# Can all hereditarily shrinking relations define sets?

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 $$ZFC$$?

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

• 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? – Noah Schweber Feb 14 at 17:50
• @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.. – Zuhair Feb 14 at 19:51

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.