Jech states the following lemma in his book:
Lemma 21.9 Let $j:V\to M=\operatorname{Ult}_U(V)$ be an elementary embedding with a critical point $\kappa$. If $\mathbb{P}\in M$ is a forcing poset such that $|\mathbb{P}|\le\kappa$ and $G$ is a $V$-generic filter over $\mathbb{P}$, then $M[G]$ is closed under $\kappa$-sequences. (i.e., $({^\kappa}M[G])^{V[G]}\subseteq M[G]$.)
(The original statement involves with $\lambda$-supercompactness, and states $M[G]$ is closed under $\lambda$-sequences under the hypothesis $|\mathbb{P}|\le\lambda$. I will only consider the measurable case for simplicity.)
He starts with the proof as follows:
It suffices to show that if $f\in V[G]$ is a function from $\kappa$ into ordinals, then $f\in M[G]$. (…)
I do not understand this point. I can see that if $M[G]$ is closed under functions $\kappa\to \mathrm{Ord}$, then the standard Mostowski argument with some coding shows $M[G]$ is closed under functions from $\kappa$ to $H_{\kappa^+}^{V[G]}$. Hence I tried to prove it in another way:
My attempt. Let $f\in V[G]$, $f:\kappa\to M[G]$. Take $p_0\in G$ such that $p_0\Vdash \dot{f}\text{ is a function from $\kappa$ to $M^\mathbb{P}$}.$
For each $\alpha<\kappa$, take
$$A_\alpha := \{p\le p_0\mid \exists \sigma\in M^\mathbb{P}[p\Vdash \dot{f}(\alpha)=\sigma]\}.$$
For each $\alpha<\kappa$ and $p\in A_\alpha$, choose $\sigma_{\alpha,p}\in M^\mathbb{P}$ that witnesses $p\in A_\alpha$. Since the choice is made over $V$, $\langle \sigma_{\alpha,p}\mid\alpha,p\rangle\in V$.
Furthermore, we can choose $g_{\alpha,p}:\kappa\to V$ such that $\sigma_{\alpha,p}=[g_{\alpha,p}]_U$.Now define a function $g_p$ as $$g_p(\alpha)(\xi) := g_{\alpha,p}(\xi)$$
if it is defined. Then $g_p$ sends $\xi$ to a partial function over $\kappa$.
Take $h_p=[g_p]_U\in M$. Then $M$ thinks $h_p$ is a partial function from $j(\kappa)$ to $M^\mathbb{P}$. Moreover, we have
- $p'\le p\implies \operatorname{dom} h_p\subseteq \operatorname{dom} h_{p'}$, and
- $p'\le p \implies p'\Vdash h_p(\alpha)=h_{p'}(\alpha)$ (Here $h_p(\alpha)$ and $h_{p'}(\alpha)$ themselves are treated as a single $M^\mathbb{P}$-name.)
Let $h(\alpha):= h_p(\alpha)$ for some $p$. Then $h$ is a partial function from $j(\kappa)$. Since $A_\alpha\cap G$ is nonempty for each $\alpha$, $h_p(\alpha)$ is defined for all $\alpha<\kappa$. Moreover, by definition of $A_\alpha$, we have $f(\alpha)=h(\alpha)$ for $\alpha<\kappa$. Hence $f=h\upharpoonright \kappa\in M[G]$.
My questions are as follows:
- Why just showing $({^\kappa}\mathrm{Ord})^{V[G]}\subseteq M[G]$ suffices to prove Lemma 21.9?
- Is my argument correct?
Thank you for any help in advance.
Best Answer
Let me give a brief answer to (1).
If $A$ is any set, fix in $M[G]$ a well-ordering of $A$, and let $\alpha$ be the order type of this well-ordering. If $f\colon\kappa\to A$ is any sequence, then $f^*\colon\kappa\to\alpha$ given by the composition of $f$ with the isomorphism is a function in $\mathrm{Ord}^\kappa$, but then by composing again with the inverse isomorphism, which is in $M[G]$, we get $f$.