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Elliot Glazer's user avatar
Elliot Glazer's user avatar
Elliot Glazer
  • Member for 7 years, 7 months
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Is this set theory equivalent to ZFC?
@MarioCarneiro You're right, all three of those were backwards.
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Is this set theory equivalent to ZFC?
Fixing error pointed out by Mario
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Is this set theory equivalent to ZFC?
I'm using the standard definition here. They are finite transitive sets well-ordered by $\in.$
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Is this set theory equivalent to ZFC?
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Is this set theory equivalent to ZFC?
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Is this set theory equivalent to ZFC?
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Does $H\vDash AC$
It's also worth mentioning replacement can fail in $H_{\kappa}.$ For example, in the Feferman-Levy extension of $L,$ $H_{\omega_1}$ fails replacement since the $\omega_n^L$'s are definable in $H_{\omega_1}.$
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Can we make ZF − infinity + “all ordinals are finite” as strong as ZFC?
Alright. Also, I do think there's an interesting question to be asked here if power set is removed, and we look for strengthenings of ZFfin - Power set which interpret $\text{ZFC}^-.$
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Can we make ZF − infinity + “all ordinals are finite” as strong as ZFC?
@AlexMennen Take an infinite set $X.$ Then $\mathcal{P}(\mathcal{P}(X))$ has a countably infinite subset, setting $X_n=\{Y \subset X: |Y|=n\}.$ A replacement on that should yield $\omega.$
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Inner model theory without choice
Philip is saying that fails in $M,$ because if some ZFC inner model covers every Prikry sequence in $M,$ that gives us a way to well-order them.
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