8
$\begingroup$

Assume that $V\neq HOD$ and let $\kappa = \min \{\alpha\in On \mid \mathcal{P}(\alpha) \not\subseteq HOD\}$.

Clearly, $\kappa$ is a cardinal.

Question: Is it consistent that $\kappa = \aleph_\omega$?

Note that it is consistent that $\kappa$ is a regular cardinal: start with $V=L$ and force with $Add(\kappa,1)$. Since this forcing is weakly homogeneous, its generic filter is not in $HOD$. Since we don't add any bounded subsets to $\kappa$, for every $\alpha < \kappa$, $\mathcal{P}(\alpha) \subseteq L \subseteq HOD$.

Similarly, it is consistent that $\kappa$ is singular with countable cofinality. Let $\kappa$ be a measurable cardinal and let $V = L[\mu]$ ($\mu$ is a normal measure for $\kappa$), the canonical inner model for one measurable cardinal. Let $C$ be a Prikry sequence. Then $HOD^{V[C]}\cap \kappa^{<\kappa} = L[\mu]\cap \kappa^{<\kappa}\subseteq HOD$, but since the Prikry forcing is weakly homogeneous, $C\notin HOD^{V[C]}$.

$\endgroup$
  • $\begingroup$ @MohammadGolshani: It sounds good. Can you describe the tail forcing? Why is it homogeneous in the intermediate model? $\endgroup$ – Yair Hayut Mar 31 '15 at 7:28
  • $\begingroup$ I think that you also need to have that the Prikry forcing is definable in $L[\mu]$, not just weakly homogeneous. $\endgroup$ – Asaf Karagila Mar 31 '15 at 7:41
  • $\begingroup$ @MohammadGolshani: I'll take a look at these papers. Thank you. $\endgroup$ – Yair Hayut Mar 31 '15 at 7:45
  • $\begingroup$ @AsafKaragila The forcing is definable at least in intermediate submodel, and that's enough. $\endgroup$ – Mohammad Golshani Mar 31 '15 at 7:46
  • $\begingroup$ @Mohammad: I thought that for a weakly homogeneous $HOD^{V[G]}\subseteq HOD(\Bbb P)^V$. So you need to have $HOD(\Bbb P)=HOD$. Which is indeed the case for $L[\mu]$. $\endgroup$ – Asaf Karagila Mar 31 '15 at 7:56
8
$\begingroup$

Assume $GCH$ and let $\kappa$ be $(\kappa+2)-$strong. Force with extender based Prikry forcing $P$ with interleaved collapses to make $\kappa=\aleph_\omega$ and $2^{\aleph_\omega}=\aleph_{\omega+2}.$ Call the resulting extension $V[H].$ Also let $V[G]$ be an intermediate submodel, which just adds the Prikry sequence related to the normal measure and adds collapses, so that the following holds:

(1) $V[G] \subset V[H]$ have the same cardinals and bounded subsets of $\kappa=\aleph_\omega,$

(2) $V[G] \models GCH.$

Note that there are many new $\omega$-sequences through $\kappa=\aleph_\omega$ in $V[H]\setminus V[G].$

A much stronger version of the following lemma will appear in my paper "$HOD, V$ and the $GCH$" (where extender based Prikry forcing is replaced by extender based Radin forcing):

Homogeneity lemma. Assume $p,q\in P$ are such that $\pi(p)$ is compatible with $\pi(q),$ where $\pi$ is the projection map. Then there are $p' \leq p, q' \leq q$ and an isomorphism $\Phi: P/p' \simeq P/q'.$

It follows from the above lemma that $HOD^{V[H]} \subseteq V[G].$

Now force over $V[H]$ to code any bounded subset of $\aleph_\omega$ into $HOD$ using a homogeneous forcing, call the extension $V[H][K].$

Now $V[G] \subset V[H][K],$ and any new $\omega-$sequence cofinal in $\kappa$ which is in $V[H]\setminus V[G]$ witnesses $\min\{\alpha: P(\alpha) \nsubseteq HOD\}=\aleph_\omega$.

Remark. In fact it seems we need much weaker assumption. All we need is to be able to add a new cofinal $\omega$-sequence in $\kappa$ which is not in $V[G],$ and it seems to me that for this just a measurable is sufficient.

$\endgroup$
  • $\begingroup$ Thank you. I don't understand why the quotient forcing is (cone) homogeneous. The usual proof for the weakly homogeneity of the Prikry forcing uses the fact that if $p, q$ has the same stem then they are compatible. This is not the case in the quotient forcing (since in order for a conditions to be compatible the intersection of the large sets must contain an element that can be projected to the next element in the normal Prikry sequence from $G$). $\endgroup$ – Yair Hayut Oct 19 '15 at 19:12
  • $\begingroup$ @YairHayut First, in fact the whole forcing is cone homogeneous. Second, by homogeneity of quotient forcing I essentially mean the following: If p, q are in the extender based forcing $P$, and they have the same projection, then there are $p^* \leq p, q^* \leq q$ and an isomorphism from $P/p^*$ onto $P/q^*.$ This is sufficient for homogeneity argument. $\endgroup$ – Mohammad Golshani Oct 21 '15 at 12:17

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.