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


1 Answer 1


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.

  • $\begingroup$ Thanks for your reply. I accept it (although it does not answer the whole question) because it is not of interest for me any more (too long ago and I don't understand it anymore). $\endgroup$
    – Justus87
    Commented Jan 4, 2018 at 10:23
  • $\begingroup$ 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.. $\endgroup$ Commented Jan 5, 2018 at 2:45
  • $\begingroup$ ...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 $M$... $\endgroup$ Commented Jan 5, 2018 at 2:58
  • $\begingroup$ ...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. $\endgroup$ Commented Jan 5, 2018 at 3:02
  • $\begingroup$ If you have other questions about set theory don't hesitate to ask! $\endgroup$ Commented Jan 5, 2018 at 3:03

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