Cantor-Bendixson Theorem

Question

I know two different proofs of the Cantor-Bendixson theorem, however both explicitly construct the perfect set and the countable set directly, without using the fact that closed sets have the perfect set property. I was wondering, assuming that we know that a closed set has the perfect set property (let's say by the Gale-Steward theorem), is there an easier way to show the Cantor-Bendixson theorem?

Answer 1

I think the complexity is in showing that the other part is countable.

Let $X$ be a closed subset of $\Bbb{R}$. Let $I$ be the subset of isolated points of $X$. Since $I$ is a set of isolated points of $X$ it is countable (there's a rational in each interval isolating a point in $I$.)

Note that since $X$ is closed in $\Bbb{R}$, taking a closure of a subset of $X$ relative to $X$ is the same as taking a closure in $\Bbb{R}$.

Now let $X_1 = \overline{I}$ and let $X_2=\overline{X\setminus X_1}=\overline{\operatorname{int}_X(X\setminus I)}$.

The set $X_2$, if non-empty is perfect. To show this, let $G=\operatorname{int}_X(X\setminus I)$. If $y\in X_2\setminus G$ then $y$ is a limit point of $G$, so $y$ is not isolated. If $y\in G$, assume by contradiction that $y$ is isolated in $X_2$. Then there is an open set $U$ (of $\Bbb{R})$ such that $U\cap X_2=\{y\}$. Since $G$ is relatively open in $X$, there is an open set $V$ (of $\Bbb{R}$) such that $V\cap X=G$. Then $y\in U\cap G=U\cap V\cap X$ and $U\cap V\cap X=U\cap G\subseteq U\cap X_2=\{y\}$. Hence $U\cap V\cap X=\{y\}$, and $U\cap V$ is an open subset of $\Bbb{R}$. This proves that $y$ is an isolated point of $X$ - hence $y\in I$, but $I\cap G=\emptyset$ - a contradiction.

What's left to prove is that the set $X_1$ - a closure of a countable discrete subset of $\Bbb{R}$ is countable - and that's where the complexity lies. This doesn't have to be true outside of $\Bbb{R}$. For example, $\Bbb{N}$ is discrete and dense in the Stone-Cěch compactification $\beta\Bbb{N}$.

What the Cantor-Bendixon proof does is instead of focusing on $I$, it focuses on $X_2$ - using a transfinite process to chop $X_1$ away a countable piece at each iteration taking the derived set. Since the iteration provably terminates at a countable ordinal the total number of points of $X_1$ chopped away before reaching $X_2$ is countable. If the transfinite process did not stop at a countable ordinal then $X_1$ could hypothetically be an uncountable set.