# What large cardinals are limits of Ramsey cardinals?

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 $$< \kappa$$.

(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.

• Regarding the last paragraph: Woodin cardinals are not measurable, but are an inaccessible limit of measurable cardinals. Commented Apr 25, 2020 at 0:53
• (And, again, congratulations!) 🥳 Commented Apr 25, 2020 at 0:54
• @AndrésE.Caicedo: Thanks! :) Commented Apr 25, 2020 at 7:18
• What does "has measure $1$" mean? Does it just mean there are $\kappa$ Ramsey cardinals below $\kappa$? Commented Aug 6 at 2:08
• @Lucenaposition: It means that the set of Ramsey cardinals is in a normal measure on $\kappa$. Indeed, in any normal measure. So it is in fact stationary, and quite large at that. That there are $\kappa$ of them simply follows from the conjunction of $\kappa$ is regular and it is the limit of Ramsey cardinals. Commented Aug 6 at 5:22