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What is the consistency strength of the following statement:

$\kappa$ is a strong cardinals and it is indestructible under $Col(\kappa, <\theta),$ where $\theta> \kappa$ is some fixed inaccessible cardinal.

Remark 1. By Laver's result, the above statement is consistent relative to the existence of a supercompact cardinal (and an inaccessible above it).

Remark 2. By results of Mitchell, Schimmerling and Steel, ``the covering lemma up to a Woodin cardinal'', if $\kappa$ remains measurable after collapsing $\kappa^+$ to $\kappa,$ then there is an inner model with a Woodin cardinal.

Remark 3. Apter's paper ``Strong cardinals can be fully indestructible'' have some positive consistency results related to the question.

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Apter and I proved that the only way to force this situation by any forcing that resembles anything like the Laver preparation is to begin with a supercompact cardinal (or a partially supercompact cardinal for $\theta$ fixed). Specifically, in our article A. W. Apter and J. D. Hamkins, “Indestructible weakly compact cardinals and the necessity of supercompactness for certain proof schemata,” MLQ Math. Log. Q., vol. 47, iss. 4, pp. 563-571, 2001, we prove:

Theorem 2. If after forcing with a closure point below $\kappa$ the weak compactness of $\kappa$ becomes indestructible by the forcing to collapse cardinals to $\kappa$, then $\kappa$ was supercompact in the ground model.

Theorem 3. Suppose $\kappa$ is weakly compact in a forcing extension that admits a closure point below $\kappa$ and that collapses $(2^{\theta^{<\kappa}})^V$ to $\kappa$. Then $\kappa$ was $\theta$-supercompact in $V$.

The proof really uses only that the universe has the $\delta$-approximation and cover properties over the ground model, for some $\delta<\kappa$.

This result does not settle the exact consistency strength, because of the assumption on the forcing, but provides fairly strong evidence that a supercompact cardinal is required for arbitrary collapses and a $\theta$-supercompact is required for collapsing $2^{\theta^{<\kappa}}$.

Meanwhile, as in your remark 2, I am given to understand from the inner model theory experts that having a weakly compact cardinal that is indestructible by collapse forcing to $\kappa$ outstrips the inner model theory, going past Woodin cardinals and more. But we'll have to hear the details from the inner model theory experts.

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  • $\begingroup$ Thanks a lot. It essentially answers what I had in mind. $\endgroup$ – Mohammad Golshani Jul 31 '15 at 2:57
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If κ is weakly compact in VCol(κ,κ+), then there is a non-domestic premouse, see Theorem 0.4 of https://ivv5hpp.uni-muenster.de/u/rds/stacking_mice.pdf

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