3 dense uncountable pairwise disjoint subsets of real line

Question

Can we find three dense uncountable pairwise disjoint subsets of $\mathbb{R}$? If so, what are these three sets?

I feel like it's not possible.

I was trying to counter it using Baire category and trying to construct such sets using the idea of construction of Cantor set (setting elements in trinary system). But I failed in both ways. I'm not even sure if answer would be affirmative or not.

Answer 1

Fix three disjoint dense subsets of $\mathbb{Q}$, call them $D_0, D_1$ and $D_2$.

Let $I_0=(0,1)\backslash \mathbb{Q}$, $I_1=(1,2)\backslash \mathbb{Q}$, $I_2=(2,3)\backslash \mathbb{Q}$ and set $J_k = I_k \cup D_k.$

Then, the sets $J_0, J_1$, and $J_2$ are disjoint, uncountable and dense.

Answer 2

Let $A$ be the set of all real numbers $x$ such that $|x|$ has a decimal expansion in which all but finitely many digits are in $\{1,2,3\}$.

Let $B$ be the set of all real numbers $x$ such that $|x|$ has a decimal expansion in which all but finitely many digits are in $\{4,5,6\}$.

Let $C$ be the set of all real numbers $x$ such that $|x|$ has a decimal expansion in which all but finitely many digits are in $\{7,8,9\}$.

Now $A$, $B$, and $C$ are three disjoint uncountable dense subsets of $\mathbb R$, and $D=\mathbb R\setminus(A\cup B\cup C)$ is a fourth one.

Answer 3

Hint for an uncountable partition of $\mathbb{R}$ into uncountable dense subsets.

Take first the binary fractions ( rationals with denominator a power of $2$) and leave them aside for now (a countable dense subset).

On the complement in $\mathbb{R}$ consider for each real number the part of the binary expansion after the dot $ \sum a_n/2^{n+1}$. For any $\alpha \in [0,1]$ let \begin{eqnarray}A_{\alpha}=\{ x \ \mid \ \lim \frac{\sum_{k=0}^n a_k}{n}= \alpha \} \end{eqnarray}

Now $A_{\alpha}$ are disjoint ( clear), uncountable ( not that hard to show), and dense. Let also $B$ formed by the binary fractions and the complement of the union of $A_{\alpha}$'s. We get our desired partition.