If $P$ is a forcing notion, $A \subseteq P$ (usually an antichain), $q\in P$, then I write $A\cap q$ for the set of all conditions in $A$ which are compatible with $q$.

For a proper forcing notion $P$, and countable elementary submodel $N$ (containing everything necessary, in particular $P$), a condition $q$ is $N$-generic iff for all maximal antichains $A\subseteq P$ with $A\in N$ the inclusion $A\cap q\subseteq N$ holds.

I will call a generic condition $q$ $N$-super-generic (just for the moment, until somebody gives me a better name) if for all maximal antichains $A\subseteq P$ with $A\in N$ the set $A\cap q $ is finite.

It is an exercise to show that the following conditions are equivalent:

1. $P$ is proper and $\omega^\omega$-bounding.
2. For every $N$, every $p\in P\cap N$ there is an $N$-super-generic $q$ stronger than p.
3. Every $N$-generic $q$ can be strengthened to a super-generic $q'$.

(Proof sketch of 1$\Rightarrow$3: enumerate $P\cap N$ as $\{p_0,p_1,\ldots\}$, and let $\{A_0,A_1,\ldots \}$ be the set of all maximal antichains that are elements of $N$. A generic filter $G$ defines a function $g:\omega\to \omega$ such that $A_n\cap G = \{ p_{g(n)}\}$. If $q$ forces that $f\in V$ bounds $g$, then clearly $A_n \cap q \subseteq \{ p_i:i\le f(n)\}$.)

Proof sketch of 2$\Rightarrow$1: If $h\in N$ is the name of a function in $\omega^\omega$, then there is a family $(H_n:n\in \omega)$ such that $H_n$ is a maximal antichain deciding the value of $h(n)$; a super-generic condition will force a bound to $h$.)

Question: What is the right (or "proper", or perhaps already established) name for "super-generic"?

If $P$ is a forcing notion, $A \subseteq P$ (usually an antichain), $q\in P$, then I write $A\cap q$ for the set of all conditions in $A$ which are compatible with $q$.

For a proper forcing notion $P$, and countable elementary submodel $N$ (containing everything necessary, in particular $P$), a condition $q$ is $N$-generic iff for all maximal antichains $A\subseteq P$ with $A\in N$ the inclusion $A\cap q\subseteq N$ holds.

I will call a generic condition $q$ $N$-super-generic (just for the moment, until somebody gives me a better name) if for all maximal antichains $A\subseteq P$ with $A\in N$ the set $A\cap q $ is finite.

It is an exercise to show that the following conditions are equivalent:

1. $P$ is proper and $\omega^\omega$-bounding.
2. For every $N$, every $p\in P\cap N$ there is an $N$-super-generic $q$ stronger than p.
3. Every $N$-generic $q$ can be strengthened to a super-generic $q'$. (EDIT: and $P$ is proper.)

(Proof sketch of 1$\Rightarrow$3: enumerate $P\cap N$ as $\{p_0,p_1,\ldots\}$, and let $\{A_0,A_1,\ldots \}$ be the set of all maximal antichains that are elements of $N$. A generic filter $G$ defines a function $g:\omega\to \omega$ such that $A_n\cap G = \{ p_{g(n)}\}$. If $q$ forces that $f\in V$ bounds $g$, then clearly $A_n \cap q \subseteq \{ p_i:i\le f(n)\}$.)

Proof sketch of 2$\Rightarrow$1: If $h\in N$ is the name of a function in $\omega^\omega$, then there is a family $(H_n:n\in \omega)$ such that $H_n$ is a maximal antichain deciding the value of $h(n)$; a super-generic condition will force a bound to $h$.)

Question: What is the right (or "proper", or perhaps already established) name for "super-generic"?

If $P$ is a forcing notion, $A \subseteq P$ (usually an antichain), $q\in P$, then I write $A\cap q$ for the set of all conditions in $A$ which are compatible with $q$.

For a proper forcing notion $P$, and countable elementary submodel $N$ (containing everything necessary, in particular $P$), a condition $q$ is $N$-generic iff for all maximal antichains $A\subseteq P$ with $A\in N$ the inclusion $A\cap q\subseteq N$ holds.

I will call a generic condition $q$ $N$-super-generic (just for the moment, until somebody gives me a better name) if for all maximal antichains $A\subseteq P$ with $A\in N$ the set $A\cap q $ is finite.

It is an exercise to show that the following conditions are equivalent:

1. $P$ is proper and $\omega^\omega$-bounding.
2. For every $N$, every $p\in P\cap N$ there is an $N$-super-generic $q$ stronger than p.
3. Every $N$-generic $q$ can be strengthened to a super-generic $q'$.

(Proof sketch of 1$\Rightarrow$3: enumerate $P\cap N$ as $\{p_0,p_1,\ldots\}$, and let $\{A_0,A_1,\ldots \}$ be the set of all maximal antichains that are elements of $N$. A generic filter $G$ defines a function $g:\omega\to \omega$ such that $A_n\cap G = \{ p_{g(n)}\}$. If $q$ forces that $f\in V$ bounds $g$, then clearly $A_n \cap q \subseteq \{ p_i:i\le f(n)\}$.)

Proof sketch of 3$\Rightarrow$1: If $h\in N$ is the name of a function in $\omega^\omega$, then there is a family $(H_n:n\in \omega)$ such that $H_n$ is a maximal antichain deciding the value of $h(n)$; a super-generic condition will force a bound to $h$.)

Question: What is the right (or "proper", or perhaps already established) name for "super-generic"?

If $P$ is a forcing notion, $A \subseteq P$ (usually an antichain), $q\in P$, then I write $A\cap q$ for the set of all conditions in $A$ which are compatible with $q$.

For a proper forcing notion $P$, and countable elementary submodel $N$ (containing everything necessary, in particular $P$), a condition $q$ is $N$-generic iff for all maximal antichains $A\subseteq P$ with $A\in N$ the inclusion $A\cap q\subseteq N$ holds.

I will call a generic condition $q$ $N$-super-generic (just for the moment, until somebody gives me a better name) if for all maximal antichains $A\subseteq P$ with $A\in N$ the set $A\cap q $ is finite.

It is an exercise to show that the following conditions are equivalent:

1. $P$ is proper and $\omega^\omega$-bounding.
2. For every $N$, every $p\in P\cap N$ there is an $N$-super-generic $q$ stronger than p.
3. Every $N$-generic $q$ can be strengthened to a super-generic $q'$.

(Proof sketch of 1$\Rightarrow$3: enumerate $P\cap N$ as $\{p_0,p_1,\ldots\}$, and let $\{A_0,A_1,\ldots \}$ be the set of all maximal antichains that are elements of $N$. A generic filter $G$ defines a function $g:\omega\to \omega$ such that $A_n\cap G = \{ p_{g(n)}\}$. If $q$ forces that $f\in V$ bounds $g$, then clearly $A_n \cap q \subseteq \{ p_i:i\le f(n)\}$.)

Proof sketch of 2$\Rightarrow$1: If $h\in N$ is the name of a function in $\omega^\omega$, then there is a family $(H_n:n\in \omega)$ such that $H_n$ is a maximal antichain deciding the value of $h(n)$; a super-generic condition will force a bound to $h$.)

Question: What is the right (or "proper", or perhaps already established) name for "super-generic"?