证明或否证: 存在 [0,1] 上取值为无理数的严格增实函数

elim

题：证明或否证: 存在\(\,[0,1]\,\)上取值为无理数的严格增实函数.

这个题目大致要求数学专业本科两年级以上程度...... 在罗马尼亚一个数学网站上看到的. 还没有人贴出解答.

王守恩

题目看不懂。
与下面的题目有关系吗？
可以把（）里的3个数是看作1个数。
\((\frac{1}{1}-\frac{1}{2}+\frac{1}{3})+(\frac{1}{5}-\frac{1}{4}+\frac{1}{7})+(\frac{1}{9}-\frac{1}{6}+\frac{1}{11})+(\frac{1}{13}-\frac{1}{8}+\frac{1}{15})+......=\frac{\ln(8)}{2}\)

elim

王守恩 发表于 2020-9-17 16:34
题目看不懂。
与下面的题目有关系吗？
可以把（）里的3个数是看作1个数。

判断是否存在一个定义域是 [0,1], 取值均为无理数的严格增函数.

小fisher

没有权限发链接，只好截图发上来了

elim

回 小fisher 先生, \(\mathbb{Q}\) 在\(\,\mathbb{R}\,\)中稠密直观上很明显, 严格的证明确实要用到实数域的阿基米德性.
但是我们考虑的严格增加函数 f 不必是连续的, 所以 f([0,1]) 不必含任何区间.  

elim

题：证明或否证:存在\([0,1]\)上取值为无理数的严格增实函数.
解：任取定二无理数\(\,\alpha,\color{grey}{\scriptsize(<)\,}\beta,\)双射\(\,r:\mathbb{N}^+\hspace{-1pt}{\small\longleftrightarrow}(\mathbb{Q}\cap(\alpha,\beta))\),令
\(\qquad\,d=\beta-\alpha,\;d_n=d(\frac{1}{2}+\frac{1}{2^{n+1}}),\;f_0(x)=(1-x)\alpha+x\beta,\)
\(\qquad{\small\ E=\{p+\lfloor q\alpha\rfloor: p,q\in\mathbb{Z},q>0.\}},\;x_0=0=x'_1,\;x_1=f_0^{-1}(r_1)\).
\(\qquad\)由\(\small\,E\subset(\mathbb{R}\cap\mathbb{Q}^c)\subset\overline{E}=\mathbb{R}\)及\(\small\,E\,\)可数\((\exists u:\mathbb{N}^+\hspace{-2pt}{\small\longleftrightarrow}E)\),取\(\,m_k=\)
\(\qquad\,\min\{m\mid u(m)\in{\small[\alpha+(x_k-x'_k)d_k,\alpha+(x_k-x'_k)d_{k-1})\cap E}\}\,\)
\(\qquad\)令\(\;\;v_k=u(m_k),\;w_k(x)=\large\frac{(x_k-x)f_{k-1}(x'_k+)+(x-x'_k)v_k}{x_k-x'_k},\)
\(\qquad\quad\;\;f_1=f_0+\mathbf{1}_{(x_k',x_k]}(w_1-f_0)\;\;(\mathbf{1}_S\,\)是集\(S\)上的特征函数\()\).
\(\qquad\)易见\(\;r_1\not\in f_1([0,1]),\;{\small\underline{D}f_1=}\inf\big\{\frac{f_1(t)-f(s)}{t-s}\mid 0\le s< t\le 1\big\}\ge d_1.\)
\(\qquad\)设\(\,x_0,x_1,\ldots,x_k,\;x'_1,\ldots,x'_k,\;v_1,\ldots,v_k,\;f_1,\ldots,f_k\) 已取定,
\(\qquad\)使\(\;f_{m-1}(x_j)=f_m(x_j)\;(j<m\le k),\;\;\underline{D}f_k\ge d_k\),
\(\qquad\)取\(\,x_{k+1}=f_k^{-1}(r_{k+1}),\,x'_{k+1}=\max\{x_j\mid x_j< x_{k+1},\,j\le k\},\)
\(\qquad\,v_{k+1}=u(m_{k+1}),\quad f_{k+1}=f_k+\mathbf{1}_{(x'_{k+1},x_{k+1}]}(w_{k+1}-f_k),\)易见
\(\qquad\;f_{m-1}(x_j)=f_m(x_j)\;{\small(j<m\le k+1),\;\;\underline{D}f_{k+1}\ge\,}d_{k+1}\underset{\,}{,}\)且对降极限
\(\qquad\,f={\displaystyle\lim_{k\to\infty}f_k}\,\)有\(\,\underline{D}f\ge d/2>0,\;f([0,1])\subset[\alpha,\beta]\cap\mathbb{Q}^c.\) 即\(\,f\,\)是
\(\qquad\,[0,1]\,\)上严格增,取无理数值.\(\quad\small\square\)

luyuanhong

楼上 elim 的帖子很好！已收藏。

elim

问题其实不需要太多技巧, 却需要较强的基础. 其理论重要性超过实用性, 咱们国家的短板是基础理论.

题： 证明或否证:存在\([0,1]\)上取值为无理数的严格增实函数.
解： 对满射\(\,\sigma：\small\mathbb{N}^+\to A,\;A\cap B\ne\varnothing,\,\)定义\(\,\underline{\sigma}{\small(B)=}\sigma(\min\sigma^{-1}{\small(A\cap B)})\).
\(\qquad\)任取定二无理数\(\,\alpha,\color{grey}{\scriptsize(<)\,}\beta,\;d=\beta-\alpha,\;d_n=d(\frac{1}{2}+\frac{1}{2^{n+1}})\)
\(\qquad\)令\(\,{\small\ E=\{p+q\alpha-\lfloor q\alpha\rfloor: p,q\in\mathbb{Z},\,q>0\}}.\;\)则\(\small\,(\overline{E}\supset\mathbb{R}\cap\mathbb{Q}^c\supset)E\,\)可数.
\(\qquad\)记\(\,I=[0,1],\;\)取定双射\(\,r:{\small\mathbb{N}^+\to(\mathbb{Q}\cap(\alpha,\beta))},u:\small\mathbb{N}^+\to E.\)
\(\qquad\)令\(\;f_0(x)=(1-x)\alpha+x\beta,\;x_0=0=x'_1,\;x_1=f_0^{-1}(r_1)\).
\((^*)\quad\)已知\(\;f_{k-1},\;x_k,{\scriptsize\color{grey}{(>)}}x'_k,\,\)定义
\(\qquad\qquad v_k=\underline{u}([\alpha+(x_k-x'_k)d_k,\alpha+(x_k-x'_k)d_{k-1}))\)
\(\qquad\qquad \omega_k(x)=\large\frac{(x_k-x)f_{k-1}(x'_k+)+(x-x'_k)v_k}{x_k-x'_k},\)
\(\qquad\qquad f_k=f_{k-1}+(\omega_k-f_0)\mathbf{1}_{(x_k',x_k]}\;\;(\mathbf{1}_S\,\)是集\(S\)上的特征函数\()\).
\((\dagger)\quad{\small\underline{D}}\varphi:=\inf\{\frac{|\varphi(t)-\varphi(s)|}{t-s}\mid s,t\in{\small\text{Domain}}(\varphi),\,s< t\}\)
\(\qquad\)易见\(\;r_1\not\in f_1([0,1]),\;{\small\underline{D}f_1=}\inf\big\{\frac{f_1(t)-f(s)}{t-s}\mid 0\le s< t\le 1\big\}\ge d_1.\)
\(\qquad\)设\(\,x_0,x_1,\ldots,x_k,\;x'_1,\ldots,x'_k,\;v_1,\ldots,v_k,\;f_1,\ldots,f_k\) 已取定,
\(\qquad\)使\(\;f_{m-1}(x_j)=f_m(x_j)\;(j<m\le k),\;\;{\small\underline{D}}f_k\ge d_k\),
\(\qquad\)取\(\,x_{k+1}=f_k^{-1}(\underline{r}(f_k(I))),\,x'_{k+1}=\max\{x_j\mid x_j< x_{k+1},\,j\le k\},\)
\(\qquad\)则由\((^*),\;[\alpha,\beta]-\{r_1,\ldots,r_{m-1}\}\supset f_{m-1}(I)-{r_m}\supset f_m(I)\)
\(\qquad\qquad\qquad f_{m-1}(x_j)=f_m(x_j),\quad{\small\underline{D}}f_{k+1}\ge\,d_{k+1}\;{\small(j<m\le k+1)}\)
\(\qquad\)故对降极限\(\;f={\displaystyle\lim_{k\to\infty}f_k}\,\)有\(\,{\small\underline{D}}f\ge d/2>0,\;f\small([0,1])\subset[\alpha,\beta]\cap\mathbb{Q}^c.\)
\(\qquad\)即\(\,f\,\)是\(\,[0,1]\,\)上严格增,取无理数值.\(\quad\small\square\)

luyuanhong

楼上 elim 的帖子很好！已收藏。

王守恩

王守恩 发表于 2020-9-18 07:34
题目看不懂。
与下面的题目有关系吗？

\(\Table\bigg[N\bigg[\displaystyle\sum_{k=1}^{9}\frac{1}{k^{(n+1)/n}}\bigg], (n, 1, 9)\bigg]\)
{1.53977, 1.96371, 2.18474, 2.31704, 2.40462, 2.46674, 2.51305, 2.54889, 2.57744}
\(\Table\bigg[N\bigg[\displaystyle\sum_{k=1}^{99}\frac{1}{k^{(n+1)/n}}\bigg], (n, 1, 9)\bigg]\)
{1.63488, 2.41187, 2.95353, 3.32862, 3.59905, 3.80194, 3.95929, 4.08469, 4.18688}
\(\Table\bigg[N\bigg[\displaystyle\sum_{k=1}^{999}\frac{1}{k^{(n+1)/n}}\bigg], (n, 1, 9)\bigg]\)
{1.64393, 2.54911, 3.30089, 3.88371, 4.33551, 4.69169, 4.97802, 5.21246, 5.40758}
\(\Table\bigg[N\bigg[\displaystyle\sum_{k=1}^{9999}\frac{1}{k^{(n+1)/n}}\bigg], (n, 1, 9)\bigg]\)
{1.64483, 2.59237, 3.46169, 4.19511, 4.79913, 5.29654, 5.70962, 6.05640, 6.35080}
\(\Table\bigg[N\bigg[\displaystyle\sum_{k=1}^{99999}\frac{1}{k^{(n+1)/n}}\bigg], (n, 1, 9)\bigg]\)
{1.64492, 2.60605, 3.53630, 4.37018, 5.09158, 5.70854, 6.23603, 6.68914, 7.08094}
\(\Table\bigg[N\bigg[\displaystyle\sum_{k=1}^{999999}\frac{1}{k^{(n+1)/n}}\bigg], (n, 1, 9)\bigg]\)
{1.64493, 2.61038, 3.57094, 4.46862, 5.27610, 5.98922, 6.61487, 7.16362, 7.64625}
\(\Table\bigg[N\bigg[\displaystyle\sum_{k=1}^{\infty}\frac{1}{k^{(n+1)/n}}\bigg], (n, 1, 9)\bigg]\)
{1.64493, 2.61238, 3.60094, 4.59511, 5.59158, 6.58922, 7.58752, 8.58624, 9.58525}
\(\Table\bigg[\displaystyle\sum_{k=1}^{\infty}\frac{1}{k^{(n+1)/n}}, (n, 1, 9)\bigg]\)
{\[pi]^2/6, Zeta[3/2], Zeta[4/3], Zeta[5/4], Zeta[6/5], Zeta[7/6], Zeta[8/7], Zeta[9/8], Zeta[10/9]}