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# Cantor-Bendixson derivative

Let $A$ be a subset of a topological space $X$. Its *Cantor-Bendixson
derivative* $A^{{\prime}}$ is defined as the set of accumulation points of $A$. In
other words

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

Through transfinite induction, the Cantor-Bendixson derivative can be
defined 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 limit
ordinal, then $A^{{(\lambda)}}=\bigcap_{{\alpha<\lambda}}A^{{(\alpha)}}$.
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 $A$. This implies 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 above properties, if $A$ has Cantor-Bendixson rank $\alpha$, the sets

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

form a strictly decreasing chain of closed sets.

## Mathematics Subject Classification

54H05*no label found*03E15

*no label found*

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