$\Bbb R^n∖\Bbb Q^n$和$(\Bbb R∖\Bbb Q)^n$的连通性

hbghlyj

James Dugundji - Topology (1978)
page 117 Section 2 Question 5
a)和c)是相同的,不妨设$x_n\in\Bbb R∖\Bbb Q$,只需要把$(π,0,0,\cdots,0)$用直线段连接到$(π,x_2,0,\cdots,0)$再连接到$(π,x_2,x_3,\cdots,0)$⋯连接到$(π,x_2,x_3,\cdots,x_n)$最后连接到$(x_1,x_2,x_3,\cdots,x_n)$即可把$(x_1,\cdots,x_n)$连接到$(π,0,\cdots,0)$,每段都是具有一个无理数坐标的,所以每段包含于$\Bbb R^n∖\Bbb Q^n$,所以整个路径包含于$\Bbb R^n∖\Bbb Q^n$,证毕.

b) 如何证明呢?

这里的证明不对: The fact that $\mathbb{R}^2$ is connected tells us that $U' \cap V'$ is not empty. 这句话不对
而且假如它的证明是对的话, 按照下面就能证明a)是不连通的了($n=2$):

$A$ is the set of points with one coordinate rational and the other one irrational. Suppose that $A=U∪V$ where $U,V$ are disjoint open sets in $A$.
By Lemma 4.5.1. there are open sets $U',V'$ in $ℝ^2$ such that $U = U'∩A$ and $V = V'∩A$. The fact that $ℝ^2$ is connected tells us that $U'∩V'≠∅$. Take $x=(x_1,x_2)∈U'∩V'$.
$U',V'$ are open, so $U'∩V'$ is open, so $∃ϵ>0$ such that $B(x,ϵ)⊂U'∩V'$, so $B(x,ϵ)∩A=∅$.
[All the above are same, but the below will be slightly different:]
$U'∩V'∩A=(U'∩A)∩(V'∩A)=U∩V=∅$, so $x∉A$, so either $x_1,x_2∉ℚ$ or $x_1,x_2∈ℚ$.
If $x_1,x_2∉ℚ$, take $q∈(x_1-ϵ,x_1)∩ℚ$; if $x_1,x_2∈ℚ$, take $q∈(x_1-ϵ,x_1)∖ℚ$.
Then $y=(q,x_2)∈A$, and $d(x,y)=x_1-q< ϵ$, so $y∈B(x,ϵ)∩A$, contradicting with $B(x,ϵ)∩A=∅$, so $A$ is connected.

Czhang271828

那就是证明 $(\mathbb R\setminus \mathbb Q)^n$ 在 $\mathbb R^n$ 的通常拓扑下不连通.

直接根据定义来, 记 $\mathbb H_+:=\mathbb R^{n-1}\times \mathbb R_{>0}$ 为右半平面, $\mathbb H_-:=\mathbb R^{n-1}\times \mathbb R_{<0}$ 为左半平面, 则 $\mathbb H_+$ 与 $\mathbb H_-$ 均是 $\mathbb R^n$ 中的开集. 那么根据子空间拓扑定义, $(\mathbb R\setminus \mathbb Q)^n\cap \mathbb H_+$ 与 $(\mathbb R\setminus \mathbb Q)^n\cap \mathbb H_-$ 都是子空间 $(\mathbb R\setminus \mathbb Q)^n$ 中的开集.

显然 $(\mathbb R\setminus \mathbb Q)^n$ 是上述两个开集的无交并, 根据定义知不连通.

hbghlyj

整理了一下:

a) Let $A⊆ℝ^2$ be the set of all points with at least one rational coordinate. $A$ is connected.
b) Let $A⊆ℝ^2$ be the set of all points with at least one irrational coordinate. $A$ is connected.
c) Let $A⊆ℝ^2$ be the set of all points with exactly one rational coordinate. $A$ is disconnected.
d) Let $A⊆ℝ^2$ be the set of all points with two rational coordinates. $A$ is disconnected.
e) Let $A⊆ℝ^2$ be the set of all points with two irrational coordinates. $A$ is disconnected.
Proof.
a) For any $(x,y)∈A$, wlog let $x$ be rational. Consider the polyline Γ from $(0,0)$ to $(x,0)$ to $(x,y)$. Any point on Γ has a rational coordinate, so $Γ⊂A$, so $A$ is a star domain, so $A$ is path connected, so $A$ is connected.
b) For any $(x,y)∈A$, wlog let $x$ be irrational. Consider the polyline Γ from $(0,π)$ to $(x,π)$ to $(x,y)$. Any point on Γ has an irrational coordinate, so $Γ⊂A$, so $A$ is connected.
c) Let $U=\{(x,y)∈ℝ^2∣x>y\}$ and $V=\{(x,y)∈ℝ^2∣x< y\}$. No element of $A$ lies on the line $x=y$, so $A⊂U∪V$. Since $U$ and $V$ are open sets, and $U∩V∩A=∅$, by lemma 7.1.3 $A$ is disconnected.
d) Let $U=\{(x,y)∈ℝ^2∣x>π\}$ and $V=\{(x,y)∈ℝ^2∣x< π\}$. No element of $A$ lies on the line $x=π$, so $A⊂U∪V$. Since $U$ and $V$ are open sets, and $U∩V∩A=∅$, by lemma 7.1.3 $A$ is disconnected.
e) Let $U=\{(x,y)∈ℝ^2∣x>0\}$ and $V=\{(x,y)∈ℝ^2∣x< 0\}$. No element of $A$ lies on the line $x=0$, so $A⊂U∪V$. Since $U$ and $V$ are open sets, and $U∩V∩A=∅$, by lemma 7.1.3 $A$ is disconnected.

还有一个问题:
$(\mathbb R\setminus \mathbb Q)^n∪\mathbb Q^n$连通吗

Czhang271828

hbghlyj 发表于 2022-11-3 22:18
整理了一下:

a) Let $A⊆ℝ^2$ be the set of all points with at least one rational coordinate. $A$ is ...

$n=1$ 显然成立. 记集合 $T_n:=(\mathbb R\setminus \mathbb Q)^n\cup \mathbb Q^n$.

$n\neq 1$ 时, 经过原点的有理直线族 $\mathcal R_n:=\{(r_1t,r_2t,\ldots, r_nt),t\in \mathbb R\mid \prod r_i\neq 0\}$ 在 $\mathbb R^n$ 中稠密, 这族曲线的并刚好是 $T_n$ 去掉坐标轴($n-1$ 维超平面). 假设存在 $\mathbb R^n$ 中开集 $U_1$ 与 $U_2$, 使得 $(U_1\cap T_n, U_2\cap T_n)$ 是一个 $T_n$ 的划分. 从测度论的显然事实得知一定存在一条直线 $l\in \mathcal R_n$, 使得其同时属于 $U_1\cap T_n$ 和 $U_2\cap T_n$. 从而 $(U_1\cap l,U_2\cap l)$ 是 $l$ 的一个划分, 矛盾!