We know that each perfect set can be written as a continuum many pairwise disjoint many perfect set. This will rely on the well know theorem which says:

Let $X$ be a nonempty perfect polish space. Then there exists an embedding of $C$ (cantor set) into $X.$

My question is how can I write a perfect set as a countable many pairwise disjoint perfect sets ? Does anyone have an idea?

Thank you in advance

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    $\begingroup$ @bof Do you mean $2^{\aleph_0}$ in your first comment? To write $[0,1]$ as a disjoint union of $\mathfrak c$ many perfect sets consider the Peano curve $[0,1]\to [0,1]^2$ and look at the preimages of the vertical segments. Surely they are $\mathfrak c$ many disjoint closed sets. To show that they are perfect note that you can split $[0,1]^2$ into $4^n$ squares each of which is mapped on a segment of length $1/4^n$ in $[0,1]$ by the inverse of the Peano curve, and then as $n\to\infty$ their diameters go to zero $\endgroup$ – Alessandro Codenotti Jun 30 at 9:10
  • $\begingroup$ @AlessandroCodenotti Thank you. I will have to think about your last sentence. I guess it depends on a specific construction of the curve? $\endgroup$ – bof Jun 30 at 9:25
  • $\begingroup$ @bof Sorry, I misremembered the fractal I wanted, looking at the images it's actually the Hilbert curve, if you look at the coloured gif on wikipedia you can see the $4^n$ squares I'm talking about after a few iterations. Also my comment about $2^{\aleph_0}$ is wrong, the ZFC result is that $[0,1]$ is not the union of less than $\mathfrak d$ disjoint closed sets. $\endgroup$ – Alessandro Codenotti Jun 30 at 9:31
  • $\begingroup$ @AlessandroCodenotti I think I see how to partition $[0,1]$ into $\frak c$ many disjoint perfect sets by transfinite induction. If my thinking is right it also shows how to partition $[0,1]\setminus S$ into $\frak c$ disjoint perfect sets whenever $|S|\lt\frak c$. $\endgroup$ – bof Jun 30 at 17:05
  • $\begingroup$ @AlessandroCodenotti I found another proof of the fact that $[0,1]$ can be partitioned into $\mathfrak c$ copies of the Cantor set $\mathbb C$ in Thearem 1.14 of Bankston & McGovern, Topological partitions, General Topology Appl, 10 (1979),215-229. First they partition $[0,1]$ into countably many copies of $\mathbb C$ and one copy of $\mathbb P$, the space of irrational numbers. Then $\mathbb P$ is easily partitioned into $\mathfrak c$ copies of $\mathbb C$ since $\mathbb P$ is homeomorphic to $\mathbb P\times\mathbb C$. $\endgroup$ – bof Jul 1 at 9:55

Theorem 1. The perfect set $[0,1]$ can not be written as the union of a pairwise disjoint family $\mathcal F$ of nonempty closed sets with $1\lt|\mathcal F|\le\aleph_0$.

Proof. Since $[0,1]$ is connected, we need only consider the case where $\mathcal F$ is countably infinite. Let $\mathcal F=\{F_1,F_2,F_3,\dots\}$. It is easy to see that there is a closed interval $I_1\subset[0,1]$ such that $I_1\cap F_1=\emptyset$ while $I_1\cap F_n\ne\emptyset$ for infinitely many $n$. Similarly there is a closed interval $I_2\subset I_1$ which is disjoint from $F_2$ but still meets infinitely many $F_n$. Continuing in this way we get a nested sequence of closed intervals whose intersection contains a point of $[0,1]$ which belongs to no $F_n$.

Theorem 2. If $X$ is a nonempty zero-dimensional metric space with no isolated points, then $X$ can be written as the union of $\aleph_0$ pairwise disjoint nonempty closed sets with no isolated points.

Proof. Choose a point $a\in X$. Construct a sequence of nonempty clopen sets $A_1,A_2,A_3,\dots$ converging to $a$; every neighborhood of $a$ contains all but finitely many of the $A_n$. Partition $X$ into the closed sets $A_2,A_4,A_6,\dots$ and $X\setminus(A_2\cup A_4\cup A_6\cup\cdots)$.

Corollary. If $X$ is a nowhere dense perfect subset of $\mathbb R$, then $X$ can be written as the union of $\aleph_0$ pairwise disjoint perfect sets.

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  • $\begingroup$ @bf , Thank you. But as you perfect might not be zero dimensional in general. How we can do the countable partition $\endgroup$ – 00GB Jun 30 at 13:31
  • $\begingroup$ @00GB bof's answer shows that a countable partition is impossible in some cases, such as the perfect set $[0,1]$ $\endgroup$ – Alessandro Codenotti Jun 30 at 13:37
  • $\begingroup$ @AlessandroCodenotti, you are right. My perfect set is algebraically independent and I know it can be written as countable pairwise disjoint of perfect sets . But how? $\endgroup$ – 00GB Jun 30 at 14:09
  • $\begingroup$ @00GB It sounds like you asked the wrong question. If you don't know how to write your perfect set as a countable union of disjoint perfect sets, what makes you so sure it can be done? $\endgroup$ – bof Jun 30 at 17:09
  • $\begingroup$ @00GB By the way, a totally disconnected subset of $]mathbb R$ is zero-dimensional. Is your perfect set zero-dimensional? $\endgroup$ – bof Jun 30 at 17:11

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