_{n} \\ x_{n}-\delta \leqslant \inf \limits_{k \geqslant n} x_{k}=a_{n} \leqslant b_{n}=\sup\limits _{k \geqslant n} x_{k} \leqslant x_{n}+\delta, \delta=? \end{gathered}\right. \end{gathered}$$

$$\begin{gathered} b_{n}-a_{n} \leqslant 2 \delta<\varepsilon \Rightarrow \delta=\frac{\varepsilon}{3} \\ \ \\ x_{n}-\frac{\varepsilon}{3}<x_{k}<x_{n}+\frac{\varepsilon}{3} \\ \ \\ \left|A-x_{k}\right|<\varepsilon \end{gathered}$$

$\textbf{Weierstrass}$
$\text{上确界的表示方式, }s = \sup \limits_{n \in \mathbb{N}} x_n$

$$\forall \varepsilon>0 \quad \exists x_{N} \in \left\{ x_n \right\} \quad s-\varepsilon<x_{N} \leqslant s$$

$$\Rightarrow \forall n>N, \quad s-\varepsilon<x_{N} \leqslant x_{n} \leqslant s \Rightarrow s-\varepsilon <x_{n}$$

$$\left|s-x_{n}\right|=s-x_{n}<\varepsilon$$

$\textbf{不减数列有上确界} \ s = \sup\limits_{n \in \mathbb{N}} x_n$
$\textbf{不增数列有下确界} \ s = \inf\limits_{n \in \mathbb{N}} x_n$