MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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? – Ali Enayat Mar 20 '13 at 2:05
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. – Tim Mercure Mar 20 '13 at 18:00
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. – 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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.