family of pairwise disjoint sets in the complement of meager. set

Question

Let $M$ be a meager subset of $\mathbb R$. I want to construct the following family in $\mathbb R\setminus M $ $$F:= \{A_{r}^\xi\colon r\in\mathbb R \ \&\ \xi<c\}$$ Such that all elemnets of $F$ are pairwise disjoint dense sets and $\bigcup F$does not contain any perfect sets. I want to try to construct by using transfinite induction. My first challenge is how can I do the transfinite induction for $r$ and $\xi$. If they are two ordinal number like $A_{\xi}^{\eta}$ then we can define well ordering like $\leq$ on $c\times c$ ($c$ is the cardinality of $\mathbb R$ by putting $<\alpha,\beta>\leq<\gamma,\delta>$ if and only if $$max\{\alpha,\beta\}=\{\gamma,\delta\}$$ OR $$max\{\alpha,\beta\}<\{\gamma,\delta\}\ and \ \alpha=\gamma \ \ OR \ \ max\{\alpha,\beta\}<\{\gamma,\delta\}, \ \alpha<\gamma, \ and \ \beta=\delta.$$ But in my case I have $r\in\mathbb R.$ What should I do? My thought process for the proof is enumerate all open intervals and all perfect sets which there are continuum many of them and construct by picking one element of each open interval and avoid element of perfect set at the same time. but the challenge when I reach the inductive step I have no control how can I keep the pairwise disjoint since I am doing this proof in ZFC. I like transfinite induction but sometimes is challenging for me. Any hint or idea how this can be done it will be useful. Thank you in advance

Answer 1

Significantly revised to correct an oversight and for greater clarity.

Let $\mathscr{B}$ be a base for $\Bbb R\setminus M$, let $\mathscr{C}$ be the family of uncountable closed subsets of $\Bbb R\setminus M$, and let $\mathscr{S}=\mathscr{B}\cup\mathscr{C}$; note that $\mathscr{C}$ contains all of the perfect subsets of $\Bbb R\setminus M$. Let $T=\Bbb R\times\mathscr{S}\times\mathfrak{c}$; $|T|=\mathfrak{c}$, so we can index $T$ as $T=\{\langle r_\xi,S_\xi,\alpha_\xi\rangle:\xi<\mathfrak{c}\}$. Now you can construct your sets $A_r^\alpha$ by straightforward transfinite recursion on $\xi<\mathfrak{c}$.

(Note: Transfinite induction is a proof technique: what you’re doing here is transfinite recursion.)

Suppose that $\eta<\mathfrak{c}$, and you’ve already chosen points $x_\xi,y_\xi\in S_\xi$ for $\xi<\eta$ so that all of these points are distinct. You’ve chosen fewer than $\mathfrak{c}$ points, and $|S_\eta|=\mathfrak{c}$, so you can choose distinct

$$x_\eta,y_\eta\in S_\eta\setminus\big(\{x_\xi:\xi<\eta\}\cup\{y_\xi:\xi<\eta\}\big)$$

and continue the recursion. (You do need to know here that every member of $\mathscr{S}$ has cardinality $\mathfrak{c}$.)

Now for each $\langle r,\eta\rangle\in\Bbb R\times\mathfrak{c}$ let $A_r^\alpha=\{x_\xi:r_\xi=r\text{ and }\alpha_\xi=\alpha\}$.

• For each $B\in\mathscr{B}$ there is a $\xi<\mathfrak{c}$ such that $\langle r_\xi,S_\xi,\alpha_\xi\rangle=\langle r,B,\alpha\rangle$, and $x_\xi\in A_r^\alpha\cap B$, so $A_r^\alpha$ is dense in $\Bbb R\setminus M$.
• For each $C\in\mathscr{C}$ there is a $\xi<\mathfrak{c}$ such that $\langle r_\xi,S_\xi,\alpha_\xi\rangle=\langle r,C,\alpha\rangle$; by construction $x_\xi\in A_r^\alpha\cap C$ and $y_\xi\in C\setminus A_r^\alpha$, so $A_r^\alpha$ is a Bernstein set and contains no perfect set.
• The points $x_\xi$ for $\xi<\mathfrak{c}$ are all distinct, so $\{A_r^\alpha:\langle r,\alpha\rangle\in\Bbb R\times\mathfrak{c}\}$ is a pairwise disjoint family.