TITLE: Can we find uncountably many disjoint dense measurable uncountable subsets of $[0,1]$?
QUESTION [10 upvotes]: Can we find uncountably many disjoint dense measurable uncountable subsets of $[0,1]$? Obviously we may as well assume all the subsets have measure $0$. If I didn't specify the subsets were uncountable, we could easily find uncountably many dense disjoint measurable subsets $X_\alpha$ (all countable) by taking a Hamel basis $H$ of $\mathbb{R}$ as a vector space over $\mathbb{Q}$, where $H$ contains a non-zero rational number, and then set $X_\alpha = (\mathbb{Q} + \alpha) \cap [0,1]$ for every $\alpha \in H$.
But what about if we want uncountable measurable dense subsets? I think it should still be that there are uncountably many that can be found that are disjoint, according to my intuition, but I don't know an "easy" way to show it like I can in the case of countable dense subsets.

REPLY [4 votes]: It's enough to do this for $\mathbb R,$ since $(0,1)$ and $\mathbb R$ are homeomorphic. Let $c$ denote the cardinality of $\mathbb R.$
Claim: There are $c$ many disjoint countable dense subsets of $\mathbb R.$
Proof: Let $K$ be the classical Cantor set. By "a Cantor set" I'll mean any set of the form $a+bK,$ where $a\in \mathbb R$ and $b>0.$
Let $I_1,I_2,\dots$ be the open intervals with rational end points. Then we can inductively choose disjoint Cantor sets $K_n\subset I_n$ for $n=1,2,\dots.$ (This is  pretty straightforward. I'll leave it for now; ask questions if you like.)
For each $n,$ write $K_n = a_n + b_nK,$ and let  $f_n: K \to K_n$ be the map $t\to a_n +b_nt.$ For $t\in K,$ define the sets $D_t = \{f_n(t): n=1,2,\dots \}.$ Then the sets $D_t$ are disjoint. Each $D_t$ is dense in $\mathbb R,$ simply because $D_t$ contains a point in each of the intervals $I_n.$ Because $K$ has cardinality $c,$ the claim follows.
To prove the full result, observe that there is a continuous surjective map $g:K\to [0,1]\times [0,1].$ (For example $g =P\circ C,$ where $P$ is the Peano space filling curve, and $C$ is the Cantor function.) For $t\in K,$ let $K_t = \{t\} \times [0,1].$ Then $\{ g^{-1}(K_t): t \in K\}$ is a disjoint collection of $c$ many sets, each of which is compact and of cardinality $c.$
Finally, consider the sets $D_t\cap (\mathbb R\setminus K) \cup g^{-1}(K_t)$ for $t\in K.$ These sets are disjoint, and each such set is dense in $\mathbb R$ and has cardinality $c.$ Since there are $c$ many such sets, we're done.