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Why not $\sf ZFC+[V=HOD]$ as the standard set theory?

It implies the existence of a definable global choice and well-order, and it is compatible with all large cardinal axioms extending $\sf ZFC$, so per maximality standpoint we are not losing anything, and more we are gaining a stronger notion of choice and more theorization and expressibility. Also, $\sf V=HOD$ is equivalent to saying that the parameter free definable sets forms an elementary substructure of the universe of sets, so it is saying that whatever is done by sets can be done by using parameter free definable sets. It's only among the latter ones that concrete sets can thrive, and accordingly are of the most convincing kind of sets intuitively speaking, so seeing that all work done in set theory can be done using them is a very strong feature to hold to, intuitively speaking. If the exposition of $\sf V=HOD$ is seen as being complex, it can be expressed metatheoretically in a simple manner by adding to $\sf ZFC$ an $\omega$-rule of definability that is:

$\textbf{Definability: }$ if $\phi_1,\phi_2, \phi_3,...$ are all $\sf FOL(=,\in)$ formulas in which only symbol "$y$" occurs free, and "$y$" never occur bound, and that doesn't use the symbol "$x$", and $\psi$ is a formula in which only symbol "$x$" occurs free, and "$x$" never occur bound; then:

$\underline {[i=1,2,3,...; \\ \forall x \, (x=\{y \mid \phi_i\} \to \psi)]} \\ \forall x: \psi$

In English: if a parameter free formula holds for all parameter free definable sets, then it holds for all sets.

This was proved by Hamkins to be equivalent over $\sf ZFC$ to the set theoretic axiom $\sf V=HOD$

So, it meets all intuitive and technical qualifications for it to be there!?

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    $\begingroup$ Strength is not the sole thing which makes something a desireable foundational theory. First and foremost it should be considered "natural" by the bulk of the community, and I highly doubt any form of V=HOD is gonna fly by that requirement. $\endgroup$
    – Wojowu
    Commented Apr 13 at 9:13
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    $\begingroup$ Axioms of foundations are supposed to be clearly true. Not a formal gimmick. $\endgroup$
    – Emil Jeřábek
    Commented Apr 13 at 9:34
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    $\begingroup$ @JoshuaZ Not sure I agree with your remark, since of coruse replacement was absent from Zermelo's original theory, but this was seen as something to be addressed, and this is how we got ZFC. $\endgroup$
    – Joel David Hamkins
    Commented Apr 13 at 13:34
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    $\begingroup$ I disagree with "parameter free definable sets ... are the most concrete of sets." The set $\{n\in\omega:2^{\aleph_n}=\aleph_{n+1}\}$ is parameter free definable, but I wouldn't call it concrete. We don't know, even for a single $n$, whether it's in this set. $\endgroup$
    – Andreas Blass
    Commented Apr 13 at 13:45
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    $\begingroup$ @JoelDavidHamkins That's an excellent point and completely refutes my claim. $\endgroup$
    – JoshuaZ
    Commented Apr 13 at 14:30

1 Answer 1

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What does it mean to be a "standard" theory?

By any account, the theory ZFC + V=HOD already is one of the "standard" theories. The axiom V=HOD is intensely studied by set theorists; it appears as a hypothesis in innumerable theorems; it is seen as expressing a certain regularity phenomenon for the universe of sets; it is true in the constructible universe and all the canonical inner models of large cardinals; it is forceable by class forcing over any model of ZFC; it is a central concern in the model theory of set theory, in light of the fact that many theorems holding of models of PA in the arithmetic realm generalize to models of ZFC+V=HOD in the set-theoretic realm; and it is seen as saturated with philosophical significance.

We are already working out the nature of this theory and its relation with all manner of set theoretic ideas. What else would one want for this theory?

I see that you refer to "the" standard theory, but if this is meant to suggest that somehow everyone would agree to work only in that theory, then of course there is no standard theory in this sense.

The closest thing would be ZFC, of course, but even so, people routinely work in theories other than ZFC. Whole parts of the subject investigate failures of the axiom of choice, and in certain parts of the subject one routinely works in (often not completely specified) large-cardinal extensions of ZFC, or with determinacy assumptions, and so forth. Certainly it is not true that there is unanimous agreement to work exclusively in ZFC set theory.

Meanwhile, there are certain reasons not to take V=HOD as expressing a fundamental truth of set theory. In my multiverse paper, I explain how our understanding of HOD proceeds mainly by analyzing it in outer models where V=HOD fails:

We understand the coquettish nature of HOD, for example, by observing it to embrace an entire forcing extension, where sets have been made definable, before relaxing again in a subsequent extension, where they are no longer definable.

In other words, the axiom V=HOD is not robust with respect to inner and outer models—it is not always preserved by forcing; it's negation is not preserved by forcing. These features can be seen as a mark of fragility, perhaps a mark against it as a fundamental principle.

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  • $\begingroup$ By standard axiomatic set theory I mean the one that set theorists are expected to mention in any chapter they'd write on introducing axiomatic set theory, it is the most accepted set of axioms about sets. I see ZFC having that status, but I see that ZFC+ V=HOD is ought to be considered also as such. There can be no unanimous consensus about any of the axioms of set theory even the most innocents ones like Extensionality. But, from your answer here, I'm even more motivated to consider ZFC+V=HOD to be in the canonical set of axioms about sets. $\endgroup$
    – Zuhair Al-Johar
    Commented Apr 13 at 13:29
  • $\begingroup$ @ZuhairAl-Johar It is a purely sociological question which set theory is "standard" in your sense of the term. We now live in a "post-standards" world, and there's little hope of dislodging ZFC from its conventional role (well, there may be some hope of adding infinitely many inaccessibles, since that is assumed by the most popular proof assistants). Of course, people continue to argue about axioms, such as V=L. But the result of such arguments is predictable. $\endgroup$
    – Timothy Chow
    Commented Apr 14 at 11:38
  • $\begingroup$ @TimothyChow, I do concede that there is some sociological aspect to coining of standards. But, I do also think that it is not merely so. There must be some reasonably presented argument for motivating such development, and by reasonable I mean not merely being sociological, but something that touches the very nature of the manipulated concept. It is this true core touch that I'm into here. So, it is YES, there is a human factor into it, and NO, it is not only that human factor, and it is the NO that is most important! $\endgroup$
    – Zuhair Al-Johar
    Commented Apr 14 at 19:11
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    $\begingroup$ @ZuhairAl-Johar Certainly, back when ZF was first proposed, "reasonably presented arguments" were indeed given. But you seem to implicitly assume that an open dialogue is continuing until this very day, with the choice of foundation remaining "up for grabs" to the most persuasive debater. While a minority of people still think that way, it's a sociological fact that the majority regards the choice of foundation as largely a matter of convenience and convention. The attitude is, if you have a favorite axiom, by all means investigate it, but don't expect us to switch foundations. $\endgroup$
    – Timothy Chow
    Commented Apr 14 at 19:44
  • $\begingroup$ Similarly, I think you exaggerate the extent to which the other axioms, such as extensionality, are in flux. Some standard weakenings of ZFC are widely studied, such as ZF, ZF_fin, KP, ZF-power, ZFC-power, as well as alternatives such as ZFA, ZFU, ZFCU, etc., and these can all be considered "standard" theories. Dropping extensionality, although studied, is at best a curiosity. $\endgroup$
    – Joel David Hamkins
    Commented Apr 14 at 19:57

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