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As noted in this answer by Eric Wofsey, as well as the Wikipedia page on the Constructible Universe, we have that $L_\alpha$ is strictly smaller than $V_\alpha$ for any $\alpha > \omega$ unless $\alpha = \omega_\alpha$. This is apparently a general statement that is true in ZFC even if $V = L$.

I am somewhat curious how this works in the minimal standard model of ZFC, where the axiom of constructibility holds. Apparently, even within the model, we are supposed to be able to prove that $L^M_{\omega+1} \subsetneq V^M_{\omega+1}$, where $L^M$ and $V^M$ refer to the model's idea of what $L$ and $V$ are. That is, although the model (and even the metatheory) agree on what $L^M_{\omega}$ and $V^M_{\omega}$ are, the model will somehow think that the definable $M$-powerset of $V^M_{\omega}$ (and, I guess, of $\Bbb N$) is strictly smaller than its true $M$-powerset.

Thus, it would seem that the minimal standard model includes at least some undefinable sets, or at least, enough to make $\mathbf{Def}^M(\Bbb N)$ strictly smaller than $\mathbf{P}^M(\Bbb N)$ within the model.

I, on the other hand, had always envisioned the minimal standard model as basically this thing which pretends that the true powerset is the definable powerset. My reasoning: the first-order axiom schema of comprehension only requires definable subsets to exist, so the definable powerset would seem to be the "thinnest possible powerset" that satisfies these requirements. I figured we'd just iterate that in $V$-like fashion, pretending that this minimalist powerset is the true powerset, and keep going until we get, at some interesting countable ordinal, a model of ZFC. But really what I was envisioning is something like the statement $L^M_\alpha = V^M_\alpha$ for all $\alpha$ in the model -- much stronger than $V=L$ -- but above we have that $L^M_{\omega+1} \subsetneq V^M_{\omega+1}$.

Questions:

  1. Am I understanding all this correctly?
  2. If this is really right, what are the elements that are in, for instance, the model's idea of $\mathbf{P}(\Bbb N)$ which are "missing" from $\mathbf{Def}(\Bbb N)$?
  3. In general, what does the minimal standard model think that powersets are? That is, if the model thinks the "true powerset" of $V^M_\omega$ is strictly larger than the "definable powerset" of it, then the model's "true powerset" equals the set of definable subsets of $V^M_\omega$, plus _____?
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    With regards to your first paragraph, where does it say that in either source? If V=L, then there are a proper class of $\kappa$ at which $V_\kappa=L_\kappa$ (namely, any $\beth$ fixed point). – spaceisdarkgreen Apr 23 '23 at 01:12
  • See https://math.stackexchange.com/a/2546720/28111. – Noah Schweber Apr 23 '23 at 02:23
  • Thanks, forgot to add that clarification; I was mostly focused on successor ordinals (e.g. which are power sets of something). I've added that although it doesn't seem to affect the meat of the question being asked. – Mike Battaglia Apr 23 '23 at 03:02
  • @MikeBattaglia Yeah, I gathered that was tangential to the question. (And I wasn't the downvoter here... apologies if my brusque correction somehow contributed to that.) – spaceisdarkgreen Apr 23 '23 at 03:08
  • @NoahSchweber Having forgotten whatever paltry amount I may have once learned about this subject, it seems like an easy corollary of what I wrote below is that there are no new constructible reals added between $\omega_1^M$ and $Ord(M)$, where $M$ is the minimal model. Do I have that right? – spaceisdarkgreen Apr 23 '23 at 03:13
  • @spaceisdarkgreen Yup, that's condensation (relativized to $M$) + the absoluteness of the $L$-hierarchy. Building $L$ within $M$ looks just like building $L$ within $V$ up to $Ord(M)$ (and this is true for all transitive models $M$, not just the minimal one). – Noah Schweber Apr 23 '23 at 03:50

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I think you need to understand why $V_{\omega+1}\ne L_{\omega+1}$ in $L$ before thinking about the minimal model, because the situation in the minimal model is not really any different.

The basic results are that for any ordinal $\alpha,$ $L_\alpha^L = L_\alpha$ and $V_\alpha^L = V_\alpha\cap L$ and for any $x\in L,$ $P^L(x) = P(x)\cap L.$ So, $V_{\omega+1}^L$ is all the constructible subsets of $V_\omega$ whereas $L_{\omega+1}^L$ are all of the subsets of $V_{\omega}$ that are definable with parameters in $V_\omega.$ These can't be the same, just on cardinality grounds: the former manifestly has cardinality continuum in $L$ whereas the latter is countable since the number of definitions with parameters are countable.

What's "missing" from $L_{\omega+1}$ is all the subsets of $V_\omega$ that will be added to the constructible hierarchy at later stages, e.g. there may be some subset of $V_\omega$ that is definable with parameters in $L_{\omega+5}$ that wasn't definable with parameters in $L_{\omega+n}$ for $n=0$ through $4.$

In the minimal model the idea is the same. The minimal model is $L_\xi$ for some countable ordinal $\xi,$ and for $\alpha<\xi,$ $L_\alpha^M=L_\alpha$ and $V_\alpha^M = V_\alpha\cap M$ and for any $x\in M,$ $P^M(x)= P(x)\cap M.$ In the minimal model, $L_{\omega+1}$ is exactly the same thing it was in $L$ (or in $V$ for that matter). $V_{\omega+1}^M$ isn't all constructible subsets of $V_\omega,$ but rather it's all subsets of $V_\omega$ with constructibility rank less than $\xi.$

So it's the same thing... what's "missing" from $L_{\omega+1}^M$ is all the subsets of $V_\omega$ that are constructible at some higher rank. There are fewer higher ranks that can contribute now since we cut off at $\xi$ (actually, per Godel's results, only ranks up to $\omega_1^M<\xi$ contribute) but that doesn't really make a difference for the question.

  • Aha! This is way subtler than I thought. Ok, I get it now. What do you mean by "actually, per Godel's results, only ranks up to $ω^M_1 < ξ$ contribute"? – Mike Battaglia Apr 23 '23 at 03:42
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    @MikeBattaglia The proof that CH holds in $L$ is done by showing every constructible real (or equivalently constructible subset of $V_\omega$) appears in $L_{\omega_1^L}$, i.e. there are no new reals added after $\omega_1^L.$ Then since $|L_\alpha|=|\alpha|,$ we have $|P(\omega)|\le \aleph_1$ in $L$. My remark is just this relativized to $M.$ (By "contribute" I mean there are new subsets added at that rank... though I should have said "can contribute" since some. but not all of them do.) – spaceisdarkgreen Apr 23 '23 at 03:49
  • Thanks. Regarding the notation $L_{\omega_1^L}$: do you mean that $\omega_1^L$ is different in $L$ from $\omega_1$ in $V$, in general? Because I thought that $L$, as an inner model of $V$, had the same ordinals as $V$. – Mike Battaglia Apr 23 '23 at 03:56
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    @MikeBattaglia Same ordinals, possibly different cardinals. The thing $L$ thinks is $\omega_1$ might actually (= in $V$) be countable. – Noah Schweber Apr 23 '23 at 04:01