Timeline for Minimal Generalized Continuum Hypothesis & Axiom of Choice
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Sep 10, 2015 at 16:20 | comment | added | Thomas Benjamin | @JoelDavidHamkins: Very interesting. So one could actually have a model of $ZF$ where $GCH$ held below an $I_0$ cardinal but have $AC$ fail above it? | |
| Nov 30, 2013 at 14:38 | comment | added | Andreas Blass | @bof I'm not aware of any general convention as to what "GCH" should mean in the absence of AC. Of course, the issue comes up only in contexts (like the present one) where one can't take for granted that (just about any form of) GCH implies AC. So, in such situations, I would recommend not using "GCH" without some clarification of which version is intended. (The same recommendation applies to CH, even more strongly since CH doesn't imply AC.) | |
| Nov 30, 2013 at 11:25 | comment | added | bof | @JoelDavidHamkins I always thought that, in choiceless set theory, the continuum hypothesis is not $2^{\aleph_0}=\aleph_1$ but the weaker assertion that there are no cardinals between $\aleph_0$ and $2^{\aleph_0}$; and the generalized continuum hypothesis is the same assertion for all infinite cardinals, not just alephs; and that "GCH implies AC" refers to an old Lindenbaum-Tarski-Sierpiński proof which does not require foundation. I wonder if I always had that wrong, or if I used to have it right and the fashions changed on me. (Oops, never mind--I just noticed Asaf Karagila's answer.) | |
| Nov 30, 2013 at 11:00 | comment | added | Adam Epstein | @bdf Regarding Foundation, just last night I was reading the proof that "general GCH implies AC" and I started wondering about the relevance of Foundation. Glad to hear otherwise. | |
| Nov 30, 2013 at 5:50 | comment | added | bof | @AndreasBlass Thanks. When did "generalized continuum hypothesis" come to mean $\forall\alpha)\,\text{CH}_\alpha$ instead of "no cardinals between $m$ and $2^m$ unless $m$ is finite"? | |
| Nov 30, 2013 at 5:37 | comment | added | Andreas Blass | @bof It is indeed the case that the proof of AC from $(\forall\alpha)\,\text{CH}_\alpha$ requires the axiom of foundation. If you begin with a ZFC universe satisfying GCH and build a Fraenkel-Mostowski-Specker permutation model over it, then that model will still satisfy $(\forall\alpha)\,\text{CH}_\alpha$. (Specker is important here, to make the atoms violate foundation rather than extensionality.) | |
| Nov 30, 2013 at 4:59 | comment | added | Joel David Hamkins | I had in mind just the usual argument in ZF, which uses foundation. | |
| Nov 30, 2013 at 4:52 | comment | added | bof | Does that "standard argument" for AC from GCH depend on the axiom of foundation? Is it true that, in the absence of the axiom of foundation, $2^{\aleph_{\alpha}}=\aleph_{\alpha+1}$ is not enough to prove AC, but you need the general GCH which says that there are no infinite cardinals $m$ and $n$ such that $m\lt n\lt2^m$? | |
| Nov 30, 2013 at 2:53 | vote | accept | CommunityBot | ||
| Nov 30, 2013 at 1:55 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |