TITLE: Partition $\Bbb{R}$ into a family of sets each one homeomorphic to the Cantor set
QUESTION [9 upvotes]: It is known that there is no (nontrivial) partition of $\Bbb{R}$ into a countable number of closed set. But is there a partition of $\Bbb{R}$ into sets, each one homeomorphic to the cantor ternary set?

REPLY [18 votes]: Let $f$ be the x-coordinate of Hilbert's space-filling curve,
whose graph is shown here:

Then the sets $\{f^{-1}(t)\}_{t\in [0,1]}$
form a partition of the interval [0,1] into Cantor sets.
An easy variation of the above construction produces a partition of the reals (take the point-preimages of $F$, where $F:\mathbb R\to \mathbb R$ is the periodic extension given by $F(t):=\lfloor t\rfloor+f(t-\lfloor t\rfloor)$).