# Thinning chains of elementary extensions

I'm bumping this question, since I'm still curious regarding the answer but this question seems to have gone unnoticed.

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:

1. $$\gamma_0 = \kappa$$
2. $$\gamma_1 > \theta$$
3. For every $$0 \leq i \leq f'(\kappa)$$, $$\gamma_i \in X$$
4. 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 $$V_\kappa \vDash \textrm{Ord is Mahlo}$$?

• "iff, for every ordinal $\theta$, $\kappa\in\textrm{dom}(f')$ and there is a ...". What if we replace this with "iff $\kappa\in\textrm{dom}(f')$ and, for every ordinal $\theta$, there is a ...", i.e. moving the $\kappa\in\textrm{dom}(f')$ outside the scope of $\theta$'s quantification?
– C7X
May 12 at 15:54
• Good point. Done. May 13 at 16:53