I would like to start by showing that there exists a Bernstein set $B$ that is additive group. To see that, it is an easy transfinite induction to construct a Bernstein set that is a linear independent over $\mathbb Q$, say $B_1$. Let $B$ be a linear space that generated by $B_1.$
Claim , The the set $B$ is as needed. Indeed, $B$ intersects each perfect set since $B_1$ does so. It remains to show $B$ does not contain any perfect set, Consdier the quotient space $\mathbb R\setminus B$, for contradiction, assume $P\subset B$ where $P$ is a perfect set. Then, any nonzero cost $x+B$ where $x\notin B$ would conatin $x+P$ and $x+P$ is perfect set disjoint with $B.$ This proves the claim.
Now, assume we have a meager set $M\subset R$ with cardinality $\mathfrak c.$ My question, I want to construct a Bernstein set $B$ that is an additive group sucht $R\setminus (B\cdot M)$ has condinality $\mathfrak c$, where $B\cdot M=\{bm\colon b\in B \ \& \ m\in M\}.$
I do not know how I can start. I got a hint that said this can be done by using Martian axioms. I have a little knowledge about Martin axioms, actually, I know the standard applications of Martin axioms.
Any help will be appreciated greatly.an