Does a generic normal measure extend the club filter? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T10:46:59Z http://mathoverflow.net/feeds/question/61077 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/61077/does-a-generic-normal-measure-extend-the-club-filter Does a generic normal measure extend the club filter? Amit Kumar Gupta 2011-04-08T16:14:41Z 2011-04-10T19:05:27Z <p>This question is related to <a href="http://mathoverflow.net/questions/61075/normal-measures-on-p-kappa-lambda-extend-the-club-filter" rel="nofollow">this one</a>. The setup is as follows:</p> <p>In $V$, $\kappa$ is supercompact and $\mathbb{P} = \mathrm{Coll}(\kappa, \aleph_2)$ is the Levy collapse. $G$ is $(V,P)$-generic, and in $V[G]$, $C \subset P_{\kappa}(\lambda)$ is club. Let $j : V \to M$ be an embedding witnessing $\lambda$-supercompactness of $\kappa$. Lift this embedding to $j^{\ast} : V[G] \to M[G \times H]$ where we may assume $H$ is generic over $V[G]$ for the poset $\mathrm{Coll}^{V[G]}(j(\kappa), \aleph_2)$ (note $\aleph_2 ^{V[G]} = \kappa$). Now in $V[G \times H]$ we define $U \subset P_{\kappa}^{V[G]}(\lambda)$ by:</p> <p>$x \in U$ iff $j[\lambda] \in j^{\ast}(x)$.</p> <p>This $U$ belongs to $V[G \times H]$, but $V[G]$ "would" think its a normal measure on $P_{\kappa}^{V[G]}(\lambda)$, i.e. it's $\kappa$-complete if we restrict to $&lt;\kappa$-sequences in $V[G]$ and it's normal if we restrict to regressive functions $f : P_{\kappa}^{V[G]}(\lambda) \to \lambda$ where $f \in V[G]$. My question:</p> <p><b>Does $U$ also extend the club filter, if we restrict to clubs in $V[G]$?</b></p> http://mathoverflow.net/questions/61077/does-a-generic-normal-measure-extend-the-club-filter/61237#61237 Answer by Amit Kumar Gupta for Does a generic normal measure extend the club filter? Amit Kumar Gupta 2011-04-10T19:05:27Z 2011-04-10T19:05:27Z <p>With the help of Jason's answer to the question I linked to, I think I might be able to solve this one. The key is to show that $P_{\kappa}^{V[G]}(\lambda) \in M[G\times H]$. Let's simply denote this set by $X$. An element $x$ of $X$ can be regarded as a function $x : \omega_1 \to \lambda$, which will be a subset of $\omega_1 \times \lambda$. A nice name for such a subset is a map, in $V$, from $\omega_1 \times \lambda$ to the collection of antichains in $\mathbb{P}$. Since $\mathbb{P}$ has the $\kappa$-chain condition and $\mathbb{P} \in M$, $M$ correctly knows the set of antichains of $\mathbb{P}$. Since $M^{\lambda} \subset M$, $M$ correctly knows the set of nice $(V,\mathbb{P})$-names for subsets of $\omega_1 \times \lambda$, let's call this set $Y$.</p> <p>Now $\mathbb{P}$ names are $j(\mathbb{P})$ names (since $\mathbb{P} \subset j(\mathbb{P})$), so we get that</p> <p>$X = \{ \dot{x}^{G\times H}\ |\ \dot{x}^{G\times H} : \omega_1 \to \lambda, \dot{x} \in Y \}$ </p> <p>This is since $\dot{x}^{G \times H} = \dot{x}^G$ for nice $(V,\mathbb{P})$-names. So $X \in M[G \times H]$ as desired.</p> <hr> <p>Recall, we want to show that $U$ extends the club filter, so take $C \in V[G]$ club in $X$. Following the hint in the previous question, we want to show that </p> <ul> <li>$D = \{j^{\ast}(x)\ |\ x \in C\}$ belongs to $M[G \times H]$</li> <li>$D$ has size less than $j(\kappa)$ in $M[G \times H]$, and</li> <li>$\bigcup D = j''\lambda$. </li> </ul> <p>Then, since $D$ is a directed subset of $j^{\ast}(C)$, elementarity will give us that $j''\lambda \in j^{\ast}(C)$, as desired.</p> <p>Since $M^{\lambda} \subset M$, we know that $g := j\upharpoonright \lambda \in M$. So for $x \in X$ (and in particular for $x \in C$), $j^{\ast}(x) = j''x = g''x$. Using this it's not hard to see that $\bigcup D = j''\lambda$. It also implies that</p> <p>$j^{\ast}$ $''X$ $= \{j^{\ast}(x)\ |\ x \in X\}$ $= \{j''x\ |\ x \in X\} = \{g''x\ |\ x \in X\}$ </p> <p>belongs to $M$. Now if $h : X \to C$ is a surjection in $V[G]$, then $j^{\ast}(h)\upharpoonright j^{\ast}$ $''X$ is belongs to $M[G\times H]$ and its range is $D$, so $D \in M[G\times H]$. </p> <p>It remains to show $M[G\times H] \vDash |D| &lt; j(\kappa)$. Since $j(\kappa)$ is inaccessible in $M$, there's some bijection $i: \alpha \to Y$ in $M$ for some $\alpha &lt; j(\kappa)$. This gives a surjection $k : \alpha \to X$ in $M[G\times H]$. We can obtain a bijection $l : X \to j^{\ast}$ $''X$ via $l(x) = g''x$. And $j^{\ast}(h)\upharpoonright j^{\ast}$ $''X$ is a surjection onto $D$. So: </p> <p>$M[G\times H] \vDash |D| \leq |j^{\ast}$ $''X| = |X| \leq \alpha &lt; j(\kappa)$.</p>