Is there a subset $A\subseteq \mathbb{R}$, not having the BP (Baire property)?

Question

Let $\mathcal{I}$ be the $\sigma$-ideal of meager sets of a topological space $X$.

We say, that a set $A\subseteq X$ has the Baire property (BP) iff $A=^*U$ for some set $U\subseteq X$. Here $A=^*B$ means, that the symmetric differene $A\bigtriangleup B=(A\setminus B)\cup(B\setminus A)\in\mathcal{I}$.

Is there a subset $A\subseteq \mathbb{R}$, not having the Baire property?

Answer 1

Yes, assuming the axiom of choice. (Actually, one needs only the weaker Boolean prime ideal theorem, though it’s not known whether it suffices for the following example.) Vitali’s construction produces an example. For $x,y\in\Bbb R$ write $x\sim y$ if and only if $x-y\in\Bbb Q$; it’s straightforward to verify that $\sim$ is an equivalence relation on $\Bbb R$. Let $A$ contain exactly one member of each $\sim$-equivalence class. (In group-theoretic terms $A$ contains one member of each coset of $\Bbb Q$, where $\Bbb Q$ is viewed as an additive subgroup of $\Bbb R$.)

For $x\in\Bbb R$ let $A+x=\{a+x:a\in A\}$. If $0\ne q\in\Bbb Q$, $A+q$ is disjoint from $A$: if there were some $a\in(A+q)\cap A$, then we’d have $a\in A$ and $a-q\in A$, but $a-q\sim a$, so $A$ cannot contain both $a$ and $a-q$. Thus, $\{A+q:q\in\Bbb Q\}$ is a countable partition of $\Bbb R$.

Clearly the sets $A+q$ for $q\in\Bbb Q$ are homeomorphic, so $A$ can’t be meagre. Suppose that $A$ has the Baire property, and let $U$ be an open set and $M$ a meagre set such that $U\mathrel{\triangle}M=A$. Evidently $U\ne\varnothing$, so there are $a,b\in\Bbb R$ such that $a<b$ and $(a,b)\subseteq U$. Let $J=(a,b)$ and $r=b-a$, and suppose that $x\in\Bbb R$ with $|x|<r$. Then $(J+x)\cap J$ is an open interval, and $M\cup(M+x)$ is meagre, so

$$\big((J+x)\cap J\big)\setminus\big(M\cup(M+x)\big)\ne\varnothing\;.$$

Clearly $$(A+x)\cap A\supseteq\big((J+x)\cap J\big)\setminus\big(M\cup(M+x)\big)\;,$$ so $(A+x)\cap A\ne\varnothing$ whenever $|x|<r$. Now let $x$ be a positive rational less than $r$ to get a contradiction.

There are models of $\mathsf{ZF}$, however, in which every subset of $\Bbb R$ has the Baire property; obviously $\mathsf{AC}$ fails in such models.

Answer 2

The answer is yes.

Bernstein set :$A$ Bernstein set $\mathcal{B}$ is a subset of $\Bbb{R}$ that intersect every non empty perfect set or any uncountable closed set but contains none of them.

If $\mathcal{B}$ is Bernstein, then so it's complement $\mathcal{B}^c$.

Lemma: $A\subset \mathcal{B}$ ( any Bernstein set) has the property of Baire implies $A$ is of first category.

Now it is clear that a Bernstein set doesn't have the property of Baire. If a Bernstein set has the property of Baire, then by the previous lemma, it would be first category and so it's complement and that contradict $\Bbb{R}$ is a Baire space ( a Baire space must be second category in itself).

For another approach see There exists a measurable subset of $\Bbb{R}$ without the property of Baire and a non measurable set with the property of Baire.

Note :Although we need the Axiom of Choice here but a weaker versions of choice are sufficient(to prove the existence of a set without B.P ) : the Boolean prime ideal theorem implies that there is a nonprincipal ultrafilter on the set of natural numbers; each such ultrafilter induces, via binary representations of reals, a set of reals without the Baire property (see here).