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I have been looking at a proof of a technical forcing lemma, and I have a couple of questions.

Here is the setup: $(N, \epsilon) \prec (\mathbf{H}( \chi ), P, \epsilon ) $ is a countable submodel, $P$ is a notion of forcing.

Lemma: The following conditions are equivalent:
1. $q \in P$ is $N$-generic, and
2. $\forall \dot{\alpha} \in N, ( N \models$ "$\dot{\alpha}$ is a $P$-name for an ordinal" $\rightarrow q \Vdash \dot{\alpha} \in N)$

To prove $2 \rightarrow 1$, let $A \in N$, $N \models $"A is a maximal antichain." In $N$, we enumerate $A = \{ a_{\xi} : \xi < \kappa \}$. Then, we define $\dot{\alpha} = \{ ( \xi, a_\xi ) : \xi < \kappa \}$.

The argument then claims that $\dot{\alpha}$ is a $P$-name for an ordinal. But, isn't it actually a $P$-name for a set containing a single ordinal? So how could we apply 2?

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I have to say that there is not enough set up. Is $P$ a notion of forcing in $N$? Is it a notion of forcing in $V$ as well? Since $N$ is countable $P\cap N$ is countable what is the relationship between $P\cap N$ and $N$? – Asaf Karagila Apr 20 '12 at 23:07
@user27974 If $\kappa$ is uncountable then $A\not\subset N$. – azarel Apr 20 '12 at 23:11
@azarel: I suspect that $N$ thinks that $\kappa$ is uncountable but it really just a countable ordinal (well, at least if $N$ is transitive...) – Asaf Karagila Apr 20 '12 at 23:51
Where is the lemma from? – Asaf Karagila Apr 22 '12 at 19:35
@AsafKaragila I have added the complete setup now, as well as the statement of the Lemma. This is from the book Baroszynski and Judah, Set Theory of the Real Line. – Kuhndog Apr 22 '12 at 19:39

Your objection is correct, but the problem is easily fixed. Define $\dot\alpha$ to be $\{(\check\beta,a_\xi):\beta<\xi\}$. Then, because the $a_\xi$ form a maximal antichain, each $a_\xi$ will force $\dot\alpha=\check\xi$, and $\dot\alpha$ is forced (by all conditions) to be an ordinal. Furthermore, the name $\dot\alpha$ is in $N$ (because it's defined from the antichain $A$ and it enumeration, which are in $N$). By statement 2, $q$ forces $\dot\alpha\in\check N$. Some extension $r$ of $q$ forces a specific value, say $\xi\in N$, for $\dot\alpha$. So $r$ can't be compatible with $a_\eta$ for any $\eta\neq\xi$ (because $a_\eta$ forces $\dot\alpha$ to equal $\check\eta$). Therefore $r$ is an extension of $a_\xi$ (by maximality of the antichain and separativity of the forcing notion), which means $q$ is compatible with $a_\xi$. Finally, since both the enumeration of $A$ and the ordinal $\xi$ are in $N$, $a_\xi$ is also in $N$, and that completes the proof of statement 1.

A few minor comments: In case you don't require separativity as part of the definition of "notion of forcing", either pass to the separative quotient first, or note that even though my argument won't make $r$ an extension of $a_\xi$, it will make them compatible, and this suffices.

One of the comments entertained the notion that $N$ might be transitive. It won't be, except in trivial cases. Usually $\chi$ is large, and then every definable element of $H(\chi)$, for example $\omega_1$, will be in $N$, whereas a countable $N$ can't contain all the elements of $\omega_1$.

Since a structure and an elementary substructure must have the same language, the setup at the start of the question should have $(N,P,\in)$ in place of $(N,\in)$. By elementarity, $P$ will be a notion of forcing in $N$ iff it is one in $H(\chi)$, and that is equivalent to being a notion of forcing in $V$ if $\chi$ is big enough.

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