4
$\begingroup$

I am employing a large cardinal notion that has been used explicitly before, and I am wondering if someone has given it a good succinct name.

A cardinal $\kappa$ is huge if there is an elementary $j : V \to M$ with $M^{j(\kappa)} \subseteq M$. $\kappa$ is 2-huge if the same but $M^{j^2(\kappa)} \subseteq M$. There are provably intermediate notions where one just asserts for some particular cardinal $\lambda \in (j(\kappa), j^2(\kappa))$, $M^\lambda \subseteq M$. This has a characterization in terms of the existence of a normal ultrafilter concentrating $\{ x \subseteq \lambda \mathrel: \text{$\lvert x\rvert < j(\kappa)$ and $\operatorname{ot}(x \cap j(\kappa)) = \kappa$} \}$.

Did anyone name these cardinals?

Improve this question
$\endgroup$
12
  • 2
    $\begingroup$ Actually I like it. $\lambda$-sesquihuge. Your comment was good, phantom commenter. $\endgroup$
    – Monroe Eskew
    Commented Apr 23, 2015 at 10:19
  • 4
    $\begingroup$ $\lambda$-sesquihuge is interesting, but inevitably someone will want to look at similar things between $n$- and $(n+1)$-huge cardinals and I don't think our Latin forefathers gave us enough prefixes for all of these. $\endgroup$
    – Miha Habič
    Commented Apr 23, 2015 at 13:06
  • 2
    $\begingroup$ I like the fanciful names. But meanwhile, there is a notion "tall with closure $\lambda$", and in analogy with that, you could say that $\kappa$ is 1-huge with closure $\lambda$. For example, $n$-huge with closure $\lambda$ would mean $j:V\to M$ is $n$-huge and also $M^\lambda\subset M$. So the interesting case arises when $j^n(\kappa)<\lambda$. $\endgroup$
    – Joel David Hamkins
    Commented Apr 23, 2015 at 13:15
  • 2
    $\begingroup$ In Magidor and Shelah's paper, "The tree property at successors of singular cardinals," a cardinal $\kappa$ is called $\tau$-huge if there is an elementary embedding $j:V \rightarrow M$ with critical point $\kappa$ such that $\kappa < \tau < j(\kappa) < j(\tau)$ and $M^{j(\tau)} \subseteq M$. $\endgroup$
    – Chris Lambie-Hanson
    Commented Apr 24, 2015 at 9:05
  • 1
    $\begingroup$ @ChrisLambie-Hanson, this answers the question but it looks like there is also some inconsistency in terminology. Woodin says "$\kappa_0$ is $\kappa_1$-huge'' to mean there is a huge embedding $j$ with critical point $\kappa_0$ and $j(\kappa_0) = \kappa_1$. I think Joel's terminology sounds the most clear and robust: "crit = x, target = y, closure = z..." $\endgroup$
    – Monroe Eskew
    Commented Apr 24, 2015 at 9:09

1 Answer 1

Reset to default
1
$\begingroup$

If there is a $\mu$ such that $j(\mu)=\lambda$, your property is called 2-fold $\mu$-supercompact (Sato 2007 - Double helix in large large cardinals and iteration of elementary embeddings) or $\mu$-hyperhuge (Usuba 2017 - The downward directed grounds hypothesis and very large cardinals).

Improve this answer
$\endgroup$

You must log in to answer this question.

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