# Construct $X$ so that $X$ is not meager and for any non-empty open set $O$, $O\setminus X$ is not meager

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 $$X$$.

• The construction by @bof totally works. But can it be any more elegant?
– QED
Oct 20 '20 at 14:15

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 $$\mathcal W$$.

• 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?
– QED
Oct 20 '20 at 13:39
• 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.
– bof
Oct 20 '20 at 13:46
• @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. Oct 20 '20 at 23:47
• @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?
– bof
Oct 20 '20 at 23:57
• @bof: Absolutely; thanks much! Oct 20 '20 at 23:58