# What's the consistency strength of this strengthening of weakly superstrong cardinals?

Recall that a cardinal $$\kappa$$ is weakly superstrong if, for every $$A \subseteq V_\kappa$$, there is a cardinal $$\lambda$$ and a set $$A^* \subseteq V_\lambda$$ such that $$\langle V_\kappa, \in, A \rangle \prec \langle V_\lambda, \in, A^* \rangle$$. We can define the stronger notion of a weakly superstrong cardinal with common target by requiring that the same $$\lambda$$ works for every $$A \subseteq V_\kappa$$.

Weakly superstrong cardinals with common target is a stronger notion than strongly uplifting, for by theorem 3 of the paper linked above, there are, for every $$A \subseteq V_\kappa$$, a club of cardinals $$\mu \lt \lambda$$ such that $$\langle V_\kappa, \in, A \rangle \prec \langle V_\mu, \in, A^* \rangle$$ for some $$A^* \subseteq V_\mu$$, so $$V_\lambda \vDash \text{"\kappa is strongly uplifting"}$$.

For an upper bound, if $$\delta$$ is weakly ineffable, it is a limit of cadinals that are weakly superstrong with common target $$\delta$$. To see this, define a function $$f$$ such that for every $$\kappa \lt \delta$$ that is not weakly superstrong with common target $$\delta$$, $$f(\kappa)$$ is a set of ordinals less than $$\kappa$$ that does not contain 0 coding the truth predicate for some $$\langle V_\kappa, \in, A \rangle$$ that has no elementary extention of the form $$\langle V_\lambda, \in, A^* \rangle$$, and if $$\kappa$$ is weakly superstrong with common target $$\delta$$, $$f(\kappa) = \{0\}$$. Since $$\delta$$ is weakly ineffable, there is an unbounded homogeneous set for $$f$$. Suppose for contradiction that there is a homogeneous set $$H$$ of ordinals that are not weakly superstrong with common target $$\delta$$. Then $$\cup_{\kappa \in H} f(\kappa)$$ codes (the truth predicate for) a set $$A^*$$ such that $$\langle V_\kappa, \in, A \rangle \prec \langle V_\delta, \in, A^* \rangle$$ for every $$A$$ coded by $$f(\kappa)$$, contradicting the definition of $$f(\kappa)$$. Thus, the homogeneous set consists of cardinals that are weakly superstrong with common target $$\delta$$.

Thus there remains one question: Is weakly superstrong with common target a stronger or weaker notion than subtle? More precisely, is subtle cardinals limits of weakly superstrong cardinals with common targets or are weakly superstrong cardinals with common targets limits of subtle cardinals? Is the answer independent, and if so, what is the answer for the constructible universe?

• Is this definition of weak superstrongness correct? It looks equivalent to the elementary substructure characterization of inaccessibility, and it's different than the one in the paper.
– C7X
May 5, 2022 at 13:11
• @C7X This definition is a weakening of the definition of strongly unfoldable cardinals. By theorem 5 of the paper I linked, it is equivalent to weakly superstrong. It differs from the characterization of inaccessible cardinals by involving a common $\kappa$ while $\lambda$ may vary, whereas the the characterization of inaccessible cardinals involve a common $\lambda$ while $\kappa$ will vary. May 5, 2022 at 21:58