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$\sf V=HOD$ is stated as:

$\forall X \, \exists \theta \, \exists \alpha < \theta \, \exists \varphi : X=\{y \in V_\theta\mid V_\theta\models\varphi(y,\alpha)\}$

This use two ordinal parameters (other than the code for $\varphi$) $``\theta; \alpha"$.

Can we do with just ONE parameter (other than the code for $\varphi$)? that is:

$\forall X \, \exists \alpha \, \exists \varphi: X= \{y \in V_\alpha \mid V_\alpha \models \varphi(y)\}$

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    $\begingroup$ But actually, one should $\varphi$ also as a parameter here, so you have two ordinal parameters: $\alpha$ and $\varphi$. I updated my answer to explain this. $\endgroup$
    – Joel David Hamkins
    Commented Jun 26, 2023 at 18:52
  • $\begingroup$ @JoelDavidHamkins, ah OK, I should edit then. $\endgroup$
    – Zuhair Al-Johar
    Commented Jun 26, 2023 at 18:56
  • $\begingroup$ @JoelDavidHamkins, on second thought. when we have a formula $\varphi(y,\alpha_1,..,\alpha_n)$ what is meant by parameters is usually $\alpha_1,..,\alpha_n$ and not the predicate $\varphi$ nor the set builder variable $y$, although in reality $\varphi$ is indeed a parameter (especially when coded by an object), but this is not usually counted among parameters in the context of definability. And besides we can code $\varphi$ by non-ordinals. So, along the usual spirit of definability, here we do have a single ordinal parameter. The predicate code shouldn't enter the count. $\endgroup$
    – Zuhair Al-Johar
    Commented Jun 29, 2023 at 13:40
  • $\begingroup$ In order to formulate HOD in ZFC, you must have codes for the formulas $\varphi$, and in an $\omega$-nonstandard model, these will include nonstandard formulas. It is important in that to know that the code is a (possibly nonstandard) natural number, in order to know that the set is still ordinal definable. That is, ultimately one must treat $\varphi$ as a parameter, and to know one has succeeded in capturing the notion of ordinal definable, it is important that the codes of formulas are ordinals. $\endgroup$
    – Joel David Hamkins
    Commented Jun 29, 2023 at 13:49
  • $\begingroup$ For example, to interpret your $\exists \varphi$ in a nonstandard model, this will include nonstandard $\varphi$, and the set might not be definable in $V_\theta$ using any standard formula. But it is definable in $V_{\theta+1}$ using also $\varphi$ as a parameter. $\endgroup$
    – Joel David Hamkins
    Commented Jun 29, 2023 at 13:51

1 Answer 1

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Yes, because there is a definable ordinal pairing function.

Specifically, if you want to get the set $\{y\in V_\theta\mid V_\theta\models\varphi(y,\alpha)\}$, then let $\beta=\langle\theta,\alpha\rangle$ be the ordinal coding the pair, and then look at $V_{\beta+1}$. Inside this structure, we have $\beta$ as the largest ordinal, and the model can decode $\beta$ as the pair $\langle\theta,\alpha\rangle$, and then get the set $\{y\in V_\theta\mid V_\theta\models\varphi(y,\alpha)\}$. So this set is definable inside $V_{\beta+1}$ without any extra parameter.

Ultimately one should view $\varphi$ also as a parameter (and formulas can be coded by finite numbers, hence by ordinals, a fact that is important when proving that HOD as defined internally is the same as the external concept of definable-from-ordinal parameters, since the internal definition might use nonstandard $\varphi$, but this is still OK since the code of a nonstandard formula is still a number, hence an ordinal). But this is fine, since we can consider $\beta=\langle\theta,\alpha,\ulcorner\varphi\urcorner\rangle$ and then proceed as I explained.

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  • $\begingroup$ Thanks. Why the first version is often used? $\endgroup$
    – Zuhair Al-Johar
    Commented Jun 26, 2023 at 18:22
  • $\begingroup$ In Zuhair's definition you don't act any function on $α$, that is you can't use the pairing function on $α$ and use the resulting ordinals as parameters $\endgroup$
    – Holo
    Commented Jun 26, 2023 at 18:22
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    $\begingroup$ @Holo Your objection is answered by my second paragraph. But actually, one can get the pair inside $V_\beta$ even without $\beta$ as an element. That is, the model can identify which pair it woud be, and so the $\beta+1$ subterfuge is not actually necessary. $\endgroup$
    – Joel David Hamkins
    Commented Jun 26, 2023 at 18:24
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    $\begingroup$ @ZuhairAl-Johar It often happens in mathematics that when one has two different statements that are equivalent, then one is more useful for some purposes and the other for other purposes. For example, to validate that a given model satisfies V=HOD, the more generous version is often more convenient, since no coding is required. But to prove a consequence of V=HOD, the second version can be more convenient, since it states a seemingly stronger property. I have seen both used in just this way. $\endgroup$
    – Joel David Hamkins
    Commented Jun 26, 2023 at 18:31
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    $\begingroup$ The reason some people prefer the first version is that it is closer to the original intention of OD as "definable with an ordinal parameter," i.e. $\exists\phi\exists\alpha\,x=\{y:\phi(y,\alpha)\}$. That isn't a definition in ZF, so one has to bring in the additional parameter $\theta$ and invoke the reflection principle. After all that, one notices that, as long as one needs $\theta$ anyway, one can use it to encode the original parameter $\alpha$. That's technically useful, but it's getting rather far from the original idea. $\endgroup$
    – Andreas Blass
    Commented Jun 26, 2023 at 19:19

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