4
$\begingroup$

Suppose $M$ is a countable transitive model for ZFC. Suppose $X\subset M\cap \textbf{ON}$ is bounded in $M\cap \textbf{ON}$. Why does $M[X]$ (the least ctm $N$ such that $M \subset N$ and $X \in N$) exists? Furthermore, is it true that there exists a forcing notion $P \in M$ and a $P$-generic filter over $M$, $G$, such that $M[G]=M[X]$?

I have looked for this information in some books such as Kunen, Jech and Kanamori's but I have found nothing.

Edit: @Elliot Glazer's answer shows that $M[X]$ is not always a generic extension. So I am changing the question a bit.

Suppose that there exists a forcing notion $Q$ and a generic filter $H$ over $M$ such that $X \in M[H]$ is a subset of $M\cap \mathbf{ON}=M[H]\cap \mathbf{ON}$. Is it true that there exists a is a forcing notion $P \in M$ and a $P$-generic filter over $M$, $G$, such that $M[G]=M[X]$?

Improve this question
$\endgroup$
4
  • 3
    $\begingroup$ No, $M[X]$ is not necessarily a model of anything interesting, and even if it is a model of set theory, it is not necessarily a forcing extension. For example, $X$ could be $0^\sharp$, seen as a subset of $\omega$, or it could be a subset of $\omega$ coding a well-ordering of $\omega$ in order type an ordinal larger than $\mathsf{ORD}^M$. $\endgroup$
    – Andrés E. Caicedo
    May 15, 2017 at 0:44
  • $\begingroup$ You are right @Andrés E. Caicedo. I edited the question, changing it a bit. $\endgroup$
    – user46747
    May 15, 2017 at 0:57
  • $\begingroup$ I suppose $M\cap\mathbf{ON}=M[G]\cap\mathbf{ON}$ is really meant to be $M\cap\mathbf{ON}=M[H]\cap\mathbf{ON}$, right? In this form, the result is true, and we can say a bit more about the nature of the forcing poset $P$. $\endgroup$
    – Andrés E. Caicedo
    May 15, 2017 at 1:13
  • 2
    $\begingroup$ See Lemma 15.43 in Jech's set theory book (2002). $\endgroup$
    – Andrés E. Caicedo
    May 15, 2017 at 1:19

2 Answers 2

Reset to default
3
$\begingroup$

Concerning your modified question, as it is commented by Andres Caicedo, the answer is yes; in fact we have the following theorem:

Theorem 1. Assume $V[G]$ is a set forcing extension of $V$ by a complete Boolean algebra $\mathbb{B}$ and suppose $N$ is a model of $ZFC$ with $V \subseteq N \subseteq V[G]$. Then for some regular complete subalgebra $\mathbb{C}$ of $\mathbb{B}$, such that $N=V[G \cap \mathbb{C}]$.

On the other hand, if we extend a bit the notion of genericity, then we have the following theorem of Mack Stanley (I don't recall the paper in which the result is stated):

Theorem 2. Suppose $V \subseteq W$ are countable transitive models of $ZFC$ with the same ordinals. Then $W$ is a class generic extension of $V$ (with respect to this extended notion of genericity).

Improve this answer
$\endgroup$
3
$\begingroup$

It is not true in general that $M[X]$ is a generic extension of $M,$ even in the case $X \subset \omega.$ For example, such an $X$ could code a well-ordering of $\omega$ of type $o(M),$ in which case $M[X]$ has more ordinals than $M$ and is thus not a generic extension.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thanks for the answer @Elliot Glazer. I edited the question. $\endgroup$
    – user46747
    May 15, 2017 at 1:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.