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In general I want to prove that $\mathbb{P}=\left(P,\leq\right)$ and $\mathbb{Q}=\left(Q,\leq\right)$ are forcing notions and there is a dense embedding $h:P\longrightarrow Q$ , then $\mathbb{Q\approx P}$.

Being more specific if $G\subseteq P$ is $\mathbb{P}$-generic over $\mathbf{V}$ , then the set $H=\left\{ q\in Q:\exists p\in G\left(q\leq h\left(p\right)\right)\right\}$ is $\mathbb{Q}$ -generic over $\mathbf{V}$ and $\mathbf{V}\left[G\right]=\mathbf{V}\left[H\right]$. Conversely, if a set $H\subseteq Q$ is $\mathbb{Q}$ -generic over $\mathbf{V}$ , then the set $G=\left\{ p\in P:h\left(p\right)\in H\right\}$ is $\mathbb{P}$ -generic over $\mathbf{V}$ and $\mathbf{V}\left[H\right]=\mathbf{V}\left[G\right]$ .

The first parts about being $\mathbb{Q}$-generic I have an idea how to prove, but with $\mathbf{V}[G]=\mathbf{V}[H]$ I have been through many trouble without any result.

P.S. Here forcing is a pre-order with the minimal element.

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  • $\begingroup$ It's about time. $\endgroup$ – Asaf Karagila Mar 8 '16 at 8:21
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I assume that you only need help to prove $V[G]=V[H]$ (- please correct me, if I'm wrong).

The following Theorem is key: Let $\mathbb P$ be a forcing and let $G$ be $\mathbb P$-generic. Then $V[G]$ is the minimal transitive (class) model $M$ of $\operatorname{ZFC}$ such that $\{G\} \cup V \subseteq M$.

The proof of this is basically trivial. Since $\{G\} \cup V \subseteq M$, we have $\tau \in M$ for every $\mathbb P$-name $\tau$. Furthermore $M$ satisfies $\operatorname{ZFC}$ and $G \in M$. This allows us to build $\tau^G$ in $M$ (, where $\tau^G$ is the $G$-interpretation of $\tau$). Thus $\tau^G \in M$ for every $\mathbb P$-name $\tau$ and therefore $V[G] \subseteq M$.

Now let $G$ be $\mathbb P$-generic and let $H := \{q \in \mathbb Q \mid \exists p \in G \colon h(p) \le q \}$. I'll prove that $V[G] \subseteq V[H]$ and leave the other cases as an exercise to you (should you get stuck, feel free to ask for additional help).

By the Theorem above it suffices to show that $G \in V[H]$: Work in $V[H]$ and let $G' := \{ p \in \mathbb P \mid \exists q \in H \colon h(p) \ge q \}$. Clearly $G \subseteq G'$. Conversely, let $p' \in G'$. In $V$, consider the set $D := \{ p \in \mathbb P \mid p \perp p' \vee p' \le p \}$. This is a dense set in $\mathbb P$ and since $G$ is $\mathbb P$-generic, we may fix some $p \in D \cap G$. Suppose that $p \perp p'$. Then $h(p) \perp h(p')$, contradicting the fact that $h(p),h(p') \in H$.

Thus $p' \le p$. Since $p \in G$ and $G$ is a filter, this yields $p' \in G$ and thus $G' \subseteq G$. Hence $G = G' \in V[H]$.

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  • $\begingroup$ I usually work with the convention that $p \le q$ iff "$p$ is stronger than $q$" and while typing the answer above, I switched at least one time between this convention and yours. Please let me know, if the above still contains an error. $\endgroup$ – Stefan Mesken Mar 8 '16 at 11:55
  • $\begingroup$ Thanks, I haven't found any error of this type. Maybe you know a proof which does not need the theorem about modeling ZFC? $\endgroup$ – Kesha Mar 8 '16 at 14:50
  • $\begingroup$ @Kesha You only need to know that $M$ is closed under the function that maps names to their interpretation to make this argument work. However, the fact that $V[G]$ models $\operatorname{ZFC}$ is of huge importance for the general theory. $\endgroup$ – Stefan Mesken Mar 8 '16 at 22:14

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