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 $a-\delta<r<a<s<a+\delta$. 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, $C \subseteq \bigcup\{A \cap (r,s) \mid r,s \in \Bbb Q \text{ and } r<s\}$, 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.

  • $\begingroup$ Yeah you got my points. Have you found any mistake in my proof? $\endgroup$ – Abstract Analysis Feb 14 '19 at 23:22
  • $\begingroup$ @LeAnhDung the first part is superfluous. Not an error. The rest seems ok $\endgroup$ – Henno Brandsma Feb 15 '19 at 4:55
  • $\begingroup$ Thank you so much for your verification! $\endgroup$ – Abstract Analysis Feb 15 '19 at 4:56

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