I'm bumping this question, since I'm still curious regarding the answer but this question seems to have gone unnoticed. Bumped again.
This is a follow-up to this question, regarding a stronger variant of upliftingness.
For a function $f$, we set $f' = \{(\alpha,\delta): (\alpha,\alpha+\delta) \in f\}$. For example $(\lambda \alpha.\alpha+1)'$ is the constant function with output $1$. Now, we extend the notion of iterative upliftingness to be with respect to a function:
Let $f$ be a function $\textrm{Ord} \to \textrm{Ord}$ and $X$ be a class of ordinals. $\kappa$ is called $f$-uplifting onto $X$ iff $\kappa \in \textrm{dom}(f')$ and, for every ordinal $\theta$, and there is a monotonically increasing sequence $(\gamma_i)_{0 \leq i \leq f'(\kappa)}$ such that:
- $\gamma_0 = \kappa$
- $\gamma_1 > \theta$
- For every $0 \leq i \leq f'(\kappa)$, $\gamma_i \in X$
- For every $0 \leq i < j \leq f'(\kappa)$, $(V_{\gamma_i}, \in) \prec (V_{\gamma_j}, \in)$ is a proper elementary extension.
And we thin it out:
Let $f$ be a function $\textrm{Ord} \to \textrm{Ord}$, $X$ be a class of ordinals and $\alpha$ be an ordinal. $\kappa$ is called $0$-thinly $f$-uplifting onto $X$ iff it is $f$-uplifting onto $X$ above $\theta$. For $\alpha > 0$, $\kappa$ is called $\alpha$-thinly $f$-uplifting onto $X$ above $\theta$ iff $S(f'(\kappa)) \cap \kappa$ is stationary in $\kappa$, where $S: \textrm{Ord} \to \mathcal{P}(X)$ is defined as $S(0) = X$, $S(\beta+1) = \{\xi \in X: \textrm{ for every } \gamma < \alpha, \kappa \textrm{ is } \gamma \textrm{-thinly } (\lambda \delta.\delta+\xi)\textrm{-uplifting onto } S(\beta)\}$ and $\beta \in \textrm{Lim} \rightarrow S(\beta) = \bigcap_{\gamma < \beta} S(\gamma)$.
Then, how does thinned upliftingness compare to shrewdness and iterative upliftingness, knowing that the latter is below Mahloness but above $V_\kappa \vDash \textrm{Ord is Mahlo}$?
