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Zuhair Al-Johar
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Can we have an inverted iterative hierarchy?

Let stratified $\sf Z$ have all axioms of $\sf Z$-$\sf Reg.$ with Infinity stipulated by existence of $\omega$, and with Separation restricted to use only stratified formulas, where those are defined after Quine's New Foundations. AC is not assumed.

Define: $y = \mathcal P^{-1}(x) \iff x= \mathcal P(y)$

Define: $x= \bigcap_{s \in S} s \iff x=\{y \mid S \neq \varnothing \land \forall s \in S (y \in s)\}$

Now, lets try coin what we mean by rank or stage of the inverted iterative hierarchy, which we'd label as inverted rank.

$X$ is an inverted rank, if and only if, there is an ordinal $\alpha$ and there is a bijective function $f$ whose domain is $\alpha$ that sends $0$ to $X$ and that sends each $\beta+1 \in \alpha$ to $ \mathcal P^{-1}(f(\beta))$, and sends each limit $\lambda \in \alpha$ to $ \bigcap_{\kappa < \lambda} f(\kappa)$.

We phrase that as: $X$ is an inverted rank of degree $\alpha$.

For brevity "inverted rank" shall be denoted "i.rank".

Axiom I: $X \text { is an i.rank } \land Y \text{ is an i.rank} \to X \subseteq Y \lor Y \subseteq X$

Axiom II: $\forall x \exists Y: Y \text{ is an i.rank } \land x \in Y$

Axiom III: $\alpha \text { is an ordinal } \to \exists X: X \text{ is an i.rank of degree } \alpha$

Axiom VI: $\forall A \exists x \forall y \, (y \in x \leftrightarrow y \in A \land y \text { is an i.rank } \}$

Now, can this theory be consistent? I suspect the last axiom might cause problems, but even if so, can the theory without it be consistent? I mean this theory goes against the norms of the known cumulative hierarchy, where no descending power sequence like that is possible. So, does it stand a chance of being consistent whether with or without the last axiom.

If the above is consistent can we have an external automorphism from an inverted rank to its predecessor inverted rank?

Zuhair Al-Johar
  • 11.3k
  • 1
  • 13
  • 47