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Post-Doctoral 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 Gitik-Magidor 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 Gitik-Magidor 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 Scott-Solovay 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.