Decomposition of a closed set into a perfect set and an at most countable set is unique.

Question

Exercise $4.7$ in Chapter $10$ of "Introduction to Set Theory" (Hrbacek & Jech, $3^{\text{rd}}$ edition), is the following:

The decomposition of a closed set $F$ into $P \cup C$ where $P \cap C = \emptyset$, $P$ is perfect, and $C$ is at most countable, is unique; i.e., if $F = P_{1} \cup C_{1}$ where $P_{1}$ is perfect and $P_{1}$ and $C_{1}$ are disjoint and $\vert C_{1} \vert \leq \aleph_{0}$, then $P_{1} = P$ and $C_{1} = C$. [Hint: Show that $P$ is the set of all condensation points of $F$.]

(Note that the results in this section of the book are proved specifically for $\mathbb{R}$)

Using the hint, I've been able to show that the set of condensation points of $F$ (points containing uncountably many points of F in every neighbourhood) is included in $P$. The other direction I've found to be more difficult. I've tried doing a proof by contradiction (i.e. assuming that there is an $x \in P$, and a $\delta \gt 0$ such that $(x - \delta, x + \delta) \cap P$ is at most countable), but I've gotten no farther than showing that $(x - \delta, x + \delta) \cap P$ cannot be finite. Since $P$ is perfect, $P = P^{(\omega_{1})} = \bigcap_{\alpha \lt \omega_{1}} P^{(\alpha)}$ (where $P = P^{(0)} = (P^{(0)})^{\prime}$, and $P^{(\alpha + 1)} = (P^{(\alpha)}))^{\prime}$ is the derived set (the set of accumulation points) of $P^{(\alpha)})$; maybe one can use this fact to construct an uncountable sequence of distinct points in any $(x - \delta, x + \delta) \cap P$?

Any help would be appreciated.

Answer 1

For a set $A$ denote by $A_c$ the set of its condensation points. You need to use the following fact: If $A$ is perfect, then $A = A_c$. Proving this is similar to proving that a perfect set is uncountable. Going back to your problem, this will give you that every point of $P$ is a condensation point of $F$.

Here is a sketch of the claim. Let $x \in A$ and $\varepsilon > 0$. Take any countable subset $B$ of $[x - \varepsilon, x + \varepsilon] \cap A$ which we can enumarate as $\{x_1,x_2, ...\}$. We will show that there is a point $y \in [x - \varepsilon, x + \varepsilon] \cap A$ not in $B$. This will prove our claim. We start by taking a closed interval $I_0 \subset [x - \varepsilon, x + \varepsilon]$ which intersects $A$. Then we can find a closed subinterval $I_1 \subset I_0$ which also intersects $A$ but doesn't contain $x_1$. Similarly we can find a closed subinterval $I_2 \subset I_1$ intersecting $A$ but not containing $x_2$ and so on. In the end we will have a nested family of closed intervals $\{I_n\}$ such that $I_n \cap A \neq \emptyset$ and $x_n \notin I_n$. By properties of nested compact sets the intersection $\bigcap_{n=0}^\infty I_n\cap A$ is not empty and is disjoint from $B$.