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This question already has an answer here:

Fix the language of set theory $\mathcal{L}=\{\in\}$. Let $\langle M,\in\rangle$ be a set or proper class model of ZFC (e.g. $M$ could be $L$, $HOD$, $V_{\kappa}$ for some inaccessible cardinal $\kappa$, etc.) consider the complete theory of this model in the language $\mathcal{L}$ that is $T_{M}:=Th(\langle M,\in\rangle)=\{\sigma\in\mathcal{L}~|~\langle M,\in\rangle\models \sigma\}$. My questions are about the possible number of countable models of this complete theory up to isomorphism $I(T_{M},\aleph_0)$ when $M$ varies over different models of $ZFC$.

Question 1: What are $I(T_{L},\aleph_0)$ and $I(T_{HOD},\aleph_0)$, $I(T_{V_{\kappa}},\aleph_0)$ (for $\kappa$ inaccessible)?

Question 2: Is there a set or proper class model $M$ of ZFC such that $I(T_{M},\aleph_0)=\lambda$ for each cardinal $\lambda\in\{1,3,4,\cdots,\aleph_{0},\aleph_1,2^{\aleph_{0}}\}$?

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marked as duplicate by Andrés E. Caicedo, colormegone, Cookie, Adam Hughes, Tunk-Fey Jul 29 '14 at 2:42

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    $\begingroup$ My feeling is that any completion of ZFC (assuming it is consistent) would be complicated enough to have $2^{\aleph_0}$ types and hence $2^{\aleph_0}$ countable models. $\endgroup$ – Levon Haykazyan Jun 17 '14 at 11:23
  • $\begingroup$ I think $T_L$ is clear enough: we begin with a fixed (class) model $V$ of ZFC, and $T_L$ is the set of sentences in the language of ZFC satisfied by $L$ inside $V$. Of course $T_L$ will vary depending on $V$. This can all be formalized in e.g. MK set theory. $\endgroup$ – Carl Mummert Jun 17 '14 at 11:57
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    $\begingroup$ If you don't get responses here in a reasonable time frame, you could try asking this on our sister site, mathoverflow.net. This question is at a sufficiently advanced level that it would also be on-topic there. $\endgroup$ – Carl Mummert Jun 17 '14 at 13:49
  • $\begingroup$ Duplicate. Joel's answer in essentially the same as the answer here. $\endgroup$ – Andrés E. Caicedo Jul 29 '14 at 0:02
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Levon Haykazyan has the right idea with his comment. Any consistent completion of ZFC (call it ZFC$^*$) has $2^{\aleph_0}$ countable models up to isomorphism.

Note that relative to ZFC, each finite ordinal is definable without parameters. $0$ is definable by $\phi_0: \forall y\, \lnot y\in x$, and $n+1$ is definable by $\phi_{n+1}:\forall y\,(y\in x)\leftrightarrow \lor_{i\leq n} \phi_i(y)$.

Now for each $S\subseteq \omega$ (our $\omega$!), consider the partial type $$p_S(x) = \{\exists y\, \phi_n(y)\land y\in x\mid n\in S\}\cup\{\lnot \exists y\, \phi_n(y)\land y\in x\mid n\notin S\}.$$ This is a family of $2^{\aleph_0}$ pairwise contradictory partial types over $\emptyset$, all of which are consistent with ZFC$^*$. Each of them can be realized in a countable model of ZFC$^*$, but any countable structure only realizes countably many types over $\emptyset$, and any two isomorphic structures realize the same types. So ZFC$^*$ must have $2^{\aleph_0}$ countable models up to isomorphism.

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