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2 votes

Does $\mathsf{ZF}$ prove $\operatorname{Col}(\lambda,\kappa)$ preserves cardinals below $\lambda$?

First of all, note that if $\lambda$ is a singular limit cardinal, then the forcing is terribly behaved already in $\sf ZFC$. Even if $\lambda$ is a successor, you'd still find that the same argument ...
Asaf Karagila's user avatar
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4 votes

Does $\mathsf{ZF}$ prove $\operatorname{Col}(\lambda,\kappa)$ preserves cardinals below $\lambda$?

Re the first question: No. Suppose $\lambda$ is a singular cardinal of cofinality $\omega$. Then the forcing collapses all cardinals $\leq\lambda$. For let $\left<A_n\right>_{n<\omega}\in V$ ...
Farmer S's user avatar
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1 vote

An equivalent of the axiom of choice?

Yes. The problem's statement can be reformulated as follows. Suppose that for any set $X$ there exists a function $f: (2^X\setminus \{\emptyset\})\to X$ mapping any nonempty subset (of escapees) to ...
Станислав Крымский's user avatar
2 votes

An equivalent of the axiom of choice?

With such shared/common knowledge problems, even when AC isn't involved, I think a crucial first step is to get away from the "story" version. For example, since this is only interesting ...
Noah Schweber's user avatar
8 votes

Is Morse-Kelley set theory with Class Choice bi-interpretable with itself after removing Extensionality for classes?

No, because the latter theories has parametrically definable automorphisms that swap two equivalent classes, but the former theory is definably rigid (no need for class choice). If the theories were ...
Joel David Hamkins's user avatar
21 votes

Are Berkeley cardinals easier to refute in ZFC than Reinhardt cardinals?

Yes, it is easier to refute Berkeleys than Reinhardts. There is a very simple refutation of Berkeleys in ZFC that is due to Woodin. It is part of the motivation for his contention that Berkeley ...
Gabe Goldberg's user avatar

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