Is it the case that every transitive definable set is also ordinal definable?

Formal definition of the former would be:

$TD^M=\{u\in M: \exists \tau_1,...,\tau_n\in Trs^M, \varphi\in Formula (M\models\forall x(\varphi(x,\tau_1,...,\tau_n)\leftrightarrow x=u))\}$

A element $u$ of $M$ is transitive definable in $M$ iff $u \in TD^M$

where $TD^M$ is the class of all transitive definable sets in $M$. $Trs^M$ is the class of all transitive sets in $M$.

IF the answer to the above question is to the negative, then is the class of all hereditarily transitive definable sets a model of ZFC?


Every set is transitive definable, in ZF, even.

To see why, note that $x$ is the unique maximal element of $\operatorname{tcl}(\{x\})$. Therefore the formula $\varphi(x,p)$ defined as $$x\in p\land\forall y(y\in p\to y=x\lor\exists z(z\in p\land y\in z))$$ defines $x$ with the parameter $\operatorname{tcl}(\{x\})$. The formula simply states that $x$ is the unique element of $p$ which is not an element of any other member of $p$, which is easily the case if we use the right parameter.

So the class you define is just $V$ itself, even if choice is not assumed.

| cite | improve this answer | |
  • $\begingroup$ I understand the argument, but I don't understand the formula. I think your formula must be $x \in p \wedge \not \exists y (y \in p \wedge x \in y)$ $\endgroup$ – Zuhair May 21 '19 at 4:56
  • $\begingroup$ Surely there can be more than one way to express this... Yes, yours works too. $\endgroup$ – Asaf Karagila May 21 '19 at 8:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.