7
$\begingroup$

Suppose $\chi$ is a regular cardinal, and $N$ is an elementary submodel of the structure $\langle H(\chi),\in\rangle$, where $H(\chi)$ is those sets hereditarily of cardinality less than $\chi$.

Can we characterize when the transitive collapse of $N$ is $H(\lambda)$ for some $\lambda<\chi$?

The motivation is idle curiosity: I just ran across an argument where such $N$ were used in a critical manner, and so I'm wondering if there's a relatively simple way to detect when a given $N$ has this property.

$\endgroup$

1 Answer 1

8
$\begingroup$

This hypothesis is equivalent to an instance of $\kappa$, the least point not in $N$, having some degree of extendibility. After all, if $j$ is the inverse of the collapse, then we have an elementary embedding $j:H(\lambda)\to H(\chi)$ with critical point $\kappa$.

A cardinal $\kappa$ is $\eta$-extendible if there is an elementary embedding $j:V_{\kappa+\eta}\to V_\theta$ for some ordinal $\theta$. Although extendibility is usually defined this way in terms of the von Neumann hierarchy, there is a parallel characterization in terms of the hereditary-size hierarchy, and the two hierarchies are interleaved. For example, $V_{\kappa+\eta}$ amounts essentially to $H(\beth_{\kappa+\eta})$, since the von Neumann hierarchy takes the power set each step. There is a good case to be made that the original definition of extendibility should perhaps have referred to the $H(\theta)$ hierarchy instead of the $V_\theta$ hierarchy, since $H(\theta)$ generally satisfies a better theory and has better closure properties. A similar issue affects the strong cardinals, which are also defined by reference to the $V_\theta$-hierarchy.

But meanwhile, if one is in a situation where the least cardinal $\kappa$ not in $N$ is known not to have extendibility properties (which are quite strong in the large cardinal hierarchy), then you know $N$ does not collapse to an $H(\lambda)$. Conversely, when $\kappa$ does have nontrivial extendibility, then you can expect to find such $N$, where $\kappa$ is the least ordinal missing from $N$, and the existence of these elementary submodels for every $\chi$ is equivalent to $\kappa$ being an extendible cardinal.

$\endgroup$
1
  • $\begingroup$ Thanks, Joel! Extendibility had never even crossed my mind as an underlying issue. $\endgroup$ Commented Jul 2, 2013 at 14:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .