11
$\begingroup$

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

$\endgroup$

migrated from mathoverflow.net Aug 9 '13 at 12:40

This question came from our site for professional mathematicians.

  • 1
    $\begingroup$ Have you tried replacing $[0,1]$ with $\mathbb{R}$? (You can then use cosets of subgroups...) $\endgroup$ – Alain Valette Aug 9 '13 at 11:01
  • $\begingroup$ What does it mean that a set is uncountably dense? $\endgroup$ – Emil Jeřábek Aug 9 '13 at 12:05
  • $\begingroup$ @EmilJeřábek I believe he meant to say "uncountably many dense, uncountable subsets". $\endgroup$ – Nick Peterson Aug 9 '13 at 13:25
  • $\begingroup$ hi Anyone here can give some input on: [What is the Probability that a Knight stays on chessboard after N hops][1] [1]: math.stackexchange.com/questions/464049/… $\endgroup$ – Arnab Dutta Aug 10 '13 at 10:43
  • 1
    $\begingroup$ 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. $\endgroup$ – Dave L. Renfro Jan 3 '14 at 20:47
10
$\begingroup$

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

$\endgroup$
5
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ The $U_x$ are not dense. $\endgroup$ – Andrés E. Caicedo Jan 8 '14 at 8:13
  • $\begingroup$ You are right, I miss that point! $\endgroup$ – Emanuele Paolini Jan 8 '14 at 10:08
4
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.