3
$\begingroup$

This question is about the argument for Lemma 3.7 in Forcing axioms and stationary sets (MSN) by Boban Veličković.

He defines a game $G_\alpha$ between two players, playing objects in $H_\kappa$, depending on a function $F : [\kappa]^{<\omega} \to \kappa$. At stage $n$, Player I plays an interval of ordinals $I_n$ below $\kappa$ and an ordinal $\xi_n \in I_n$. Player II plays an ordinal $\mu_n$. $\min(I_{n+1})$ must be above $\mu_n$. Player I wins if the closure $M$ of $\{ \xi_n : n < \omega \} \cup \alpha$ under $F$ is contained in $\bigcup_n I_n$, and $M \cap \omega_1 = \alpha$.

Now I understand the argument for why player I has a winning strategy when we ignore the requirement on $\alpha$. For the $\alpha$-indexed games, he says: Let $A_F$ be the set of $\alpha<\omega_1$ such that player II has a winning strategy $\sigma_\alpha$. "Since $A_F$ has cardinality $\leq \aleph_1$, there is a strategy $\sigma$ which dominates all the $\sigma_\alpha$. It follows that $\sigma$ is a winning strategy for II in $G_\alpha$ for every $\alpha \in A_F$."

What does it mean to say that a strategy "dominates" a lot of other ones (for the same player), and why does such $\sigma$ exist?

$\endgroup$
5
  • $\begingroup$ @EmilJeřábek, thank you for your edit. I cut and pasted the author's name directly out of MathSciNet, so I'm not sure why it came out funny. Since it showed up correctly on Safari, I thought it might just be a Firefox thing …. $\endgroup$
    – LSpice
    Commented Mar 9, 2022 at 18:58
  • 2
    $\begingroup$ @LSpice The name in MathSciNet is wrong: the correct diacritic is not breve, but caron. Though these would look almost indistinguishable when typeset right. The actual reason it looks weird in some browser/system/etc setup is most likely due to the fact that there is no “c with breve” precomposed character in Unicode (as this combo does not appear in any human language), thus it is written as U+0063 (LATIN SMALL LETTER C) followed by U+0306 (COMBINING BREVE). Then the positioning of the combining diacritic over ... $\endgroup$ Commented Mar 10, 2022 at 7:35
  • 2
    $\begingroup$ ... the “c” depends on instructions specified in whatever font the browser happens to use, and since this is not a natural character pair, these may be inaccurate or missing. In contrast, there is a precomposed character U+010D (LATIN SMALL LETTER C WITH CARON) for the correct “č”. $\endgroup$ Commented Mar 10, 2022 at 7:40
  • 1
    $\begingroup$ Small correction: “c with breve” is not entirely nonexistent – it turns out to be used in the ISO 9 transliteration of the Abkhaz Cyrillic letter ҽ. But anyway it is a very rare combination, does not have a precomposed Unicode character, and, apparently, does not have a good font support. $\endgroup$ Commented Mar 10, 2022 at 7:57
  • $\begingroup$ @EmilJeřábek, thanks for clarifying! I was wondering why I couldn't find the character. $\endgroup$
    – LSpice
    Commented Mar 10, 2022 at 11:06

1 Answer 1

4
$\begingroup$

What constitutes a partial play doesn't depend on $\alpha$. And $\kappa$ is assumed to have cofinality strictly larger than $\aleph_1$. So he means: define $\sigma$ applied to a partial play to just be some ordinal below $\kappa$ that is above all the outputs of the $\aleph_1$-many strategies $\sigma_\alpha$.

(And it is easy to see that for any fixed $\alpha$, if $\sigma_\alpha$ is a winning strategy for II in the game $G_\alpha$, then so is any function from partial runs into $\kappa$ that is $\ge \sigma_\alpha$.)

$\endgroup$
1
  • $\begingroup$ Thanks. I was about to post a way to get around my question. Just run the argument with the sequence $\{ \sigma_\alpha : \alpha < \omega_1 \}$ in the model $N_0$. Then find the countable $z$ with $z \cap \omega_1 = \alpha \in A_F$. Noting that $\sigma_\alpha \in N_{\xi_n}$ for each $n$, we construt a run of $\sigma_\alpha$ in which Player I wins. $\endgroup$ Commented Mar 10, 2022 at 8:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .