# Is the universe of ZFA rigid? The pairing axiom implies that even atoms have a unitary set which discern them from all other atoms. So, is it rigid?

Although any permutation of atoms induces an automorphism of the whole universe, atoms seem to be indiscernible only within the permutation models. Can a permutation model be extended to a rigid structure? We can suppose that AC holds.

In ZFA, every definable permutation $$\pi:A\to A$$ of the class of atoms extends uniquely to an automorphism of the entire set-theoretic universe, simply by defining it on sets $$y$$ as follows: $$\pi(y)=\{\pi(x)\mid x\in y\}.$$ It is easy to see that $$x\in y\iff \pi(x)\in \pi(y)$$, and so this is an isomorphism. (One needs to know that $$\pi$$ is definable in order to know that the extension is definable, which is needed in order to know that $$\pi(y)$$ is a set, etc.)
If there are at least two atoms, then we can define $$\pi$$ so as to swap them, fixing all other atoms, and this map will extend to a definable automorphism of $$V$$. So in ZFA, if there are at least two atoms, then the universe is not rigid.
• In ZFA atoms are indiscernible in the sense that every assertion $\varphi$ expressible in the language of set theory that holds of one atom $\varphi(a)$ holds of all atoms $\varphi(b)$. You cannot tell them apart in the language of set theory. This is because there is an automorphism swapping $a$ and $b$, and automorphisms preserve truth. Feb 9 at 12:58