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

The title says it all: Is there a partition of $[0,1]$ into uncountably many dense uncountable subsets ?

## migrated from mathoverflow.netAug 9 '13 at 12:40

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• Have you tried replacing $[0,1]$ with $\mathbb{R}$? (You can then use cosets of subgroups...) – Alain Valette Aug 9 '13 at 11:01
• What does it mean that a set is uncountably dense? – Emil Jeřábek Aug 9 '13 at 12:05
• @EmilJeřábek I believe he meant to say "uncountably many dense, uncountable subsets". – Nick Peterson Aug 9 '13 at 13:25
• hi Anyone here can give some input on: [What is the Probability that a Knight stays on chessboard after N hops] : math.stackexchange.com/questions/464049/… – Arnab Dutta Aug 10 '13 at 10:43
• My 20 January 2003 answer to a similar question: Aww come on, let's be really greedy --> Partition the reals into $c = 2^{{\aleph}_{0}}$ many pairwise disjoint sets $\{P_{i}: i < c\}$ such that, for each $P_i$ and for each open interval $(a,b),$ the outer Lebesgue measure of $P_{i} \cap (a,b)$ is $b-a.$ Luzin and Sierpinski constructed such a collection back in 1917. For nicer sets, $F_{\sigma}$ in fact, we can get positive Hausdorff dimension intersction with every open interval. – Dave L. Renfro Jan 3 '14 at 20:47

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

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).

• The $U_x$ are not dense. – Andrés E. Caicedo Jan 8 '14 at 8:13
• You are right, I miss that point! – Emanuele Paolini Jan 8 '14 at 10:08

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.

Here is another solution: An $x\in[0,1]$ is irrational iff its continued fraction is infinite. Recall that the continued fraction of $x$, $$x=\frac1{a_0+\frac1{a_1+\frac1{\dots}}},$$ is obtained by letting $a_0$ be the integer part of $1/x$, then letting $a_1$ be the integer part of $\displaystyle \frac1x-a_0$, etc, so each $a_i$ is a positive integer.

Recall now that the set $(\mathbb Z^+)^{\mathbb N}$ of infinite sequences of positive integers is uncountable (in fact, $|(\mathbb Z^+)^{\mathbb N}|=|\mathbb R|$). Each such sequence has the form $(b_0,b_1,b_2,\dots)$ with each $b_i$ a positive integer.

Now, for your partition: Given $t\in(\mathbb Z^+)^{\mathbb N}$, put $x\in[0,1]$ in $A_t$ iff, with the $a_i$ as above, we have that $(a_0,a_2,a_4,\dots)=t$. Note that each $A_t$ is closed, uncountable, and they are pairwise disjoint. The union of all the $A_t$ is $[0,1]\setminus\mathbb Q$. Pick countably many $t$, and add an element of $[0,1]\cap\mathbb Q$ to each of these $A_t$. The resulting sets are again closed.

To further satisfy the requirement that the sets are dense, let me modify the construction somewhat: Say that two such sequences $(a_0,a_1,\dots)$ and $(b_0,b_1,\dots)$ are $E$-equivalent iff there are $n,m$ such that $b_{n+k}=a_{m+k}$ for $k=0,1,\dots$, that is, iff, by chopping off initial segments from each (possibly of different length) we obtain the same sequence. This is an equivalence relation on $(\mathbb Z^+)^{\mathbb N}$.

Noting that each equivalence class is countable, it follows from the fact that $|(\mathbb Z^+)^{\mathbb N}|=|\mathbb R|$ that there are as many equivalence classes as there are reals. Pick from each equivalence class a representative, and call $V$ the set of representatives so chosen, so $V\subseteq (\mathbb Z^+)^{\mathbb N}$ meets each equivalence class in exactly one point.

Now, for your partition: Given $t\in V$, put $x\in[0,1]$ in $B_t$ iff, with $$x=\frac1{a_0+\frac1{a_1+\frac1{\dots}}}$$ as above, we have that $(a_0,a_2,a_4,\dots)\mathrel{E} t$. Note that each $B_t$ is $F_\sigma$, uncountable, and dense, and again they are pairwise disjoint. The union of all the $B_t$ is again $[0,1]\setminus\mathbb Q$. Since $[0,1]\cap\mathbb Q$ is itself an $F_\sigma$ set, we can pick a $t_0\in V$ and add to $B_{t_0}$ the set $[0,1]\cap\mathbb Q$, and the resulting set is once again $F_\sigma$. This is of course optimal in the sense that we cannot obtain such a partition using closed sets, because of the density requirement.

A space $X$ is called maximally resolvable when it is a pairwise disjoint union of $\Delta(X)$ many dense subsets, where $\Delta(X)$ is the minimal cardinality of all non-empty open subsets of $X$.

In this paper the author proves in theorem 3 that a space $X$ with $\aleph_0 \le \chi(X) \le \Delta(X)$ is maximally resolvable. For the definition of $\chi(X)$ see wikipedia.

As $[0,1]$ has $\chi(X) = \aleph_0, \Delta(X) = \mathfrak{c}$ we know it is maximally resolvable, so it has a partition into $\mathfrak{c}$ many dense sets.