Timeline for Forall exists formula
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 1, 2018 at 14:07 | comment | added | tomasz | @NoahSchweber: I've found the proof. It's just that I've never considered that before, and it was the first time I've heard of it, so I had just assumed that it's just independent of ZFC. | |
| Jun 30, 2018 at 7:51 | vote | accept | CommunityBot | ||
| Jun 30, 2018 at 7:51 | |||||
| Jun 29, 2018 at 21:18 | comment | added | Noah Schweber | @tomasz As a warmup, consider the ultraproduct (over $\mathbb{N}$, with respect to nonprincipal $\mathcal{U}$) of the family of sets $X_i$ defined recursively as $X_1=\{1, 2\}$, $X_{i+1}=\{1, ..., 2\cdot (\vert \prod_{j\le i}X_j \vert)\}$; there is a natural map $\pi$ from $2^\omega$ to $\prod_{i\in\mathbb{N}}X_i$ such that $\rho(f)$ and $\rho(g)$ disagree cofinitely often whenever $f\not=g$. Since $\mathcal{U}$ is nonprinciple, this gives an injection of $2^\omega$ into $\prod_{i\in\mathbb{N}} X_i/\mathcal{U}$, hence the ultraproduct has cardinality at least continuum. | |
| Jun 29, 2018 at 20:06 | comment | added | Maxime Ramzi | @tomasz : I know it needs to be proved but I didn't feel it was necessary to write it down since it is a classical result. If you wish I can add a proof of this fact ? (And indeed, I had forgotten about quantifier elimination for a minute) | |
| Jun 29, 2018 at 20:03 | comment | added | tomasz | @Max: completeness is also an easy consequence of quantifier elimination. I think this is older than the concept of categoricity (but probably not older than the observation that algebraic closure of a field is unique up to isomorphism, which implies categoricity). In any event, you need to show that ultraproduct has cardinality of the continuum, which is not trivial, as Alex said. | |
| Jun 29, 2018 at 19:56 | comment | added | tomasz | @EmilJeřábek: I know. I just didn't see that the ultraproduct has the right cardinality | |
| Jun 29, 2018 at 16:58 | comment | added | Alex Kruckman | @Max Yes indeed. | |
| Jun 29, 2018 at 16:57 | comment | added | Maxime Ramzi | @AlexKruckman : Indeed, but it's a standard fact (unless I'm mistaken) that a nontrivial ultraproduct of countable structures has cardinality $2^{\aleph_0}$ | |
| Jun 29, 2018 at 16:49 | comment | added | Alex Kruckman | The fact that the ultraproduct has size $2^{\aleph_0}$ is not completely trivial. The trick is finding a continuum family of sequences $(x_p)_{p\in \mathbb{P}}$ which you know to be distinct in the ultraproduct, i.e. such that $\{p\in \mathbb{P}\mid x_p = y_p\}$ is finite whenever $(x_p)$ and $(y_p)$ are distinct sequences in the family. | |
| Jun 29, 2018 at 15:26 | comment | added | Maxime Ramzi | @tomasz your argument is perhaps simpler, but the proof that the theory algebraically closed fields of characteristic zero is complete goes through (at least the proof I know) its $\kappa$-categoricity, for $\kappa$ uncountable, and since this field and $\mathbb{C}$ have the same cardinal... | |
| Jun 29, 2018 at 14:38 | comment | added | Emil Jeřábek | @tomasz There is only one algebraically closed field of characteristic $0$ and cardinality $2^\omega$. | |
| Jun 29, 2018 at 14:35 | comment | added | tomasz | I don't see why the resulting field would be isomorphic to the complex field (in absence of CH). But it will be an algebraically closed field of characteristic zero, which is what is important here. | |
| Jun 29, 2018 at 11:18 | history | answered | Maxime Ramzi | CC BY-SA 4.0 |