In their paper "Pointwise Definable Models of Set Theory" Hamkins, Linetsky, and Reitz prove the following theorem:

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

Let's consider, as our ground model, a c.t.m.--call it M-- that satisfies CH. By the theorem stated above, M has a pointwise definable class forcing extension M[G]. Can one provide a relatively simple example of such an M[G] where CH is false? Are there any limitations (apart from Koenig's Theorem) as to how such an M[G] can violate CH? Also, as regards set theorists living (so to speak) in M[G], do they believe that the language of set theory that describes their set-theoretic universe M[G] is a countable language?

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    $\begingroup$ I'm not sure what you mean by your last question. The language of set theory is a singleton, the single relation $\in$. $\endgroup$ – Miha Habič Jun 20 '14 at 11:15
  • $\begingroup$ @Miha: I have always understood that a formal language was to have constant symbols, variable symbols, and relation/function symbols, formation rules for well-formed formulas. In his answer to my question regarding the axiom schemata of Separation and Replacement on this website Asaf wrote that the language of set theory (ZFC in the case of my previous question and this question) had only countably many formulas. In their paper, Hamkins, Linetsky, and Reitz write, " ...if the ZFC axioms of set theory are are consistent then there are models of ZFC in which...every real number... $\endgroup$ – Thomas Benjamin Jun 20 '14 at 14:01
  • $\begingroup$ is uniquely definable without parameters. The authors also state that in any fixed structure $\mathcal M$ in a countable language the math-tea argument 'seems' to hold. So when one has a model of ZFC where CH fails and that model is pointwise definable, the number of formulas defining the reals must be the value of the continuum. What am I missing? $\endgroup$ – Thomas Benjamin Jun 20 '14 at 14:10
  • $\begingroup$ I agree that in the sense you specify the language of ZFC is countable. Still, the fact that it is countable is highly absolute, if that answers your question. $\endgroup$ – Miha Habič Jun 20 '14 at 14:30
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    $\begingroup$ Regarding the definability of the reals, this is precisely the issue with the math-tea argument they want to address. CH is irrelevant here because, regardless of whether it holds or not, there seem to be too many reals fro them to be definable. The paper's point is that (pointwise) definability is a notion external to the model. The model will be unaware that all of it's elements are definable, even if it has all of the formulas required, because it simply cannot talk about satisfaction over itself. $\endgroup$ – Miha Habič Jun 20 '14 at 14:34

The argument in the paper proceeds in two steps: adding a special predicate $U$ which makes the structure $(M,\in,U)$ pointwise definable and then forcing over this to code the whole structure into the GCH pattern. The first step doesn't depend at all on the GCH pattern of $M$ and the second step can be taken to have arbitrarily high closure and so it can be made to preserve arbitrarily long initial segments of the GCH pattern of $M$. So to get a pointwise definable $M[G]$ with a specified violation of CH we just start with $M$ satisfying CH, force over it to get the desired value of $2^{\aleph_0}$ and then perform the argument from the paper above this value. In particular, any value for the continuum which is consistent with König's theorem is also consistent with the universe being pointwise definable.


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