# Relative consistency of $\mathsf{ZFC}-$Ext$+\neg$Ext+“every set has a unique powerset”

Let $$T$$ be the $$\mathcal L_\in$$-theory whose axioms are the axioms of $$\mathsf{ZFC}$$, with extensionality replaced by its negation, and an additional axiom specifying that every set has a unique powerset, formally

$$\forall x\forall y\forall z((\forall v(v\in y\leftrightarrow v\subseteq x)\land\forall w(w\in z\leftrightarrow w\subseteq x))\rightarrow y=z),$$

where $$a\subseteq b$$ is to be expanded as an $$\mathcal L_\in$$-formula in the usual way.

Is $$T$$ consistent relative to $$\mathsf{ZFC}$$? The motivation for this question comes from the fact that $$V\setminus V_\omega$$ models $$\mathsf{ZFC}$$ with extensionality replaced by its negation, but powersets are not unique: for every $$x\in V\setminus V_\omega$$ both $$\mathcal P(x)^V\setminus V_\omega$$ and $$\{s\in V\setminus V_\omega\mid s\setminus x\subseteq V_\omega\}$$ fulfill the definition of powerset.

• ZFC with urelements? – Noah Schweber Apr 18 at 13:21
• I'm not familiar with how urelements are added to ZFC precisely, can you elaborate? @Noah – Alessandro Codenotti Apr 18 at 14:21

As Noah commented. ZFC + Ur-elements is known to be consistent relative to ZFC. It is usually referred to by "ZFA", i.e. ZF with atoms. In one formulation of that theory ONLY empty objects violate Extensionality! So we do have many empty objects (the Ur-elements). However all non-empty sets are extensional! That is, "No distinct non-empty sets are co-extensional (i.e. have the same elements)". Clearly this theory satisfy all of your conditions.

I'll post this as a separate answer, because I'm not quite sure of it, although I'm sure of the above accepted answer.

I'll present a rather trivial way of getting $$ZFA$$ from $$ZF$$.

Define an alternative membership relation $$\in^*$$ as:

$$y \in^* x \iff y \in x \wedge [x \text { is an initial segment of } N \to y \neq max(x)]$$

where naturals are defined after Zermelo as the empty set or the iterated singletons of the empty set. An initial segment of N is a finite set of naturals closed under predecessor function for non zero elements; and $$max(x)$$ is the maximal element of $$x$$.

So in English: $$y$$ is an alternative member of $$x$$ is the same as the epsilon member of $$x$$ if $$x$$ is not an initial segment of $$N$$, but when $$x$$ is an initial segment of N then $$y$$ is an alternative member of $$x$$ only if $$y$$ is a non-maximal member of $$x$$.

This way we'll have the empty set $$0$$ being $$\in^*$$ empty, and also the set $$\{0\}$$ set being $$\in^*$$ empty as well. So we have a single violation of Extensionality. I think it is provable in $$ZF$$ that all rules of $$ZFA$$ would hold for this new membership relation $$\in^*$$ over any domain of $$ZF$$. Also you can get $$ZFCA$$ from $$ZFC$$ under the same interpretation.