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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.

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You have answered your own question.

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.

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    $\begingroup$ This doesn't seem to answer the question. The OP was asking about extensions that can impose rigidity, he suggested AC, but I think this is not enough, we need GC to make a theory with atoms rigid. $\endgroup$ Feb 8 at 22:07
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    $\begingroup$ Yes, I had answered the question asked in the title "Is the universe of ZFA rigid". The OP seems to have asked a different question in the body of the post. But actually, I am unsure what that question in the body is asking. The claim that atoms are indiscernible only within the permutation models is incorrect. Also, under AC every set-sized structure can be trivially extended to a rigid structure, but it is unclear whether that is the kind of answer that is sought. $\endgroup$ Feb 8 at 22:22
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    $\begingroup$ Thank you, Joel and Zuhair for the answers. Perhaps I was not clear in my question. In other words: under AC in ZFA, does Joel and Palumbo's result that every set has a rigid relation also holds, that is, even for atoms? I guess that even a permutation model, which is a structure in ZFA, can be extended to a rigid structure. Is this correct? $\endgroup$ Feb 8 at 23:44
  • $\begingroup$ Sorry to insist, but I am in doubt about something you don't find in the books and seems to require the interpretation of a more experienced set theorist. In ZFA we have the pairing axiom, so we can form the unitary set {a} even if a is an atom. Why this fact would not tell us that the atoms can always be discerned from one another? Fraenkel took a collection of "distinct" atoms for his "cells". Then, under extensions of permutations, the atoms "are made" indiscernible within the model. But I think that they are not for someone living outside the Model. Any mistake in this reading? $\endgroup$ Feb 9 at 11:30
  • $\begingroup$ 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. $\endgroup$ Feb 9 at 12:58

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