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Let me point out that meeting all pre-dense sets is not generally the same as meeting all dense classes. Consider the forcing $\mathbb{P}$ that adds a generic function from $\text{Ord}$ to $V$. (One can use this forcing to force global choice — see Victoria Gitman's account.) That is, conditions are functions $p:\alpha\to V$ for some ordinal $\alpha$, and the order is extension of functions. If a filter $G\subset\mathbb{P}$ meets all dense classes, then it is easy to see that $\cup G$ is a surjective function total function from $\text{Ord}$ to $V$: for any set $x$, it is dense that the conditions have $x$ in their range, and for any ordinal $\alpha$, it is dense that the conditions are defined at $\alpha$.

But that argument breaks down completely if one tries to use only predense sets: the reason is that the forcing $\mathbb{P}$ has no pre-dense sets at all! (except for sets that contain the empty function) For any set $B$ of nontrivial conditions, there is a condition $p$ that is incompatible with every element of $B$. So it turns out that every filter vacuously meets all pre-dense sets, and the forcing technology would not be doing what we want.

You might reply that this example is because we are using the partial order, rather than the Boolean algebra. But that reply has problems, because in a case like this, there is no way to formalize the Boolean completion without going to meta-classes, as the antichains are generally proper classes here. So we cannot so easily go to the Boolean completion of the forcing.

One can transform $\mathbb{P}$ into forcing that also adds sets, for example, by using the generic filter to determine what happens next, for example, to code $G$ into the GCH pattern.

Let me point out that meeting all pre-dense sets is not generally the same as meeting all dense classes. Consider the forcing $\mathbb{P}$ that adds a generic function from $\text{Ord}$ to $V$. (One can use this forcing to force global choice — see Victoria Gitman's account.) That is, conditions are functions $p:\alpha\to V$ for some ordinal $\alpha$, and the order is extension of functions. If a filter $G\subset\mathbb{P}$ meets all dense classes, then it is easy to see that $\cup G$ is a surjective function total function from $\text{Ord}$ to $V$: for any set $x$, it is dense that the conditions have $x$ in their range, and for any ordinal $\alpha$, it is dense that the conditions are defined at $\alpha$.

But that argument breaks down completely if one tries to use only predense sets: the reason is that the forcing $\mathbb{P}$ has no pre-dense sets at all! (except for sets that contain the empty function) For any set $B$ of nontrivial conditions, there is a condition $p$ that is incompatible with every element of $B$. So it turns out that every filter vacuously meets all pre-dense sets, and the forcing technology would not be doing what we want.

You might reply that this example is because we are using the partial order, rather than the Boolean algebra. But that reply has problems, because in a case like this, there is no way to formalize the Boolean completion without going to meta-classes, as the antichains are generally proper classes here. So we cannot so easily go to the Boolean completion of the forcing.

One can transform $\mathbb{P}$ into forcing that also adds sets, for example, by using the generic filter to determine what happens next, for example, to code $G$ into the GCH pattern.

Let me point out that meeting all pre-dense sets is not generally the same as meeting all dense classes. Consider the forcing $\mathbb{P}$ that adds a generic function from $\text{Ord}$ to $V$. (One can use this forcing to force global choice.) That is, conditions are functions $p:\alpha\to V$ for some ordinal $\alpha$, and the order is extension of functions. If a filter $G\subset\mathbb{P}$ meets all dense classes, then it is easy to see that $\cup G$ is a surjective function total function from $\text{Ord}$ to $V$: for any set $x$, it is dense that the conditions have $x$ in their range, and for any ordinal $\alpha$, it is dense that the conditions are defined at $\alpha$.

But that argument breaks down completely if one tries to use only predense sets: the reason is that the forcing $\mathbb{P}$ has no pre-dense sets at all! (except for sets that contain the empty function) For any set $B$ of nontrivial conditions, there is a condition $p$ that is incompatible with every element of $B$. So it turns out that every filter vacuously meets all pre-dense sets, and the forcing technology would not be doing what we want.

You might reply that this example is because we are using the partial order, rather than the Boolean algebra. But that reply has problems, because in a case like this, there is no way to formalize the Boolean completion without going to meta-classes, as the antichains are generally proper classes here. So we cannot so easily go to the Boolean completion of the forcing.

One can transform $\mathbb{P}$ into forcing that also adds sets, for example, by using the generic filter to determine what happens next, for example, to code $G$ into the GCH pattern.

Let me point out that meeting all pre-dense sets is not generally the same as meeting all dense classes. Consider the forcing $\mathbb{P}$ that adds a generic function from $\text{Ord}$ to $V$. (One can use this forcing to force global choice — see Victoria Gitman's account.) That is, conditions are functions $p:\alpha\to V$ for some ordinal $\alpha$, and the order is extension of functions. If a filter $G\subset\mathbb{P}$ meets all dense classes, then it is easy to see that $\cup G$ is a surjective function total function from $\text{Ord}$ to $V$: for any set $x$, it is dense that the conditions have $x$ in their range, and for any ordinal $\alpha$, it is dense that the conditions are defined at $\alpha$.

But that argument breaks down completely if one tries to use only predense sets: the reason is that the forcing $\mathbb{P}$ has no pre-dense sets at all! (except for sets that contain the empty function) For any set $B$ of nontrivial conditions, there is a condition $p$ that is incompatible with every element of $B$. So it turns out that every filter vacuously meets all pre-dense sets, and the forcing technology would not be doing what we want.

You might reply that this example is because we are using the partial order, rather than the Boolean algebra. But that reply has problems, because in a case like this, there is no way to formalize the Boolean completion without going to meta-classes, as the antichains are generally proper classes here. So we cannot so easily go to the Boolean completion of the forcing.

One can transform $\mathbb{P}$ into forcing that also adds sets, for example, by using the generic filter to determine what happens next, for example, to code $G$ into the GCH pattern.

Let me point out that meeting all pre-dense sets is not generally the same as meeting all dense classes. Consider the forcing $\mathbb{P}$ that adds a generic function from $\text{Ord}$ to $V$. (One can use this forcing to force global choice.) That is, conditions are functions $p:\alpha\to V$ for some ordinal $\alpha$, and the order is extension of functions. If a filter $G\subset\mathbb{P}$ meets all dense classes, then it is easy to see that $\cup G$ is a surjective function total function from $\text{Ord}$ to $V$: for any set $x$, it is dense that the conditions have $x$ in their range, and for any ordinal $\alpha$, it is dense that the conditions are defined at $\alpha$.

But that argument breaks down completely if one tries to use only predense sets: the reason is that the forcing $\mathbb{P}$ has no pre-dense sets at all! For any set $B$ of conditions, there is a condition $p$ that is incompatiable with every element of $B$. So it turns out that every filter vacuously meets all pre-dense sets, and the forcing technology would not be doing what we want.

You might reply that this example is because we are using the partial order, rather than the Boolean algebra. But that reply has problems, because in a case like this, there is no way to formalize the Boolean completion without going to meta-classes, as the antichains are generally proper classes here.

One can transform $\mathbb{P}$ into forcing that also adds sets, for example, by using the generic filter to determine what happens next, for example, to code $G$ into the GCH pattern.

Let me point out that meeting all pre-dense sets is not generally the same as meeting all dense classes. Consider the forcing $\mathbb{P}$ that adds a generic function from $\text{Ord}$ to $V$. (One can use this forcing to force global choice.) That is, conditions are functions $p:\alpha\to V$ for some ordinal $\alpha$, and the order is extension of functions. If a filter $G\subset\mathbb{P}$ meets all dense classes, then it is easy to see that $\cup G$ is a surjective function total function from $\text{Ord}$ to $V$: for any set $x$, it is dense that the conditions have $x$ in their range, and for any ordinal $\alpha$, it is dense that the conditions are defined at $\alpha$.

But that argument breaks down completely if one tries to use only predense sets: the reason is that the forcing $\mathbb{P}$ has no pre-dense sets at all! (except for sets that contain the empty function) For any set $B$ of nontrivial conditions, there is a condition $p$ that is incompatible with every element of $B$. So it turns out that every filter vacuously meets all pre-dense sets, and the forcing technology would not be doing what we want.

You might reply that this example is because we are using the partial order, rather than the Boolean algebra. But that reply has problems, because in a case like this, there is no way to formalize the Boolean completion without going to meta-classes, as the antichains are generally proper classes here. So we cannot so easily go to the Boolean completion of the forcing.

One can transform $\mathbb{P}$ into forcing that also adds sets, for example, by using the generic filter to determine what happens next, for example, to code $G$ into the GCH pattern.