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Oct 6, 2018 at 6:33 comment added Keith Millar Hmm... that's interesting. If we require $\mu<j(\kappa)$, does this become more 'large-cardinal-like?' More specifically, would it also hold for $\lambda<\mu$ if it held for $\mu$?
Oct 5, 2018 at 15:15 comment added Keith Millar If you close $M$ under $\kappa$-sequences, and you make $\mu<j(\kappa)$ (like strong compactness) then it turns out to be equiconsistent to strongness; this is because every $\mu+2$-strong cardinal would have this property at $\mu$, and every such cardinal would be $\mu$-tall. Since tallness is equiconsistent to strongness, it's equiconsistent to both.
Oct 5, 2018 at 10:23 history edited Yair Hayut CC BY-SA 4.0
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Oct 5, 2018 at 10:22 comment added Yair Hayut You're right, this is true if we further assume that $M$ is closed under $\kappa$-sequences, but without this assumption I'm not certain what is the exact consistency strength.
Oct 4, 2018 at 22:24 comment added Keith Millar Thanks! Why exactly does it imply that $o^K(κ)≥κ^{++}$?
Oct 4, 2018 at 14:00 vote accept Keith Millar
Oct 4, 2018 at 11:08 history answered Yair Hayut CC BY-SA 4.0