Complement of the product of Bernstein set and meager set Assume $M\subset\mathbb R$ be a meager set with cardinality $\mathfrak c.$ I want to construct a Bernstein set $B$ such that $\mathbb R\setminus(B\cdot M)$ has cardinality $\mathfrak c$ , $B\cdot M=\{b\cdot m\mid b\in B,  m\in M\}$. It is an easy transfinite induction to construct a Bernstein set but I want an extra condition.
I do not know if I need a set-theoretical assumption for that or not. More precisely, It is possible or not. I hope I can find an answer or hint since I spent  plenty of time but I did not get any thing at least  possible or not.
 A: Eventually, I answered my question. Surprisingly, the answer is very simply even though  I spent a lot of time thinking to find such  function. There answer is
No Bernstein can be found with this property
Let $M$ be compact nowhere dense prefect set. For contradiction, let $B$ be a Bernstein set such that $\mathbb R\setminus(BM)$ has cardinality $mathfrak c.$ Now,  pick $x\notin BM$. Notice that $xM^{-1}$ is uncountable closed set( cantinas a perfect set) and $$xM^{-1}\cap B=\emptyset$$ which is a contradiction   since $B$ is a Bernstein set.
A: My idea wold be to do this, following up on my comments: assume MA. We are given a meagre set $M$ and want to construct a Bernstein set $B$ such that $\Bbb R\setminus M\cdot B$ is size continuum.
Enumerate all relevant perfect subsets of $\Bbb R$ as $\{P_\xi: \xi < \mathfrak{c}\}$.
We will construct points $\{(x_\zeta, y_\zeta, z_\zeta)\mid \zeta < \mathfrak{c}\}$  by transfinite recursion; at stage $\xi < \mathfrak c$ we pick distinct
$$x_\xi, y_\xi \in P_\xi \setminus (\{x_\zeta \mid \zeta < \xi\} \cup \{y_\zeta \mid \zeta < \xi\} \cup \{0\})\tag{1}$$
while ensuring
$$z_\xi \notin \bigcup\{ M \cdot x_\zeta\mid \zeta < \xi\}\tag{2}$$
$(1)$ follows the standard construction of a Bernstein set and works for the same reasons, while $(2)$ can be done as the sets $M \cdot x_\zeta$ are also meagre and so fewer than $\mathfrak{c}$ cannot cover $\Bbb R$ by MA.
In the end $B:=\{x_\xi: \xi < \mathfrak{c}\}$ will be Bernstein, and $(2)$ ensures the condition about $\Bbb R\setminus M \cdot M$ as the $z_\xi$ witness the size of it (I think); I see that I haven't built in assurances for them to be all distinct yet... So this is almost OK, but not quite.
If a condition like MA is at all necessary I do not know yet. It's late and I'll call it a night.
