# Assuming Axiom of Choice, every uncountable closed set contains a perfect subset

Assuming Axiom of Choice, 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 any $$A \subseteq R$$, we call $$a \in \Bbb R$$ a condensation point of $$A$$ $$\iff$$ for all $$\delta>0$$, the set $$\{x \in A \mid |x-a|<\delta\}$$ is uncountable. Clearly, $$a$$ is a condensation point of $$A$$ $$\implies$$ $$a$$ is a limit point of $$A$$. We denote the set of all condensation points of $$A$$ by $$A^c$$.

1. For all $$A \subseteq R$$, $$A^c$$ is closed

Assume that $$a \in \Bbb R$$ is a limit point of $$A^c$$. Then for all $$\dfrac{\delta}{2}>0$$, there exists $$y \in A^c$$ such that $$y \neq a$$ and $$|y-a|<\dfrac{\delta}{2}$$. Also, $$y \in A^c$$ implies the set $$\left \{x \in A \,\middle\vert\, |x-y|<\dfrac{\delta}{2} \right \}$$ is uncountable.

We have $$|x-a| = |(x-y)+(y-a)| \le |x-y| +|y-a|<\dfrac{\delta}{2}+\dfrac{\delta}{2}=\delta$$.

Hence $$\left \{x \in A \,\middle\vert\, |x-y|<\dfrac{\delta}{2} \right \}$$ is uncountable $$\implies$$ $$\{x \in A \mid |x-a|<\delta\}$$ is uncountable.

It follows that $$a$$ is a condensation point of $$A$$ and thus $$a \in A^c$$.

1. For all $$A \subseteq R$$, then $$C=A-A^c$$ is countable

If $$a \in C$$, then $$a \in A$$ is not a condensation point of $$A$$. So there exists $$\delta>0$$ such that the set $$\{x \in A \mid |x-a|<\delta\}=A \cap (a-\delta,a+\delta)$$ is countable.

Since $$\Bbb Q$$ is dense in $$\Bbb R$$, there exist $$r,s \in \Bbb Q$$ such that $$a-\delta. This shows that, for each $$a \in C$$, there exists an open interval $$(r,s)$$ such that $$r,s \in \Bbb Q$$ and $$a \in A \cap (r,s)$$ and $$A \cap (r,s)$$ is countable.

As a result, $$C \subseteq \bigcup\{A \cap (r,s) \mid r,s \in \Bbb Q \text{ and } r, which is a countable union of countable sets. By Axiom of Countable Choice, this set is countable. Consequently, $$C$$ is countable.

1. If $$A \subseteq \Bbb R$$ is uncountable and closed, then $$A^c$$ is perfect

Since $$A$$ is closed, $$A^c \subseteq A$$.

It follows from 1. that $$A^c$$ is closed.

We have $$A^c=A -(A-A^c)$$ where $$A$$ is uncountable and $$A-A^c$$ is countable (from 2.). Then $$A^c$$ is uncountable and thus non-empty.

Next we prove that $$A^c$$ contains no isolated points of itself.

Assume the contrary that $$a \in A^c$$ is an isolated point of $$A^c$$. Then there exists $$\delta > 0$$ such that $$x \neq a$$ and $$|x-a|<\delta$$ implies $$x \notin A^c$$. Thus $$x \in A$$ and $$|x-a|<\delta$$ implies $$x=a$$ or $$x \in A -A^c$$. We know that $$A-A^c$$ is countable from 2., so the set $$\{x \in A \mid |x-a|<\delta\}$$ is countable. This contradicts the fact that $$a \in A^c$$ is a condensation point of $$A$$.

As a result, $$A^c \subseteq A$$ is perfect.

The non-condensation points ($$C$$) of $$A$$ are open in $$A$$: this follows from your proof of 2: the sets $$A \cap (r,s)$$ are relatively open subsets of $$A$$ and they precisely cover $$C$$. So the condensation points themselves are closed in $$A$$ and hence closed in $$\mathbb{R}$$ when $$A$$ is closed in $$\mathbb{R}$$. No need for the proof under 1. anymore.
A point is a condensation point of $$A$$ iff every neighbourhood of it in $$A$$ is uncountable. So in particular it's never a singleton in the condensation points too, because we only throw out at most countably many points to go from $$A$$ to its condensation set. So the condensation set is perfect. This is also in essence your argument, I think.