Cantor-Bendixson Theorem Proof (existence of an ordinal)

Question

Let $A$ be a topological space, denote the set of accumulation points of $A$ by $A'$.

If $\alpha$ is an ordinal, we define $A^{(\alpha)}$ by transfinite induction:

$A^{(0)} = A$, $\, \,A^{(\alpha+1)} = (A^{(\alpha)})'\,\,$ and $A^{(\alpha)} = \bigcap_{\beta < \alpha} A^{(\beta)}\,\,$ if $\alpha$ is a nonzero limit ordinal.

Note that $A^{(\beta +1)} \subseteq A^{(\beta)}$ for any ordinal $\beta$.

1) How can we say that there is an ordinal $\alpha$ such that $A^{(\alpha+1)} = A^{(\alpha)}$?

2) Is it true that if $L=\bigcap_{\alpha} A^{(\alpha)} \neq \emptyset$, then $L=L'$?

Answer 1

If there is no such ordinal, then for every $a\in A$, map it to the least $\alpha$ such that $a\notin A^{(\alpha+1)}$. This is now a surjection onto the class of ordinals. But since $A$ is a set, this is impossible.

For the second question, note that if $x\in L\setminus L'$, then $x\notin L^{(1)}$, so it cannot be in the intersection.