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, 2019 at 16:33
  • $\begingroup$ As noted in a different Mathoverflow post, a definition that also requires $j(f)\kappa=f(\beta)$ whenever $\beta \lt \kappa$ would be easier to handle. On the other hand Gabe Goldberg's proof that Shelah cardinals are weakly Shelah doesn't seem to work with that definition. $\endgroup$ Sep 2, 2021 at 9:28

2 Answers 2


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.

  • $\begingroup$ 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, 2019 at 18:31
  • $\begingroup$ You're right, it was a typo, I meant limits of weakly Shelahs $\endgroup$ Aug 17, 2019 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, 2019 at 18:36
  • 1
    $\begingroup$ 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.) $\endgroup$ Aug 17, 2019 at 21:28

I think any measurable Woodin cardinal is a limit of weakly Shelah cardinals.

To see this, note that, if $\kappa$ is a Woodin cardinal, for any $f : \kappa \to \kappa$, $\kappa$ is a limit of cardinals $\nu$ that are $\lt \kappa$-strong for $f$ (that's probably a standard result but I know it from theorem 9.9. of "Double helix in large large cardinals and iteration of elementary embeddings" by Kentaro Sato), which means that for any $\gamma \lt \kappa$, there is an elementary embedding $\bar i : V \to \bar N$ such that $V_{f(\gamma) +1} \subset \bar N$ and $\bar i(f) \upharpoonleft (\gamma +1) = f \upharpoonleft (\gamma +1)$.

If $\kappa$ is additionally measurable, there is an elementary embedding $j : V \to M$ with critical point $\kappa$. By elementarity, there's an elementary embedding $i : M \to N$ with critical point $\nu$ such that $V_{f(\kappa) +1}^M \subset N$ and $i(f) \upharpoonleft (\gamma +1) = j (f) \upharpoonleft (\gamma +1)$. Since this works for any $f : \kappa \to \kappa$, this shows that $\kappa$ is weakly Shelah in $M$ (we even have the strengthening described in this question) and thus a limit of weakly Shelah cardinals.

By a similar argument using the elementary embedding characterization of weakly compact cardinals, we can probably prove that any weakly compact Woodin cardinal is a limit of weakly Shelah cardinals. (Added September 17, 2022) No, that doesn't work because, if $j: P \to M$ is a weakly compact embedding with critical point $\kappa$, $M$ may contain functions that are not in $P$. However, the argument does work for weakly Ramsey cardinals.

(Added September 17, 2022) A weakly superstrong Woodin cardinal is weakly Shelah. By theorem 5 of the paper that defined weakly superstrong cardinals, a cardinal $\kappa$ is weakly superstrong if and only if for any $A \subseteq V_\kappa$, there are $\lambda$ and $A^* \subseteq V_\lambda$ such that $\langle V_\kappa, \in, A \rangle \prec \langle V_\lambda, \in, A^* \rangle$. If $\kappa$ is Woodin and weakly superstrong, then, for any $f : \kappa \to \kappa$, there is a cardinal $\nu \lt \kappa$ that is $\lt \kappa$-strong for $f$, and an elementary extension $\langle V_\kappa, \in, A \rangle \prec \langle V_\lambda, \in, f^* \rangle$. By elementarity, $\nu$ is $\lt \lambda$-strong for $f^*$. Since this works for any $f : \kappa \to \kappa$, this shows that $\kappa$ is weakly Shelah and also satisfies the strengthening mentioned above.


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