Can the first ordinal in which $V\neq HOD$ be $\aleph_\omega$?

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, $C\notin HOD^{V[C]}$.

• @MohammadGolshani: It sounds good. Can you describe the tail forcing? Why is it homogeneous in the intermediate model? – Yair Hayut Mar 31 '15 at 7:28
• I think that you also need to have that the Prikry forcing is definable in $L[\mu]$, not just weakly homogeneous. – Asaf Karagila Mar 31 '15 at 7:41
• @MohammadGolshani: I'll take a look at these papers. Thank you. – Yair Hayut Mar 31 '15 at 7:45
• @AsafKaragila The forcing is definable at least in intermediate submodel, and that's enough. – Mohammad Golshani Mar 31 '15 at 7:46
• @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 $L[\mu]$. – Asaf Karagila Mar 31 '15 at 7:56

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.

• 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 $G$). – Yair Hayut Oct 19 '15 at 19:12
• @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. – Mohammad Golshani Oct 21 '15 at 12:17