Timeline for Consistency of many Erdos cardinals

5 events

when toggle format	what		by	license	comment
Aug 2, 2013 at 14:17	answer	added	Philip Welch		timeline score: 6
Mar 30, 2013 at 0:51	comment	added	Bill Mitchell		Of course if $\kappa$ is Ramsey then $V_\kappa$ satisfies that every $\kappa(\alpha)$ exists, so a Ramsey cardinal is enough (and much more than enough) to get this. I suspect that "$\kappa(\alpha)$ for every $\alpha$ is is not equivalent to anything with a simpler statement.
Mar 20, 2013 at 18:00	comment	added	Tim Mercure		Sort of, ultimately I'm curious about the strength of various sharps existing (or all sharps existing). I was hoping that if there was a paper out there looking at a class of Erdos cardinals it would be a good starting point, since having $\kappa(\alpha)$ for all alpha means that every set of ordinals has a sharp. But this seems like overkill: if $a\subseteq \alpha$ we only need $\kappa \to (\omega_1)^{< \omega}_{2^{\vert \alpha \vert}}$ for $a^\sharp$ which seems much weaker than $\exists \beta \enspace \kappa(\beta)>2^{\vert \alpha \vert}$, though perhaps the universal assertions coincide.
Mar 20, 2013 at 2:05	comment	added	Ali Enayat		Since every (uncountable) measurable cardinal is an Erdős cardinal, a safe upper bound to "$\kappa(\alpha)$ exists for each $\alpha$ is "the measurable cardinals are cofinal in the ordinals". Jech's text, as well as Kanamori's has the full details. Are you looking for sharper upper bounds?
Mar 19, 2013 at 18:05	history	asked	Tim Mercure	CC BY-SA 3.0