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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 $V_\kappa \vDash \textrm{Ord is Mahlo}$?

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  • $\begingroup$ "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? $\endgroup$
    – C7X
    May 12 at 15:54
  • $\begingroup$ Good point. Done. $\endgroup$
    – Binary198
    May 13 at 16:53

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