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Is it consistent with $\sf ZFC + \text{ countable models of } ZFC \text { exist}$, that every countable model of $\sf ZFC$ is a subset of some parameter free definable pointwise-definable model of $\sf ZFC$?

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The answer is in the positive. Since by the completeness theorem the existence of a countable model of ZFC is equivalent to Con(ZFC), the argument below works with the assumption that ZFC + Con(ZFC) is consistent.

Wojowu's answer already points out that the proof of the following fact (S) can be found in the paper Pointwise Definable Models of Set Theory (JSL 2013) by Hamkins, Linetsky, and Reitz.

(S) Every countable model of ZFC has a (class) generic extension to model of ZFC that is pointwise definable.

[Historical note: (S) was asserted with a proof outline in my paper Models of set theory with definable ordinals, which includes a detailed proof of the weaker statement: every countable model of ZFC has a generic extension in which all the ordinals are parameter-free definable. The paper of Hamkins-Linetsky-Reitz presents a detailed proof of not only (S), but also of a corresponding result for countable models of GBC].

Note that (S) is a theorem of ZFC (indeed, it is provable in much weaker systems , but that is a different story). It is easy to see that if the theory ZFC + Con(ZFC) is consistent, then so is $T$ = ZFC + Con(ZFC) + V = L. The fact that $T$ includes the statement V = L implies that $T$ has a global well-ordering of its ambient universe and therefore $T$ has definable Skolem functions. This immediately implies that $T$ has a pointwise definable model $M$. Since (S) is a theorem of ZFC, it holds in $M$, and moreover EVERYTHING in $M$ is parameter-free definable. Thus $M$ is a model in which every countable $N$ model of ZFC has a (class) generic extension $N[G]$ such that (1) $N[G]$ is pointwise definable, and (2) $N[G]$ is parameter-free definable in $M$.

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This is not only consistent, it is provably true. From Hamkins-Linetsky-Reitz's Pointwise Definable Models of Set Theory:

Theorem 11. Every countable model of ZFC has a pointwise definable class forcing extension.

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  • $\begingroup$ What is the proof that these forcing extensions are parameter free definable? Or that at least one of them should be? $\endgroup$
    – Zuhair Al-Johar
    Commented Jun 22, 2023 at 12:21
  • $\begingroup$ @ZuhairAl-Johar I slightly misunderstood your question. The property you ask for is not itself expressible in the language of ZFC, but it will hold in any pointwise definable model of ZFC. $\endgroup$
    – Wojowu
    Commented Jun 22, 2023 at 16:06
  • $\begingroup$ I don't think so, because you have continuum many of them $\endgroup$
    – Zuhair Al-Johar
    Commented Jun 22, 2023 at 16:35
  • $\begingroup$ To complete your proof you need to add the proof that at least one of those forcing extensions is parameter free definable, like proving that it is an element of some point-wise definable model of ZFC, or something similar to that. $\endgroup$
    – Zuhair Al-Johar
    Commented Jun 22, 2023 at 17:19
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    $\begingroup$ Looking back at the question, and taking it as face value, the answer is trivially yes, since there are models of ZFC in which Con(ZFC) fails, and therefore sentences of the form "every model of ZFC has such-and-such property" are automatically true. $\endgroup$
    – Ali Enayat
    Commented Jun 23, 2023 at 13:55

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