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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?

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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.

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  • $\begingroup$ 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 at 4:56
  • $\begingroup$ Surely there can be more than one way to express this... Yes, yours works too. $\endgroup$ – Asaf Karagila May 21 at 8:29

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