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In this answer, Professor Hamkins gives a proof that for models $M$ of ZF, $M$ being a model of $\text{ZFC} + V = \text{HOD}$ is equivalent to there being a definable well-ordering of the universe:

https://mathoverflow.net/a/180734

His argument easily extends to an equivalence of these properties to $M$ having a well-ordering of the universe definable with respect to an ordinal parameter. So, if there is well-ordering of the universe with respect to some parameter $p,$ but there is not a well-ordering of $V$ definable without parameters, then necessarily $p \not \in \text{OD}$. Is this situation possible? My intuition is that it shouldn't be possible, since I don't think a non-ordinal parameter should be able to define something so fundamental when an ordinal cannot do the same.

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  • $\begingroup$ I'd say welcome to M.SE, Elliot, but it seems you've been here for a while. But it's good to see you utilizing this website! $\endgroup$ – Asaf Karagila Jan 10 '16 at 22:28
  • $\begingroup$ Ay, I've been meaning to for a long time but never got around to it till now. $\endgroup$ – Elliot Glazer Jan 11 '16 at 12:46
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Of course.

If you start with a model where there is a definable well-ordering, say $V=L$, and you add a single Cohen real $r$ you have that:

  1. $V=L[r]$, so there is a definable well-ordering with a parameter $r$ (e.g. given two sets in $L[r]$ ask which one has a name appearing first in the order of the ground model, here $L$, that when interpreted with $r$ as the generic give you the two sets).
  2. Since the Cohen forcing is homogeneous, $L=\mathsf{HOD}^L=\mathsf{HOD}^{L[r]}\neq L[r]$.

So while there is a well-ordering definable from a parameter, which in this case is a real number (read: a subset of $\omega$), there is no such well-ordering which is definable without parameters or from ordinals.

(This argument shows that any set forcing over a model with global well ordering from parameters will also have a global well ordering definable from a parameter. You can violate that with a class forcing, though.)

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I just noticed this question, which I find quite interesting.

I thought I'd mention the following related result, which some readers may find interesting:

Theorem. The following are equivalent.

  1. The universe is HOD of a set: $\exists b\ (V=\text{HOD}(b))$.
  2. The axiom V=HOD is forceable.
  3. Somewhere in the generic multiverse, the universe is HOD of a set.
  4. Somewhere in the generic multiverse, the axiom V=HOD holds.

The proof is contained in my blog post, Being HOD-of-a-set is invariant throughout the generic multiverse.

In particular, it follows that the axiom V=HOD is a switch, in models for which $V=\text{HOD}(b)$, since it can be forced on and then off again as much as you like. If $V=\text{HOD}$ holds, then you can do the forcing in Asaf's answer, adding a Cohen real, and $V\neq\text{HOD}$ in the extension $V[c]$, but then you can force $V=\text{HOD}$ again in a further forcing extension. And furthermore, whenever $V=\text{HOD}(b)$, then you can force $V=\text{HOD}$, and the assertion $\exists b\ V=\text{HOD}(b)$ is invariant by forcing.

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  • $\begingroup$ Actually, I see that I must have noticed it earlier, since I have evidently upvoted both the question and Asaf's answer already. $\endgroup$ – JDH Aug 8 '16 at 1:37
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    $\begingroup$ I agree that, although now standard, the notation is not great. What is your preferred notation? $\endgroup$ – JDH Aug 8 '16 at 11:17
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    $\begingroup$ The thing is that HOD and L notations should agree, and they don't. I don't mind which way is what, just that they agree. $\endgroup$ – Asaf Karagila Aug 8 '16 at 11:21
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    $\begingroup$ This may be obvious, but what exactly is the relationship of $\exists b\,V=\mathrm{HOD}(b)$ to the existence of a well-ordering of the universe definable with a parameter? Are these equivalent? $\endgroup$ – Emil Jeřábek Jul 12 '17 at 12:51
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    $\begingroup$ Yes, those are equivalent, and with the same parameter $b$. Some authors write $\text{HOD}_{\{b\}}$, to indicate that we are taking the set $b$ as a parameter, not the set of elements of $b$. Given the well-order, any set is the $\alpha^{th}$ with respect to that order, and if $V=\text{HOD}(b)$, then there is a canonical order in terms of the definitions and ordinal parameters that are used. $\endgroup$ – JDH Jul 12 '17 at 13:01

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