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Feb 19, 2014 at 22:43 comment added Rachid Atmai ...the "$\Sigma_1$ amenability" would still characterize a (possibly) weaker large cardinal than Ramsey. I am trying to make this idea precise.
Feb 19, 2014 at 22:40 comment added Rachid Atmai I want to say something like: the ultrafilter $U$ is boldface $\Sigma_1$ definable from parameter in $M$, or we can find a $\Sigma_1$ formula $\phi$ such that $X_{\alpha} \in U$ iff $\phi(\alpha, x)$, $x$ some parameters in $M$. When $U$ is weakly amenable then $L_{\alpha}$ and its ultrapower have the same subsets of $\kappa$, and when $U$ is $\Sigma_1$-amenable then $L_{\alpha}$ and its ultrapower should have the same subsets of $\kappa$ which are $\Sigma_1$ definable. One property can hold without the other (IIRC). I was curious to see that even if one property can hold without the other...
Feb 19, 2014 at 13:18 comment added Victoria Gitman Carlo, can you clarify what you mean by $\Sigma_1$-amenable?
Feb 19, 2014 at 4:04 comment added Rachid Atmai Call $U$ $\Sigma_1$-amenable with respect to $M$ if $U$ is $\boldsymbol \Sigma_1(M)$. Is it possible to characterize a large cardinal using $\Sigma_1$-amenability and countable completeness?
Feb 18, 2014 at 22:41 history edited Victoria Gitman CC BY-SA 3.0
added 87 characters in body
Feb 18, 2014 at 21:26 history answered Victoria Gitman CC BY-SA 3.0