When comparing between V=L and AC, one of the things that gets my attention is that, if we switch to an external perspective and don't care about first-order expressibility, in models of V=L we have a fixed first-order formula $\varphi(u,v)$ such that, for each $\alpha$, the set $\{(x,y)\in V_\alpha:\varphi(x,y)\}$ is a well-ordering; on the other hand, in models of AC, for each level $\alpha$ we can construct a formula giving a sequence of well-ordering $\{<_\beta :\beta < \alpha\}$ in an extending manner using DC$_\alpha$. However, when trying to "pack up" this result in a single sentence and generalize it to the whole universe - i.e. turning $$\forall\alpha\exists\varphi_\alpha(\varphi_\alpha\text{ is a well order})$$ to $$\exists\varphi\forall\alpha(\varphi_\alpha\text{ is a well order})$$ I find it quite challenging. Is such a uniform well-ordering of the universe only probable in L?
- 1$\begingroup$ The existence of a definable well ordering is equivalent to $V=\rm HOD$, or if you allow a parameter to the well ordering, then add one to $\rm HOD$ as well. $\endgroup$Asaf Karagila– Asaf Karagila ♦2024-06-26 21:02:52 +00:00Commented Jun 26, 2024 at 21:02
- $\begingroup$ @AsafKaragila Muahahaha beat you to it! :P $\endgroup$Noah Schweber– Noah Schweber2024-06-26 21:06:22 +00:00Commented Jun 26, 2024 at 21:06
- 1$\begingroup$ @Noah: That may very well be, but I got an English Cherry end grain cutting block. So, I win. $\endgroup$Asaf Karagila– Asaf Karagila ♦2024-06-26 21:54:40 +00:00Commented Jun 26, 2024 at 21:54
- $\begingroup$ @AsafKaragila Yes that does sound knife, definitely a cut above the usual. And appropriately, English Cherry is a hodwood. $\endgroup$Noah Schweber– Noah Schweber2024-06-26 22:15:23 +00:00Commented Jun 26, 2024 at 22:15
1 Answer
A note on fonts: below I use mathsf for sentences and mathit for objects. So e.g. $\mathcal{M}\models\mathsf{V=L}$ iff $\mathcal{M}=L^\mathcal{M}$.
Depending on whether you allow parameters in your fixed formula, you're either looking for "$\mathsf{V=HOD}$" or "$\exists xV=\mathsf{HOD}[x]$." For simplicity I'll focus on the parameter-free situation.
Here $\mathsf{HOD}$ is the class of hereditarily ordinal definable sets. (See also this old answer of mine.)
It's easy to show that $\mathsf{V=HOD}$ is equivalent to the existence of a parameter-freely-definable well-ordering of the universe. The subtle feature is that $\mathsf{V=HOD}$ is in fact expressible in the language of set theory! This is a beautiful trick, so below I've spoilered the key one-word hint:
Reflection.
As to the connection with the axiom of constructibility $\mathsf{V=L}$, this is in fact vastly weaker. A good starting point is to understand just how flexible the construction $\mathcal{M}\leadsto\mathit{HOD}^\mathcal{M}$ as compared with $\mathcal{M}\leadsto L^M$; for example, any (countable!) $\mathcal{M}\models\mathsf{ZFC}$ has a forcing extension $\mathcal{M}[G]$ such that $\mathit{HOD}^{\mathcal{M}[G]}=\mathcal{M}$ (this is due to Vopenka if I recall correctly). Of course this doesn't help us build models of $\mathsf{V=HOD}$, but it's a good indicator of how weak that principle is likely to be (e.g. it has no bearing on "arithmetic" issues like $\mathsf{GCH}$, or even - to the best of my knowledge - large cardinal structure as far as we understand currently).
- $\begingroup$ Ah, thank you, that's exactly what I'm looking for! I can't believe that somehow I managed to leave it in a corner of my memory and never bothered looking at that spot :) $\endgroup$Raczel Chowinski– Raczel Chowinski2024-06-26 21:07:24 +00:00Commented Jun 26, 2024 at 21:07