It is slightly hard to disprove the hypothesis that “all triangles are isosceles” by using a triangle that _actually is_ isosceles. 🙂

I had to disabuse my son of this particular argument.

]]>Yes, that is how I understood your argument in the end, and this is how I have presented it in the revised blog post above (with thanks to you).

]]>Sorry for checking your reply lately. My initial statement is somewhat misleading: it should be as “if $\kappa$ is a totally otherworldly and $\lambda$ is a $\Sigma_2$-correct cardinal larger than $\kappa$, then $V_\lambda$ sees $\kappa$ is totally otherworldly.” As you pointed out, this statement itself implies no results about consistency strength.

If $\lambda$ is also worldly, then $V_\lambda$ is a model of ZFC with a totally otherworldly cardinal. You proved that a totally otherworldly cardinal is $\Sigma_2$-correct worldly cardinal, so we have the desired result.

]]>For question 1, I guess you mean downward absolute to transitive inner models? If so, the answer is negative, and I think it is a little easier even than in the worldly case. Suppose $\kappa$ is otherworldly to $\lambda$. This is preserved by the forcing of the GCH, so we may assume GCH. Now force violations to the GCH with Easton forcing at every successor. This also preserves otherworldliness. But if we stop this latter forcing at $\kappa$, then this would be an inner model where $\kappa$ is no longer otherworldly, since the GCH holds above $\kappa$ but fails unboundedly often below.

(I made a few edits to this comment.)

For question 2, I agree.

]]>Aw 🙁 that’s sad to hear.

If anyone’s looking for a static archive, most (all?) of the site seems to be on the Wayback Machine, and the MathJax rendering works: http://web.archive.org/web/20191104130451/http://cantorsattic.info:80/Cantor%27s_Attic

]]>Oh, I see now. The Sigma_2 correct cardinal you intend to use is the other totally otherworldly cardinal. This seems to work perfectly and it resolves my issue. Thanks! I’ll update the post tomorrow.

]]>1. Are otherworldly cardinals downward absolute (we know that worldlies, in general, are not)?

2. We always knew that strength and size didn’t go hand in hand (e.g. strong vs. superstrong cardinals). But this is really crazy how big the least totally otherworldly is in the presence of other large cardinals. I guess then we then have “there exists a measurable/huge/whatever and a totally otherworldly” is stronger than “there exists a measurable/huge/whatever and an inaccessible” (since in the latter, the inaccessible is redundant).

]]>Since ZFC proves that the $\Sigma_2$-correct cardinals are unbounded, we cannot expect to prove in ZFC that if there is an totally otherworldly cardinal with a $\Sigma_2$-correct cardinal above, then Con(totally otherworldly), since this would violate the incompleteness theorem.

]]>Your comment was garbled a little, and I tried to edit it, but now I have become confused about your argument; perhaps I have made a mistake with editing—I apologize. We don’t know that $V_\lambda$ is worldly, so how does your conclusion work?

]]>Let $\kappa$ be a totally otherworldly cardinal and $\lambda>\kappa$ be a $\Sigma_2$-correct cardinal. Take $\alpha<\lambda$ and $\beta>\alpha$ such that $V_\kappa\prec V_\beta$.

Since $V_{\beta+1}\models (\beta>\alpha\land V_\kappa\prec V_\beta)$, we have $V_{\beta+1}\models \exists\xi (\xi>\alpha\land V_\kappa\prec V_\xi)$.

By $\Sigma_2$-correctness of $\lambda$, there is $\gamma<\lambda$ with $\gamma>\alpha\land V_\kappa\prec V_\gamma)$. Hence $V_\lambda\models (\exists\xi (\xi>\alpha\land V_\kappa\prec V_\xi)$. Since $\alpha$ is arbitrary, $V_\lambda$ thinks $\kappa$ is totally otherworldly.

Since every totally otherworldly cardinal is $\Sigma_2$-correct, the existence of two totally otherworldly cardinals implies the consistency of the existence of a totally otherworldly cardinal.

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