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PostDoctoral Research Fellow, School of Mathematics, Institute for Research in Fundamental Sciences (IPM).
My Shelah number is close to one.
16h

comment 
Can the first ordinal in which $V\neq HOD$ be $\aleph_\omega$?
I think the above idea does not work, the whole forcing is enough homogeneous so it seems $HOD$ of final extension is just the original core model $K.$ 
16h

answered  Antichain on $\mathcal{P}(\omega)/fin$ of cardinality $2^{\aleph_0}$? 
21h

comment 
Can the first ordinal in which $V\neq HOD$ be $\aleph_\omega$?
@AsafKaragila The forcing is definable at least in intermediate submodel, and that's enough. 
21h

comment 
Can the first ordinal in which $V\neq HOD$ be $\aleph_\omega$?
In fact we don't need to add many Prikry sequences, it suffices to add two different ones, such that they are not constructible from each other. This is possible if we consider the $\alpha$th Prikry sequence added, where $\alpha>\kappa$ is $\kappa_1$ as defined in Def. 2.1 of Merimovich's paper "Prikry on extenders revisited". 
21h

comment 
Can the first ordinal in which $V\neq HOD$ be $\aleph_\omega$?
Consider the paper "Supercompact extender based Prikry forcing" by Merimovich. His forcing is not difficult to see has enough weak homogenerity. By extender based Prikry forcing, I essentially mean a revised version of it, which works like GitikMagidor forcing, but has a much simpler presentation. So unlike in Merimovich paper, in this version we just add subsets of $\kappa,$ and preserve all cardinals. We mix it with collapses as in GitikMagidor paper. The tail essentially adds the rest of Prikry sequences. 
21h

comment 
Can the first ordinal in which $V\neq HOD$ be $\aleph_\omega$?
What about the following idea: Start with a core model for a strong cardinal $\kappa$. Do extender based Prikry forcing to turn $\kappa$ into $\aleph_{\omega},$ and make $2^{\aleph_\omega}=\aleph_{\omega+2}.$ Consider an intermediate submodel in which the normal Prikry sequence and the collapses are added, so that $\kappa$ becomes $\aleph_\omega$ in it. The tail forcing is enough homogeneous, and it adds no bounded subsets of $\kappa=\aleph_\omega,$ so the tail forcing extension is as required 
1d

comment 
Notions of infinity in $\mathsf{ZF}$ without choice
Now the question has changed and I removed my comment. With this version, the answer is no, as is shown in Asaf's answer. 
1d

comment 
Proofs of the uncountability of the reals.
Welcome to Mathoverflow. 
2d

awarded  Nice Question 
Mar 28 
revised 
Reference for proof that consistency of $\omega_1$Erdos cardinal implies Con(Chang's Conjecture)
added 21 characters in body 
Mar 28 
answered  Reference for proof that consistency of $\omega_1$Erdos cardinal implies Con(Chang's Conjecture) 
Mar 27 
asked  Applications of set theory in physics 
Mar 24 
awarded  Necromancer 
Mar 23 
awarded  Popular Question 
Mar 23 
awarded  Nice Answer 
Mar 16 
comment 
Characterising subsets of the reals as ordered spaces
The paper "Separable linear orders and universality" is related. 
Mar 16 
awarded  Nice Answer 
Mar 16 
answered  If $\kappa$ is weakly inaccessible and $A\subset\kappa$, can $L[A]$ violate $\kappa^{\lt\kappa}=\kappa$? 
Mar 16 
comment 
Does there exist a supercompactness theorem?
@ThomasBenjamin I have no idea, but maybe check to see if it is possible to extend the above proof. 
Mar 14 
comment 
ScottSolovay unpublished paper on ``Boolean valued models of set theory''
I am also interested to hear Prof. Solovay's reasons for not completing (and publishing) the work. 