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Let $\text{BP}(X)$ denote $\sigma$-algebra of subsets of $X$ with the Baire Property BP and $\text{MGR}(X)$ denote the $\sigma$-ideal of meager sets in $X$.

Assume $X$ is second countable Baire space.

Question: There is no uncountable subset $\mathcal{A} \subseteq \text{BP}(X)$ such that $A \notin \text{MGR}(X)$ for any $A \in \mathcal{A}$ and $A\cap B \in \text{MGR}(X)$ for anytwo distinct $A,B \in \mathcal{A}$.

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Suppose I have uncountably many non-meager sets $\mathcal{A}=\{A_\eta\}_{\eta\in\omega_1}$ with the Baire property. Let $\mathcal{U}$ be a countable base for $X$.

Let $\mathcal{U}$ be a countable base for the space $X$. Given any non-meager $A\in\mathcal{A}$, there is some open set $V\in\mathcal{U}$ and some meager set $D$ such that $V\Delta D\subseteq A$: $A$ has the property of Baire, so differs from some open set by a meager set, and I'm just restricting my attention to a small piece of that open set. Without loss of generality, I'll assume each $A\in\mathcal{A}$ actually equals $V\Delta D$ for some $V\in\mathcal{U}$ and some meager $D$.

Now since $\mathcal{U}$ is countable there is some nonempty open set $O\in \mathcal{U}$ and uncountable $\mathcal{B}\subseteq\mathcal{A}$ such that for each $B\in\mathcal{B}$, $B\Delta O$ is meager. In particular, since $\mathcal{B}$ is uncountable, it has at least two elements. Pick some $B_0, B_1$ in $\mathcal{B}$. We have $B_0\cap B_1\supseteq O-(M_0\cup M_1)$ for meager sets $M_0, M_1$. But since $X$ has the Baire property, this intersection is non-meager.

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  • $\begingroup$ Good point - what I'd written wasn't quite right. Hopefully fixed now. $\endgroup$ Mar 5, 2015 at 14:53
  • $\begingroup$ OK so in the 2nd paragraph you're saying an almost open set almost contains a basic open set, fine... how about the 3rd paragraph, how are you using (un)countability? Sorry for being "dense" :) $\endgroup$ Mar 5, 2015 at 16:37
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    $\begingroup$ He's just using it to say that $\mathcal B$ contains at least two elements. $\endgroup$ Mar 5, 2015 at 17:05
  • $\begingroup$ Jay is right - hopefully I've clarified things. $\endgroup$ Mar 5, 2015 at 17:37
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    $\begingroup$ Oh and also a pigeonhole principle application... there are uncountably many $A$, each one has one of the countably many $V$, so there is some $V$ that is had by uncountably many $A$. Great. $\endgroup$ Mar 5, 2015 at 18:02

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