Every uncountable closed set contains a perfect subset.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
For $A \subseteq \Bbb R$, the derived set of $A$, denoted by $A'$, is the set of all limit points of $A$. It follows that if $A$ is closed, then $A' \subseteq A$.
Lemmas:
Every closed set $A \subseteq \Bbb R$ has countably many isolated points.
$A'$ is a closed set for all $A \subseteq \Bbb R$.
There exists a unique transfinite sequence $\langle A_\alpha \rangle_{\alpha \in \rm{Ord}}$ such that
$$\begin{align} A_0 &= A \subseteq \Bbb R\\ A_{\alpha+1} &= (A_{\alpha})' \space \text{ for all ordinal } \alpha \\ A_{\alpha} &= \bigcap_{\xi < \alpha} A_{\xi} \text{ for all limit ordinal } \alpha \end{align}$$
Let $F \subseteq \Bbb R$ be a closed set. By Lemma 3, there exists a unique transfinite sequence $\langle F_\alpha \rangle_{\alpha \in \rm{Ord}}$ such that
$$\begin{align} F_0 &= F \subseteq \Bbb R \\ F_{\alpha+1} &= (F_{\alpha})' \space \text{ for all ordinal } \alpha\\ F_{\alpha} &= \bigcap_{\xi < \alpha} F_{\xi} \text{ for all limit ordinal } \alpha \end{align}$$
By Lemma 2 and Transfinite Induction Theorem, $F_\alpha$ is closed for all ordinal $\alpha$. Next we prove that there exists a countable ordinal $\Theta$ such that
For all $\alpha < \Theta$, the set $F_\alpha -F_{\alpha+1}$ is non-empty and countable.
$F_{\Theta+1} = F_\Theta$.
$F_\Theta$ is a perfect set.
$F - F_\Theta$ is countable.
$F_\alpha -F_{\alpha+1}$ is the set of all isolated points of $F_\alpha$ and thus is countable by Lemma 1. Clearly, $F_{\alpha+1} \subseteq F_\alpha$ for all ordinal $\alpha$.
- There exists $\alpha\in \rm{Ord}$ such that $F_\alpha=F_{\alpha+1}$
If not, $F_\alpha\neq F_{\alpha+1}$ and thus $F_\alpha - F_{\alpha+1} \neq \emptyset$ for all ordinal $\alpha$. Moreover, $(F_{\alpha} -F_{\alpha+1}) \cap (F_{\beta} -F_{\beta+1}) = \emptyset$ for all ordinals $\alpha,\beta$ such that $\alpha \neq \beta$.
Let $\gamma$ be the Hartogs number of $\mathcal P(F)$ (the existence of Hartogs number require Axiom of Replacement).
Define a function $\phi:\gamma \to \mathcal P(F)$ by $$\phi(\alpha) = F_{\alpha} -F_{\alpha+1}$$
It follows that $\phi$ is an injection from the Hartogs number of $\mathcal P(F)$ into $\mathcal P(F)$. This is a contradiction.
Let $\Theta = \min \{\alpha \in \rm Ord \mid F_\alpha = F_{\alpha+1}\}$. Thus $(F_\Theta)' =F_{\Theta+1} = F_\Theta$. Hence $F_\Theta$ is perfect.
- $F - F_\Theta$ is countable
Let $C=F - F_\Theta$ and $\langle (r_n,s_n) \rangle_{n \in \Bbb N}$ be an enumeration of all open intervals with rational endpoints.
Let $\psi:C \to \Theta$ be a function that maps $a \in C$ to the unique ordinal $\alpha$ such that $a \in (F_\alpha -F_{\alpha+1})$. It follows that $a \in C$ is an isolated point of $F_{\psi(a)}$. Thus there exists an open interval $(r,s)$ such that $r,s \in \Bbb Q$ and $\{a\} = (r,s) \cap F_{\psi(a)}$.
We define a function $\mathcal G: C \to \Bbb N$ by $$\mathcal G(a)= \min \{n \in \Bbb N \mid \{a\} = (r_n,s_n) \cap F_{\psi(a)}\}$$
Let $\mathcal G(a)=\mathcal G(b) =n'$. Then $\{a\} = (r_{n'},s_{n'}) \cap F_{\psi(a)}$ and $\{b\} = (r_{n'},s_{n'}) \cap F_{\psi(b)}$.
If $\psi(a) = \psi(b) = \alpha$, then clearly $a=b$.
If $\psi(a) \neq \psi(b)$, then WLOG we assume $\psi(a) < \psi(b)$. Clearly, $a \notin F_{\psi(b)}$ and thus $a \neq b$. We have $b\in (r_{n'},s_{n'}) - \{a\}$ $\implies b \notin F_{\psi(a)} \implies b \notin F_{\psi(b)}$. This is a contradiction. Hence this case is impossible to happen.
As a result, $\mathcal G$ is injective and thus $|C| \le |\Bbb N| = \aleph_0$. Hence $C$ is countable.
- $\Theta$ is countable
We define a function $\mathcal F:\Theta \to \Bbb N$ by $$\mathcal F(\alpha) = \min \{\mathcal G(a) \mid a \in (F_\alpha -F_{\alpha+1})\}$$
Clearly, $\mathcal F$ is injective and thus $|\Theta| \le |\Bbb N| = \aleph_0$. Hence $\Theta$ is countable.
- $F_\Theta$ is uncountable
We have $F$ is uncountable and $F - F_\Theta$ is countable. Thus $F_\Theta$ is uncountable.