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The following concept is due to Shelah and I have some issues with a claim using this notion: Suppose that $\nu$ is a limit ordinal and that $P_\nu$ is an iteration of forcing notions. We say that a $P_\nu$ name $\dot{\alpha}$ of an ordinal is $prompt$ iff the following two things hold:

  1. $\Vdash_\nu \dot{\alpha} \le \nu$

  2. If $p \Vdash_\nu "\dot{\alpha} = \xi"$ then even $p \upharpoonright \xi ^\smallfrown 1_\nu \upharpoonright [\xi, \nu) \Vdash_\nu \dot{\alpha} = \xi$ ( $1_\nu$ should be the largest element of the iteration, and $\xi$ is the hacek name of an ordinal though I refused to write the hacek)

Then the following two things should hold:

  1. If $\dot{\alpha}$ is prompt $\eta \le \nu$ and if $p \Vdash_\nu \eta \le \dot{\alpha}$ then $p \upharpoonright \xi ^\smallfrown 1_\nu \upharpoonright [\xi, \nu) \Vdash_\nu \eta \le \dot{\alpha}$
  2. If $\dot{\alpha_i}$ are prompt then so is the supremum $sup$ and the minimum $min$

I have problems proving those two assertions so any help would be highly appreciated. Thank you!

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closed as too localized by Asaf Karagila, Amzoti, Chris Godsil, Henry T. Horton, Guess who it is. Jun 1 '13 at 5:39

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By $p|\xi\frown 1_\nu | [\xi,\nu)$ you mean the condition $q$ which agrees with $p$ up to $\xi$ and it has constant value the largest element after $\xi$. –  azarel Feb 8 '12 at 18:01
I would suggest that nontrivial questions about forcing are better placed on mathoverflow. (And this would seem especially true for questions about iterated forcing.) –  JDH Feb 8 '12 at 19:37
It was suggested in a moderator flag that this question be migrated to MathOverflow. Just wanted to explain that MathOverflow is not a part of the SE 2.0 network (or at least, not yet), so question migration is not possible - it would have to be reposted by hand. –  Zev Chonoles Feb 9 '12 at 7:31
Ok, its now posted on MO –  Stefan Hoffelner Feb 9 '12 at 10:09
Posted on MO: mathoverflow.net/questions/87976/… –  Asaf Karagila Feb 9 '12 at 10:26

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