I am trying to prove the following statement and i think i have reached a proof but i'm not sure, is my proof correct or have i missed something?
If $M$ is a countable transitive model of ZFC and $\mathbb{P} \in M$ is an infinite notion of forcing then there exists an $H \subset \mathbb{P}$ such that $M[H]$ is not a model of ZFC. Hint: Fix $f\in M$ such that $f$ maps $\omega \times \omega$ one-to-one into $\mathbb{P}$. Choose $H$ such that $f^{-1}(H)$ is a well-order of type $\alpha \gt o(M) = \text{Ord} \cap M$.
My proof: Fix $H$ as in the hint. Since we construct $M[H]$ using $\mathbb{P}$-names and we don't nessecarily need $H$ to be $\mathbb{P}$-generic over $M$ for the construction, we know that $M[H]$ is a transitive set and that $\text{Ord} \cap M = \text{Ord} \cap M[H]$. Now if we suppose $M[H]$ is a model of ZFC, since $H\in M[H]$ and $f \in M[H]$ then $f^{-1}(H) \in M[H]$ and it is a theorem of ZFC that every well-ordered set is isomorphic to a unique ordinal, so $f^{-1}(H) \cong \beta$ for some $\beta \in \text{Ord}^{M[H]}$. But looking from the outside we see that $\beta \cong \alpha$ for the $\alpha$ in the hint, so $\beta = \alpha$. Now since $\text{Ord} \cap M = \text{Ord} \cap M[H]$ we have $\alpha \in o(M)$ which is a contradiction.
Since this is my first exercise in forcing i am trying to be very careful, i would be very happy if you could point out any error, even small ones.
Thanks for your patience.
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1$\begingroup$ That looks good to me. The only thing you may want to explicitly mention is that all transitive models of $\mathrm{ZFC}$ do compute the correct order type, i.e. $\mathrm{otp}^{M}(f^{-1}(H)) = \alpha$. $\endgroup$– Stefan MeskenCommented Aug 29, 2018 at 12:43
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$\begingroup$ @StefanMesken, Is it enough to say that, the correct computation of the order type your saying follows from the absolutness of the notion of an ordinal for transitive models? $\endgroup$– LddddCommented Aug 29, 2018 at 13:10
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1$\begingroup$ @ShervinSorouri Your claimed characterization of order type is incorrect. Here is a correct one ($R$ is a well-order): $\mathrm{otp} (R) = \alpha \iff (\exists f( f \colon (\mathrm{dom}(R);R) \to (\alpha; \in) \wedge \alpha \in \mathrm{Ord} \wedge f \text{ is an isomorphism}) \iff \forall f ( \ldots)$ showing that it is $\Delta_{1}$. (The $\ldots$ is for you to fill in.) $\endgroup$– Stefan MeskenCommented Aug 29, 2018 at 20:21
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1$\begingroup$ @StefanMesken, Oh i see i have used otp a little bit awkwardly, by $\text{otp}X$ i meant the order type of the relation on the set $X$. For the $\Pi_1$ part can we say $\forall f \forall \beta((\beta \text{ is an ordinal} \wedge f:(\text{dom}(R), R) \rightarrow (\beta, \in) \wedge f \text{is an isomorphism}) \rightarrow \alpha = \beta)$ ? $\endgroup$– LddddCommented Aug 29, 2018 at 20:33
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1$\begingroup$ @ShervinSorouri Yes, that works. $\endgroup$– Stefan MeskenCommented Aug 29, 2018 at 20:41
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