6
$\begingroup$

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 $\mathfrak{sd} = \omega_1$?

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ Cute problem! $~~~~$ $\endgroup$ – hot_queen May 5 '15 at 13:41
2
$\begingroup$

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 $\mathfrak{sd} < \mathfrak{d}$?

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.