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?

• What Baire property are we talking about here. Just being Baire (i.e. not being a union of countably many nowhere dense sets), or something else? – Henno Brandsma Jul 1 '16 at 16:38
• Or, $A$ has the property of Baire iff there is some open set $U$ such that $A \Delta U$ is meagre? – Henno Brandsma Jul 1 '16 at 17:12
• The filter to be considered will be identified by a subset of $2^\omega$. And concerning the definition of B.P., we used the second one. – peer Jul 1 '16 at 17:18
• @BrianM.Scott Corrected – peer Jul 1 '16 at 23:01
• What is the origin of this problem. (I'd say including source of the problem is useful in general.) – Martin Sleziak Jul 7 '16 at 16:42

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.

• I was wondering about the last part: why is it true that a filter on $\omega$ with the Baire property (i.e. almost open) must be meagre? I was looking in the book this came from (which did prove the existence of the $y_F$ and $f_F$ in a separate theorem) and the author did not mention it near this theorem. That was the part that had me confused. – Henno Brandsma Jul 2 '16 at 4:58
• @Henno: You caught me just as I was adding that bit, having realized that it wasn’t quite so easy to see (for most of us, anyway!) as Talagrand implies in the paper in which he proved this result. – Brian M. Scott Jul 2 '16 at 5:05
• I think that I saw it mentioned that Sierpiński showed this first. – Henno Brandsma Jul 2 '16 at 5:16
• @Henno: It’s possible; this result is just part of a result by Talagrand, Theorem $21$ in this paper, which is where I saw it. – Brian M. Scott Jul 2 '16 at 5:24
• Only the last observation was due to Sierpiński, IIRC. The whole theorem is indeed credited to Talagrand in the book (Set theory, structure of the real line). – Henno Brandsma Jul 2 '16 at 5:36