9
$\begingroup$

Recall that an inaccessible cardinal $\kappa$ is a Woodin cardinal if for every $A\subseteq V_\kappa$ there is an unbounded set in $\kappa$ of $\lambda$ such that $V_\kappa\models\lambda$ is $A$-strong.

This raises the question, of whether or not intermediate notions have been defined in the literature, and what sort of results can they be used to obtain?

For example, $\Sigma^n_m$-Woodin is an inaccessible cardinal $\kappa$ such that for every $A$ which is $\Sigma^n_m$-definable over $V_\kappa$, there is an unbounded set of $\lambda<\kappa$ such that $V_\kappa\models\lambda$ is $A$-strong.

It seems probable that there is such hierarchy of definable Woodin-ness, and that it might be used to obtain partial determinacy (and saturation) results.

Do these notions exist in the literature, and if so do they have interesting consequences?

Improve this question
$\endgroup$
1
  • 4
    $\begingroup$ Yes, partial Woodinness is common in arguments in inner model theory. Accordingly, you obtain determinacy results addressing specific pointclasses (typically, well beyond projective). $\endgroup$
    – Andrés E. Caicedo
    Commented May 27, 2017 at 20:55

2 Answers 2

Reset to default
7
$\begingroup$

(As I pointed out in a comment) yes, partial Woodinness is common in arguments in inner model theory. Accordingly, you obtain determinacy results addressing specific pointclasses (typically, well beyond projective). To illustrate this, let me "randomly" highlight two examples:

  • See here for $\Sigma^1_2$-Woodin cardinals and, more generally, the notion of a cardinal $\delta$ being Woodin with respect to a family $\frak A\subseteq \mathcal P(\delta)$.
  • See here for $\Gamma$-Woodin cardinals (and coarse mice) for $\Gamma$ a good pointclass.

There are also more recent examples, of course.

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ As you see from the examples, rather than directly considering notions of Woodinness restricted by the standard $\Sigma^m_n $-definability requirements, typically the fragments under consideration are defined descriptive set theoretically. $\endgroup$
    – Andrés E. Caicedo
    Commented May 28, 2017 at 1:54
8
$\begingroup$

There is a useful notion of "weak Woodinness" according to which every uncountable regular cardinal is weakly Woodin :) See https://ivv5hpp.uni-muenster.de/u/rds/fabiana_ralf.pdf

Improve this answer
$\endgroup$
1
  • $\begingroup$ Welcome to MathOverflow! I see you're hitting with full force. $\endgroup$
    – Asaf Karagila ♦
    Commented Sep 14, 2017 at 9:44

You must log in to answer this question.

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