Timeline for Normal measures on $P_{\kappa }(\lambda )$ extend the club filter

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Apr 10, 2011 at 15:40	comment	added	Amit Kumar Gupta		Thanks for the response! I think the argument in the second last paragraph can be simplified a little: If $g:P_{\kappa}(\lambda)\to C$ is a surjection in V, then $j(g)\upharpoonright j''P_{\kappa}(\lambda)$ belongs to $M$ by your observation that $j''P_{\kappa}(\lambda)$ does, and it surjects onto D. Also, by your observation that $P_{\kappa}^M(\lambda) = P_{\kappa}^V(\lambda)$, we can define a surjection $e : P_{\kappa}(\lambda) \to j''P_{\kappa}(\lambda)$ by $e(x) = h''x$. Thus in M, $\|D\| \leq \|P_{\kappa}(\lambda)\| = \lambda^{<\kappa} < j(\kappa)$ since $j(\kappa)$ is inaccessible in $M$.
Apr 9, 2011 at 15:48	vote	accept	Amit Kumar Gupta
Apr 9, 2011 at 6:31	comment	added	Jason		Small omission: should say range of $j(g)$ restricted to $j''P_{\kappa}\lambda$, which is exactly $j''\lambda^{{<}\kappa}$.
Apr 9, 2011 at 4:45	history	undeleted	Jason
Apr 9, 2011 at 4:44	history	edited	Jason	CC BY-SA 3.0	Elaborated on proof
Apr 9, 2011 at 2:50	history	deleted	Jason
Apr 9, 2011 at 0:40	history	edited	Jason	CC BY-SA 3.0	$M$ will not necessarily exhibit this closure if $j: V \rightarrow M$ wasn't an ultrapower embedding
Apr 8, 2011 at 22:47	history	answered	Jason	CC BY-SA 3.0