The following definition of a large cardinal property combines parts of the definitions of "Shelah cardinal" and "Woodin cardinal":

A cardinal $\kappa$ is weakly Shelah if for all $f : \kappa \to \kappa$ there is some $\alpha < \kappa$ that is closed under $f$ and there is some elementary embedding $j : V \to M$ (where $M$ is a transitive class) such that $\operatorname{crit}(j) = \alpha$ and $j(\alpha) > \kappa$ and $V_{j(f)(\kappa)} \subset M$.


  1. Is every weakly Shelah cardinal Shelah?

  2. Is the least weakly Shelah cardinal Shelah?

  3. Is every weakly Shelah cardinal measurable?

  4. What is the consistency strength of ZFC + "there is a weakly Shelah cardinal"? In particular, how does it relate to weakly hyper-Woodin cardinals as defined by Schimmerling?

Here's what I know:

If $\kappa$ is a Shelah cardinal, then $\kappa$ is weakly Shelah. To see this, let $f : \kappa \to \kappa$. By the Shelah property applied to $f+1$, there is an elementary embedding $j : V \to M$ such that $\operatorname{crit}(j) = \kappa$ and $V_{j(f)(\kappa)+1} \subset M$. Because Shelah cardinals are Woodin, some cardinal $\alpha < \kappa$ is $\mathord{<}\kappa$-$f$-strong in $V$ and therefore $\mathord{<}j(\kappa)$-$j(f)$-strong in $M$ by elementarity. In particular $\alpha$ is $(j(f)(\kappa)+1)$-$j(f)$-strong in $M$, and therefore also in $V$ because $V_{j(f)(\kappa)+1} \subset M$. It's not hard to see that $\alpha$ witnesses the weakly Shelah property of $\kappa$ in $V$ with respect to $f$.

On the other hand, if $\kappa$ is weakly Shelah, then $\kappa$ is a Woodin limit of Woodin cardinals and there are Woodin cardinals above $\kappa$. Clearly the definition implies $\kappa$ is Woodin. Then by considering various $f$ we can obtain cofinally many $\alpha < \kappa$ as in the definition, and $j(\alpha) > \kappa$ implies $\alpha$ is a limit of Woodin cardinals, so $\kappa$ is a limit of Woodin cardinals. Now applying the definition to the function $f : \kappa \to \kappa$ where $f(\alpha)$ is the successor of the least Woodin cardinal above $\alpha$ shows that there is a Woodin cardinal above $\kappa$.

  • 7
    $\begingroup$ This is one of the reasons to avoid naming cardinals after people. Shelah is Woodin, Shelah is strong, and Shelah is inaccessible are all weird propositions with varying degrees of truth in real life. Although from his stories one can easily conclude that Woodin is worldly! :-) $\endgroup$ – Asaf Karagila Aug 15 '19 at 16:33

To answer the first three questions negatively, the key is to show that measurable weakly Shelah cardinals are limits of weakly Shelah cardinals.

To see this, suppose that $\kappa$ is weakly Shelah and measurable. Let $j : V\to M$ be an elementary embedding with critical point $\kappa$. We claim that $\kappa$ is weakly Shelah in $M$. We will show that $\kappa$ has the weak Shelah property with respect to any increasing function $f : \kappa\to \kappa$. (It suffices to handle increasing functions.) In other words, we will find a cardinal $\nu < \kappa$ and an elementary embedding $i : M\to N$ definable over $M$ such that $\text{crit}(i) = \nu$, $i(\nu) > \kappa$, and $V_{i(f)(\kappa)}\cap M\subseteq N$. Since $\kappa$ is weakly Shelah in $V$, there is a cardinal $\nu < \kappa$ and a definable elementary embedding $\bar i : V\to \bar N$ such that $\text{crit}(\bar i) = \nu$, $\bar i(\nu) > \kappa$, and $V_{\bar i(f)(\kappa)}\subseteq \bar N$. Let $i = j(\bar i)$ and let $N = j(\bar N)$. Thus $i$ (resp. $N$) is defined over $M$ by the same formula as $\bar i$ (resp. $\bar N$) but with the parameters shifted via $j$.

We now verify that $i$ is as desired. It's pretty easy to see $i(\nu) = \bar i(\nu) > \kappa$. Moreover by the elementarity of $j$, $V_{j(\bar i(f)(\kappa))}\cap M= j(V_{\bar i(f)(\kappa)})\subseteq j(\bar N) = N$. To show $V_{i(f)(\kappa)}\cap M\subseteq N$, it therefore suffices to show $j(\bar i(f)(\kappa)) \geq i(f)(\kappa)$, which is where we use that $f$ is increasing: $$j(\bar i(f)(\kappa)) = j(\bar i)(j(f))(j(\kappa))\geq j(\bar i)(j(f))(\kappa) = i(j(f))(\kappa) = i(f)(\kappa)$$

This shows $\kappa$ is weakly Shelah in $M$. Hence by the standard reflection argument, $\kappa$ is a limit of weakly Shelah cardinals. We can conclude that the least weakly Shelah cardinal is not measurable, so in particular it is not Shelah.

As for consistency strength, it is not even completely obvious from this that the existence of a Shelah cardinal implies the consistency of a weakly Shelah cardinal, but this too is true. Proving this involves examining the witnessing ordinal of (weakly) Shelah cardinals. The witnessing ordinal of a (weakly) Shelah cardinal $\kappa$ is the least cardinal $\lambda$ such that $\kappa$ is witnessed to be (weakly) Shelah by extenders in $V_\lambda$. The main observation about $\lambda$ is that $\kappa < \text{cf}(\lambda) \leq 2^\kappa$. It follows pretty easily that the witnessing ordinal for the first weakly Shelah is below the second Shelah cardinal. (Incidentally, it seems one must go through this kind of calculation just to show that two Shelahs are stronger than one.)

Still thinking about the weakly hyper-Woodin question.

| cite | improve this answer | |
  • $\begingroup$ Wait, how is this possible? I thought Shelah cardinals are measurable. No? That means that every Shelah cardinal is a measurable weakly Shelah, so it is the limit of Shelah cardinals. Therefore the ordinals are not well-founded... $\endgroup$ – Asaf Karagila Aug 17 '19 at 18:31
  • $\begingroup$ You're right, it was a typo, I meant limits of weakly Shelahs $\endgroup$ – Gabe Goldberg Aug 17 '19 at 18:35
  • $\begingroup$ Oh. Shame. It would have been nice to see all very large cardinals rendered inconsistent on MathOverflow... :-P $\endgroup$ – Asaf Karagila Aug 17 '19 at 18:36
  • $\begingroup$ I see, thanks! On further thought, it seems like $\kappa$ Woodin and $\Sigma_3$-reflecting implies $\kappa$ weakly Shelah: given $f:\kappa\to\kappa$, use Woodinness to get some $\alpha < \kappa$ that is $\mathord{<}\kappa$-$f$-strong. Then the desired conclusion (on existence of $j$) holds for cofinally many $\bar{\kappa} < \kappa$ in place of $\kappa$. Formulating this in terms of extenders, $\Sigma_3$-reflection implies the desired conclusion holds for cofinally many $\bar{\kappa} \in \operatorname{Ord}$ in place of $\kappa$, and therefore for $\kappa$ itself (since $f$ is increasing.) $\endgroup$ – Trevor Wilson Aug 17 '19 at 21:28

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.