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

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?

• It should be "open set $U \subseteq X$" right? May 29, 2023 at 7:07

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.

• It's not clear at the moment whether or not BPI implies the existence of a Vitali set. So while you correct the open parenthesis of your first remark, you might want to edit that to make it seem less like BPI implies Vitali sets exists (unless you have a proof for that, in which case I'm eager to read it). Feb 28, 2015 at 19:11
• @Asaf: I remember thinking at the time that I needed to go back and clarify that, and then I forgot. Thanks. Feb 28, 2015 at 20:53
• How do you prove the existence of a set without the Baire property using Boolean prime ideal theorem instead of AC? Sep 5, 2022 at 2:41
• @Michael: The details require a fair bit of work, but the basic idea is contained in the note at the end of Sourav Ghosh’s answer here and the discussion at the bottom of the page here. The category analogue of the zero-one law mentioned there is Theorem 4 of this paper. Sep 5, 2022 at 4:03

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).