Master-condition characterizations

Fix $D\in M$ dense. If $D\cap M$ is predense below $q$, then conditions below $q$ that extend a condition of $D\cap M$ are dense below $q$; F2 gives $q⊩\dot{G}\cap D\cap M\mathrm{\ne }\mathrm{\varnothing }$. Conversely, if some $r\le q$ were incompatible with every member of $D\cap M$, then $r$ would force that intersection empty. This proves (1) if and only if (2).

Assume (1). The inclusion $M\subseteq M[G]\cap V$ follows from check names. For the reverse inclusion, let $r\le q$ force that a name $\dot{x}\in M$ equals a ground object $x$. Define $D_{\dot{x}}$ to contain (a) every $s$ for which some ground object $y$ satisfies $s⊩\dot{x}=\check{y}$, and (b) every $s$ below which no condition has property (a). This set is dense: from any condition, either an extension has property (a), or the original condition has property (b). By definability of forcing it belongs to $M$. Since $D_{\dot{x}}\cap M$ is predense below $q$, some $s\in D_{\dot{x}}\cap M$ is compatible with $r$. It cannot have property (b), because a common extension with $r$ would force $\dot{x}=\check{x}$ while admitting no ground-value extension. Hence $s$ has property (a); by elementarity its witness may be taken as some $y\in M$. A common extension of $r$ and $s$ forces both $\dot{x}=\check{x}$ and $\dot{x}=\check{y}$, so $x=y\in M$. Thus $q$ forces every ground member of $M[\dot{G}]$ to lie in $M$, proving (3). Statement (3) immediately implies (4), since ordinals are ground objects and check names give the opposite inclusion.

Assume (4), and let $A\in M$ be a maximal antichain. In $M$, use A1 to fix a bijection $e:\xi \to A$ from an ordinal $\xi$, and form by mixing the name $\dot{\beta }$ for the unique index of the member of $A\cap \dot{G}$. Then $\dot{\beta }\in M$ and $q⊩\dot{\beta }\in M\cap \mathrm{O}\mathrm{r}\mathrm{d}$ by (4). Consequently $q$ forces $e(\dot{\beta })\in A\cap M\cap \dot{G}$, so $A\cap M$ is predense below $q$. Every dense $D\in M$ contains, by elementarity and A1, such a maximal antichain $A\in M$; hence $D\cap M$ is predense below $q$ and (1) follows.

The all-model definition immediately gives the club formulation, since the countable elementary submodels of a fixed well-ordered $H_{\mu }$ structure form a club by F4 and A1. Conversely, suppose the good models contain a club in $[H_{\mu }{]}^{\omega }$, where $\mu >2^{\mathrm{\mid }P\mathrm{\mid }}$. Represent a subclub as the models closed under a function $F:H_{\mu }^{<\omega }\to H_{\mu }$. Choose $\lambda >\mu$ and a well-order ${<}_{\lambda }$ so that $F$ may be taken as the ${<}_{\lambda }$-least such witness. Every countable $M\prec (H_{\lambda },\in ,{<}_{\lambda },P,\dots )$ is then closed under $F$, so $N=M\cap H_{\mu }$ is a good club model. Every subset of $P$, and hence every dense set or maximal antichain in $M$, belongs to $H_{\mu }$; therefore $D\cap N=D\cap M$. An $(N,P)$-master below $p\in P\cap M$ is thus also an $(M,P)$-master. This proves the all-model formulation and completes both claimed equivalences. AC is used exactly for A1; no countable transitive model or generic filter is selected.