# Why are strong cardinals tall?

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.

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

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.
• 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 $x_\alpha=g_\alpha(b_\alpha)$. Commented Sep 13, 2023 at 6:20
• What do you mean by $[b_\alpha, f_\alpha]^V_E$ and why does it follow that $x_\alpha=j(f_\alpha)(b_\alpha)$? Commented Oct 11, 2023 at 22:41
• That is the object represented by the pair $(b_\alpha,f_\alpha)$ in the ultrapower of $V$ by $E$. Here we may assume that $f_\alpha:[\kappa]^{|b_\alpha|}\to V$. Commented Oct 12, 2023 at 15:13