The constructible universe $L$ is too thin for large cardinals greater than measurable. To build $L$-like inner models for large cardinal, it is natural to think about "adding" the evidences into the model. For example, $L[U]$ is an inner model for measurable cardinal. However, the situation becomes much more complicated beyond measurable. One reason is that $L[U]$ can contain only one measurable cardinal not two, which I knew. And I was told somewhere that

if $\kappa$ is a some strong cardinal witnessed by an/some elementary embedding(s) $j$, and $E$ is an extender generated from $j$, Then $L[E]=L[U]$ where $U=E_{\{\kappa\}}$.

Is the statement true? If true, how to prove it? If not, how to argue that it is necessary to develop much more complicated technique to build inner model for larger cardinals?


2 Answers 2


One point to make is that no one embedding witnesses that a cardinal is strong (that is unless it is witnessing that $\kappa$ is something much stronger, like a supercompact). The definition requires that for every ordinal $\alpha$ there is an embedding $j_\alpha$ with critical point $\kappa$ witnessing that $V_\alpha$ is the $V_\alpha$ of the target model.

I am trying to guess at what you may have seen: it is certainly true that if you take the "$\kappa+1$"-extender $E$ derived from any such embedding $j_\alpha$ for $\alpha > \kappa +1$, then this is essentially just a measure on ${\cal P}(\kappa)$, and then, yes $L[E] =L[U]$ by Kunen's analysis.

To get an inner model for, e.g., a strong cardinal, one will need some methodology that allows you to build a class-sized predicate: one will need to have encoded somehow, for a proper class of $\alpha$, those $j_\alpha$ on to the predicate. It is this methodology that makes the matter complicated; to get a fine structural inner model, such as $L[U]$ the construction is even more delicate.

The place to read about this is in Martin Zeman's book: "Inner Models and Large Cardinals"

  • $\begingroup$ Thank you, Philip. You have mentioned Kunen's analysis, which is exactly what I want to know. Do you know where can I find the argument? Is it also found in Zeman's book? $\endgroup$ Commented Nov 17, 2011 at 20:39
  • $\begingroup$ I believe Philip is referring to Kunen's paper "Some applications of iterated ultrapowers in set theory," Ann. Math. Logic 1 (1970) 179-227. Kunen shows there (among many other things) that the universe constructed from a measure on $\kappa$ depends only on $\kappa$, not on the particular measure. $\endgroup$ Commented Nov 18, 2011 at 19:44
  • $\begingroup$ Exactly, Andreas, that is the one - thank you for filling that in. It is also given as Thm. 76 in Jech's "Set Theory" (1st Ed.) $\endgroup$ Commented Nov 21, 2011 at 11:57

A fairly complete answer to this question appears in Woodin's "In search of Ultimate $L$" at the beginning of the section on Martin-Steel extender sequences. Woodin omits the proofs, so I'll fill in some of the details.

If $E$ is a short extender, then $L[E] = L[U]$ where $U$ is the normal measure of $E$. To see this, let $j : V\to M$ be the ultrapower of $V$ by $E$. Kunen's analysis shows that $j\restriction V_{\kappa+1}\cap L[U]$ coincides with the unique iterated ultrapower embedding $i : L[U]\to L[j_E(U)]$. (Every element of $V_{\kappa+1}\cap L[U]$ is $\Sigma_2$-definable in $L[U]$ from parameters in $\kappa\cup \Gamma$ where $\Gamma$ is the class of common fixed points of $i$ and $j$; since $i$ and $j$ are elementary and agree on these parameters, they agree on $V_{\kappa+1}\cap L[U]$.) Since $i$ is definable over $L[U]$, $E\cap L[U] = \{(a,X)\in L[U] : X\subseteq [\kappa]^{<\omega}, a\in i(X)\}$ is in $L[U]$. Since $L[E]$ is the minimum proper class model $N$ of ZFC such that $E\cap N \in N$, $L[E]\subseteq L[U]$. The reverse inclusion is obvious.

If $E$ is a long extender, then $L[E]$ can be larger than $L[U]$. For example, if $U$ and $W$ are measures on distinct measurable cardinals and $E$ is the extender of $j_W\circ j_U$, then $L[E] = L[U,W]$ has two measurable cardinals. Woodin's Theorem 4.20 asserts that if $E$ is a nice enough long extender (that is not "too long"), then $L[E]$ will not contain an inner model with a Woodin cardinal. This is probably pretty hard. Woodin also claims that dropping the niceness condition, consistently, one can code arbitrary sets into long extenders, so these do not have an inner model theory. There are still some open questions here, but these results together argue that inner model theory needs more than one extender.

  • 2
    $\begingroup$ Is there a proof of Theorem 4.20 available somewhere? $\endgroup$ Commented Jul 17, 2021 at 21:09
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    $\begingroup$ No, I think it's just one of those never-to-be-published theorems... $\endgroup$ Commented Jul 18, 2021 at 22:48

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