# The notations $M[X]$ and $V[A]$ and intermediate models

I am confused about Jech's notations $M[X]$ and $V[A]$.

Let us begin with exercise 13.34 (see [Jec02, p. 199]).

Let $\mathbf{M}$ be a transitive model of $\mathsf{ZF}$ containing all the ordinals and let $X$ be a subset of $\mathbf{M}$. Then there is a least model $\mathbf{M}[X]$ of $\mathsf{ZF}$ such that $\mathbf{M} \subseteq \mathbf{M}[X]$ and $X \in \mathbf{M}[X]$ and, if $\mathbf{M} \models \mathsf{AC}$, then $\mathbf{M}[X] \models \mathsf{AC}$.

One must realize that this is really a theorem schema as we are talking about proper classes: In fact, for each formula $\phi(x) \equiv x \in \mathbf{M}$ we have to find another formula $\psi(x, X) \equiv x \in \mathbf{M}[X]$ such that ...

Q1: Can someone give me a hint more helpful than Jech's one?

This leads to my next question. As far as I understand, forcing over $V$ and the notation $V[G]$ (where $G \notin V$ is a generic filter for some forcing poset $P \in V$) is just a sloppy notation. More precisely, one should always use a large enough finite fragment $\mathsf{ZFC}^*$ and some c.t.m. $M$ for $\mathsf{ZFC}^*$ (i.e. a set model) as $G \notin V$ is nonsense from a formalist's point of view.

Q2: So, how can I formalize Jech's results Lemma 15.40 up to Lemma 15.43 (see [Jec02, p. 247]) using the c.t.m. approach?

I am particularly puzzled as the analogue of exercise 13.34 for set models seems to fail.

[Jec02] Thomas Jech: "Set Theory: The Third Millennium Edition, revised and expanded". Springer, 2002

It is way too late to post an answer but let me take a crack at it anyways. So let $$M$$ be an inner model of $$\text{ZF}$$. We want to show that there exists a least inner model $$M[X]$$ of $$\text{ZF}$$, where $$X\subseteq M$$ and such that $$M\subseteq M[X]$$ and $$X\in M[X]$$.
We use relative constructibility over sets. For every $$\alpha\in \text{ORD}$$, define $$L_0((V_{\alpha}\cap M) \cup \{X\})=(V_{\alpha}\cap M)\cup \{X\}$$. At successor stages let $$L_{\alpha+1}((V_{\alpha}\cap M) \cup \{X\})=Def_{\mathcal{P}}(L_{\alpha}((V_{\alpha}\cap M) \cup \{X\}))$$ (definable power set). At limit stages $$\gamma$$, let $$L_{\gamma}((V_{\alpha}\cap M) \cup \{X\})=\bigcup_{\alpha<\gamma}((V_{\alpha}\cap M) \cup \{X\})$$. Then let $$L((V_{\alpha}\cap M) \cup \{X\})=\bigcup_{\gamma\in\text{ORD}}L_{\gamma}((V_{\alpha}\cap M) \cup \{X\})$$.
Finally define $$M[X]=\bigcup_{\alpha\in\text{ORD}}L((V_{\alpha}\cap M) \cup \{X\})$$. Doing this ensure that $$M\subseteq M[X]$$, because $$M[X]$$ contains every chunk of $$M$$, as these are contained in the $$L((V_{\alpha}\cap M) \cup \{X\})$$. Since also $$\{X\}\subseteq M[X]$$ then $$X\in M[X]$$. $$M[X]$$ satisfies $$\text{ZF}$$. If $$M\models \text{AC}$$, then wellorder $$X$$. Then for every $$\alpha, L((V_{\alpha}\cap M) \cup \{X\})\models \text{AC}$$ (which is not true in general). Feel free to add to this, I did it very quickly.
• I re-read your question. Sorry for not answering the second question. Let me try to give you some quick intuition for your second question. In forcing theory, people who follow the $V$ and $V[G]$ notation realize that there is no real meaning in writing $V[G]$. What this notation really is, is just a shorthand for writing what one can write with the forcing language and using the definability of the forcing in the ground model $V$. In this context what one must then do is define a new forcing relation $\Vdash^*$ by recursion without referring to any specific model. This is the approach taken.. Jan 5, 2018 at 2:45
• ...by Kunen in his book. With regards to your comment following Joel David Hamkins answer on the fact that it is consistent that there may not be a smallest, let me add the following observation: assume that $M$ is a countable transitive model of $ZFC$ and let $\mathbb{P}=Fn(\omega,2)$, also referred as $Add(\omega,1)$, the forcing notion for a Cohen real. Then there is a filter $G$ on $\mathbb{P}$ such that there is no transitive set model $M\subseteq N$ such that $G\in N$ and $N\models ZF-Power Set Axiom$ and $o(N)=o(M)$ where $o(M)$ is the ordinal height of $M$... Jan 5, 2018 at 2:58
• ...the reason why there cannot be be any such $N$ is the same as the arguments given by Joel David Hamkins. $G$ codes a wellorder of $\omega$ of order type larger $M$, revealing that all ordinals in $M$ are countable. Why this does not apply to Jech's exercise that you mentioned above is because one important assumption in Jech's exercise is to have $X\in M[X]$. But in this counterexample (as in Hamkins' counterexample) we have $G\notin M[G]$, $G$ cannot be generic. Jan 5, 2018 at 3:02