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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.

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  • $\begingroup$ Not sure yet about the first question. The second question is trivial: It's clear that every $C_r$ has measure zero. (To get a Cantor set of positive measure you need to take a variable ratio at each stage...) $\endgroup$
    – David C. Ullrich
    Jun 5, 2018 at 14:59
  • $\begingroup$ Oops. Yes, your $C_r$ has positive measure - I read the definition wrong. $\endgroup$
    – David C. Ullrich
    Jun 5, 2018 at 16:33

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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.

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