Timeline for Unbounded set in $V[G]$ has an unbounded subset in $V$?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 25, 2021 at 6:15 | vote | accept | Clement Yung | ||
| Aug 23, 2021 at 14:30 | comment | added | Gabe Goldberg | I remember this from the introduction to Magidor's paper on the dual Jensen covering lemma, which shows that if a set of ordinals has some weak closure properties (primitive recursive closure) and $0^\#$ doesn't exist, then the set is the union of countably many constructible sets. This shows that under anti-large cardinal principles, with some reasonable constraints on the set in the extension, the answer to your question becomes yes. | |
| Aug 23, 2021 at 14:22 | comment | added | Gabe Goldberg | If you force to add a subset $G$ of an infinite cardinal $\kappa$ using finite conditions (which is equivalent to adding $\kappa$-many Cohen reals), then $G$ is unbounded in $\kappa$ but contains no infinite set from the ground model. | |
| Aug 23, 2021 at 9:11 | comment | added | Corey Bacal Switzer | Another example of a forcing adding an unbounded $Y \subseteq \kappa$ with no unbounded ground model set, somewhat similar to a Matthias real, is the Prikry forcing to add an $\omega$ cofinal sequence to a measurable cardinal $\kappa$ while preserving $\kappa$. The generic $\omega$-sequence you add to the measurable can't contain an infinite ground model set since $\kappa$ is regular in $V$. | |
| Aug 23, 2021 at 6:02 | answer | added | LCollapse | timeline score: 4 | |
| Aug 23, 2021 at 4:25 | history | asked | Clement Yung | CC BY-SA 4.0 |