# 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 uncountable $$\mathcal{F}_{\sigma}$$ set.

$$B\subset \Bbb{R}$$ is said to have the property of Baire if $$B=U\triangle M$$ where $$U$$ is open and $$M$$ is meager.

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 $$F$$ defined above ?

Edit $$1$$ : $$F$$ intersects every uncountable $$F_{\sigma}$$ set iff F intersects every uncountable closed set.

Edit $$2$$: If $$f:\Bbb{R}\to \Bbb{R}$$ is additive i.e $$f(x+y) =f(x) +f(y)$$ then $$f$$ is $$\Bbb{Q}$$-linear.In other words $$f$$ is a linear function if we consider $$\Bbb{R}$$ as a vector space of $$\Bbb{Q}$$.

1. $$f(x+y) =f(x) +f(y)$$

2. $$f(qx) =qf(x)$$

$$\forall x, y\in\Bbb{R}$$ and $$\forall q\in\Bbb{Q}$$

$$\underline{ \text{Case } 1}$$ : $$( f \text{ is } \Bbb{R} \text{ linear})$$

Then $$f(x) =ax$$ for some $$a\in \Bbb{R}$$.

Hence clearly $$f$$ is additive onto map $$($$ and moreover $$f$$ is a linear homeomorphism / topological isomorphism $$)$$.

But $$f$$ fails to hold the third property mentioned above. For an example ,if we take $$F=\mathcal{B}( \text{Bernstein set})$$ then $$f(F) =a\mathcal{B}$$ doesn't have the property of Baire.

$$\boxed{\text{ Required function can't be \Bbb{R} -linear}}$$

$$\underline{\text{Case }2}$$: $$f$$ is $$\mathbb{Q}$$-linear but not $$\Bbb{R}$$-linear.

Points to be considered:

1. If $$g:\Bbb{R}\to \Bbb{R}$$ defined by $$g(x) =qx , q\in\Bbb{Q}$$ is continuous then $$g=f$$ on $$\Bbb{Q}$$ implies $$g=f$$ on $$\Bbb{R}$$ i.e any continuous $$\Bbb{Q}$$-linear map extends linearly on $$\Bbb{R}$$.

$$\boxed{ \text{ So g can't be a continuous on \Bbb{Q}}}$$

1. A linear map is completely determined by it's action on the basis. So our task is reduced to construct a discontinuous linear map from the vector space $$\Bbb{R}_{\Bbb{Q}}$$ to $$\Bbb{R}$$.

2. A non-continuous solution of an additive function ( called ugly function) is non-measurable.

Conjecture $$1$$: $$f:\Bbb{R}\to \Bbb{R}$$ is $$\Bbb{Q}$$-linear and non-measurable function. Then $$\exists F\subset \Bbb{R}$$ closed uncountable set such that $$f(F)$$ doesn't have the property of Baire.

1. There exists a discontinuous additive function $$f:\Bbb{R}\to\Bbb{R}$$ satisfying Darboux property ( It can be shown by defining $$f$$ linealy on a hamel basis of $$\Bbb{R}_{\Bbb{Q}}$$ and then extending additively on $$\Bbb{R}$$ )

Conjecture $$2$$ : The Darboux property of $$f$$ ( discontinuous $$\Bbb{Q}$$-linear function) is sufficient to conclude that $$f(F)$$ have property of Baire for every closed uncountable set $$F$$.

1. Any second category (non meager) subset of $$\Bbb{R}$$ contains a set that fails to have the property of Baire. $$[$$ Suppose $$A\subset \Bbb{R}$$ non meager. Then $$A\cap \mathcal{B}$$ or $$A\cap \mathcal{B}^c$$ atleast one of the set doesn't have the B.P otherwise both would be meager and eventually $$A$$ would be meager $$]$$

2. $$f:\Bbb{R}\to \Bbb{R}$$ is an additive onto function such that $$f$$ is not injective then $$f^{-1}(y)$$ is dense in $$\Bbb{R}$$ for all $$y\in\Bbb{R}$$.

Question: 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 $$F$$ defined above ?

Here is the MO post of this question.

• Isn't it true that "$F$ intersects every uncountable $F_\sigma$ set" iff "$F$ intersects every uncountable closed set"? Jun 11, 2022 at 20:27
• If additive function $f$ is continuous, $f(x) = ax$, and then it does not have your property. One method of building discontinuous additive functions uses a Hamel basis. So perhaps an example can be constructed using transfinite recursion, either starting with a Hamel basis, or constructing the Hamel basis as you go. Jul 9, 2022 at 15:04
• One thing to note is that a nonlinear additive function cannot be measurable. Most likely, it cannot be Baire either. Jul 15, 2022 at 19:04
• An observation: if $F$ intersects every uncountable $F_\sigma$, then the same is true for every superset of $F$. So given $f$, it would suffice to find such a set $F$ for which $f(F)$ is not comeager. For then there exists a set $A \supset f(F)$ which does not have the BP, and by surjectivity, there is a set $G \supset F$ with $f(G) = A$. Jul 31, 2022 at 21:33
• @NateEldredge I feel that observation could be useful but I don't know how to find $F$ such that $f(F)$ is non meager ( or comeager is enough as $\Bbb{R}$ is Baire space)? Can you provide any further simplification? Aug 1, 2022 at 14:50