Does anyone have a reference for the consistency strength of saying that the Erdos cardinal $\kappa(\alpha)$ exists for all $\alpha$? Or of various weakenings about how far such cardinals extend into $\text{Ord}$?

This comes up while looking at the consistency strength of sharps of sets of ordinals, so any reference on that front would also be welcome.

  • $\begingroup$ 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? $\endgroup$ – Ali Enayat Mar 20 '13 at 2:05
  • $\begingroup$ 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. $\endgroup$ – Tim Mercure Mar 20 '13 at 18:00
  • $\begingroup$ 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. $\endgroup$ – Bill Mitchell Mar 30 '13 at 0:51

I agree with the last sentence of Bill Mitchell's comment. But here is something closer than the cardinals mentioned. In [1] the notion of "almost Ramsey" cardinal was coined. Such a cardinal $\kappa$ is required to be $\alpha$-Erdos for all $\alpha<\kappa$. Then $V_\kappa$ is a model of what you are after (but in fact this is still not exact as there are many other $\gamma <\kappa$ for which this is true too. Almost Ramseys also get an outing in [2].

If you are interested in sharps alone, then, eg just for sharps for reals, $\kappa \rightarrow (\omega_1)^{<\omega}_{2}$ is already overkill: one really just needs for any function $f$ homogeneous sets of arbitrarily large but countable length, all of which have the same "type". This has been investigated closely in [3]. Similar considerations would hold for sharps of other sets of ordinals. The least inner model in which every set has a sharp, $L^\#$ say, is too thin to contain any Erdos cardinals (other than trivially $\kappa(\delta)$ for $\delta< \omega_1^{L^\#}$).

[1] J. Vickers & P.Welch "Elementary Embeddings of an inner model into the Universe" JSL vol 66, 2001.

[2] A. Apter & P. Koepke "Making All cardinals almost Ramsey" Archive for Math. Logic, vol 47, 2008.

[3] J. Baumgartner & F.Galvin "Generalized Erdos cardinals and $0^\#$", Ann. of Math. Logic, vol. 15 , 1978.


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