4
$\begingroup$

What large cardinal axiom, if any, implies that there are unboundedly many Ramsey cardinals below the large cardinal?

That is, is there a large cardinal axiom, such that if $\kappa$ is that large cardinal there are unboundedly many Ramsey cardinals $< \kappa$.

$\endgroup$
0
8
$\begingroup$

(I removed the past version of the answer, since I read "unboundedly many Ramsey cardinals" as "proper class of Ramsey cardinals", which is perhaps the more natural reading for a set theorist.)


If $\kappa$ is measurable then it is the limit of Ramsey cardinals, and in fact it is a measure $1$ limit of Ramsey cardinals. To see that, note that (1) measurable is Ramsey; and (2) if $M$ is an inner model that agrees with $V$ up to $V_{\alpha+1}$, then $M$ agrees with $V$ on whether or not $\alpha$ is a Ramsey cardinal, since being Ramsey is a 2nd order property.

Combine the two facts, and we see that if $j\colon V\to M$ witnessing that $\kappa$ is measurable, then $V_{\kappa+1}\subseteq M$, so $\kappa$ is Ramsey in $M$. But then the set of Ramsey cardinals below $\kappa$ has measure $1$.

 

Of course, we can just look at "worldly limit of Ramsey cardinals" or "inaccessible limit of Ramsey cardinals", etc. Much weaker axioms.

$\endgroup$
3
  • $\begingroup$ Regarding the last paragraph: Woodin cardinals are not measurable, but are an inaccessible limit of measurable cardinals. $\endgroup$ Apr 25 '20 at 0:53
  • 1
    $\begingroup$ (And, again, congratulations!) 🥳 $\endgroup$ Apr 25 '20 at 0:54
  • $\begingroup$ @AndrésE.Caicedo: Thanks! :) $\endgroup$
    – Asaf Karagila
    Apr 25 '20 at 7:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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