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Let $R$ be a relation definable in the language of set theory. We call $R$ as inherently proper class relation over domain $D$ if it and its complementary output over domain $D$ are proper classes, i.e. for each $x$ the class of all $y \in D$ such that $y R x$ is a proper class, and the class of all $y \in D$ such that $ y \not R x $ is also a proper class.

We recursively define $\in^n$ as:

$y \in^0 x \leftrightarrow y=x$

$y \in^{n+1} x \leftrightarrow \exists z (z \in^n x \land y \in z)$

Define $R^n$ recursively as:

$y R^0 x \leftrightarrow y=y$

$y R^{n+1} x \leftrightarrow y R^n x \land \forall z \in^n y (z R x)$

We say a relation is hereditarily shrinking if and only if for each $n>0$:$\ R^n$ is inherently proper class relation over its prior domain. That is, for each $x$ the class of all $y$ such that $y R^{n-1} x \land y R^n x$ is a proper class, and the class of all $y$ such that $y R^{n-1} x \land \neg y R^n x$ is a proper class too.

Is the following provable in $ZFC$?

$R \text { is hereditarily shrinking } \to \forall x \exists s (s=\{y|\forall n \in \omega (y R^n x)\}) $

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  • $\begingroup$ In your definition of $R^n$, should "$zRx$" be "$zR^nx$" at the end? Also, do you have an example of a hereditarily shrinking relation? $\endgroup$ – Noah Schweber 2 days ago
  • $\begingroup$ @NoahSchweber, I think the defintion I gave is OK. Examples of hereditarily shrinking relations are: "is singleton or empty","is subnumerous to", "is almost a subset of", the last is what I call as "quasi-subset" and it entails that there can maximally be one element of the quasi-subset that is not an element of the mother set, etc.. $\endgroup$ – Zuhair 2 days ago
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The above schema is inconsistent with ZF.

Take the $x R y$ to be $x=x \land (y \text{ is singleton} \lor y\text{ is a set of singletons})$

Clearly $R$ is hereditarily shrinking relation, yet the class of all sets $y$ hereditarily satisfying $R$ (for whatever $x$) is indeed a proper class.

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