My thesis is related with the Cantor sets. I was reading a lot of papers, blogs, etc, in order to look for the mean properties of these sets.
In one blog a read a proposition.
''Every perfect set contains a Cantor set''.
This proposition was without proof, so I wonder if this proposition is true or if you have any ideas of how to prove or papers or something.
All your ideas are very welcome!!
On
If $P$ is a perfect subset of the real numbers, show that there exist $a < b < c < d$ such that $P \cap [a,b]$ and $P \cap [c,d]$ are both perfect. Let $P_1 = (P \cap [a,b]) \cup (P \cap [c,d])$. Repeat this process with each of $P \cap [a,b]$ and $P \cap [c,d]$, obtaining $P_2 \subset P_1$ consisting of intersections of $P$ with 4 intervals, $P_3 \subset P_2$ consisting of intersections of $P$ with 8 intervals, .... and making sure the lengths of the intervals approach $0$. The intersection of all the $P_i$ will be a Cantor set.
I doubt you can prove this in the most general case. Take any finite set $S$ with the indiscrete topology. Then $S$ will be a perfect subset, indeed a perfect space. Now define a Cantor Set to be a set that is topologically isomorphic to the usual construction of removing thirds of the real line. Every Cantor set must then be uncountable, so there can be no Cantor set inside of $S$, since $S$ is finite.
Edit: Just re-read the title, but you can also do the same thing with a finite or countable subset of the reals if you allow the indiscrete topology. So unless you mean also the usual Euclidean/open ball topology on the reals, the general claim will also fail. But I think you should be able to prove it assuming that topology.