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We work in $\mathsf{ZFC}$ + "Every set is an element of a transitive model of $\mathsf{ZFC}$."


Suppose $M$ is a transitive set model of $\mathsf{ZFC}$. Say that $M$ is locatable iff there is some transitive set $N$ such that:

  • $N\models\mathsf{ZFC}$,

  • $M\in N$, and

  • $M$ is parameter-freely-definable in $N$.

The requirement that $N\models\mathsf{ZF}$ prevents coding $M$ into $N$ in a "silly" way. To see how this can be an issue, let $\alpha=\min\{\gamma: L_\gamma\models\mathsf{ZFC}\}$. If $N$ is a transitive model of $\mathsf{ZFC}$ with $L_\alpha\in N$, then $N$ sees that $L_\alpha$ is countable and so contains lots of (say) reals which are Cohen-generic over $L_\alpha$. Suppose $r,s$ are such reals; how is $N$ to distinguish between $L_\alpha[r]$ and $L_\alpha[s]$ in a parameter-free way? This seems to be an obstacle in general to the locatability of $L_\alpha[c]$ for $c$ Cohen-generic over $L_\alpha$. However, such an $L_\alpha[c]$ is nonetheless locatable: force over $L_\beta$, where $\beta=\min\{\gamma: \gamma>\alpha, L_\gamma\models\mathsf{ZFC}\}$, to code $c$ into the continuum pattern appropriately.

Indeed, such coding tricks give:

Every countable $M$ is locatable.

Note that the locating model is very simple - it's an element of $L(M)$. For uncountable $M$s this seems like a nontrivial further condition, and in particular none of the absoluteness theorems I'm familiar with seem to directly apply.

Similarly, we have:

If $L_\theta\models\mathsf{ZFC}$ then $L_\theta$ is locatable, even if $\theta$ is uncountable.

(For example, every ordinal $\theta$ is parameter-freely-definable in the $(\vert\theta\vert^++\theta+1)$th level of $L$ satisfying $\mathsf{ZFC}$ since $(i)$ that level of $L$ can "count" the $\mathsf{ZFC}$-satisfying levels below it, and $(ii)$ we can recover $\theta$ from $\vert\theta\vert^++\theta$ as there is a unique way to write the latter in the form $\kappa+\gamma$ for $\kappa$ a cardinal and $\gamma<\kappa$.)

However, the general situation is not clear to me:

Question: is every transitive set model of $\mathsf{ZFC}$ locatable?

Of course it is quite plausible that this will depend on additional set-theoretic hypotheses; for example, I think an affirmative answer is more likely assuming $\mathsf{V=L}$ than assuming the existence of large cardinals. But I don't immediately see how to show any of this.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Xander Henderson
    Jan 31, 2021 at 13:42
  • $\begingroup$ the fact that $L$ is unchanged if you use the parameter-free-definable powerset at successor stages might be relevant, although I don't immediately see how. $\endgroup$ Jan 31, 2021 at 23:48
  • $\begingroup$ @JasonZeshengChen I don't see how that would be useful if $M$ is not a level of $L$ (or similar to a level of $L$). $\endgroup$ Feb 1, 2021 at 0:15
  • $\begingroup$ I was thinking that it might be useful towards a consistently positive answer by working in $L$. But that was a wild guess anyways. $\endgroup$ Feb 1, 2021 at 0:17
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    $\begingroup$ @JasonZeshengChen Ah, the problem is that even if $\mathsf{V=L}$ there may be models of $\mathsf{ZFC}$ which are highly non-$L$-like. E.g. even in $L$ there could be a transitive set model of $\mathsf{ZFC}$ + "There is a measurable cardinal." $\endgroup$ Feb 1, 2021 at 0:31

3 Answers 3

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Great question! It has a positive answer in general:

Let $M$ be a transitive set-sized model of $\mathrm{ZFC}$. We may now find some regular cardinal $\kappa$ and some $X\subseteq\kappa$ so that $M\in L_\kappa[X]$. From now on we will work in $L[X]$. Let $\beta$ be the ordinal so that $M$ is the $\beta$-th set according to $<_X=:$ the canonical wellorder on $L[X]$. If both $X$ and $\beta$ were parameter-free definable then we could define $M$ without parameters. Unfortunately, neither of this is necessarily the case so we will try to fix that. Note that $\mathrm{GCH}$ holds in $L[X]$ at and above $\kappa$ and by our assumption on $V$, there are arbitrarily large $\lambda$ so that $L_\lambda[X]\models\mathrm{ZFC}$.

Claim: There is some $\kappa<\lambda<\kappa^+$ with

  1. $X\in L_\lambda[X]\models\mathrm{ZFC}$
  2. $L_\lambda[X]^{<\kappa}\subseteq L_\lambda[X]$

Proof: Let $\theta>\kappa^+$ so that $L_\theta[X]\models\mathrm{ZFC}$. Construct a continuous increasing sequence $\langle X_\alpha\mid \alpha\leq\kappa\rangle$ of elementary submodels of $L_\theta[X]$ of size $\kappa$. Make sure thate $\kappa\cup\{X\}\subseteq X_0$ and that $X_\alpha^{<\kappa}\subseteq X_{\alpha+1}$ for any $\alpha<\kappa$. Then the collapse $L_\lambda[X]$ of $X_\kappa$ is as desired.$\square$

We aim to find a generic extension of $L_\lambda[X]$ (inside $L[X]$) so that $M$ is definable there w/o parameters.

In $L_\lambda[X]$, let $\mathbb P$ be the/an ${<}\kappa$-closed Easton-style forcing that codes $\{X, \beta\}$ (observe that $\beta<\kappa<\lambda$) into the $\mathrm{GCH}$-pattern on the successor cardinals in the interval $(\aleph_\kappa, \aleph_{\kappa+\kappa})^{L_\lambda[X]}$ and also makes $\mathrm{GCH}$ fail at $\aleph_{\kappa+\kappa+1}^{L_\lambda[X]}$. Now the following holds:

  • $\mathbb P$ is ${<}\kappa$-closed in $L_\lambda[X]$
  • $L_\lambda[X]$ has size $\kappa$
  • $L_\lambda[X]^{<\kappa}\subseteq L_\lambda[X]$

Thus we can construct a filter $G\subseteq\mathbb P$ (in $L[X]$!) that is generic over $L_\lambda[X]$.

Finally, $N=L_\lambda[X][G]\in L[X]\subseteq V$ is a transitive model of $\mathrm{ZFC}$ that locates $M$: First of all $\kappa$ is definable w/o parameters in $N$ since $$N\models \aleph_{\kappa+\kappa+1}\text{ is the largest cardinal at which }\mathrm{GCH}\text{ fails}$$ Using $\kappa$, $N$ can decode $X$ and $\beta$ from the $\mathrm{GCH}$-pattern on the successors in the interval $(\aleph_\kappa, \aleph_{\kappa+\kappa})^N$ and so can define $M$ as the $\beta$-th set according to $<_X$.

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  • $\begingroup$ how can one see that there are stationary many $\alpha<\kappa^+$ such that $L_\alpha[X]^{<\kappa}\subseteq L_\alpha[X]$? $\endgroup$
    – ikrto
    Feb 1, 2021 at 17:27
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    $\begingroup$ Look at the map $f:\kappa^+\rightarrow\kappa^+$ that sends $\alpha$ to the least $\gamma$ so that $L_\alpha[X]^{<\kappa}\subseteq L_\gamma[X]$. The set $C$ of closure points is a club and the set $S$ of ordinals $<\kappa^+$ of cofinality $\kappa$ is stationary. Now $C\cap S$ is stationary and any $\alpha\in C\cap S$ satisfies $L_\alpha[X]^{<\kappa}\subseteq L_\alpha[X]$. $\endgroup$ Feb 1, 2021 at 17:38
  • $\begingroup$ (I edited the argument to find $\lambda$ as the claim "there are club many $\alpha<\kappa^+$ so that $L_\alpha[X]\models\mathrm{ZFC}$" is not true) $\endgroup$ Feb 1, 2021 at 19:20
  • $\begingroup$ This is really neat! The genuine existence of generics over non-cardinal-indexed levels of $L$ is something I always forget about. $\endgroup$ Feb 2, 2021 at 0:07
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A partial positive answer belatedly occurs to me, motivated by Jason Zesheng Chen's comments above (although I'm not sure this is what they had in mind):

Suppose $\mathsf{V=L}$ and $M\models\mathsf{ZFC}$. Let $M$ be the $\theta$th set in the $L$-ordering, and let $\delta$ be the $(\vert\theta\vert^++\theta+1)$th index of a level of $L$ satisfying $\mathsf{ZFC}$. Then $M\in L_\delta$ and $\theta$ is parameter-freely definable in $L_\delta$ (by the argument that every level of $L$ is locatable); since the $L$-ordering is absolute to $L_\delta$, this gives the parameter-free-definability of $M$ in $L_\delta$.

So assuming $\mathsf{V=L}$, then every transitive set model of $\mathsf{ZFC}$ is locatable. (And I suspect the same trick will work for similar fine-structural models.)


Interestingly, at a glance this doesn't seem to generalize to $\mathsf{V=HOD}$. The $L$-ordering is playing two roles in the above argument, both getting sets from ordinals and building arbitrarily "big" models which recognize a given ordinal (the map $\theta\mapsto L_{\vert\theta\vert^++\theta+1}$), and while the former is handled by $\mathsf{HOD}$ the latter seems genuinely problematic to me. But I could be missing something.

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  • $\begingroup$ This is simpler than what I had in mind. I was thinking something like this: if $M\in L^-_{\beta+1}$, where $L^-_{\beta+1}$ is the $\beta+1$th level of the parameter-free $L$, then $M$ is a parameter-free definable subset of $L^-_\beta$. Now move to a ZFC-modelling level of $L$ where $\beta$ is parameter-free definable, define $L^-_{\beta}$, and define $M$ in it. $\endgroup$ Feb 1, 2021 at 1:19
  • $\begingroup$ Maybe a silly comment, but does your argument carry over $\mathsf{V=HOD}$ with a proper class of $\Sigma_2$-correct worldly cardinals? If $\kappa$ is a $\Sigma_2$-correct worldly cardinal, then $V_\kappa$ thinks $\mathsf{V=OD}$ and it also reflects the well-ordering over $\mathsf{OD}$. $\endgroup$
    – Hanul Jeon
    Feb 1, 2021 at 5:48
  • $\begingroup$ @HanulJeon I think that works, yes, by exactly the same argument. $\endgroup$ Feb 1, 2021 at 5:51
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Here's another case where you get a positive answer, one which is quite far from $V=L$. The idea is, with the right hypothesis the strategy Noah outlined for coding countable $M$ into the continuum pattern of some larger model works even if $M$ is uncountable. There's some annoying details to make it work, but thats the core of it.

Work over ZFC + the Inner Model Hypothesis. For the sake of the reader who is unfamiliar, let me say a bit about the IMH, introduced by Sy Friedman. (The reader who already knows can skip to the next paragraph.) The IMH is a sort of width-reflection principle, saying that if something happens in an outer model you can locate it in an inner model. Specifically, it states that if a parameter-free assertion $\varphi$ is true in an inner model of a width-extension of $V$, then it is true in an inner model of $V$. The IMH implies there are inner models with measurable cardinals of arbitrarily large Mitchell order, so in particular it implies $V=L$ badly fails. On the other hand, it implies there are no large cardinals in $V$. For consistency, you can get a model of the IMH from a Woodin cardinal with an inaccessible above.

You might worry about how you actually formulate the IMH, given that it involves quantifying over arbitrary outer models. But for the purposes of this answer, it's enough to just look at extensions from a (tame) class forcing. This is formalizable as a single assertion in second-order set theory. Call this principle the CIMH.

There is a class-forcing extension $V[H]$ of $V$ where we add the following: for each transitive model $M$ of ZFC and each $N \in V$ with $M \in N$, an $N$-generic $G_{N,M}$ for the forcing which codes $M$ into the continuum pattern of an extension of $N$, starting at, say, $\aleph_{13}^N$. Let me note that the point of doing all the coding at once is that to use the CIMH we need the $\varphi$ we are reflecting to not use parameters, and so we cannot just do the coding for one pair $(M,N)$.

By the CIMH, we have an inner model $W$ which thinks that for all appropriate pairs $(M,N)$ that $G_{N,M}$ exists. This is because that is a parameter-free assertion satisfied by $V[H]$. It's now immediate that if a pair $(M,N)$ as above is in $W$ then $M$ is locatable in $V$. This is because $W$ must have $N[G_{N,M}]$ as an element, and this model can define $M$ by looking at the continuum pattern. But of course then $N[G_{N,M}]$ must be in $V$, so $V$ thinks $M$ is locatable.

We are not done yet, however, since we have only seen that every $M \in W$ is locatable, not every $M \in V$. This takes a small bit more.

Suppose otherwise. Then we have a least-rank counterexample, and the rank $\alpha$ of the least-rank counterexample is definable without parameters. So consider the following forcing extension. First ensure that some appropriate $M$ of rank $\alpha$ is coded into the continuum pattern in $V[k]$ at a definable location, then force over $V[k]$ to get $V[k][H]$ like above, in such a way that we don't destroy the coding of $M$. We now reflect this property of $V[k][H]$ down to an inner model $W'$, by the CIMH. That is, some transitive model $M$ of ZFC of rank $\alpha$ must be in $W'$ and so $M$ is locatable in $V$, just like the $W$ argument. This is the desired contradiction.

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    $\begingroup$ Excellent, I love the inner model hypothesis(/es) and didn't think to try and use it here! $\endgroup$ Feb 1, 2021 at 3:40

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