Normal measures on $P_{\kappa }(\lambda )$ extend the club filter - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T20:05:23Z http://mathoverflow.net/feeds/question/61075 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/61075/normal-measures-on-p-kappa-lambda-extend-the-club-filter Normal measures on $P_{\kappa }(\lambda )$ extend the club filter Amit Kumar Gupta 2011-04-08T15:58:21Z 2011-04-09T04:44:41Z <p>This is (a variation on) exercise 20.4 in Jech's "Set Theory." Let $j : V \to M$ witness $\lambda$-supercompactness of $\kappa$, and consider the normal measure $U$ on $P_{\kappa}(\lambda)$ consisting of those $x$ such that $j[\lambda] \in j(x)$. (How do you make the left quotation mark symbol to denote 'j-image-of-lambda'?) </p> <p><b>We want to show that this measure extends the club filter.</b></p> <p>This hint is as follows: Suppose $C$ is club. Then define $D = j[C]$. Then: </p> <ol> <li>$D$ is a directed subset of $j(C)$.</li> <li>$D$ has size $|C| \leq \lambda ^{&lt; \kappa} &lt; j(\kappa )$.</li> <li>Therefore $\bigcup D \in j(C)$.</li> <li>$\bigcup D = j[\lambda ]$</li> </ol> <p>I'm fine with 1. I'm not sure about 2 - where is the argument taking place, in $V$ or in $M$, or both? For 3, it appears the underlying argument is this:</p> <p>$V \vDash \forall E \subset C\ (E$ directed and $|E| &lt; \kappa \Rightarrow \bigcup E \in C)$</p> <p>and so</p> <p>$M \vDash \forall E \subset j(C)\ (E$ directed and $|E| &lt; j(\kappa) \Rightarrow \bigcup \in j(C))$</p> <p>I can accept this assuming that 2 means "$D \in M$ and $M \vDash |D| &lt; j(\kappa )$." I'm having trouble with 4 as well - I believe that $j[\lambda] \subseteq \bigcup D$, but why does the reverse inclusion hold, i.e. why is it that $x \in C, \beta \in j(x) \Rightarrow \beta \in j[\lambda]$?</p> http://mathoverflow.net/questions/61075/normal-measures-on-p-kappa-lambda-extend-the-club-filter/61101#61101 Answer by Jason for Normal measures on $P_{\kappa }(\lambda )$ extend the club filter Jason 2011-04-08T22:47:56Z 2011-04-09T04:44:41Z <p>Suppose $\kappa$ is $\lambda$-supercompact for some $\lambda \geq \kappa$, and let $j: V \rightarrow M$ be an elementary embedding with critical point $\kappa$ such that $j(\kappa) > \lambda$ and $M^{\lambda} \subseteq M$ for some inner model $M$. First, observe that $V$ and $M$ agree on $P_{\kappa}\lambda$ because $M$ is closed under ${&lt;}\kappa$ sequences. In particular, this means that $\lambda^{{&lt;}\kappa} \leq (\lambda^{{&lt;}\kappa})^M$ since $M \subseteq V$. But this then means that $j(\kappa) > (\lambda^{{&lt;}\kappa})^M \geq \lambda^{{&lt;}\kappa}$ because $j(\kappa)$ is inaccessible in $M$ and $j(\kappa)$ is greater than both $\lambda$ and $\kappa$. Next, note that any $x \in P_{\kappa}\lambda$ will be a subset of $\lambda$ having size less than the critical point $\kappa$ so that $j(x) = j''x \subseteq j''\lambda$.</p> <p>[Specifically, if for some $\alpha &lt; \kappa$, we have a bijection $f: \alpha \rightarrow x$, then $j(f)$ will be a bijection between $j(\alpha) = \alpha$ and $j(x)$. So every element of $j(x)$ is of the form $j(f)(\beta)$ for some $\beta &lt; \alpha$, but $j(f)(\beta) = j(f(\beta)) \in j''x$ since $\beta$ is also below the critical point.]</p> <p>Also, $M$ will contain $h = j \upharpoonright \lambda$ by its closure under $\lambda$ sequences. Therefore, $M$ will have $j''P_{\kappa}\lambda = \{j(x)| x \in P_{\kappa}\lambda\} = \{j''x| x \in P_{\kappa}\lambda\} = \{h''x| x \in P_{\kappa}\lambda\}$. Now letting $g: P_{\kappa}\lambda \rightarrow \lambda^{{&lt;}\kappa}$ be a bijection in $V$, we will have a bijection $j(g) \upharpoonright j''P_{\kappa}\lambda: j''P_{\kappa}\lambda \rightarrow j''\lambda^{{&lt;}\kappa}$ in $M$. Therefore, $M$ will have the range of $j(g)$, which is exactly $j''\lambda^{{&lt;}\kappa}$. Now, since $C$ has size at most $\lambda^{{&lt;}\kappa}$ (in $V$), we may let $e: \lambda^{{&lt;}\kappa} \rightarrow C$ be a surjection. Then $j(e) \upharpoonright j''\lambda^{{&lt;}\kappa}: j''\lambda^{{&lt;}\kappa} \rightarrow j''C$ is a surjection in $M$ so similarly, its range, $D = j''C$, will be in $M$. But $M$ will also know that $j''\lambda^{{&lt;}\kappa}$ has size $\lambda^{{&lt;}\kappa} &lt; j(\kappa)$ because $M$ can construct $j \upharpoonright \lambda^{{&lt;}\kappa}$ from $j''\lambda^{{&lt;}\kappa}$ by virtue of $j$ being order-preserving. Therefore $\bigcup D \in j(C)$ by the ${&lt;}j(\kappa)$-directed closure of $j(C)$ in $M$, as you mention.</p> <p>Also, if $x \in C$, then $|x| &lt; \kappa$ and $x \subseteq \lambda$ so $j(x) = j''x \subseteq j''\lambda$. Therefore, $\bigcup D = \bigcup j''C \subseteq j''\lambda$.</p>