Equivalent definitions of M-genericity. - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T13:01:59Z http://mathoverflow.net/feeds/question/35825 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/35825/equivalent-definitions-of-m-genericity Equivalent definitions of M-genericity. unknown (google) 2010-08-17T01:48:35Z 2010-08-17T02:15:38Z <p>I'm trying to learn about forcing, and have heard that there are several equivalent ways to define genericity. For instance, let M be a transitive model of ZFC containing a poset (P, &le;). Suppose G &sube; P is such that q &isin; G whenever both p &isin; G and q &ge; p. Suppose also that whenever p,q &isin; G then there is r &isin; G such that r &le; p and r &le; q. Then the following are equivalent ways to say that G is generic:</p> <p>(1) G meets every element of M dense in P. That is, for all D &isin; M, if for all p &isin; P there is q &isin; D such that q &le; p, then G &cap; D is nonempty.</p> <p>(2) G is nonempty and meets every element of M dense below some p &isin; G. That is, for all p &isin; G and all B &isin; M, if for each q &le; p there is r &isin; B such that r &le; q, then G &cap; B is nonempty.</p> <p>Proving this equivalence seemed like it would be an easy exercise, but I think I'm missing something. Can someone point me toward a source where I can find a proof? I hope this is an acceptable question; this is my first time posting.</p> <p>EDIT: Typo and omission fixed.</p> http://mathoverflow.net/questions/35825/equivalent-definitions-of-m-genericity/35828#35828 Answer by Joel David Hamkins for Equivalent definitions of M-genericity. Joel David Hamkins 2010-08-17T01:58:48Z 2010-08-17T02:15:38Z <p>If \$G\$ satisfies (1), then it satisfies (2) because if \$p\$ is in \$G\$ and \$D\$ is dense below \$p\$, then let \$D'\$ be the set of conditions \$q\$ which are either in \$D\$ or incompatible with \$p\$. This is dense in \$P\$ since any condition that is compatible with \$p\$ will have elements of \$D\$ below it, and any condition incompatible with \$p\$ is already in \$D'\$. But \$G\$ cannot meet \$D'\$ in something incompatible with \$p\$, by your assumption on \$G\$, and so it must meet it in \$D\$, as desired. </p> <p>Conversely, if \$G\$ satisfies (2), then it will satisfy (1) because if \$D\$ is dense, then it is dense below any \$p\$, and so \$G\$ will meet it.</p>