Does there exist $f:\Bbb{R}\to \Bbb{R}$ additive onto function such that $f(F) \subset \Bbb{R}$ has the property of Baire for every $F$?

Question

Let $F\subset \Bbb{R}$ intersect every closed uncountable subsets of $\Bbb{R}$.

Does there exist $f:\Bbb{R}\to \Bbb{R}$ additive onto function such that $f(F) \subset \Bbb{R}$ has the property of Baire for every $F$ ?

I have explained my thoughts here on MSE.

Answer 1

Yes, there exists such a function: Consider the real line as a linear space over the field $\mathbb Q$ and find a linearly independent Cantor set $C\subseteq \mathbb R$ (using the Kuratowski-Mycielski Theorem 19.1 in Kechris' "Classical Descriptive Set Theory"). Identifying $C$ with $C\times C$, we can write $C$ and the union $\bigcup_{\alpha\in\mathbb R}C_\alpha$ of continuum many uncountable compact sets. Since the set $C$ is linearly independent, there exists an additive function $f:\mathbb R\to\mathbb R$ such that $C_\alpha\subseteq f^{-1}(\alpha)$ for every real number $\alpha\in \mathbb R$. This function $f$ has the required property: any Bernstein set $F\subset\mathbb R$ has non-empty intersection with each set $C_\alpha$, $\alpha\in\mathbb R$, and hence $f[F]=\mathbb R$, so $f[F]$ has the Baire property in $\mathbb R$.