For a given $\kappa > \omega$, define the game $d(\kappa)$ that runs for $\omega$ stages as follows: At stage $n$, player I chooses a sequence of elements of $\omega$, $g_n$ of length $\kappa$, and player II picks a natural number $f(n)$. Define $h_{\alpha} (n) = g_n (\alpha)$.

Player I wins the game if exists $\alpha \in \kappa$ such that $h_\alpha$ dominates $f$.

Let $\mathfrak{sd} = \min \left\lbrace \kappa : \text{player I has a winning strategy in } d(\kappa) \right\rbrace$.

$\mathfrak{sd}$ is uncountable by a diagonal argument and $\leq \mathfrak{d}$ since player I can play a dominating set.

Is $\mathfrak{sd} = \omega_1$?

  • $\begingroup$ Intuitively, it seems it could be possible to show $\mathfrak{b}\le\mathfrak{sd}$. The existence of a bounding function $g_\mathcal{D}\in\omega^\omega$ for every bounded $\mathcal{D}\subseteq\omega^\omega$ of size $\kappa$ would seem to imply that player I has no winning strategy in $d(\kappa)$, since if player I "threatens" to end with $\mathcal{D}$, then player II could "threaten" to end with the function $n\mapsto g_\mathcal{D}(n)+1$. But I'm not sure how proofs involving two player games and winning strategies should go (even if this idea is plausible). $\endgroup$ – user52534 May 5 '15 at 0:01
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    $\begingroup$ Cute problem! $~~~~$ $\endgroup$ – hot_queen May 5 '15 at 13:41

Notice that player I has a winning strategy for the game $d(\kappa)$ iff there exists $\langle s_i : i < \kappa \rangle$ such that the following hold.

(1) Each $s_i: \omega^{\omega} \to \omega^{\omega}$.

(2) For every $n$, knowing $n$ bits of input gives you $n+1$ bits of output - i.e., for every $x, y \in \omega^{\omega}$, if $x \upharpoonright n = y \upharpoonright n$ then $s_i(x) \upharpoonright (n+1) = s_i(y) \upharpoonright (n+1)$.

(3) For every $x \in \omega^{\omega}$ there is some $i < \kappa$ such that $s_i(x)$ dominates $x$.

Now it is easy to check that for any $s_i$ as above, $\{x \in \omega^{\omega} : (\exists^{\infty} n)(s_i(n) = x(n))\}$ is comeager. Hence if $\kappa <$ cov(Meager), then player I does not have a winning strategy on $d(\kappa)$. In particular, $\mathfrak{sd} > \omega_1$ is consistent (add $\omega_2$ Cohen reals).

Although this answers your question, it is now natural to ask: Can $\mathfrak{sd} < \mathfrak{d}$?

  • $\begingroup$ I cannot construct such $s_i$. I was trying to build those by adding asking the winning strategy $\sigma$ what would she do if player II plays each finite sequence: $s_i^\varphi (n) = \sigma (n) (i)$, this gives (1) and (2), however, (3) can be made by the stronger condition $s_\alpha (n) \geq n$. (I am thinking $(s_i (x))(n)$ is $s_i (x(b))$ in (3)). Under this, I cannot prove that the set you asserted is comeager it actually comeager unless the family $s_i$ is already a dominating family. $\endgroup$ – Karv May 6 '15 at 20:13
  • $\begingroup$ Hi Karv, Could you explain why you do not get (3) from a winning strategy? $\endgroup$ – hot_queen May 16 '15 at 0:00

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