5
$\begingroup$

As someone unfamiliar with set theory, I was surprised not to find any information online about the following question, which seems natural to me. I might be missing the right search keywords, and would appreciate being pointed to relevant literature.

Broadly speaking: What can be said about the transfinite sequence of first-order structures $(⟨V_α; ∈⟩)_{α∈\operatorname{Ord}}$ viewed up to elementary equivalence?

Some more specific questions:

  • I see that many large cardinal properties have to do with elementary embeddings between the $V_α$, but the relationship to elementary equivalence is not clear to me. Can we have $α < β$ such that $V_α$ does not elementarily embed into $V_β$ yet is elementarily equivalent to it?

  • Is there some $α$ such that $V_α$ is elementarily equivalent to the whole $V$? (Needs NBG to formulate.)

  • On this Wikipedia page, one finds that if $κ$ is (strongly) inaccessible then there are club-many $α < κ$ such that $V_α$ elementarily embeds into $V_κ$. Is there a weaker assumption than inaccessibility (e.g., what about weak inaccessibility?) which makes this true?

  • There are set-many ($2^{\aleph_0}$) first-order theories in the language of set theory, so some theory must appear class-many times across this sequence of theories. Do we have a somewhat concrete example of this? For example, is there some well-known large cardinal property such that all the $α$ with this property have elementarily equivalent $V_α$s?

$\endgroup$
1
  • 1
    $\begingroup$ FWIW, those $\kappa$ such that $V_\kappa \prec V$ are known as “correct” cardinals. But again, this is about being an elementary submodel, not [the weaker condition of] being elementarily equivalent. Still, this may be a keyword to search for. $\endgroup$ Commented Sep 10 at 11:04

1 Answer 1

Reset to default
7
$\begingroup$

Here is the basic rundown.

  1. As you correctly identify, there are only set-many theories (indeed, continuum at most) for the various $V_\alpha$s, and therefore of $\{V_\alpha\mid\alpha<(2^{\aleph_0})^+\}$ there will already be a large set of those with the same theory, and therefore elementary equivalent. This is an easy proof in $\sf ZFC$, and in fact you don't really need choice if we replace $(2^{\aleph_0})^+$ by the Hartogs number of $\Bbb R$.

  2. If $V_\alpha\prec V_\beta$, then we can show that both are models of $\sf ZFC$, and so this is already quite beyond the reach of $\sf ZFC$ itself, as it gives us not just models of $\sf ZFC$, and not just transitive models of $\sf ZFC$, but in fact worldly cardinals. Indeed, if $\alpha$ is the least worldly cardinal, then there is no $V_\beta$ which is elementarily equivalent to it. So in this situation we actually have quite a lot of worldly cardinals in the universe to begin with.

    However, this is not as strong as it may seem. If $\kappa$ is any worldly cardinal whose cofinality is uncountable, then $\{\alpha<\kappa\mid V_\alpha\prec V_\kappa\}$ is a club in $\kappa$. So all of this happens well before the least worldly cardinal with an uncountable cofinality, and certainly well before the least inaccessible cardinal.

  3. If, however, the elementary embedding $V_\alpha\prec V_\beta$ is nontrivial, namely some $\kappa<\alpha$ is actually moved by the embedding, we again make a very significant jump in the strength, as this is now starting to get into the territory of very strong compactness axioms. Specifically, if $\kappa$ is moved by the embedding, $j\colon V_\alpha\to V_\beta$, then $j(V_\kappa)=V_{j(\kappa)}$, since $\kappa<\alpha$ and $j(\kappa)<\beta$, this generally means that $j$ itself is inside $V_\beta$, and so $V_\beta$ thinks that $\kappa$ is a superstrong cardinal.

So, to your question, yes, we can certainly have two ordinals $\alpha$ and $\beta$, indeed well below $\beth_2$, such that $V_\alpha\equiv V_\beta$, but there is no elementary embedding between them (not even the identity). And assuming some large cardinals, we can certainly have a situation where $V_\alpha\prec V_\beta$, but the only embedding is the identity.

$\endgroup$
3
  • $\begingroup$ In 2, why does $V_\alpha\prec V_\beta$ imply both are models of $\mathsf{ZFC}$? As long as $\kappa$ has uncountable cofinality (doesn't have to be worldly), $\{\alpha<\kappa\mid V_\alpha\prec V_\kappa\}$ will be a club. Also, it seems possible for $\alpha,\beta$ to be successor ordinals, although that probably has pretty high large cardinal strength. $\endgroup$ Commented Sep 20 at 23:15
  • 3
    $\begingroup$ That's actually a great exercise. Try and think about it. $\endgroup$ Commented Sep 21 at 8:05
  • 2
    $\begingroup$ See the other worldly cardinals. jdh.hamkins.org/otherwordly-cardinals $\endgroup$ Commented Sep 27 at 19:38

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.