# Baire property and $F_\sigma$ sets

I am currently reading a proof for a theorem which concludes that a set A does not have the Baire property, if (1) is correct. So, $(1)\Rightarrow$A does not have the Baire property is shown. The proof is by contradiction and starts like this: Let $F=\cup_{n\in\omega}F_n$ be any meager set of type $F_\sigma$. My question is, if every set which does have the Baire property is a $F_\sigma$ set and if the converse statement does hold .

Every $F_\sigma$ set has the property of Baire; however, the converse is extremely false. Indeed, every Borel set (even every analytic or co-analytic set) has the property of Baire, and $F_\sigma$ sets represent only the first two levels of the Borel hierarchy, which has uncountably many levels.
It doesn't stop there, either. For example, it is consistent with ZF+DC (assuming large cardinals are consistent) that every set has the property of Baire. A way to bring this into the ZFC-world is: it follows from large cardinals that every set of reals in $L(\mathbb{R}$) has the property of Baire ($L(\mathbb{R})$ is a bit technical; roughly, this means that any set you can define in terms of real numbers has the property of Baire - in particular, every set you've ever heard of is in $L(\mathbb{R})$).
• The set A is a filter on $\omega$ and the filter can be considered as a subset of $2^\omega$ as stated in my question here: math.stackexchange.com/questions/1842206/…. (1) actually is the following statement: For every partition $\{I_n:n\in\omega\}$of $\omega$ into finite sets, there exists $X\in\mathcal{F}$ s.t. $X\cap I_n=\emptyset$ for infinitely many $n\in\omega$. – peer Jun 29 '16 at 14:55
• But then, i don't understand the proof. The author should have started like this: Suppose $A$ does have the Baire property... This is a contradiction to ... – peer Jun 29 '16 at 14:59
• Re: your second comment, I can't comment on the proof without seeing more of it. But if you're trying to prove that $A$ does have the Baire property by contradiction, then you do not start out by assuming it does - you assume it doesn't. (Or was that a typo?) Can you give a link to the proof you're reading? – Noah Schweber Jun 29 '16 at 15:19