It is well known that if $P$ is a totally disconnected perfect compact subset of reals, then $P$ is homeomorphic to the Cantor set. My question is that whether $P$ always contain a non empty perfect subset of measure zero?
A more particular question is the following: Let $0<r<1.$ Similar to the Cantor set, the fat Cantor set $C_r$ is obtained from the unit interval $[0,1]$ by inductively removing middle third open intervals of size $r/3, r/9$ and so on. I am interested in the collection of $r\in (0,1)$ such that the set $C_r$ contains an uncountable null perfect subset.
The answer to your first question is yes.
Let $P$ be a subset of $\mathbb R$ which is homeomorphic to the Cantor set $C=\{0,1\}^\omega.\ $ Since $C$ is homeomorphic to $C\times C,$ it follows that $P$ can be partitioned into continuum many nonempty perfect sets, most of which will have Lebesgue measure zero.
Since every uncountable Borel set contains a homeomorph of the Cantor set, it follows that every uncountable Borel set contains a nonempty perfect set of Lebesgue measure zero.