Equivalent definitions of $(M,P)$-genericity - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T17:19:43Z http://mathoverflow.net/feeds/question/43836 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/43836/equivalent-definitions-of-m-p-genericity Equivalent definitions of $(M,P)$-genericity Stefan Hoffelner 2010-10-27T17:08:08Z 2010-10-27T19:39:39Z <p>Hi!</p> <p>I started to read the chapter 31 in Jechs book about proper forcing. Unfortunately it is written in a rather sketchy way and I do have some issues in proving a lemma about two equivalent definitions of $(M,P)$-genericity. Jech gives the following definition:</p> <p>Let $(P,&lt;)$ be a fixed notion of forcing, $\lambda > 2^{|P|}$ and let $M \prec (H_{\lambda}, \in, &lt;, ..)$, where $H_{\lambda}$ is the set of all sets whose transitive closure has size less than $\lambda$, and $' &lt;'$ is a well-ordering of $H_{\lambda}$. Then a condition $q \in P$ is $(M,P)$-generic if for every maximal antichain $A \in M$, the set $A \cap M$ is predense below $q$.</p> <p>He moves on to state a small lemma, saying that its proof is a routine verification (that leads us to the assertion that I am not an old hand):</p> <p>Assume the same situation as in our definition above. Then the following are equivalent:</p> <p>(i) $q$ is $(M,P)$-generic.</p> <p>(ii) $q \Vdash \dot{G} \cap M$ is a filter on $P$ generic over $M$</p> <p>The (ii) $\Rightarrow$ (i) direction is quite easy but I couldn't find a proof for (i) $\Rightarrow$ (ii).</p> <p>Thank you</p> http://mathoverflow.net/questions/43836/equivalent-definitions-of-m-p-genericity/43859#43859 Answer by Amit Kumar Gupta for Equivalent definitions of $(M,P)$-genericity Amit Kumar Gupta 2010-10-27T19:39:39Z 2010-10-27T19:39:39Z <p>Let $G$ be a $(V,P)$-generic filter containing $q$. We want to show that $G \cap M$ is a filter on $P$ generic over $M$. $G \cap M$ will be countable so it typically won't be closed upwards in $P$, the relevant point is that it's closed upwards in $P \cap M$. So it's not hard to see that it's a filter on $P \cap M$. Next we need to see it's generic over $M$.</p> <p>Let $A \in M$ be a maximal antichain in $P$. Let $p$ be the sole element of $A \cap G$. Let $r$ extend both $p$ and $q$, which is possible because they're both in $G$ and $G$'s a filter. By (i), $A \cap M$ is predense below $q$, and $r \leq q$, so there's some $p' \in A \cap M$ compatible with $r$. But since $A$ is a maximal antichain, there's precisely one element of $A$ compatible with $r$, and we already know $p \in A$ is compatible with $r$ (in fact it's above $r$), so $p' = p$. Therefore $p \in A \cap G \cap M$, meaning $G \cap M$ meets $A$. Since $A$ was arbitrary, $G \cap M$ meets every maximal antichain belonging to $M$, hence it's generic over $M$.</p> <p>Just as genericity of a filter can be defined in terms of dense sets, open dense sets, predense sets, or maximal antichains, it can so be done for genericity of a condition. See Lemma 1.2 in my notes below; it gives three equivalent conditions for $(M,P)$-genericity. The first and third are just like your (i) and (ii) but with antichains replaced with dense sets. The second one is also in terms of dense sets but slightly different from your (i) and (ii), and it's another useful way of looking at $(M,P)$-genericity.</p> <p><a href="http://math.berkeley.edu/~akgupta/PFA.pdf" rel="nofollow">http://math.berkeley.edu/~akgupta/PFA.pdf</a></p>