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Let $X$ be a collection of cardinals such that if $\kappa,\lambda\in X$ and $\kappa<\lambda$, then there is a non-trivial elementary embedding $j:V_{\kappa+1} \to V_{\lambda+1}$ with $crit(j) = \kappa$.

What is the consistency strength of the claim that there are such $X$ for various sizes of $X$? In particular, what is the consistency strength when $X$ is a proper class of cardinals?

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First, let me point out that having a chain of $3$ such cardinals, with elementary embeddings $$V_{\kappa+1}\to V_{\lambda+1}\to V_{\eta+1},$$ already implies that $\kappa$ is $1$-inaccessible with target $\lambda$ in $V_{\eta+1}$, and so by elementarity $\kappa$ will be $1$-extendible with arbitrarily large targets below $\lambda$, and so $V_\kappa$ must also think that there are unboundedly many $1$-extendible cardinals. So even chains of length three exceed a proper class of $1$-extendible cardinals in consistency strength.

Meanwhile, a $2$-extendible cardinal is sufficient to produce very large chains:

Theorem. If $\kappa$ is $2$-extendible, then $V_\kappa$ has such a proper class chain of $1$-extendibles as you desire.

Proof. Suppose that $\kappa$ is $2$ extendible with target $\lambda$, so that there is an elementary embedding $j:V_{\kappa+2}\to V_{\lambda+2}$ with critical point $\kappa$, for which necessarily $j(\kappa)=\lambda$. It follows that $V_{\lambda+2}$ can see the embedding $j\upharpoonright V_{\kappa+1}:V_{\kappa+1}\to V_{\lambda+1}$ and therefore $\kappa$ is $1$-extendible in $V_{\lambda+2}$. It follows that there is a normal measure one set of $1$-extendible cardinals below $\kappa$. But furthermore, we can make them cohere in the way you desire in your question. Namely, since $V_{\lambda+2}$ thinks that $\lambda$ is the target of the $1$-extendibility embedding $j\upharpoonright V_{\kappa+1}$, and $\lambda=j(\kappa)$, it follows by elementarity that $V_{\kappa+2}$ thinks that $\kappa$ is the target of a $1$-extendibility embedding $j_0:V_{\delta+1}\to V_{\kappa+1}$. Thus, we've achieved a chain of length $2$. By Zorn's lemma, let us extend this to a maximal set $X\subset\kappa$ (maximal under inclusion) of $1$-extendible-with-target-$\kappa$ cardinals $\delta\lt\kappa$, which is also a chain in your sense. I claim that $X$ must have size $\kappa$. If not, then $X$ has size less than $\kappa$. Notice that $V_{\kappa+2}$ is capable of verifying the property on $X$ we have described, and so by elementarity, $V_{\lambda+2}$ will think that $j(X)=X$ is a maximal chain of $1$-extendible-with-target-$\lambda$ cardinals below $j(\kappa)=\lambda$, which is a chain in your sense. But the point is that $X\cup\{\kappa\}$ is a strictly larger set of $1$-extendible-with-target-$\lambda$ cardinals, which is a chain. This violates the maximality of $X$. Thus, $X$ must be unbounded, and so we get a proper class chain in $V_\kappa$ of such cardinals. QED

One can do a similar argument with an alternative weaker hypothesis.

Theorem. If $\kappa$ is $2^\kappa$-supercompact, then $V_\kappa$ has a $\kappa$-chain of $1$-extendible cardinals.

Proof. Let $j:V\to M$ be a $2^\kappa$-supercompactness embedding. Thus, $j\upharpoonright V_{\kappa+1}:V_{\kappa+1}\to M_{j(\kappa)+1}$ is a $1$-extendibility embedding inside $M$. So $M$ thinks $\kappa$ is $1$-extendible with target $j(\kappa)$. Let $X\subset\kappa$ be a maximal chain of $1$-extendible-with-target-$\kappa$ cardinals, as in the previous argument. If $X$ is bounded below $\kappa$, then $j(X)=X$, which would contradict maximality, since we may add $\kappa$ to have a strictly larger chain. So $X$ is unbounded, and we've found the desired chain. QED

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  • $\begingroup$ It's an interesting question. I suspect that one can formulate similar questions in terms not just of linear chains, as you have, but in terms of the rank of the partial order underlying this, where $\kappa$ is below $\lambda$ if $\kappa$ is $1$-extendible with target $\lambda$. Asserting nontrivial rank for this order drives up the consistency strength. $\endgroup$ – Joel David Hamkins Nov 27 '13 at 1:45
  • $\begingroup$ That sounds interesting. What would the assertion that the partial order has nontrivial rank look like? $\endgroup$ – Sam Roberts Nov 27 '13 at 1:58
  • $\begingroup$ The partial order is well-founded, and so it has a natural rank function. In the finite rank cases, the rank is the length of the largest linear chain in your sense. But suppose we have a model with arbitrarily large finite chains, but no infinite chain? The rank of the partial order in this case would be $\omega$, but you don't get a chain of length $\omega$. Similarly, one can formulate the hypthoses that the rank is any particular ordinal $\alpha$, or that there are arbitrarly large set-size chains (but perhaps no proper class chain). $\endgroup$ – Joel David Hamkins Nov 27 '13 at 2:07
  • $\begingroup$ Ah, I see what you mean now. I'd be very interested to see how the consistency strength is sensitive to assumptions on the rank in this way! In any case, thanks again! $\endgroup$ – Sam Roberts Nov 27 '13 at 2:17
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I think for the Theorem one can get by with a subcompact cardinal, which is weaker than a $\kappa^+$-supercompact.

( $\kappa$ is subcompact if for any $A\subseteq H_{\kappa^+}$ there are $j,\mu < \kappa, B\subseteq H_{\mu^+}$ with $j:(H_{\mu^+},\in, B) \longrightarrow_e (H_{\kappa^+},\in, A)$. ( where $ \longrightarrow_e$ means ``is an elementary map''. ) )

Let $\kappa$ be subcompact as above, and let $A = \mathop{Sat} = {Sat}_{\kappa^+}$ Then $\mathop{Sat} \subseteq H_{\kappa^+}$, and applying the definition of subcompactness there are $\mu < \kappa$, $j$, and $\overline{\mathop{Sat}}$ with

$j : (H_{\mu^+}, \overline{\mathop{Sat}}) \longrightarrow_e (H_{\kappa^+}, \mathop{Sat})$.

Then $\overline{\mathop{Sat}}= {Sat}_{\mu^+}$ can be checked easily. $j$ is easily in $H_{\kappa^+}$ but the point of having the Sat predicate is that we can express in a first order way (in the extended language) that there is a $j$ with $ran(j)$ an elementary substructure of $(H_{\kappa^+},\in)$.

Thus $(H_{\kappa^+}, {Sat}) \models$``${ran} (j) \prec_e (V, \in)$'' indeed:

$(H_{\kappa^+}, {Sat}) \models$``There is $k, \kappa_0,$ with $k : (H_{\kappa_0^+}, \in) \longrightarrow (V, \in)$ and ${ran} (k) \prec_e (V, \in) ,$ and thus $k$ is an elementary map.''

This goes down to $(H_{\mu^+}, {Sat}_{\mu^+})$. This starts out links in the chain, using the restriction of the map(s) $k$ to the domain $V_{\kappa_0 +1}$ etc. and the rest of the argument is as before.

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