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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Sep 24, 2013 at 6:07 vote accept CommunityBot
Sep 24, 2013 at 3:37 vote accept CommunityBot
Sep 24, 2013 at 6:07
Sep 21, 2013 at 15:03 history edited Joel David Hamkins CC BY-SA 3.0
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Sep 21, 2013 at 14:43 comment added Andrés E. Caicedo Joel, you may want to add a small remark to the theorem, indicating that adding a Cohen real results in a partial rather than linear order.
Sep 21, 2013 at 14:14 comment added Joel David Hamkins Yes, you are right. I should assume that $\mathbb{B}$ is atomless (and I had meant that $\mathbb{B}$ is a complete Boolean algebra).
Sep 21, 2013 at 14:00 comment added Asaf Karagila The comment seems a bit fishy. What if $\Bbb B$ is the completion of a trivial forcing which just chooses an ordinal below $\omega_1$? Certainly this has a lot of nontrivial complete subalgebras (defined by subsets of atoms) but in fact there are no inner models, because we didn't add anything to the universe.
Sep 21, 2013 at 13:57 comment added Joel David Hamkins The inner models arising in a forcing extension over $L$ by a forcing notion $\mathbb{B}$ are precisely isomorphic to the collection of complete subalgebras of $\mathbb{B}$.
Sep 21, 2013 at 13:31 history edited Joel David Hamkins CC BY-SA 3.0
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Sep 21, 2013 at 13:19 history answered Joel David Hamkins CC BY-SA 3.0