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We believe the answer to the following question, that is relevant to a joint research project with Piotr Szewczak, should be known. We would appreciate any help or pointer. Needed definitions may be found in, e.g., Blass's chapter in the Handbook of Set Theory.

For $f\in\omega^\omega$, let $K_f:=\{g\in \omega^\omega : g\le^* f\}$.

Let $\mathfrak{tmp}$ be the minimal cardinality of a set $Y\subseteq \omega^\omega$ such that the set $\bigcup_{f\in Y}K_f$ is not meager.

Since every set $K_f$ is meager, we have that $\mathfrak{b}\le \mathfrak{tmp}\le \mathrm{non}(meager)$.

Is $\mathfrak{tmp}$ (provably) equal to either of these cardinals?

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    $\begingroup$ You really aim at this to be a temporary name, eh? :-) $\endgroup$
    – Asaf Karagila ♦
    Commented Mar 26, 2015 at 23:22
  • $\begingroup$ Yes, we do. Such a clean definition - the cardinal must be known under some other (and better :) ) name. $\endgroup$
    – Boaz Tsaban
    Commented Mar 26, 2015 at 23:24
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    $\begingroup$ Note also that $\mathfrak{tmp} \leq \mathfrak{d}$. $\endgroup$
    – Ramiro de la Vega
    Commented Mar 26, 2015 at 23:40
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    $\begingroup$ So $\mathfrak{tmp}<\mathrm{non}(meager)\,$ is consistent. $\endgroup$
    – Ramiro de la Vega
    Commented Mar 26, 2015 at 23:51
  • $\begingroup$ @RamirodelaVega: Excellent point. In our application we work below $\mathfrak{d}$, so we missed this point. $\endgroup$
    – Boaz Tsaban
    Commented Mar 27, 2015 at 1:08

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For every meager $M \subseteq \omega^{\omega}$ there exists $f_M \in \omega^{\omega}$ such that for every increasing $f \in \omega^{\omega}$, $K_f \subseteq M$ implies $f \leq^{\star} f_M$. For a proof, see Theorem 2.2.2 in Bartozynski, Judah book. It follows that your invariant is the bounding number.

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  • $\begingroup$ Nice! Lyubomyr Zdomskyy points out another proof: $\omega^\omega$ is homeomorphic to $[\omega]^\omega$, and here $\mathfrak{b}$ is enough: Take a non-meager filter of character $\mathfrak{b}$ and note that each upper cone of a subset of $\omega$ is compact. $\endgroup$
    – Boaz Tsaban
    Commented Mar 27, 2015 at 14:51

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