# Smallest Transitive Model Containing Certain Sets

Let $V$ denote the underlying universe. Let $M \in V$ be a transitive set or class such that $M \models \text{ZF}$. Let $x \in V$ be some set.

One often sees the notation for $M[x]$ and $M(x)$. If $x$ is a generic filter on a poset, the forcing construction using names gives a construction of $M[x]$. When $x$ is just an arbitrary set, what should $M[x]$ and $M(x)$ mean.

Taking some of the properties from $L[x]$ and $L(x)$, it seems that one should expect $M[x]$ to be the smallest transitive model of $\text{ZF}$ in $V$ such that $M \subseteq M[x]$ and $x \cap M[x] \in M[x]$. Also $M(x)$ should be the smallest transitive model of $\text{ZF}$ such that $M \subseteq M(x)$ and $x \in M(x)$.

Is this intended meaning of these $M[x]$ and $M(x)$? Moreover, can $M[x]$ and $M(x)$ be explicitly constructed in a hierarchy like $L[x]$ and $L(x)$?

Let $$\Omega$$ denote $$\mathsf{ORD}^M$$, the height of $$M$$. This may be $$\mathsf{ORD}$$ if $$M$$ is a proper class. If $$\Omega=\mathsf{ORD}$$, let $$V_\Omega$$ simply mean $$V$$. For $$\alpha<\Omega$$, let $$M_\alpha$$ denote $$M\cap V_\alpha=V_\alpha^M$$. Finally, for $$t$$ a set, let $$\mathrm{tc}(t)$$ denote the transitive closure of $$t$$, that is, the smallest transitive set that contains $$t$$ (as a subset).
Given $$x$$ whose transitive closure belongs to $$V_\Omega$$, by $$M(x)$$ we denote $$\bigcup_{\alpha<\Omega}L\bigl(M_\alpha\cup\mathrm{tc}(\{x\})\bigr).$$ As you anticipate, this is indeed the smallest transitive model of $$\mathsf{ZF}$$ that contains $$M$$ and has $$x$$ as an element.
In this sense, our notation for forcing is incorrect, and if $$M[x]$$ is a forcing extension, instead we should write $$M(x)$$ -- though, of course, it is too late to hope this will happen. This is probably a remnant of the time when the distinction in the case of $$L(x)$$ vs $$L[x]$$ was not followed as carefully as it is now, and one can easily find papers discussing $$L[\mathbb R]$$ when what is meant is $$L(\mathbb R)$$.
The classes $$M[x]$$ and $$M(x)$$ are sometimes used in contexts where $$M$$ is $$\mathsf{OD}$$ or subclass of $$\mathsf{OD}$$ that is (internally) definable. In those instances, the internal definability of the class $$M$$ allows us to define $$M[x]$$ and $$M(x)$$ by straightforward generalizations of the $$L[x]$$ and $$L(x)$$ classes. (See for instance Chapter 13 of Jech's Set theory book.)
I've only seen the notation $$M[x]$$ used in a setting where the above is not the case, $$M[x]$$ is not a forcing extension, and $$x\notin M[x]$$. Namely, when $$M$$ itself is internally defined, just as $$L$$, through a hierarchy each level $$N'$$ of which extends the previous one $$N$$ by adding certain constants or relations definable over $$N$$. This is the case when $$M$$ is a fine structural model over some predicate. For instance, models of the form $$L(\mathbb R)[\vec E]$$ where $$\vec E$$ is a coherent sequence of extenders. In these cases, there is no ambiguity with the notation, as we just apply the same distinction as with $$L$$, and the models are anyway defined through a transfinite recipe that explicitly indicates how each level is formed. But if $$M$$ is not carefully presented through such a hierarchy, I do not see how one would make sense of the square bracket notation unambiguously. (Note in particular that what Jech calls $$M[x]$$ in exercise 13.34 is really $$M(x)$$.)
• I realize this is an old question, but would you mind adding why $M(x)\models$ Comprehension? I'm trying to mimic the reflection argument we use to show that $L\models ZF$, but I keep running into trouble. You can write $M(x)$ as a union of sets indexed by $\Omega\times ORD$, which you can then well order in type $ORD$, but you need continuity of the hierarchy at limits for the Reflection theorem to apply. Jul 21, 2020 at 23:39