# 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 $F$?

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.

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 $$\mathbb R$$.