Cantor-Bendixson derivative

Let $A$ be a subset of a topological space $X$. Its Cantor-Bendixsonderivative $A^{\prime }$ is defined as the set of accumulation points of $A$. Inother words

$$A^{\prime }=\{x\in X\mid x\in \overline{A\setminus \{x\}}\}.$$

Through transfinite induction, the Cantor-Bendixson derivative can bedefined to any order $\alpha$, where $\alpha$ is an arbitrary ordinal.Let $A^{(0)}=A$. If $\alpha$ is a successor ordinal, then$A^{(\alpha )}={\left(A^{(\alpha -1)}\right)}^{\prime }$. If $\lambda$ is a limitordinal, then .The Cantor-Bendixson rank of the set $A$ is the least ordinal$\alpha$ such that $A^{(\alpha )}=A^{(\alpha +1)}$. Note that $A^{\prime }=A$implies that $A$ is a perfect set.

Some basic properties of the Cantor-Bendixson derivative include

1. ${(A\cup B)}^{\prime }=A^{\prime }\cup B^{\prime }$,

2. ${({\bigcup }_{i\in I}{A}_i)}^{\prime }\supseteq {\bigcup }_{i\in I}{A}_i^{\prime }$,

3. ${({\bigcap }_{i\in I}{A}_i)}^{\prime }\subseteq {\bigcap }_{i\in I}{A}_i^{\prime }$,

4. ${(A\setminus B)}^{\prime }\supseteq A^{\prime }\setminus B^{\prime }$,

5. $A\subseteq B\Rightarrow A^{\prime }\subseteq B^{\prime }$,

6. $\overline{A}=A\cup A^{\prime }$,

7. $\overline{A^{\prime }}=A^{\prime }$.

The last property requires some justification. Obviously, $A^{\prime }\subseteq \overline{A^{\prime }}$. Suppose $a\in \overline{A^{\prime }}$, then every neighborhood of$a$ contains some points of $A^{\prime }$ distinct from $a$. But by definition of$A^{\prime }$, each such neighborhood must also contain some points of $A$. Thisimplies that $a$ is an accumulation point of $A$, that is $a\in A^{\prime }$.Therefore $\overline{A^{\prime }}\subseteq A^{\prime }$ and we have $\overline{A^{\prime }}=A^{\prime }$.

Finally, from the definition of the Cantor-Bendixson rank and the aboveproperties, if $A$ has Cantor-Bendixson rank $\alpha$, the sets

$$A^{(1)}\supset A^{(2)}\supset \mathrm{\cdots }\supset A^{(\alpha )}$$

form a strictly decreasing chain of closed sets.