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What is the consistency strength of "ZC + failure of $\text{CH}$ for all many membered sets"?

I know that for the case of ZFC the failure of $\text{CH}$ for every many membered set is too strong, going above measurable cardinals. Is it the same situation when we remove replacement?

Zuhair

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No. Just start with a model of GCH, force CH to fail below $\aleph_\omega$, without going above it, and consider $V_{\omega+\omega}$ of the extension.

Large cardinals only come into play when you want to also have strong limit cardinals.

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