Timeline for Do these ordinals exist?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 20, 2017 at 5:29 | vote | accept | Zetapology | ||
| Sep 20, 2017 at 5:29 | vote | accept | Zetapology | ||
| Sep 20, 2017 at 5:29 | |||||
| Sep 19, 2017 at 1:37 | comment | added | Andreas Blass | Yes, the supremum $F_\omega(\alpha)$ of a countable set $\{F_n(\alpha):n\in\omega\}$ of countable ordinals is countable. This uses the axiom of choice. I'm also using the availability of replacement for formulas involving the satisfaction predicate; without that, I couldn't guarantee that $\{F_n(\alpha):n\in\omega\}$ is a set. | |
| Sep 18, 2017 at 22:39 | comment | added | Zetapology | Just to be clear, the reason $F_\omega(\alpha)$ is countable is because $\aleph_1$ has cofinality $\omega_1$, meaning that the set of all $F_n(\alpha)$ for finite $n$ (which has cardinality $\aleph_0$) is not cofinal in $\omega_1$, and therefore does not have supremum $\omega_1$. Is this correct? | |
| Sep 18, 2017 at 15:53 | history | answered | Andreas Blass | CC BY-SA 3.0 |