Timeline for Constructing a model of $\mathrm{DCF}_0$ via forcing
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Jun 15, 2019 at 5:55 | history | bounty ended | Noah Schweber | ||
| S Jun 15, 2019 at 5:55 | history | notice removed | Noah Schweber | ||
| Jun 15, 2019 at 5:14 | vote | accept | James E Hanson | ||
| Jun 14, 2019 at 21:59 | answer | added | James E Hanson | timeline score: 4 | |
| S Jun 13, 2019 at 15:15 | history | bounty started | Noah Schweber | ||
| S Jun 13, 2019 at 15:15 | history | notice added | Noah Schweber | Draw attention | |
| May 1, 2019 at 3:14 | comment | added | James E Hanson | Ah, sorry about that. | |
| Apr 30, 2019 at 20:20 | comment | added | Jonathan Schilhan | I am very familiar with forcing. I was misreading "functions in $V$" as "functions in $V[G]$". In the sentence before that you write "the field of germs of meromorphic functions at a generic point", which implied, for me, that you mean meromorphic functions in $V[G]$. | |
| Apr 30, 2019 at 15:00 | comment | added | James E Hanson | On the other hand if I choose a non-computable complex number $a$ and look at the field of germs of (computable) meromorphic functions at that point, then this field should be algebraically closed, since the zeroes and poles of a computable meromorphic function are always computable, so given any meromorphic function $f$ we can find a neighborhood of $a$ small enough that $f$ has no zeroes or poles in that neighborhood. Then there is a meromorphic square root of this function in that neighborhood. | |
| Apr 30, 2019 at 14:57 | comment | added | James E Hanson | How familiar are you with forcing? The point $z$ is in some sense 'new'. Here's an analogous situation: Suppose I'm considering computable meromorphic functions on simple domains (like circles of rational radius with rational centers) and I want to find some field of germs at a point that is algebraically closed. This can never work if my point is computable because for any given computable complex number $a$, the function $f(z) = z-a$ cannot have a square root in any neighborhood of $a$. | |
| Apr 30, 2019 at 1:03 | history | asked | James E Hanson | CC BY-SA 4.0 |