Recall that given a finite language $\mathcal{L}$, we say that an $\mathcal{L}$-structure is *computably saturated* (or *recursively saturated*) if for any computable set $\Sigma(\bar{x},y)$ of $\mathcal{L}$-formulas in the variables $\bar{x}y$ and any $\bar{a} \in M^{\bar{x}}$, if $\Sigma(\bar{a},y)$ is finitely satisfiable in $M$, then it is satisfied in $M$. An easy inductive argument shows that every countable $\mathcal{L}$-structure has a countable computably saturated elementary extension. Another easy argument shows that any expansion of a computably saturated structure by finitely many constants is still computably saturated.


One important property of countable computably saturated structures is resplendence, which for our purposes can be defined like this: An $\mathcal{L}$-structure $M$ is *resplendent* if for any finite extensions $\mathcal{L}' \supseteq \mathcal{L}$ and any $\mathcal{L}'$-sentence $\varphi$ that is consistent with $\mathrm{Th}(M)$, there is an expansion $M'$ of $M$ that is a computably saturated model of $\mathrm{Th}(M)\cup\{\varphi\}$.


All countable computably saturated structures are resplendent. Moreover, the same is true after any expansion by any finite number of constants.


Forcing is typically defined in terms of well-founded models of $\mathsf{ZFC}$, but, as discussed in [the answers to this question](https://mathoverflow.net/q/48522/83901), forcing ultimately makes sense over arbitrary models of $\mathsf{ZFC}$: Given an model $M$ of $\mathsf{ZFC}$ and a forcing poset $\mathbb{P} \in M$, we can consider the language that contains $\in$, $\mathbb{P}$, and a unary predicate $G$. There is a single sentence $\chi(\mathbb{P},G)$ in this language that says that $\mathbb{P}$ is a forcing poset and $G$ is a generic filter on $\mathbb{P}$. As Emil Jeřábek points out in his answer to that question, $\mathsf{ZFC} \cup \{\chi(\mathbb{P},G)\}$ is a conservative extension of $\mathsf{ZFC}$. (The rest of the work of a forcing argument is showing that $(M,\mathbb{P},G)$ interprets an end extension of $M$ modeling $\mathsf{ZFC}$ in which $G$ is a set. We don't really need to worry about that for the sake of this question.)


We have, furthermore, that the theory $\mathrm{eldiag}(M) \cup \{\chi(\mathbb{P},G)\}$ is consistent. If $M$ is a countable computably saturated model of $\mathsf{ZFC}$, then, by resplendence, we have that for any forcing poset $\mathbb{P}\in M$, there is a $G \subseteq \mathbb{P}$ which is $M$-generic such that $(M,\mathbb{P},G)$ is computably saturated. (In particular, this means that the actual forcing extension $M[G]$ will be computably saturated as well.)


My question is about whether this always happens, 'usually' happens, or 'usually' doesn't happen for various choices of $G$.



> 
> **Question 1:** Let $M$ be a countable computably saturated model of $\mathsf{ZFC}$, and let $\mathbb{P} \in M$ be a forcing poset. If $G \subseteq \mathbb{P}$ is an $M$-generic filter, does it follow that $(M,\mathbb{P},G)$ is computably saturated?
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I doubt that this does in fact always hold, but I don't know how to build a counterexample. What's less certain to me is whether a 'sufficiently generic' $M$-generic filter results in a computably saturated expansion.


To state this carefully, we need the following observation: Fix a countable computably saturated model $M$ of $\mathsf{ZFC}$ and a forcing poset $\mathbb{P} \in M$. Let $A$ be the set of all elements $\alpha$ of $\mathbb{P}^\omega$ such that for each $i<\omega$, $\alpha(i+1) \leq \alpha(i)$. $\mathbb{P}^\omega$ can easily be identified with Baire space, and $A$ is a $G\_\delta$ subset of $\mathbb{P}^\omega$ and therefore itself a Polish space. It's easy to see that the set $$B = \{\alpha \in A : \alpha\text{ generates an }M\text{-generic filter}\}$$ is comeager in $A$. (This is essentially the Rasiowa–Sikorski lemma.) For any $\alpha \in A$, let $G\_\alpha = \{p \in \mathbb{P} : p \geq \alpha(i)\text{ for some }i\}$. (This is what we mean by 'generating' a filter.)



> 
> **Question 2:** Is the set $$\{\alpha \in B : (M,\mathbb{P},G\_\alpha)\text{ is computably saturated}\}$$ always comeager in $A$?
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I would also be interested in answers involving a weaker set theory than $\mathsf{ZFC}$.



How often are forcing extensions of countable computably saturated models of $\mathsf{ZFC}$ computably saturated?

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> The answer to Question 1 is positive (thus the answer to Question 2 is also positive). More explicitly, the positive answer to Question 1 follows from the following well-known facts:
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**Lemma 1.** $(M,\mathbb{P},G)$ *is parametrically definable in* $M[G]$.


**Lemma 2.** $M[G]$ *is recursively saturated*.


**Lemma 3.** *Any structure that is parametrically definable in a computably saturated structure is also computably saturated.*


Lemma 1 is due independently to Laver and Woodin, who proved that $M$ is parametrically definable in $M[G]$; see [this MO post of Hamkins](https://mathoverflow.net/questions/83203/definability-of-ground-model#:%7E:text=The%20answer%20is%20yes.,W%20using%20parameters%20in%20V.) for more detail.


Lemma 2 was proved as part of the proof of Theorem 2.6 of [this paper](https://www.researchgate.net/publication/243117837_Counting_models_of_set_theory) of mine. You can also find a proof of Lemma 2 in [this blogpost](http://kamerynjw.net/2019/08/07/recursive-saturation.html) of Kameryn Williams (see the third proposition). The same blogpost also includes a reference to another result of mine which shows that Lemma 2 can fail if $\mathbb{P}$ is a proper class notion of forcing in $M$.


Lemma 3 follows from the definitions involved.


Finally, let me point out that Lemma 2 is true for all models $M$ of ZF, but I do not know the status of Lemma 1 for a model of ZF in which AC fails.


