I want to show that given a countable transitive model $M$ a notion of forcing $\mathbb P$ and a generic filter $G$, then forcing over a dense set of $\mathbb P$, $D$ is just as forcing over $\mathbb P$, i.e., it is enough to use only conditions from the dense subset in order to create $M[G]$.

Is it true? I was trying to prove it with induction, but I got confused. do I need to go over the naming process with conditions from $D$?

  • $\begingroup$ Is $D$ a poset? $\endgroup$
    – Jonny
    Apr 20, 2015 at 19:24
  • $\begingroup$ @jonny Since it's a subset of a poset the inherited relation makes it a poset. It might not have a maximum element but that shouldn't necessarily matter. $\endgroup$
    – DRF
    Apr 20, 2015 at 19:27

1 Answer 1


Yes, this is true.

Depending on how you defined genericity, or whether or not you have seen the following equivalences, this is not very difficult:

Let $M$ be a countable model of set theory, $\Bbb P\in M$ a partial order and $G\subseteq\Bbb P$ a filter. Then $G$ is generic if and only if one of the following hold:

  1. For every $D\in M$ which is a dense subset of $\Bbb P$, $G\cap D\neq\varnothing$.

  2. For every $D\in M$ which is a dense open subset of $\Bbb P$, $G\cap D\neq\varnothing$.

  3. For every $D\in M$ which is a pre-dense subset of $\Bbb P$, $G\cap D\neq\varnothing$.

  4. For every $D\in M$ which is a maximal antichain of $\Bbb P$, $G\cap D\neq\varnothing$.

Now use the first characterization, and note that if $E\subseteq D$, $E$ is dense in $D$ and $D$ is dense in $\Bbb P$, then $E$ is dense in $\Bbb P$.

Since every $D$-name is a $\Bbb P$-name, interpreting using $G$ is not a problem. All you need to show is that every $\Bbb P$-name has an equivalent $D$-name, and that can be done by induction on the rank of the name.

If $\dot x$ is a $\Bbb P$-name, we may assume that if $\dot y$ appears in $\dot x$, then $\dot y$ is a $D$-name; now just replace the conditions in $\dot x$ by conditions in $D$ and we're done.


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