【柯】代数学引论 第1章 §5.集合与映射

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\(Page.26\\1.\ f\omega=\omega_{1}\omega\dot{}_{1}\omega_{2}\omega\dot{}_{2}···\omega_{n}\omega\dot{}_{n}\\\quad f(f\omega)=\omega_{1}\omega_{1}\omega\dot{}_{1}\omega\dot{}_{1}\omega_{2}\omega_{2}\omega\dot{}_{2}\omega\dot{}_{2}···\omega_{n}\omega_{n}\omega\dot{}_{n}\omega\dot{}_{n}\\\quad 可知，在\ f(f\omega)\ 的长度>4的任意区间内会出现以下四种情况\left\{\begin{array}{l}\omega_{i}\omega\dot{}_{i}\omega\dot{}_{i}\omega_{i}\\\omega\dot{}_{i}\omega\dot{}_{i}\omega_{i}\omega_{i+1}\\\omega\dot{}_{i}\omega_{i}\omega_{i+1}\omega\dot{}_{i+1}\\\omega_{i}\omega_{i+1}\omega\dot{}_{i+1}\omega_{i+2}\end{array}\right.\\\quad 考虑当\omega_{i}中不存在++或--，即为以-+-+-型式错开。在以上四种情况中都会再次出现++或--\)
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\(2.\ 映射\ f\ 不存在右逆(f\ 不满射)\\\quad \left\{\begin{array}{l}由规则x \mapsto int[x]定义的映射\ g:\mathbb{N}\mapsto \mathbb{N}\\由规则x \mapsto int[x+1]-1定义的映射\ h:\mathbb{N}\mapsto \mathbb{N}\end{array}\right.\quad 都可作为\ f\ 的左逆\)
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\(3.\ (1)对\forall x\in S\cup T,皆有x\in S或x\in T\\\qquad 若x\in S,则f(x)\in f(S),即f(x)\in f(S)\cup f(T)\\\qquad 若x\in T,则f(x)\in f(T),即f(x)\in f(S)\cup f(T)\\\qquad 故f(S\cup T)\subset f(S)\cup f(T)\\~\\\qquad 对\forall y\in f(S)\cup f(T),皆有y\in f(S)或y\in f(T)\\\qquad 若y\in f(S)，又S\subset S\cup T,f(S)\subset f(S\cup T),则y\in f(S\cup T)\\\qquad 若y\in f(T)，又T\subset S\cup T,f(T)\subset f(S\cup T),则y\in f(S\cup T)\\\qquad 故f(S)\cup f(T)\subset f(S\cup T)\\~\\\qquad 综上，f(S\cup T)=f(S)\cup f(T)\\\qquad QED\\~\\\quad (2)对\forall x\in S\cap T,皆有x\in S且x\in T\\\qquad 当x\in S时，有f(x)\in f(S)\\\qquad 又当x\in T时，有f(x)\in f(T)\\\qquad 即f(x)\in f(S)\cap f(T)\\\qquad 故f(S\cap T)\subset f(S)\cap f(T)\\\qquad QED\\~\\\qquad e.g.\quad f(x)=x^{2}\quad S=\{0,1\}\quad T=\{-1,0\}\)???
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\(4.\ \textrm{C}_{n}^{0}+\textrm{C}_{n}^{1}+\textrm{C}_{n}^{2}+···+\textrm{C}_{n}^{n-1}+\textrm{C}_{n}^{n}=2^{n}\)
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\(5.\ 对于\forall b_{i},b_{j}\in Imf(b_{i}\neq b_{j})\\\quad 若\exists x\in f^{-1}(b_{i})\cap f^{-1}(b_{j})\\\quad 则有b_{i}=b_{j},故f^{-1}(b_{i})与f^{-1}(b_{j})互不相交\\~\\\quad 设\exists x\in X且x\notin f_{b}\\\quad 则有b_{x}=f(x),x=f^{-1}(b_{x})\\\quad 综上,集合X是互不相交的纤维的并\\\quad QED\)
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\(6.\ 对角线法，与无穷级数的Cauchy乘积类似\)

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\(7.\ 对\forall x\in S\Delta T,皆有x\in S\backslash T或x\in T\backslash S\\\quad 又S\backslash T\Leftrightarrow S\backslash (S\cap T),T\backslash S\Leftrightarrow T\backslash (S\cap T)\\\quad 故S\Delta T\subset (S\cup T)\backslash (S\cap T)\\~\\\quad同理(S\cup T)\backslash (S\cap T)\subset S\Delta T\\\quad 故S\Delta T=(S\cup T)\backslash (S\cap T)\\\quad QED\)