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Timeline for A question on ultrapower

Current License: CC BY-SA 2.5

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Feb 11, 2010 at 1:54 comment added Joel David Hamkins I have posted a revised answer, for the case mu_1 not in ran(j_01). Here, the answer is affirmative.
Feb 11, 2010 at 0:17 history edited Joel David Hamkins CC BY-SA 2.5
Revised answer for case mu_1 not in ran(j_01).
Feb 6, 2010 at 2:08 vote accept Ant emyy Lee
Feb 5, 2010 at 10:03 vote accept Ant emyy Lee
Feb 5, 2010 at 10:03
Feb 5, 2010 at 10:02 vote accept Ant emyy Lee
Feb 5, 2010 at 10:02
Feb 3, 2010 at 13:38 comment added Joel David Hamkins Welcome to MO! Ah, I read your question more carefully now. You are right: my answer applies to the case where mu_1 IS in the range of j_01, for then mu_1=j_01(nu), and in this case, j02 is precisely the ultrapower by mu_0 x nu, since every object in M_2 would be j02(f)(kappa_0,kappa_1), and you map this to [f]_{mu_0xnu} for an isomorphism. For your case, where mu_1 is NOT in the range of j_01, then I believe the answer is Yes, but this is more complicated, and I will post an answer later.
Feb 3, 2010 at 3:20 comment added Ant emyy Lee So the model-theoretic fact you said in your answer may not make sense in this case, since $\mu_1$ is not a normal measure in $V$. How to deal with this case?
Feb 3, 2010 at 3:20 comment added Ant emyy Lee I am the student exchange you with email, you introduce me to go here, so I register this new name. In my question, I require $\mu_1$ is not in the range of $j_{0,1}$, i.e. the second step of the iterated ultrapower is done by a normal measure not in $j_{0,1}''V$. For example, suppose $\kappa$ has $\kappa$ many normal measures <U_\alpha:\alpha<\kappa>, the first step of the iterated ultrapower I use $U_0$, and the second step of the iteration, I use the \kappa-th measure in the sequence j_{0,1}(<U_\alpha:\alpha<\kappa>) .
Feb 2, 2010 at 5:14 history answered Joel David Hamkins CC BY-SA 2.5