Return to Question

EDIT: The first three questions were answered by Monroe Eskew below; the fourth, however, seems like a fundamentally different question. I've accepted Monroe's answer, and have asked the fourth question separately here.

EDIT: The first three questions were answered by Monroe Eskew below; the fourth, however, seems like a fundamentally different question. I've accepted Monroe's answer, and have asked the fourth question separately here.

That said, if we drop choice, then such filters can be ultrafilters! Let $V$ be a model of $ZF+V=L(\mathbb{R})$ + "The theory of $L(\mathbb{R})$ is absolute" - formally, this last condition is a scheme asserting that for every sentence $\varphi$ in the language of set theory with real parameters and every set forcing $\mathbb{P}$, $V\models\varphi\iff L(\mathbb{R})^{V[G]}\models\varphi$. (Informally, this should suggest that $V$ is the $L(\mathbb{R})$ of some choice model with a proper class of Woodins.)

Now consider the "labelled Levy collapse" $\mathbb{Q}=Col_{\mathbb{R}}(\omega,\omega_1)$ - a condition in this forcing is a finite sequence of reals coding well-orderings, and conditions are ordered by extension. Like the Levy collapse, this forcing makes $\omega_1$ countable, but - importantly - does so by adding a single real, $G$; in particular, the generic extension $V[G]$ satisfies "$V=L(\mathbb{R})$."

This gives us lots of absoluteness between $V$ and $V[G]$. In particular, the sentence "The club filter on $\omega_1$ is an ultrafilter" is true in $V$, and hence in $V[G]$; and in particular, this means that in $V$ the filter $\mathcal{F}_{Col_\mathbb{R}(\omega, \omega_1)}$ is an ultrafilter on $\omega_2^V$. Indeed, this filter witnesses that $\omega_2^V$ is measurable in $V$.

EDIT: The first three questions were answered by Monroe Eskew below; the fourth, however, seems like a fundamentally different question. I've accepted Monroe's answer, and have asked the fourth question separately here.

That said, if we drop choice, then such filters can be ultrafilters! Let $V$ be a model of $ZF+V=L(\mathbb{R})$ + "The theory of $L(\mathbb{R})$ is absolute" - formally, this last condition is a scheme asserting that for every sentence $\varphi$ in the language of set theory with real parameters and every set forcing $\mathbb{P}$, $V\models\varphi\iff L(\mathbb{R})^{V[G]}\models\varphi$. (Informally, this should suggest that $V$ is the $L(\mathbb{R})$ of some choice model with a proper class of Woodins.)

This gives us lots of absoluteness between $V$ and $V[G]$. In particular, the sentence "The club filter on $\omega_1$ is an ultrafilter" is true in $V$, and hence in $V[G]$; and in particular, this means that in $V$ the filter $\mathcal{F}_{Col_\mathbb{R}(\omega, \omega_1)}$ is an ultrafilter on $\omega_2^V$. Indeed, this filter witnesses that $\omega_2^V$ is measurable in $V$.

EDIT: The above argument relies on the assumption that $\omega_1$ in the new $L(\mathbb{R})$ is the old $L(\mathbb{R})$'s $\omega_2$; this seemed obvious to me at first, but now that I think about it (prompted by Asaf's answer below) I see it is wildly unjustified.

EDIT: The first three questions were answered by Monroe Eskew below; the fourth, however, seems like a fundamentally different question. I've accepted Monroe's answer, and have asked the fourth question separately here.

Of particular interest are the forcings which preserve $\omega_2$ but collapse $\omega_1$, like $Col(\omega_1,\omega)$ (call such forcings relevant). Unfortunately, the associated filters are never ultrafilters, at least in ZFC: this is because they are always $\omega_2$-closed (since $\omega_1$ is made countable, and $V^\mathbb{P}\models$ the intersection of countably many clubs is countable).

Now consider the "labelled Levy collapse" $\mathbb{Q}=Col_{\mathbb{R}}(\omega_1,\omega)$ - a condition in this forcing is a finite sequence of reals coding well-orderings, and conditions are ordered by extension. Like the Levy collapse, this forcing makes $\omega_1$ countable, but - importantly - does so by adding a single real, $G$; in particular, the generic extension $V[G]$ satisfies "$V=L(\mathbb{R})$."

This gives us lots of absoluteness between $V$ and $V[G]$. In particular, the sentence "The club filter on $\omega_1$ is an ultrafilter" is true in $V$, and hence in $V[G]$; and in particular, this means that in $V$ the filter $\mathcal{F}_{Col_\mathbb{R}(\omega_1, \omega)}$ is an ultrafilter on $\omega_2^V$. Indeed, this filter witnesses that $\omega_2^V$ is measurable in $V$.

What is $\mathcal{F}_{Col(\omega_1,\omega)}$?

I suspect the answer is "all of them," but I don't see how to prove it.

Of particular interest are the forcings which preserve $\omega_2$ but collapse $\omega_1$, like $Col(\omega,\omega_1)$ (call such forcings relevant). Unfortunately, the associated filters are never ultrafilters, at least in ZFC: this is because they are always $\omega_2$-closed (since $\omega_1$ is made countable, and $V^\mathbb{P}\models$ the intersection of countably many clubs is countable).

Now consider the "labelled Levy collapse" $\mathbb{Q}=Col_{\mathbb{R}}(\omega,\omega_1)$ - a condition in this forcing is a finite sequence of reals coding well-orderings, and conditions are ordered by extension. Like the Levy collapse, this forcing makes $\omega_1$ countable, but - importantly - does so by adding a single real, $G$; in particular, the generic extension $V[G]$ satisfies "$V=L(\mathbb{R})$."

This gives us lots of absoluteness between $V$ and $V[G]$. In particular, the sentence "The club filter on $\omega_1$ is an ultrafilter" is true in $V$, and hence in $V[G]$; and in particular, this means that in $V$ the filter $\mathcal{F}_{Col_\mathbb{R}(\omega, \omega_1)}$ is an ultrafilter on $\omega_2^V$. Indeed, this filter witnesses that $\omega_2^V$ is measurable in $V$.

What is $\mathcal{F}_{Col(\omega,\omega_1)}$?

I suspect the answer is "all of them," but I don't see how to prove it.

EDIT: The first three questions were answered by Monroe Eskew below; the fourth, however, seems like a fundamentally different question. I've accepted Monroe's answer, and have asked the fourth question separately here.