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Nov 25, 2023 at 0:24 comment added Farmer S @JoelDavidHamkins: Re $\kappa_0<\gamma_0$: I see, I was complicating it unnecessarily. And yes, got the email.
Nov 23, 2023 at 17:25 comment added Joel David Hamkins Can't we argue $\kappa_0<\gamma_0$ more simply like this: having the reflective property is $\Pi_2$ expressible. So if it exists, $\kappa_0$ is below every $\Sigma_3$-correct cardinal, and extendibles are $\Sigma_3$ correct. So $\kappa_0<\gamma_0$.
Nov 23, 2023 at 16:50 history edited Farmer S CC BY-SA 4.0
added 3415 characters in body
Nov 23, 2023 at 0:12 history edited Farmer S CC BY-SA 4.0
added 453 characters in body
Nov 22, 2023 at 19:30 history edited Farmer S CC BY-SA 4.0
Added upper bound on consistency strength
Nov 22, 2023 at 16:26 history edited Farmer S CC BY-SA 4.0
Improved answer.
Nov 22, 2023 at 16:00 history edited Farmer S CC BY-SA 4.0
Improved and simplified answer.
Nov 22, 2023 at 15:44 history answered Farmer S CC BY-SA 4.0