Bernstein set and it proof. Definition. A set $B\subset\mathbb R$ is called a Bernstein set if both $B$ and $\mathbb R\setminus B$ intersect each perfect set.
I know the construct  the of Bernstein set but I got confused with this construction that I found in a book.
Let $\mathcal F=\{P_\xi\colon \xi<c\}$ be an enumeration of all perfect subsets of $\mathbb R.$ By recursion induction on $\xi<c$ define the sequences $\{a_\xi\colon \xi<c\}$ and $\{b_\xi\colon \xi<c\}$ by choosing in step $\xi<c$ points $a_\xi\neq b_\xi$ from $\mathbb R$ such that $$ a_\xi, b_\xi\in P_\xi\setminus(\{a_\zeta\colon \zeta<\xi\}\cup \{b_\zeta\colon \zeta<\xi\})$$  clearly, this choice is possible. Now define $$B:=\{a_\xi\colon \xi<c\}$$ Then $B$ is Bernstein set.
This is not my proof but I have doubts about this proof is correct. More specifically, I did not see how $B$ will intersect each perfect at continuum many points. Am I missing some thing ?
Any help will appreciated greatly.
 A: That is in fact the standard construction of a Bernstein set. I’m not sure why you are concerned about whether $B$ intersects each perfect set in continuum many points, since that is not part of the definition of a Bernstein set. As you say in your answer, it’s clear that $B\cap P$ and $B\setminus P$ are non-empty for each perfect subset $P$ of $\Bbb R$, and by definition that makes $B$ a Bernstein set.
However, it is true that if $P$ is a perfect subset of $\Bbb R$, then $|B\cap P|=2^\omega$. One way to see this is to note that $P$ necessarily contains a Cantor set $C$, and every Cantor set contains $2^\omega$ distinct subsets that are Cantor sets.1 Each of these Cantor sets is a perfect subset of $\Bbb R$, so each of them is one of the sets $P_\xi$ in the construction of $B$. $B$ contains a point $a_\xi$ from each of these sets, so $B$ contains $2^\omega$ points of $C$ and hence $2^\omega$ points of $P$.
1 For each $n\in\omega$ let $X_n=\{0,1\}$ with the discrete topology, and let $X=\prod_nX_n$; then $X$ is homeomorphic to $C$. For each infinite $S\subseteq\omega$ let
$$X_S=\{x\in X:x_n=0\text{ for all }n\in\omega\setminus S\}\,;$$
$X_S$ is clearly homeomorphic to $\prod_{n\in S}X_n$ and hence to $X$, so $X_S$ is a Cantor set. And $\omega$ has $2^\omega$ infinite subsets $S$, so $X$, and therefore $C$, contains $2^\omega$ distinct Cantor sets $X_S$.
A: I think the proof is correct here is why $B$ is Bernstein set. For every prefect set $P$ there is a $\xi<c$ such that $P=P_\xi$. Then $$a_\xi\in P\cap  B$$ and $$b_\xi\in P\cap( \mathbb R\setminus B) $$ Thus $B$ is as needed.
