Suppose we do have a filter $\mathcal{F}$ on $\omega$ which contains the cofinite filter, so $X\in\mathcal{F}$ implies $X$ is infinite. For $X\in\mathcal{F}$, let $f_X$ be the increasing enumeration of $X$. Let $\tilde{\mathcal{F}}=\{f_X\mid X\in\mathcal{F}\}$. Let $f\in\omega^\omega$ s.t. $f$ is a bound of $\tilde{\mathcal{F}}$.

Identify $X\subseteq\omega$ with $x\in2^\omega$ defined by $x(n)=1$ if $n\in X$ and $x(n)=0$, if $n\notin X$ and consider the discrete topology on $2$ and the product topology in $2^\omega$.

I want to prove, that $A_n:=\{X\subseteq\omega\mid\forall k\geq n\ f_X(k)\leq f(k)\}$ is meager for all $n\in\omega$. Any hints?

Definition of an unbounded familiy $F$: A family of subsets of $\omega^\omega$ is unbounded, if for all $g\in \omega^\omega$ exists a $f\in F$ s.t. $\neg(f\leq^* g)$, where $f\leq^*g$ means $f(n)\leq g(n)$ for all but finitely many $n\in\omega$.

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    $\begingroup$ Meager in what space and what topology exactly? Clearly the space is not $\omega$, but it seems not to be $\omega^\omega$ either. $\endgroup$ – Asaf Karagila Jun 28 '16 at 8:21
  • $\begingroup$ In $2^\omega$ with discrete Topology in 2 and product Topology. $\endgroup$ – peer Jun 28 '16 at 11:12
  • $\begingroup$ One can identify a subset $X\subseteq\omega$ with $x\in2^\omega$ defined by $x(n)=1$ if $n\in X$ and $x(n)=0$, if $n\notin X$. $\endgroup$ – peer Jun 28 '16 at 17:17
  • $\begingroup$ Yes, I know. You should clarify this in your question. $\endgroup$ – Asaf Karagila Jun 28 '16 at 17:18
  • $\begingroup$ It's not clear to me that such $f$ exists. Why should it exist? $\endgroup$ – Asaf Karagila Jun 28 '16 at 17:27

First, note that $\mathcal{F}$ is a red herring; the real question is

For $f\in\omega^\omega$ unbounded, show that $A_f:=\{X: \forall k, f_X(k)\le f(k)\}$ is meager.

Note that I've also gotten rid of $n$; this is just for simplicity, and won't change anything important.

EXERCISE: for every $\sigma\in 2^{<\omega}$ there is some $\tau\in 2^{<\omega}$ with $\sigma\preccurlyeq\tau$ such that for any $X$, if the characteristic function of $X$ extends $\tau$, then $X$ is not in $A_f$.

If you can do this, do you see why this means $A_f$ is in fact nowhere dense (not just meager)? (The passage to meager comes when we bring $n$ back into the picture.)

By the way, this line of attack comes from the game picture of category; see e.g. my answer to Analogy for Baire categories?.

  • $\begingroup$ f unbounded means that one cannot find a M s.t. $f\leq M$ and $\sigma\preccurlyeq\tau$ means $\sigma$ is extended by $\tau$? $\endgroup$ – peer Jun 30 '16 at 22:20
  • $\begingroup$ Let $n$ be the number of $1$'s in the finite sequence $\sigma$. Then, if the chracteristic function $\chi_X$ of a $X\subseteq\omega$ extends $\sigma$, $f_X(n)$ denotes the n-th element in X (ordering $<$. Then one can define $\tau$ as $\tau(i)=\sigma(i)$ for $i=0,...,|\sigma|$ and $\tau(i)=0$ sufficently many $i>|\sigma|$ s.t. the element $f_X(n+1)$ must be greater than $f(n+1)$, if the characteristic function extends $\tau$, hence $X\notin A_f$. $\endgroup$ – peer Jul 2 '16 at 9:18
  • $\begingroup$ But i am not able to see why thiis mean $A_f$ is nowhere dense (in what topology?) $\endgroup$ – peer Jul 2 '16 at 9:20
  • $\begingroup$ @peer Such a $\tau$ determines an open set (in the usual topology on Cantor space, generated by finite strings) in which $A_f$ has no element (since no function extending $\tau$ lies in $A_f$). Moreover, what we've shown is that for every $\sigma$, there is such a $\tau$ extending $\sigma$. Think about finite strings as basic open sets: what this means is that, for every open set $U$ in Cantor space, there is some smaller open set $V$ such that $A_f\cap V=\emptyset$. This is exactly the definition of "nowhere dense." $\endgroup$ – Noah Schweber Jul 2 '16 at 9:23
  • $\begingroup$ To clarify: the topology on Cantor space used here is the standard one, generated by the sets of the form $\{r: \rho\prec r\}$ for $\rho\in 2^{<\omega}$ (these basic opens are sometimes called cylinders). As to the game picture mentioned at the end of my answer: consider the following game between two players, I and II. They alternately play finite strings of $0$s and $1$s, and I wins if the infinite string they produce (after infinitely many moves) lies in $A_f$. Clearly II has a winning strategy; it turns out that this means that the payoff set (in this case, $A_f$) is meager! $\endgroup$ – Noah Schweber Jul 2 '16 at 9:24

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