Meager set union Bernstein set with empty interior

Question

Fix a meager set $M\subset \Bbb R$. of course $\text{int} M=\emptyset$. I want to construct a Bernstein set $B\subset \Bbb R$ such that $$\text{int}(M\cup B)=\emptyset \tag{1}$$

We know it is true for any Bernstein set that $\text{int}B=\emptyset.$ I am not sure such $B$ can be found to make (1) true.

Any help will be appreciated greatly

Answer 1

In fact, we have a stronger result:

Suppose $B$ is Bernstein and $M$ is meager. Then $int(M\cup B)=\emptyset$.

In the interest of brevity, I'll use an oversized hammer for the problem: the perfect set property for $G_\delta$ sets, that every uncountable $G_\delta$ set has a perfect subset.

Suppose $int(M\cup B)\not=\emptyset$ and $M$ is meager. Let $(a,b)\subseteq M$. Then $B\cap (a,b)$ is comeager in $(a,b)$. We then observe:

• If $X$ is nowhere dense, then so is its closure $\overline{X}$; so every nowhere dense set is contained in a closed nowhere dense set.

• Consequently, every meager set is contained in a meager $F_\sigma$ set.

• And so dually every comeager set has a comeager $G_\delta$ subset.

And all of this relativizes to a nondegenerate open interval like $(a,b)$. So $B$ contains an uncountable $G_\delta$ set, and hence is not Bernstein by the perfect set property.