$mathbb R$ as continuum many of pairwise disjoint of Bernstein sets

Question

The set $Bsubsetmathbb R$ is called Bernstein set if neither $B$ nor $mathbb Rsetminus B$ contains any perfect sets.
Theorem: $mathbb R$ can be written as continuum many of pairwise disjoint Bernstein sets.
Proof:  Let $mathcal{P}$ be a family of all perfect subsets of $mathbb R$. Then  $|mathcal{P}timesmathbb R|=mathfrak c , $ so we can choose an enumeration ${langle P_xi,y_xiranglecolonxi<mathfrak c}$ of $mathcal{P} times mathbb R.$
We will construct, by induction on $xi<mathfrak c,$ a
sequence ${x_xicolon xi<mathfrak c}$ such that
$$x_xiin P_xisetminus{x_zetacolon zeta<xi}$$
Since each $x_lambdaneq x_xi$ for all $lambda<xi<mathfrak c$, we can define $f$ on ${x_xicolon xi<mathfrak c}$ such that $f(x_xi)=y_xi$ and $f(x)=0$ otherwise.
Claim : $f^{-1}(r)$ is Bernstein set for each $rinmathbb R.$
Indeed, for each $rinmathbb R$ and $Pinmathcal P$ there is an $xi<mathfrak c$ such that $langle r,Prangle = langle r_xi, P_xi rangle$,  then $x_xiin f^{-1}(r_xi)cap P_xi=f^{-1}(r) cap Pneq emptyset.$  Also the same true for the complement for $f^{-1}(r)$ (as $f^{-1}(t)$ where $tneq r$).  Then $f^{-1} (r)$ is Bernstein for all $rinmathbb R$. It is clear that $$ mathbb R=bigcup_{rinmathbb R} f^{-1} (r)$$ as we need.
I have no probelm with this proof at all. My question is Can we have the same result without  define function like that? I mean just start to construct the by transfinite induction. I would love to see completely different approache to get the same result. I am hoping for theos who have deeply understanding for transfinite induction to share their proofs.

Brian M. Scott · Answer

First, and least important, your construction is by recursion, not by induction: induction is a proof technique.
Next, the argument doesn’t really use the function $f$ at all: you could just as well define your Bernstein sets by letting $B_r={x_xi:y_xi=r}$ for each $rinBbb R$.
Finally, your construction does not ensure that the union of your Bernstein sets is all of $Bbb R$; it could, for instance, be $Bbb RsetminusBbb Q$ if you happened to choose irrational numbers for all of your points $x_xi$. If you want them to be a partition of $Bbb R$, you have to be a bit more careful in your recursion.
Enumerate $Bbb R={r_xi:xi<mathfrak{c}}$. At stage $eta$, when you’ve already chosen $x_xi$ for all $xi<eta$, let $mu=minbig{zeta<mathfrak{c}:r_zetain P_etasetminus{x_xi:xi<eta}big}$, and let $x_eta=r_mu$. This is just your construction with a little extra care taken in choosing which point of $P_etasetminus{x_xi:xi<eta}$ to use as $x_eta$, so it does give you pairwise disjoint Bernstein sets $B_r={x_xi:y_xi=r}$ for $rinBbb R$.
Suppose that $Bbb Rsetminusbigcup_{rinBbb R}B_rnevarnothing$, let $r_nuinBbb Rsetminusbigcup_{rinBbb R}B_r$ and let $X={xi<mathfrak{c}:r_nuin P_xi}$; then $|X|=mathfrak{c}$, and $x_xiin P_xi$ for each $xiin X$. For each $xiin X$ there is a $mu(xi)<mathfrak{c}$ such that $x_xi=r_{mu(xi)}$, and the points $x_xi$ are distinct, so the map $mu$ is injective. Let $M={mu(xi):xiin X}$; $|M|=mathfrak{c}$, so we can let $theta=min{muin M:nulemu}$. Note that $theta=mu(eta)$ for some $etain X$, and $x_eta=r_theta$.
By hypothesis $r_nunotinbigcup_{rinBbb R}B_r$, so there is no $xi<mathfrak{c}$ such that $x_xi=r_nu$, and therefore $theta>nu$. But this is impossible: $r_nuin P_etasetminus{x_xi:xi<eta}$, and $nu<theta$, so at stage $eta$, when we chose $x_eta$, $theta$ was not the least element of $big{zeta<mathfrak{c}:r_zetain P_etasetminus{x_xi:xi<eta}big}$, and we did not set $x_eta=r_theta$.
Added: Here are a few links to answers of mine containing constructions by transfinite recursion, and one to an old paper of mine that does so; I’ve arranged them very roughly in ascending order of complexity. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 [PDF; see Theorem $bf{5}$].

$mathbb R$ as continuum many of pairwise disjoint of Bernstein sets

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