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Let $\lambda$ be a singular cardinal. Is it consistent that there is a forcing of size $\lambda^+$ that collapses $\lambda^+$ while preserving all cardinals below $\lambda$?

(Note that even without the size requirement this implies a failure of the Jensen covering property, so such a forcing does not necessarily exist.)

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2 Answers 2

up vote 4 down vote accepted

The answer is no and it follows easily from the following theorem:

Theorem. Suppose $\kappa$ is a regular uncountable cardinal and $|P|\leq \kappa.$ Then $\Vdash_P cf(\kappa)=|\kappa|.$

For a proof of the theorem see Singularizing forcing of "small" cardinality?

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I think this is a consequence of stationary forcing. See P. Larson: The Stationary Tower, p. 60.

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Isn't the stationary tower forcing much larger than $\lambda^+$? – Monroe Eskew May 13 '13 at 15:32

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