Timeline for Can we force the existence of this function?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 23, 2014 at 1:54 | answer | added | Geoff Galgon | timeline score: 6 | |
| Jan 7, 2014 at 18:47 | vote | accept | Ioannis Souldatos | ||
| Jan 4, 2014 at 15:31 | answer | added | Péter Komjáth | timeline score: 9 | |
| Jan 4, 2014 at 7:01 | comment | added | Péter Komjáth | Miha: The f-values taken on $\omega$ of the conditions in the generic set will be cofinal in $\omega_1$, therefore collapsing $\aleph_1$. | |
| Jan 3, 2014 at 20:26 | comment | added | Miha Habič | Here's an idea: the forcing conditions consist of a finite symmetric partial function $f$ from $\aleph_2\times \aleph_2$ to $\aleph_1$ and a finite subset $A$ of $\aleph_2$. A condition $(f',A')$ is stronger than $(f,A)$ if $f'$ and $A'$ extend $f$ and $A$, respectively, and if for any (coded) pair $\langle a,b\rangle\in A$ any new values that $f'$ assigns to the same rows of the $a$-th and $b$-th columns are distinct. Taking as $F$ the union of the first components of a generic filter, it seems to me that this satisfies both of your requirements. | |
| Jan 2, 2014 at 20:28 | comment | added | Ioannis Souldatos | @WillSawin: It is not very different. If (a) is missing then (b) must include that the sets {x|F(x,a)=F(x,b)} are finite. | |
| Jan 2, 2014 at 20:13 | comment | added | Ioannis Souldatos | @WillSawin: No. | |
| Jan 2, 2014 at 19:44 | comment | added | Will Sawin | Do you know the answer without condition a? | |
| Jan 2, 2014 at 19:26 | history | edited | Noah Schweber | CC BY-SA 3.0 |
Added "distinct"
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| Jan 2, 2014 at 19:24 | history | asked | Ioannis Souldatos | CC BY-SA 3.0 |