CH implies $\omega_1$ pairwise disjoint of perfect set This is an exercise in Fundamental of real analysis by James Foran. 
Prove that the continuum hypothesis implies that each perfect subset of $\mathbb R$ can be written as the union of $\omega_1$ pairwise disjoint non-empty perfect sets.
This is not homework but I like to see the construction. I do not know how to start. Also, if you know  book or paper that have some constructions like that please let me know. Thank you  in advance.        
 A: A subset $P$ of $\mathbb{R}$ is perfect if $P=P'$, where $P'$ is the set of all limit points of $P$. Hence, our construction will focus on to make every point of a constructed family of disjoint sets a limit point.
Let $\{x_\alpha\mid \alpha<\omega_1\}$ be an enumeration of all reals. We will construct $P_{\xi,\eta}$ by induction on $\langle \xi,\eta\rangle \in \omega_1\times\omega_1$, where $\omega_1\times\omega_1$ is equipped with the following order:
$$\langle \xi,\eta\rangle < \langle \xi',\eta'\rangle\text{ iff } [\max(\xi,\eta)<\max(\xi',\eta')]\text{ or }[\max(\xi,\eta)=\max(\xi',\eta') \text{ and } \xi<\xi']\\ \text{ or } [\max(\xi,\eta)=\max(\xi',\eta') ,\,\xi=\xi'\text{ and } \eta<\eta'].$$
We can see that $<$ is a well-order over $\omega_1\times \omega_1$. Moreover, every initial segment under this order is countable.
Then we can recursively construct a family of sets $\{P_{\xi\eta}\mid \xi,\eta<\omega_1\}$ satisfying the following conditions:


*

*$P_{\xi,0}\subseteq P_{\xi,1} \subseteq \cdots$,

*$P_{\xi,\eta}\cap P_{\xi',\eta}=\varnothing$ if $\xi\neq \xi'$, 

*Every point of $P_{\xi,\eta}$ is a limit point of $P_{\xi,\eta+1}$,

*$|P_{\xi,\eta}|\le\aleph_0$ for all $\xi,\eta<\omega_1$, and

*$\bigcup_{\xi,\eta}P_{\xi\eta} = \mathbb{R}.$
Now let $P_\xi=\bigcup_{\eta<\omega_1} P_{\xi\eta}$. Then $P_\xi$s from a partition of $\mathbb{R}$ whose members are perfect.

Here it is a construction of a desired family: 
Consider the case $\eta=0$. If $\xi=0$, set $P_{0,0}=\{x_0\}$. Now assume that $P_{\mu,\nu}$ is given for all $\langle\mu,\nu\rangle<\langle \xi,\eta\rangle$. If $\xi=0$, set 
$$P_{\xi,0} := \begin{cases}\{x_\eta\} &\text{if } x_\eta \notin \bigcup_{\langle\mu,\nu\rangle<\langle \xi,\eta\rangle}P_{\mu,\nu}, \\
\{\text{any other point of }\mathbb{R}-\bigcup_{\langle\mu,\nu\rangle<\langle \xi,\eta\rangle}\}&\text{otherwise.}\end{cases}$$
Now consider the case $\eta\neq 0$. Let us also assume that we have constructed $P_{\mu,\nu}$ for all $\langle\mu,\nu\rangle<\langle \xi,\eta\rangle$. Let $S$ be a collection of all isolated points of 
$$T=\bigcup \{P_{\xi,\nu} \mid \nu<\eta \text{ and } \langle \xi,\nu\rangle<\langle \xi,\eta\rangle\}.$$
Then $S$ is countable; so enumerate it as $\{y_i\mid i<\omega\}$. For each $i$, inductively, choose a sequence $\langle z^i_n\mid n<\omega\rangle$ converging to $y_i$ and also satisfies $z^i_n\notin \bigcup_{\langle \mu,\nu\rangle<\langle \xi,\eta\rangle} P_{\mu,\nu} \cup\{z^j_k\mid j<\omega,\,k<n\}$. Now set $P_{\xi,\eta}= T\cup \{z^i_n\mid i,n<\omega\}.$

My argument shows how $\mathbb{R}$ is partitioned into $\aleph_1$ perfect subsets under CH. However, the only properties of $\mathbb{R}$ used in the proof are


*

*$\mathbb{R}$ is uncountable, and

*no points of $\mathbb{R}$ are isolated points.
Therefore, the same argument works for any Polish spaces (i.e. complete metric space with a countable dense subset) with no isolated points. Especially, if $P\subseteq \mathbb{R}$ is a perfect subset, then $P$ is closed (so $P$ is a complete metric space) and has no isolated points. Therefore, my argument effectively shows every perfect subset of $\mathbb{R}$ is partitioned by $\aleph_1$ perfect subsets.
