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
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 disjoint 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.