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

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.

• For those interested, although Cantor stated the decomposition was unique in 1884, he did not actually prove this. Instead, he only proved the decomposition exists using a method that leads to a unique decomposition, and he did not prove that the decomposition itself (obtained by any method) is unique. The first proof of uniqueness was given in: Giulio Benedetto Isacco Vivanti (1859-1949), Sugli aggregati perfetti [On Perfect Sets], Rendiconti del Circolo Matematico di Palermo 13 (1899), pp. 86-88. Mar 7 at 15:40

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 $$B$$.