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A set is called meager is it can written as a countable union of nowhere dense sets. A set $S$ is said to have the Baire property if for some open set $O$, the symmetric difference $S\Delta O$ is meager.

Problem: Assuming that there is a set of real numbers which does not have the Baire property construct a set $X$ which is not meager and such that for any non-empty open set $O$, $O\setminus X$ is not meager.

My feeble attempt: Suppose $S\subseteq\Bbb R$ does not have the Baire property. Then for all nonempty open sets $O$, $S\Delta O$ is not meager. Taking $\Bbb R$ as the open set $S\Delta\Bbb R=\Bbb R\setminus S$, which cannot be meager. I thought I could take the set $X$ as $\Bbb R\setminus S$. Now for any non-empty open set $O$, $O\setminus X=O\cap S$. I have no idea how to show $O\cap S$ is not meager. In fact I think my choice for $X$ is wrong.

Please suggest how to construct such $X$.

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  • $\begingroup$ The construction by @bof totally works. But can it be any more elegant? $\endgroup$
    – QED
    Oct 20 '20 at 14:15
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Let $S$ be a set (of real numbers) that does not have the Baire property. Let $\mathcal U$ be the collection of all nonempty open sets $U$ such that $S\cap U$ has the Baire property. Let $\mathcal W$ be a maximal pairwise disjoint subcollection of $\mathcal U$ and let $W=\bigcup\mathcal W$. Since $\mathcal W$ is countable, $S\cap W$ has the Baire property, so the set $X=S\setminus W$ does not have the Baire property.

Since $X$ does not have the Baire property, $X$ is not meager. Let $O$ be a nonempty open set, and assume for a contradiction that $O\setminus X$ is meager.

Since $O\cap W$ is an open subset of the meager set $O\setminus X$, we must have $O\cap W=\emptyset$. Hence $O\cap S=O\cap(S\setminus W)=O\cap X$, which has the Baire property, since $O\setminus X$ is meager. This means that $O\in\mathcal U$, but then the fact that $O\cap W=\emptyset$ contradicts the maximality of $\mathcal W$.

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  • $\begingroup$ 1. Why did you take maximal pairwise disjoint subcollection? 2. Why is $\mathcal{W}$ countable? 3. why $S\cap W$ has Baire property imply that $S\setminus W$ does not have Bair property? $\endgroup$
    – QED
    Oct 20 '20 at 13:39
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    $\begingroup$ 1. It seemed like a good idea at the time. 2. Any collection of disjoint open subsets of the real line is countable. 3. If $S\cap W$ and $S\setminus W$ both had the Baire property then the set $S=(S\cap W)\cup(S\setminus W)$ would also have the Baire property. $\endgroup$
    – bof
    Oct 20 '20 at 13:46
  • $\begingroup$ @bof: Very possibly I’m missing the obvious, but while $O\cap X$ is clearly not co-meagre, I don’t see what ensures that it’s actually meagre. $\endgroup$ Oct 20 '20 at 23:47
  • $\begingroup$ @BrianM.Scott Oops! Thanks! I think what I meant to say was, $O\cap X$ has the Baire property, since $O\setminus X$ is meager. Does that make sense? $\endgroup$
    – bof
    Oct 20 '20 at 23:57
  • $\begingroup$ @bof: Absolutely; thanks much! $\endgroup$ Oct 20 '20 at 23:58

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