6
$\begingroup$

Let $\kappa$ be supercompact in $V$. Let $\mathbb{P}$ be one of the standard forcing notions (or an iteration of such), and for simplicity assume that $\mathbb{P}$ is ${<}\kappa$-directed closed (e.g. a $\kappa$-Hechler iteration). In particular, $\kappa$ will remain supercompact in $V^{\mathbb{P}}$ if $V$ was 'Laver prepared'.

Question: What conditions can we impose on $\mathbb{P}$ and $V$ such that $\kappa$ remains supercompact in every intermediate forcing extension, i.e. $V \vDash \forall \mathbb{Q} \triangleleft \mathbb{P} \, \colon \, \Vdash_{\mathbb{Q}} \check{\kappa} \, \text{is supercompact}$ ?

$\endgroup$
9
  • 6
    $\begingroup$ It is likely hard to say much. Results of Kunen show that if $\mathbb{P}$ is even just the forcing to add a single Cohen real to $\kappa$, there is an intermediate extension where $\kappa$ isn't even weakly compact. $\endgroup$ Nov 30, 2020 at 19:59
  • 4
    $\begingroup$ @AsafKaragila Yes, I meant Cohen subset of $\kappa$. Kunen showed that if you first add a Suslin tree to $\kappa$ and then force with the tree to kill it, the iteration is equivalent to adding a single Cohen subset to $\kappa$. Of course, in the intermediate model the Suslin tree kills weak compactness, but if $\kappa$ was sufficiently indestructible to being with, a lot of strength can be resurrected by killing the tree. I think this first appeared in his Saturated ideals paper, but it's been written up many times since then. $\endgroup$ Nov 30, 2020 at 20:46
  • 2
    $\begingroup$ Related to Kunen's result: Large cardinals need not be large in HOD $\endgroup$ Dec 1, 2020 at 14:48
  • 4
    $\begingroup$ Here’s another example. If GCH holds in the neighborhood, then for a regular $\kappa$, there is a two-step iteration $\mathbb S * \mathbb T$ that adds a $\square_\kappa$-sequence and then threads it, and the iteration is isomorphic to $\mathsf{Col}(\kappa,\kappa^+)$. $\endgroup$ Dec 1, 2020 at 18:01
  • 1
    $\begingroup$ @Johannes: That's why I asked Miha to clarify that "Cohen real" meant Cohen subset. $\endgroup$
    – Asaf Karagila
    Dec 1, 2020 at 20:07

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.