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.

  • 4
    $\begingroup$ Isn't it true that "$F$ intersects every uncountable $F_\sigma$ set" iff "$F$ intersects every uncountable closed set"? $\endgroup$
    – GEdgar
    Commented Jun 11, 2022 at 20:27
  • $\begingroup$ 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. $\endgroup$
    – GEdgar
    Commented Jul 9, 2022 at 15:04
  • 1
    $\begingroup$ One thing to note is that a nonlinear additive function cannot be measurable. Most likely, it cannot be Baire either. $\endgroup$ Commented Jul 15, 2022 at 19:04
  • $\begingroup$ 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 $f(G) = A$. $\endgroup$ Commented Jul 31, 2022 at 21:33
  • $\begingroup$ @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? $\endgroup$ Commented Aug 1, 2022 at 14:50

1 Answer 1


This question was answered on the MathOverflow cross post by Taras Banakh:

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 $\mathbb R$.


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