A partition of the unit interval into uncountably many dense uncountable subsets

Solution 1:

This is a nice problem but I think this is not a problem for MO.

Anyway, the coset trick mentioned by @Alain Valette is nice.

As another way to approach a solution, consider the function $f : [0,1]\longrightarrow \Bbb{R}$ with $f(x) = \limsup_n \frac{x_1+x_2+\cdots+x_n}{n}$ where $0.x_1x_2\cdots$ is the non-terminating binary expansion of $x$. Then it is not hard to show that the family $\lbrace f^{-1}(\lbrace r \rbrace) \; | \; r \in \Bbb{R}\rbrace$ is a partition of $[0,1]$ into uncountably many dense uncountable subsets.

You may want to look at HERE

Solution 2:

Yes. Of course $[0,1]^2$ can be written as $\bigcup_{x\in [0,1]} \{x\}\times[0,1]$ and there is a bijection between $[0,1]^2$ and $[0,1]$.

added. An explixit solution is the following. Define the set $U_x$ as the set of all numbers $y\in[0,1]$ such that if you write $y$ in binary and remove the digits in odd position, you get the digits of $x$.

added bis Unfortunately the $U_x$ defined above are not dense (they are Cantor like).

Solution 3:

Define an equivalence relation $\sim$ on $\Bbb R$ by $x\sim y$ iff $x-y\in\Bbb Q$. (Check that this is an equivalence relation.) Show that the $\sim$-equivalence class of a real number $x$ is $x+\Bbb Q=\{x+q:q\in\Bbb Q\}$. Note that each $x+\Bbb Q\in\mathscr{C}$ is dense in $\Bbb R$. Let $\mathscr{C}=\Bbb R/\sim$ be the set of $\sim$-equivalence classes. Each $x+\Bbb Q\in\mathscr{C}$ is countable, so $|\mathscr{C}|=|\Bbb R|=|\Bbb R^2|$. Thus, we can index $\mathscr{C}$ by points in the plane: $\mathscr{C}=\{C_{\langle x,y\rangle}:\langle x,y\rangle\in\Bbb R^2\}$, where $C_{\langle x,y\rangle}\ne C_{\langle u,v\rangle}$ if $\langle x,y\rangle\ne\langle u,v\rangle$.

For each $x\in\Bbb R$ let $S_x=\bigcup_{y\in\Bbb R}C_{\langle x,y\rangle}$; clearly $S_x$ is dense in $\Bbb R$, and $\{S_x:x\in\Bbb R\}$ is therefore an uncountable partition of $\Bbb R$ into uncountable dense sets. Finally, let $\varphi:\Bbb R\to(0,1)$ be any surjective homeomorphism, e.g., $\varphi(x)=\frac1\pi\tan^{-1}x+\frac12$, and let

$$D_x=\begin{cases} \varphi[S_x],&\text{if }x\ne 0\\ \varphi[S_0]\cup\{0,1\},&\text{if }x=0\;; \end{cases}$$

it's easy to check that $\{D_x:x\in\Bbb R\}$ is an uncountable partition of $[0,1]$ into uncountable dense subsets.