I looked in a couple of books focusing on set theory in the real line, but I have not seen the proof or disproof of the following question. It might be something easy just eluding me, but I wonder if the answer is no for reasons similar to why a continuous surjection from the Cantor Set $C$ onto $[0,1]$ is at least two-to-one.
The question is: Can the closed interval $I = [0,1]$ be partitioned into a collection of pairwise-disjoint copies of $C$? A weaker version would be, can an uncountable union of pairwise-disjoint Cantor Sets merely contain an interval? Can the union be equal to, or maybe just contain, $[0,1] \setminus \mathbb{Q}$?
Here, by a Cantor Set I mean any space homeomorphic to the standard ternary one. So, any collection of totally disconnected, perfect, compact sets in $I$.
Thanks!
Edit: In the same paper linked below, theorem 1.3 shows that $I \setminus \mathbb{Q}$ can be partitioned by Cantor Sets. In fact, it shows that the Cantor Set can also be partitioned by homeomorphic copies of the irrationals.
Best Answer
A positive answer (that the interval can be partitioned into pairwise disjoint Cantor sets) is given by Theorem 1.14 of Paul Bankston and Richard J. McGovern, Topological partitions, General Topology and its Applications 10 (1979), 215–229.
In fact any nonempty Polish space without isolated points can be partitioned into Cantor sets, according to the answer to this question:
Partitioning a metric space into Cantor sets