# Difference between revisions of "Talk:Huge"

Eaglgenes101 (Talk | contribs) (Amateur analysis...) |
Zetapology (Talk | contribs) m (added rebuttal to amateur analysis) |
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So I thought about it a bit. The first fixed point of the elementary embedding $j$ above $crit(j)$ is the supremum of all of $crit(j)$, $j(crit(j))$, $j^2(crit(j))$, $j^3(crit(j))$, etc..., and said iterates of $j$ form a cofinal set of $j^\omega(crit(j))$. So for any $\alpha < j^\omega(crit(j))$, there is some whole number $n$ such that $j^n(crit(j)) > \alpha$. If I reasoned correctly, this means that $\omega$-almost hugeness is equivalent to $n$-hugeness for all whole number $n$, which is known to be perfectly consistent at least relative to an $I3$ cardinal. [[User:Eaglgenes101|Eaglgenes101]] ([[User talk:Eaglgenes101|talk]]) 08:26, 29 April 2019 (PDT) | So I thought about it a bit. The first fixed point of the elementary embedding $j$ above $crit(j)$ is the supremum of all of $crit(j)$, $j(crit(j))$, $j^2(crit(j))$, $j^3(crit(j))$, etc..., and said iterates of $j$ form a cofinal set of $j^\omega(crit(j))$. So for any $\alpha < j^\omega(crit(j))$, there is some whole number $n$ such that $j^n(crit(j)) > \alpha$. If I reasoned correctly, this means that $\omega$-almost hugeness is equivalent to $n$-hugeness for all whole number $n$, which is known to be perfectly consistent at least relative to an $I3$ cardinal. [[User:Eaglgenes101|Eaglgenes101]] ([[User talk:Eaglgenes101|talk]]) 08:26, 29 April 2019 (PDT) | ||

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+ | I thought a bit too. You've reasoned incorrectly, because while there could be a sequence of $M_n$ for each $n$ for which $j:V\rightarrow M_n$ witness the $n$-hugeness of a cardinal $\kappa$, there may not be a single model $M$ for which $j:V\rightarrow M$ witnesses the $n$-hugeness of $\kappa$ for all $n$. This would make a cardinal which is $n$-huge for all $n$ but not $\omega$-almost huge. [[User:Zetapology|Zetapology]] ([[User talk:Zetapology|talk]]) 11:05, 9 May 2019 (PST) |

## Latest revision as of 22:06, 9 May 2019

## $\omega$-almost hugeness underexplored?

The page briefly mentions the $\omega$-almost huge cardinals, but then immediately moves on to $\omega$-huge cardinals without mentioning anything about the properties or possibility/impossibility of $\omega$-almost huge cardinals. Furthermore, none of the other pages in this wiki make any mention of $\omega$-almost hugeness. Is it consistent for them to exist, how strong are they, and is it known how they relate to other large cardinals? Eaglgenes101 (talk) 14:23, 27 April 2019 (PDT)

So I thought about it a bit. The first fixed point of the elementary embedding $j$ above $crit(j)$ is the supremum of all of $crit(j)$, $j(crit(j))$, $j^2(crit(j))$, $j^3(crit(j))$, etc..., and said iterates of $j$ form a cofinal set of $j^\omega(crit(j))$. So for any $\alpha < j^\omega(crit(j))$, there is some whole number $n$ such that $j^n(crit(j)) > \alpha$. If I reasoned correctly, this means that $\omega$-almost hugeness is equivalent to $n$-hugeness for all whole number $n$, which is known to be perfectly consistent at least relative to an $I3$ cardinal. Eaglgenes101 (talk) 08:26, 29 April 2019 (PDT)

I thought a bit too. You've reasoned incorrectly, because while there could be a sequence of $M_n$ for each $n$ for which $j:V\rightarrow M_n$ witness the $n$-hugeness of a cardinal $\kappa$, there may not be a single model $M$ for which $j:V\rightarrow M$ witnesses the $n$-hugeness of $\kappa$ for all $n$. This would make a cardinal which is $n$-huge for all $n$ but not $\omega$-almost huge. Zetapology (talk) 11:05, 9 May 2019 (PST)