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

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

proof\textbf{proof}

=(xX)(xCMX){(xX)(yY)=(xX)(xCMX)(xX)(yY)=(yY)(yCMY)\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}

(xX)(yY)=(xCMX)(yCMY){X=CMXY=CMY(x=)(y=)\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=)(Y=)(x,y), (xCMX)(yCMY)(X = \emptyset) \vee (Y=\emptyset) \Rightarrow \forall(x,y), \ (x\in C_MX) \vee (y \in C_MY)

(x,y), xCM(X×Y)(x,y), xX×Y\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,YX, Y 任意, 所以X×Y=X \times Y = \emptyset

solution2\textbf{solution2}
(A×B)(X×Y)    (AX)(BY)(A\times B)\subset (X \times Y) \iff (A \subset X) \wedge (B \subset Y)

proof\textbf{proof}

Let P=(xA)(yB), Q=(xX)(yY)PQ    (¬pQ)\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)1)

(¬p)((xX)(yY))    (¬p(xX))(¬p(yY))    (p(xX))(p(yY))\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}

xAxX    (AX)yByY    (BY)\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)2)
根据命题等价性, 右边推左边也同理

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

xX,yY,zZ, (xX)(yY)(zZ)(yY)     {(x,z)(xX)(zZ)}Y    (XZ)×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}

solution4\textbf{solution4}
(X×Y)(X×Y)=(XX)×(YY)(X \times Y) \cap (X' \times Y') = (X \cap X') \times (Y \cap Y')

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

{(u,v)(u(XX))(v(YY))}     {(u,v)(uX)(uX)(vY)(vY)}     {(u,v)[(ux)(vY)]  [(uX)(vY)]}\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}

A=X×Y, B=X×YA = X \times Y, \ B = X' \times Y'

{(u,v)(uX)(vY)}{ppA} {(u,v)(uX)(vY)}{ppB}     {p(pA)(pB)}    AB    (X×Y)(X×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}

solution5 反对称性\textbf{solution5} \ \text{反对称性}

SΔT=(S\T)(T\S) S\T=S(¬T),T\S=T(¬S) (ST)(¬T¬S)=(ST)(¬(ST)) =(ST)\(ST)\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

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

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

f(A):={yYx,(xA)(yf(x))} AB    (xAxB) {yYx,(xB)(yf(x))}=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)f(B))⇏(AB)2) \ (f(A) \subset f(B)) \not \Rightarrow (A \subset B)

{yYx,(xA)}{yYx,xB} x,(xA)x,(xB)     x, (xA)(x∉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}

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

x,xA Let x1Af(x1)=y1y1Y     {xxA}{yf(x)=y}\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}

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

f(A):={yYx (x(AB))(yf(x))} {x,x(AB)yf(x)}{x(xA)(yf(x))(xB)(yf(x)} f(A)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}

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

f(AB):={yYx,(x(AB)(yf(x)))} {x,y((xA)(xB))(yf(x))}     {x,y(xAyf(x))(xByf(x))}     f(A)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}

solution5\textbf{solution5}
ABf1(A)f1(B)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
f1(B)={xXf(x)B}f^{-1}(B)=\{x \in X | f(x) \in B\}

AB    f(x)Af(x)B    {xXf(x)A}{xXf(x)B} f1(A)f1(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}

solution6\textbf{solution6}
f1(AB)=f1(A)f1(B)f^{-1}\left(A^{\prime} \cap B^{\prime}\right)=f^{-1}\left(A^{\prime}\right) \cap f^{-1}\left(B^{\prime}\right)

{xX(f(x)A)(f(x)B)} {xXf(x)A}{xXf(x)B}     f1(A)f1(B)\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}

solution7\textbf{solution7}
f1(AB)=f1(A)f1(B)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

solution8\textbf{solution8}
f1(A/B)=f1(A)/f1(B)f^{-1}(A / B)=f^{-1}(A) / f^{-1}(B)

{xX(f(x)A)(f(x)B)}    {xXf(x)A}{xXf(x)B} f1(A)/f1(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}

solution9\textbf{solution9}
f1(CYA)=CXf1(A)f^{-1}\left(C_{Y} A\right)=C_{X} f^{-1}(A)

CXf1(A)=X{xXf(x)A} ={xXf(x)A}={xXf(x)CYA}=f1(CYA)\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}

solution10\textbf{solution10}
A, AXB, BY\begin{gathered} \forall A, \ A \subset X\\ \forall B , \ B \subset Y \end{gathered}
f1(f(A))Af^{-1}(f(A)) \supset A

需要用到的很重要的结论是

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

 i) xAf(x)f(A) ii)AXxA,xX{xXf(x)f(A)}=f1(f(x))let M={xXf(x)f(A)}\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}

    {xAxM}AM    f1(f(A))A\iff \{x \in A \Rightarrow x \in \textbf{M} \} \Rightarrow A \subset M \iff f^{-1}(f(A)) \supset A

solution11\textbf{solution11}
f(f1(B))Bf\left(f^{-1}(B)\right) \subset B

M=f1(B)={xXf(x)B} f(M)={yY,x(xM)(yf(x))} f(M)={yY,xX  (f(x)B)(yf(x))}\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xx

xi,f(xi)B {yY(f(x)B)(yf(x))}B     f(f1(B))B\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)

solution1\textbf{solution1}
R1X×Y,R2Y×X\mathcal{R}_1 \subset X \times Y, \quad \mathcal{R}_2 \subset Y \times X 满足
(R1R2=Δx)(R2R1=ΔY)\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

{R1X×YR2Y×XR2R1={y,(xR1y)(yR2x)}=Δy\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R=R2R1\mathcal{R} = \mathcal{R}_2 \circ \mathcal{R}_1

{xRy1={y1,(xR1y1)(y1R2x)}=ΔY xRy2={y2,(xR1y2)(y2R2x)}=ΔY    y1=y2\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,R2R1\mathcal{R}_2 \circ \mathcal{R}_1为函数关系

solution2\textbf{solution2}
RX2\mathcal{R} \subset X^{2}, 传递性等价于RRR\mathcal{R} \circ \mathcal{R} \subset \mathcal{R}

i)\text{i)} 根据传递性(aRb)(bRc)(aRc)(a \mathcal{R} b) \wedge (b \mathcal{R} c) \Rightarrow (a \mathcal{R} c)
aX,bX\exists a \in X, \exists b \in X, 因为R\mathcal{R} 是建立在X×XX \times X 上的关系
所以有cXc \in X

ii)\text{ii)}

RR={x,(xRx)(xRx)}{xxX×X}R     RRR\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}

solution3\textbf{solution3}
RY×X 和 RX×Y 满足 (yRx)(xRy),则 R是 R的转置关系\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}
RX2的反对称性等价于 RRΔX\mathcal{R} \subset X^2 \text{的反对称性等价于} \ \mathcal{R} \cap \mathcal{R}' \subset \Delta_X

aX,bX,(aRb)(bRa)a=b\exists a \in X, \exists b \in X, \quad(a \mathcal{R} b) \wedge(b \mathcal{R} a) \Rightarrow a=b

因为(bRa)(aRb)    (aRb)(aRb)a=b    (RR)X=X 所以有RRΔX\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}

solution4\textbf{solution4}
当且仅当RR=X2时, 集合X中的任意两个元素由RX2相联系\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}

因为RX2,a,bX,aRbXbRaX 所以(aRb)(bRa)X2     RR=X2\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}

另一方面,RR,显然RX2, RX2\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:XY是映射,yY的原像f1(y)X称为 y上的层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}

solution1\textbf{solution1}
x1X,x2X,如果f(x1)=f(x2),认为x1,x2由关系RX2相联系, 并记作x1Rx2,R是等价关系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}

i), 很显然x1Rx2=x2Rx1 ii) x1Ra:=f(x1)=f(a),aRx2:=f(a)=f(x2) (x1Ra)(aRx2)f(x1)=f(a)=f(x2)f(x1)=f(x2) 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}

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

如果XY的层相交,f1(y1)=f1(y2)=x1 f(x1)=y1f(x1)=y2,与映射定义矛盾\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}

ii)f1(y1)f1(y2)f1(yn)=x1x2xn=X\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

solution1\textbf{solution1}
映射 f:XY是满射, 当且仅当 BY,f(f1(B))=B\textbf{映射} \ f:X \to Y \text{是满射, 当且仅当 } \forall B \subset Y, f(f^{-1}(B))= B

a)xf1(B),f(x)BY xX,f(x)=i=1Bi=Y    f(X)=Y\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}

b)BY,f(f1(B))=B,保证了 f 是满射, 下面证明 f 为单射     f(x1)=f(x2)x1=x2 不妨设 f(x1)=f(x2)=pY f1(f(x1))=f1(p)={xAf(x)=p}=M1A f1(f(x2))=f1(p)=M2A f1(p)=M1=M2, 即它的层是唯一的 f(x1)=f(x2)x1=x2\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}

solution2\textbf{solution2}
如果  f:XY 与  g:YX 满足 gf=eX,eX 为中性元, \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, }
则  g 称为 f 的左逆映射,  f 称为 g 的右逆映射, 可能存在多个单侧逆映射\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}

(x1,x2,,xn)fa(a,x1,xnxn) 注意到 a 的取值可以是任意的\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}

solution3\textbf{solution3}

f:XYg:YZ 是双射, 映射  gf:XZ 是双射, 并且 (gf)1=f1g1f: 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}

i)先证满射性, XfYgZ (gf)X=Z gf为满射\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}

ii)再证单射性, (gf)(x1)=(gf)(x2)g(f(x1))=g(f(x2)) g为单射, f(x1)=f(x2) f为单射, x1=x2\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}

iii)(f1g1)(gf)=f1(g1g)f=f1eYf =f1f=eX\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}

solution4\textbf{solution4}
对于任何映射  f:XY,xF(x,f(x)) 定义的映射\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:XX×Y 是单射F: X \to X \times Y\text{ 是单射}

f(x1)=f(x2)x1=x2 x1x2,f(x1)f(x2)(x1,f(x1))(x2,f(x2)) F(x1)=F(x2)x1=x2\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

01
极限为AA,AA 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

n>N1,x1V(A1) n>N2,x2V(A2) 不妨设  A1<A2,N1<N2 此时, 使得  A1+ϵ=A2ϵ ϵ=A1A22,更一般地, 让  ϵ=A1A22\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}

solution1\textbf{solution1}

ABxnynxn(Δyn)+yn(Δxn)+ΔxnΔyn 构造 ϵ3\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}

{Δxn<min(1,ε3)1Δyn<min(1,ε3)1ΔxnΔyn<ε3\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}

{xn<A+ΔxnA+1yn<B+ΔynB+1xnΔynϵ3Δynmin(1,ε3(A+1))这样所需要的 ϵ 已经构造完毕了\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}

solution2\textbf{solution2}

ABxnynxn+ΔxnynΔynxnyn=xnyn+ynΔxnxnyn+xnΔynyn2ynΔyn =xnΔyn+ynΔxnyn211δ(yn),δ(yn)=Δynyn\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}

构造证明的第一部分, 已知条件 xn<A+Δxn,yn>BΔyn, 先凑出\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}

11δ(yn)<?\frac{1}{1-\delta\left(y_{n}\right)} < ?

Δyn<B4yn>BΔyn>B2δ(yn)<1211δ(yn)<2\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}

接下来只要证明 xnΔynyn2<ε4,ynΔxnyn2<ε4 即可\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}

i)\textbf{i)}

xn<A+1,1yn2<(2B)2=4B2 此时构造出 Δyn<116εB2(A+1),Δyn<min{116εB2(A+1),B4}\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}

ii)\textbf{ii)}

1yn<2B,ΔxnεB8Δxnmin{εB8,1}\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}

solution3对于 ϵ>0,可以求出{n,m}>N,使得 xmxn<ϵ\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

i)基本数列是收敛数列\textbf{i)} \quad \text{基本数列是收敛数列}

xmxn<xmA+xnAifk>N,xkA<ε2xmxn<ε2+ε2=ε\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}

ii)收敛数列是基本数列\textbf{ii)} \quad \text{收敛数列是基本数列}

an:=infknxkbn:=supknxk an+1:=infkn+1xkbn+1:=supkn+1xk anan+1bn+1bn 根据闭区间套, [an,bn]有公共点 A\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}

[an,bn],{Axkbnanxnδinfknxk=anbn=supknxkxn+δ,δ=?\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}

bnan2δ<εδ=ε3 xnε3<xk<xn+ε3 Axk<ε\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}

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

ε>0xN{xn}sε<xNs\forall \varepsilon>0 \quad \exists x_{N} \in \left\{ x_n \right\} \quad s-\varepsilon<x_{N} \leqslant s

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

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

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