Is $[0,1]$ the union of $2^{\aleph_0}$ perfect sets which are pairwise disjoint?

I need represent [0,1] as the union of $2^{\aleph_0}$ perfect sets which are pairwise disjoint.

I have thought about removing open disjoint sets but the number of open sets I get is countable. Thanks.

Solution 1:

Yes, one can do this. The idea is to take a suitable space-filling curve $f: [0,1]\to [0,1]^2$ and take as the partition of $I$ the preimages of, say, vertical lines under $f$. The preimages will be clearly compact, disjoint, but one needs to work (by taking a suitable function $f$) to ensure their perfectness. For the details, see:

Edgar A. Cohen Jr., A Solution of a Problem of I. P. Natanson Concerning the Decomposition of an Interval into Disjoint Perfect Sets, Advances in Pure Mathematics, Vol.4 (2014) No.5, 189-193.