命题5.4.14 给定任意两个实数\(x< y\),我们能够找到一个有理数\(q\)使得\(x<q<y\)。
证明:
设\(x=LIM_{n \to \infty}a_n,y=LIM_{n \to \infty}b_n\)
\(\because x<y\),由定义5.4.6(实数的排序)、定义5.4.3(实数的正负)得
\(y-x=LIM_{n \to \infty}(b_n-a_n)\)为正\(=LIM_{n \to \infty}c_n\)
其中\({(c_n)^{\infty}_{n=1}}\)为正远离\(0\)的柯西序列,${\forall}n \geq 1,c_n \geq c > 0,c \in \mathbb{Q} $
由定义5.3.1(实数的相等)得\((b_n-a_n)^{\infty}_{n=1}\)与\((c_n)^{\infty}_{n=1}\)等价,即
\({\forall}\varepsilon_1 > 0,{\exists}N_1 \in \mathbb{N},s.t. \vert b_n-a_n-c_n\vert \leq \varepsilon_1\)对\({\forall}n \geq N_1\)成立
由定义5.1.8(柯西序列)得
\({\forall}\varepsilon_2 > 0,{\exists}N_2 \in \mathbb{N},s.t. \vert a_n-a_{N_2}\vert \leq \varepsilon_2\)对\({\forall}n \geq N_2\)成立
\({\forall}\varepsilon_3 > 0,{\exists}N_3 \in \mathbb{N},s.t. \vert b_n-b_{N_3}\vert \leq \varepsilon_3\)对\({\forall}n \geq N_3\)成立
令\(\varepsilon_1=\varepsilon_2=\varepsilon_3=\frac{c}{4}\),\(N=max\{N_1,N_2,N_3\}\),对\({\forall}n \geq N\)
\(\vert b_n-a_n-c_n\vert \leq \frac{c}{4}\Rightarrow c-\frac{c}{4}=\frac{3c}{4} \leq c_n-\frac{c}{4} \leq b_n-a_n \leq c_n+\frac{c}{4} \Rightarrow \frac{3c}{4} \leq c_N-\frac{c}{4} \leq b_N-a_N \leq c_n+\frac{c}{4}\)
\(\vert b_n-b_N\vert \leq \frac{c}{4}\Rightarrow b_N-\frac{c}{4} \leq b_n \leq b_N+\frac{c}{4}\)
\(\vert a_n-a_N\vert \leq \frac{c}{4}\Rightarrow a_N-\frac{c}{4} \leq a_n \leq a_N+\frac{c}{4}\)
取\(q=\frac{a_N+b_N}{2}\)
\(b_n-q=b_n-\frac{a_N+b_N}{2} \geq b_N-\frac{c}{4}-\frac{a_N+b_N}{2}=\frac{b_N-a_N}{2}-\frac{c}{4} \geq \frac{3c}{8}-\frac{c}{4}=\frac{c}{8}>0\),\(n \geq N\)
\(q-a_n=\frac{a_N+b_N}{2}-a_n \geq \frac{a_N+b_N}{2}-a_N-\frac{c}{4}=\frac{b_N-a_N}{2}-\frac{c}{4} \geq \frac{3c}{8}-\frac{c}{4}=\frac{c}{8}>0\),\(n \geq N\)
令\((D_n)^{\infty}_{n=1}=\begin{equation} \left\{ \begin{array}{c} \frac{c}{8},n<N \\ b_n,n \geq N \\ \end{array} \right. \end{equation}\)与\((b_n-q)^{\infty}_{n=1}\)等价,\(y-q=LIM_{n \to \infty}(b_n-q)=LIM_{n \to \infty}D_n\)
\(\because {\forall}n \geq 1,D_n \geq \frac{c}{8}\),\(\therefore (D)^{\infty}_{n=1}\)为正远离\(0\)的柯西序列,\(y-q=LIM_{n \to \infty}D_n\)为正,即\(y>q\)
同理令\((E_n)^{\infty}_{n=1}=\begin{equation} \left\{ \begin{array}{c} \frac{c}{8},n<N \\ a_n,n \geq N \\ \end{array} \right. \end{equation}\)与\((q-a_n)^{\infty}_{n=1}\)等价,\(q-x=LIM_{n \to \infty}(q-a_n)=LIM_{n \to \infty}E_n\)
又\(\because {\forall}n \geq 1,E_n \geq \frac{c}{8}(E_n)^{\infty}_{n=1}\)为正远离0的柯西序列,\(\therefore q-x=LIM_{n \to \infty}E_n\)为正,即\(q>x\)
综上,对任意两个实数\(x<y\),总存在有理数\(q\)使得\(x<q<y\)
原文地址:https://www.cnblogs.com/puyiniao/p/11104783.html