Timeline for Forcing the consistency of $ZF$ from a fragment of $ZF$
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Jan 17, 2017 at 12:14 | comment | added | Thomas Benjamin | (cont.) $GPK^{+}_{\infty}$ serves to prove its consistency will soon begin to unravel (as arguments of this type usually do....) | |
| Jan 17, 2017 at 12:11 | comment | added | Thomas Benjamin | (cont.) (F. Colombini et al, eds.), Birkhauser, Boston 1989, pp. 473-518--my comment also] [prove] that the hyperuniverses $N_{\kappa}$ ($\kappa$ being a strongly inaccessible, weakly compact cardinal) satisfy $GPK^{+}_{\infty}$...", he seemingly implies that $GPK^{+}_{\infty}$ is itself consistent. If so, the consistency of $KMC$ (and consequently the consistency of $ZF$ and $ZFC$) follows. On the surface, this seems to be interesting, but I'll bet if one demands more rigor, the claim that $N_{\kappa}$, (where $\kappa$ is a strongly inaccessible, weakly compact cardinal) satisfies | |
| Jan 17, 2017 at 11:51 | comment | added | Thomas Benjamin | (cont.) tree with On-finite levels has an On-infinite branch".). Since it is known that $KM$ proves $CON(ZF)$, $GPK^{+}_{\infty}$ $+$ $AC_{WF}$ should prove $CON(ZFC)$. Since Esser also states in his paper that, "...The arguments of [M. Forti and R. Hinnion in their paper, "The consistency problem for positive comprehension principles." (J. Symbolic Logic 54, (1989), 1401-1418--my comment] and [M. Forti and F. Honsell in their paper, "Models of selfdescriptive set theories". In: Partial Differential Equations and the Calculus of Variations, Essays in Honor of Enni DeGiorgi, I | |
| Jan 17, 2017 at 11:09 | comment | added | Thomas Benjamin | It might also be interesting to have as the base theory $GPK^{+}_{\infty}$, since Olivier Esser shows in his paper "On the Consistency of a Positive Theory" that the theories $GPK^{+}_{\infty}$ + $AC_{WF}$ and $KMC$ $+$ "On is ramifiable" are mutually interpretable ($AC_{WF}$ is the axiom of choice defined on all well-founded sets, $KMC$ is Kelly-Morse with the Axiom of Global Choice, On is the class of all ordinals, and "On is ramifiable" is defined as follows: A class of $KMC$ is On-finite iff it is a set, On-infinite iff not. "On is ramifiable" means "Every On-infinite | |
| Jan 17, 2017 at 1:39 | vote | accept | Thomas Benjamin | ||
| Jan 12, 2017 at 16:59 | comment | added | Noah Schweber | And indeed, regardless of whether AD holds, $V_{\omega_1}$ is never a model of ZF! This is since every ordinal in $V_{\omega_1}$ is countable, hence has a real code, in $V_{\omega_1}$, so there is no set of all reals. What is true is that if $V\models AD$, then $L_{\omega_1^V}\models ZF$. And this is enough to prove that ZF+AD proves Con(ZF). Note that this is how we show that a weakly inaccessible cardinal implies the consistency of ZF: if $\kappa$ is weakly inaccessible in $V$, then it is (strongly) inaccessible in $L$, and so $L_\kappa$ (but not necessarily $V_\kappa$) is a model of ZF. | |
| Jan 12, 2017 at 16:56 | comment | added | Noah Schweber | @ThomasBenjamin ZF+AD does indeed prove Con(ZF) (and much, much more), but not quite the way you think. In my comment above, by "inaccessible" I meant "strongly inaccessible" - in order to conclude that $V_\kappa\models ZF$ in the most direct possible way, we need to show that $V_\kappa$ is closed under powersets. Now, large cardinal notions behave weirdly without choice, but there is no good sense in which $\omega_1$ is strongly inaccessible assuming AD: we will always have a surjection from $\mathbb{R}$ to $\omega_1$. (cont'd) | |
| Jan 12, 2017 at 15:49 | comment | added | Thomas Benjamin | So in $ZF$ $+$ $AD$ (since $\omega_1$ is inaccessible), one could prove the consistency of $ZF$? | |
| Jan 12, 2017 at 15:41 | comment | added | Noah Schweber | @ThomasBenjamin If you consider $\omega$ to be inaccessible, then it is not true that ZF+"there is an inaccessible" proves Con(ZF). You do indeed need the inaccessible to be uncountable. (Think about the proof - given $\kappa$ inaccessible, we argue that $V_\kappa\models ZF$. But in order for $V_\kappa$ to satisfy $ZF$, it has to satisfy in particular Infinity - so we need $\kappa>\omega$.) My understanding, in fact, is that most authors do not consider $\omega$ to be inaccessible, but I could be wrong. | |
| Jan 12, 2017 at 12:19 | comment | added | Thomas Benjamin | Since $ZF$ $+$ "there is an inaccessible cardinal" proves $CON(ZF)$, does the "inacessible cardinal" in question need to be uncountable? I am given to understand that $\omega$ is countable yet inaccessible. | |
| Jan 10, 2017 at 7:25 | comment | added | Thomas Benjamin | Considering the nature of the comments given, I am willing to delete this question. Your answer was helpful to me, but if you think your answer will help others, I will not delete. What do you wish to do? | |
| Jan 10, 2017 at 7:19 | vote | accept | Thomas Benjamin | ||
| Jan 10, 2017 at 7:23 | |||||
| Jan 9, 2017 at 18:01 | history | undeleted | Noah Schweber | ||
| Jan 9, 2017 at 17:59 | history | deleted | Noah Schweber | via Vote | |
| Jan 9, 2017 at 17:50 | history | answered | Noah Schweber | CC BY-SA 3.0 |