Cantor's Attic defined tall cardinals as follows:

A cardinal $\kappa$ is $\theta$-tall iff there is an elementary embedding $j:V \to M$ into a transitive class $M$ with critical point $\kappa$ such that $j(\kappa) \gt \theta$ and $M^\kappa \subset M$. $\kappa$ is tall iff it is $\theta$-tall for every $\theta$; i.e. $j(\kappa)$ can be made arbitrarily large.

It also claimed that every strong cardinal is tall. Why is that? It doesn't follow immediately from the definition of strong cardinals, and as far as I know, extenders can't even ensure that $M^\omega \subset M$

Most of what I know about extenders comes from the paper Double helix in large large cardinals and iteration of elementary embeddings by Kentaro Sato, whose definition 3.2(3) (which relies on definition 3.1) seems to be roughly the same as what other set theories call extenders.

  • $\begingroup$ Who is $V$ here? $\endgroup$
    – Vincent
    Commented Jul 14, 2023 at 15:36
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    $\begingroup$ @Vincent I think $V$ is the Von Neumann universe. $\endgroup$
    – David Lui
    Commented Jul 14, 2023 at 16:15

1 Answer 1


Let $\kappa$ be strong and let $\gamma$ be a strong limit cardinal of cofinality $>\kappa$. Let $E$ be a short extender of strength $\gamma$ and support $\gamma$ and critical point $\kappa$. Using that $V_\gamma$ is closed under $\kappa$-sequences, show that $M=\mathrm{Ult}(V,E)$ is also.

  • $\begingroup$ But I have no idea how to prove that. This paper proves an analogous result as part of lemma 3.6, but for extenders the restriction that indices have to be finite, so that a union of infinitely many indices is not generally an index, ruins that proof idea. $\endgroup$ Commented Jul 14, 2023 at 14:48
  • $\begingroup$ Also, the result you're claiming would also answer another question that I've asked before. $\endgroup$ Commented Jul 14, 2023 at 14:51
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    $\begingroup$ Let $\vec{x}=\left<x_\alpha\right>_{\alpha<\kappa}\subseteq M=\mathrm{Ult}(V,E)$. Let $\left<f_\alpha,b_\alpha\right>_{\alpha<\kappa}$ be s.t. $b_\alpha\in[\gamma]^{<\omega}$ and $x_\alpha=[b_\alpha,f_\alpha]^V_E$, and so $x_\alpha=j(f_\alpha)(b_\alpha)$ where $j:V\to M$ is the ultrapower map. Let $\vec{f}=\left<f_\alpha\right>_{\alpha<\kappa}$. Note $\vec{g}=j(\vec{f})\upharpoonright\kappa=\left<j(f_\alpha)\right>_{\alpha<\kappa}\in M$. But also $\vec{b}=\left<b_\alpha\right>_{\alpha<\kappa}\in V_\gamma\subseteq M$. From $\vec{g},\vec{b}$, we get $\vec{x}$, as $x_\alpha=g_\alpha(b_\alpha)$. $\endgroup$
    – Farmer S
    Commented Sep 13, 2023 at 6:20
  • $\begingroup$ What do you mean by $[b_\alpha, f_\alpha]^V_E$ and why does it follow that $x_\alpha=j(f_\alpha)(b_\alpha)$? $\endgroup$ Commented Oct 11, 2023 at 22:41
  • $\begingroup$ That is the object represented by the pair $(b_\alpha,f_\alpha)$ in the ultrapower of $V$ by $E$. Here we may assume that $f_\alpha:[\kappa]^{|b_\alpha|}\to V$. $\endgroup$
    – Farmer S
    Commented Oct 12, 2023 at 15:13

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