Let $\mathcal{M}$ be a countable transitive standard-model of ZFC.
Let $B \in \mathcal{M}$ be a boolean algebra that is complete in $\mathcal{M}$.
Further, let $\mathcal{M}^{(B)}$ be the corresponding boolean model of ZFC.

Now we consider an $\mathcal{M}$-generic ultrafilter $U$ on $B$.
According to Jech (Set Theory, p.216), the elements of the generic extension $\mathcal{M}[U]$ can be obtained via the map $\cdot^U: x \mapsto x^U := \lbrace{y^U \mid x(y) \in U \rbrace}$ where $x \in \mathcal{M}^{(B)}$. But when reading Bell (Set Theory Boolean-Valued models and Independence Proofs, p.91) the following map is given: $i(x) = \lbrace i(y) \mid \Vert y \in x\Vert \in U \rbrace$ where $x \in \mathcal{M}^{(B)}$. I assume that the following holds: $$\forall x \in \mathcal{M}^{(B)}: x^U = i(x). \ \ \ [*]$$

But at the same time I am not yet fully convinced as $u(x) \leq \Vert x \in u \Vert$ for $x \in \text{dom}(u)$.

Can anyone verify that $[*]$ (or a variant of that statement) holds (and give a short reason why)?


Simply use induction on $\mathbb{B}$-names (or the rank of $\mathbb{B}$-names, if you are not familiar with applying induction directly to $\mathbb{B}$-names.) Suppose that $i(z)=z^U$ holds for all $z\in\operatorname{dom} x$.

Since $x(z)\le \|z\in x\|$, we have $x^U\subseteq i(x)$. Conversely, assume that $\|y\in x\|\in U$. Then $$\sum_{z\in\operatorname{dom}x}x(z)\cdot \|z=y\| \in U.$$ By genericity of $U$, we can find $z\in\operatorname{dom} x$ such that $x(z)\cdot\|z=y\|\in U$. This proves

  1. $i(z)=z^U\in x^U$, and
  2. $i(z)=i(y)$.

Hence $i(y)\in x^U$. This proves $i(x)\subseteq x^U$ (since $i(y)\in i(x)$ means $\|y\in x\|\in U$.)


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