The questions:

  1. Are there any cardinals $\kappa$ such that $\mathrm{V}\setminus\{\kappa\}\prec\mathrm{V}$? If so, I think a neat name for them would be "ghost cardinals", because their existance leaves no impact on $\mathrm{V}$ and it's properties.
  2. Given a theory $\mathrm{T}$, what is the minimum cardinality of $\mathcal{M}$ such that $\mathcal{M}$ is nonempty and $\mathcal{M}\models\mathrm{T}$? If so, what is it respective to Peano Arithmetic, Z, ZF, ZFC, KM, etc.? (SOLVED)
  3. Are there any cardinals $\kappa$ larger than the smallest correct cardinal such that $\forall\lambda<\kappa(\mathrm{V}_\lambda\prec\mathrm{V}\rightarrow\mathrm{V}_{\lambda}\prec\mathrm{V}_\kappa)$? (In other words, every rank of a correct cardinal is a substructure of the rank of $\kappa$, making $\mathrm{V}_\kappa$ very similar to $\mathrm{V}$, a good name for these is hypercorrect)
  4. Given a structure $\mathcal{M}$, are there any non-singleton chains of substructures of $\mathcal{M}$ ordered by $\prec$? What is the largest suprememum of these chains' order types when $\mathcal{M}=\mathrm{V}$? What about when $\mathcal{M}=L$?

Where I have gotten to so far on them:

  1. If they do exist, they are not "definable" (i.e. there is no formula that is true for them and only them.) Every $\aleph_\alpha$ and $\beth_\alpha$ for finite $\alpha$ is not a ghost cardinal, and GCH implies every ghost cardinal is a limit cardinal.
  2. For Peano Arithmetic, the answer is clearly $\aleph_0$. For all the others, there are no finite models of Z, ZF, ZFC, or KM, and since they are all $L(\omega,\omega)$-theories, there is a countable model (if any). Thus, for all of these, $\aleph_0$.
  3. Every one of these cardinals are correct. Since not much is known on correct cardinals, this one seems hard to work on.
  4. Yes, there is an $\mathcal{M}$ where this property is true. In fact, that $\mathcal{M}$ is $\mathrm{V}$. Clearly, the existance of a correct cardinal implies that this supremum is at least $3$ (as $\{\mathrm{V}_\kappa,\mathrm{V}\}$ is well-ordered by $\prec$). It is also true that the existence of a hypercorrect cardinal implies it is at least $4$ (as $\{\mathrm{V}_\lambda,\mathrm{V}_\kappa,\mathrm{V}\}$ is well-ordered by $\prec$)

1 Answer 1

  • 1: No - since e.g. the ordinal $\kappa+1$ (remember, all cardinals are ordinals) will be a successor in $V$ but not in $V\setminus\{\kappa\}$. EDIT: a better example, given any $a\in V$, is to think about $\{a\}$. In $V$, $\{a\}$ is nonempty, but $V\setminus\{a\}$ thinks $\{a\}$ is empty; in particular, $V\setminus \{a\}$ won't satisfy the axiom of extensionality, regardless of what $a$ is. I think you're asking the wrong question here - a better question would be e.g. "Can we have an $M\prec V$ with $\kappa\not\in M$?"

    • Another question of this form, arguably more important, is : "Can we have an elementary embedding $j:V\rightarrow M$, for some inner model $M$, with $\kappa\not\in ran(j)$?" This is connected with measurable (and larger) cardinals.
  • 2: Lowenheim-Skolem.

  • For 3, the answer will depend on large cardinal assumptions - e.g. if $V$ happens to be a rank-minimal model of ZFC, then there are no $\lambda\in Ord$ with $V_\lambda\prec V$.

  • For 4, you're trying way too hard: think about (say) an infinite pure set. Or the rationals, as a linear order. As to chains of elementary substructure in $V$, $L$, or similar, my answer re: 3 applies.

For what it's worth, I think it would be a good idea to get a solid grasp of basic model theory before diving in to the model theory of set theory specifically; in particular, I think that comfort with the basic tools like compactness, Lowenheim-Skolem, and omitting types is a necessity here (to be clear, I don't think omitting types itself is specifically necessary, but comfort with its proof and use is a turning point in understanding model theory). I think it's also a good idea in the future to not ask several questions at once.

  • 1
    $\begingroup$ @Zetapology "they can be "switched" with other cardinals (via automorphism)" No, they can't - there's no nontrivial automorphism of $V$ at all. (Precisely: a well-founded model of ZF has no nontrivial automorphisms.) And regardless, there's a huge difference between "can be moved by an automorphism" and "can be omitted, while keeping every other element, without changing the theory of the structure." I think you need to think carefully about exactly what question you're trying to ask, and what the various terms you're saying mean precisely. $\endgroup$ Sep 12, 2017 at 1:44
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    $\begingroup$ @Zetapology No, it doesn't. Read the definition again. It gives a nontrivial elementary embedding from $V$ into itself; that's not the same thing at all. $\endgroup$ Sep 12, 2017 at 1:58
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    $\begingroup$ @Zetapology No. An automorphism is bijective, while an elementary embedding need not be. Let me reiterate my previous point: before diving into this stuff, you should really familiarize yourself with the basic model theory involved. $\endgroup$ Sep 12, 2017 at 1:59
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    $\begingroup$ @Zetapology I'm aware. That statement is false. I'll say it again: an automorphism is, by definition, bijective. Elementary embeddings need not be. An elementary embedding from $V$ to itself need not be an automorphism, and in fact (so long as $V$ is well-founded and the embedding is not just the identity) it never will be. $\endgroup$ Sep 12, 2017 at 2:01
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    $\begingroup$ @Zetapology Isomorphisms are bijective. Elementary embeddings need not be. I don't know how many times I have to say this. $\endgroup$ Sep 12, 2017 at 2:04

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