11
$\begingroup$

What are some examples of posets $\mathbb{P}$ which have the following properties? It's OK if the definition uses large cardinals or some other hypothesis.

  1. $\mathbb{P}$ preserves stationary subsets of $\omega_1$
  2. $\mathbb{P}$ fails to have the $\omega$-covering property. So in particular, $\mathbb{P}$ cannot be $\sigma$-distributive, or proper on any stationary set of countable models.
  3. Whenever $\kappa$ has uncountable cofinality, then this remains true in $V^{\mathbb{P}}$.

Item (3) rules out Namba forcing,Prikry forcing, and the standard stationary tower forcings with critical point at least $\omega_2 $ (the latter preserve stationary subsets of $\omega_1 $ if the height of the tower is a Woodin cardinal, but change lots of regular cardinals to cof $\omega $).

Item (2) rules out several forms of antichain sealing forcings (all variants of sealing forcings I know of are of the form "proper followed by shooting a club", which has the countable covering property).

$\endgroup$
  • 1
    $\begingroup$ This seems somewhat close to recent work of Yair Hayut and yours truly. Although there's still quite a gap, if you require that $\Bbb P$ do not add new subsets of $\omega_1$, and that it is a "tiny bit homogeneous" (in an odd, technical sense), then it not change cofinalities at all. $\endgroup$ – Asaf Karagila Jul 15 '15 at 19:55
11
$\begingroup$

Let $(\kappa_n: n<\omega)$ be an increasing sequence of measurable cardinals with limit $\kappa,$ and for each $n$ let $U_n$ be a normal measure on $\kappa_n.$ Let $\mathbb{P}$ be the Prikry type forcing notion which adds one element Prikry to each $\kappa_n:$

$p\in \mathbb{P}$ iff $p=(s, A),$ where

1)$s$ is a finite sequence,

2) $\forall i< |s|, s(i) < \kappa_i,$

3) $A=(A_n: |s| \leq n < \omega),$ and $A_n \in U_n$,

$p=(s,A)\leq q=(t, B)$ iff:

1) $s$ end extends $t$,

2) $\forall |t| \leq i < |s|, s(i)\in B_i,$

3) $\forall |s| \leq n < \omega, A_n \subseteq B_n.$

The forcing adds an $\omega$-sequence to $\kappa,$ and can not be covered by any ground model set of size $\omega$ (in fact of size $< \kappa$), it adds no bounded subsets to $\kappa,$ and does not change cofinalities.

So the forcing is as requested.

$\endgroup$
  • $\begingroup$ Do you know if the consistency strength of this violation of $\omega$-covering is exactly $\omega$ many measurable cardinals? $\endgroup$ – Yair Hayut Jul 16 '15 at 15:18
  • $\begingroup$ Unfortunately I have no idea. $\endgroup$ – Mohammad Golshani Jul 17 '15 at 4:54
  • 1
    $\begingroup$ It can be done with just a single measurable cardinal: just take a quotient of Prikry forcing by another Prikry forcing on co-infinitely many coordinates. $\endgroup$ – Bill Chen Jul 18 '15 at 22:39

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.