Filter, which does not have the Baire property I do not understand the following proof of the following theorem. It is part of Theorem 4.1.2 (Talagrand theorem) in the book Tomek Bartoszyński and Haim Judah: Set theory. On the structure of the real line.
Let $\mathcal{F}$ be a filter on $\omega$ which contains the cofinite filter. 
(1) For every partition of $\omega$ into finite sets $I_n$, there exists $X\in\mathcal{F}$ s.t. $X\cap I_n=\emptyset$ for infinitely many $n\in\omega$.
(2)$\mathcal{F}$ does not have the Baire property.
Then (1) implies (2)
Proof: Let $F=\cup_{n\in\omega}F_n$ be any meager set of type $F_\sigma$. Then there exists $x_F\in 2^\omega$ and a strictly increasing function  $f_F$ s.t
$F\subseteq \{x\in 2^\omega:\forall^\infty n \exists j\in[f_F(n),f_F(n+1))x(j)\not=x_F(j)\}$
Let $I_n=[f_F(n),f_F(n+1))$ for $n\in\omega$. Find $X\in\mathcal{F}$ s.t. $X\cap I_n=\emptyset$ for infinitely many $n\in\omega$. Define $Y\in 2^\omega$ as follows:
$Y|I_n=X|I_n$, if $X\cap I_n=\emptyset$ and $Y|I_n=x_F|I_n$ if $X\cap I_n\not=\emptyset$. Then $X\subseteq Y$ and $Y\notin\mathcal{F}$, which is a contradiction. 
q.e.d.
Why does the author starts with a meager set of type $F_\sigma$ and not with a filter, which has the Baire property?
 A: The problem is that the argument is stated very tersely. The author is assuming that the filter $\mathscr{F}$ satisfies (1) and showing that it is not contained in any meagre $F_\sigma$. Let me expand the argument a bit.
Let $F=\bigcup_{n\in\omega}F_n$, where each $F_n$ is closed and nowhere dense in $2^\omega$. For $n\in\omega$ let $$G_n=2^\omega\setminus\bigcup_{k\le n}F_n\;,$$ and let $$G=\bigcap_{n\in\omega}G_n=2^\omega\setminus F\;.$$
For $x\in 2^{<\omega}$ let 
$$B(x)=\{y\in 2^\omega:y\upharpoonright\operatorname{dom}x=x\}\;.$$
Each $G_n$ is dense and open in $2^\omega$, so we can recursively construct a strictly increasing $f:\omega\to\omega$ such that for each $n\in\omega$ and each $x\in 2^{f(n)}$ there is a $y\in 2^{f(n+1)}$ such that $y\cap f(n)=x$, and $B(y)\subseteq G_n$. For $n\in\omega$ let $I_n=[f(n),f(n+1))$, and let $X\in\mathscr{F}$ be such that $X\cap I_n=\varnothing$ for infinitely many $n\in\omega$.
Recursively construct $y_n\in 2^{f(n)}$ as follows. 


*

*If $X\cap I_n\ne\varnothing$, let $y_{n+1}=y_n\cup(X\cap I_n)$.  

*If $X\cap I_n=\varnothing$, choose $y_{n+1}\in 2^{f(n+1)}$ such that $y_n=y_{n+1}\cap f(n)$, and $B(y_{n+1})\subseteq G_n$. 


Let 
$$Y=\bigcup_{n\in\omega}y_n\in 2^\omega\;.$$
Clearly $X\subseteq Y$, so $Y\in\mathscr{F}$. Moreover, $Y\cap f(n)=y_n$ for each $n\in\omega$.
If $X\cap I_n=\varnothing$, then $Y\in B\big(Y\cap f(n+1)\big)=B(y_{n+1})\subseteq G_n$. There are infinitely many such $n\in\omega$, and the sequence $\langle G_n:n\in\omega\rangle$ is decreasing, so $Y\in G$. Thus, $\mathscr{F}\cap G\ne\varnothing$, so $\mathscr{F}\nsubseteq F$, and hence $\mathscr{F}$ is not meagre in $2^\omega$.
Finally, a filter on $\omega$ with the Baire property must be meagre in $2^\omega$, so $\mathscr{F}$ cannot have the Baire property.
Added: It occurs to me that it might be a good idea to prove that last observation. 
Suppose that $\mathscr{F}$ is a non-meagre filter with the Baire property; then there are a non-empty open set $U$ and a meagre set $A$ such that $U\mathrel{\triangle}\mathscr{F}=A$. Then $U\setminus\mathscr{F}\subseteq A$, so there is a $y\in 2^{<\omega}$ such that $\varnothing\ne B(y)\setminus\mathscr{F}\subseteq A$. $B(y)$ is homeomorphic to $2^\omega$, and $A\cap B(y)$ is meagre in $B(y)$, so $\mathscr{F}$ contains a dense $G_\delta$ subset $H$ of $B(y)$.
Let $n=\operatorname{dom}y$, and define 
$$h:B(y)\to B(y):z\mapsto y\cup\{k\ge n:k\notin z\}\;;$$
then $h$ is an autohomeomorphism of $B(y)$, so $h[H]\subseteq h[\mathscr{F}]$ is a dense $G_\delta$ subset of $B(y)$, and therefore so is $H\cap h[H]\subseteq\mathscr{F}\cap h[\mathscr{F}]$. Now $h$ is an involution, so $z\in\mathscr{F}\cap h[\mathscr{F}]$ iff $z\in\mathscr{F}$ and $h(z)\in\mathscr{F}$, in which case $z\cap h(z)\in\mathscr{F}$. But $z\cap h(z)=y\notin\mathscr{F}$, since $\mathscr{F}$ contains the cofinite filter. Thus, $\mathscr{F}$ cannot both be non-meagre and have the Baire property.
