Timeline for About Grothendieck Universe and Tarski's A and A' Axioms
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
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| Apr 17, 2011 at 21:19 | comment | added | The Mathemagician | I WILL say I totally agree with Joel that it's strange that Tarski didn't insist on transitivity.Transitive versions of the TG axioms are worth investigating for this reason-maybe the Polish master saw subtle problems with it that aren't immediately obvious from a cursory analysis. | |
| Apr 17, 2011 at 21:17 | comment | added | The Mathemagician | I recently discussed this matter with a fellow graduate student whose thesis-in-progress is on category theory.He's a devout believer in the TG set theory.We agreed the entire point of this construction is to be able to have "inaccessibly large" sets in mathematics while still preserving the essential features of the ZFC foundations. It's an extrordinarily clever construction by The Mad Scientist of Mathematics,but I'm still not sure it avoids the paradox problems of classes and type theory.I will reserve judgement until my knowledge of such matters rivals that of my friend. | |
| Jun 17, 2010 at 14:19 | comment | added | Gérard Lang | Again, thank you very much for the case ZF+ GU. Indeed, Solovay's message is building upon Blass's paper that shows four inequivalent possible definitions of an inaccessible ordinal when you do not have AC, that become equivalent in presence of AC. Concerning transitivity and axiom A of Tarski, this is taken care of inside condition A'2, pertaining to Tarski's axiom A' that he offers as equivalent to axiom A, and this is what I am finding intriguing in his assertion of the equivalence of the sixteen axioms B deriving from mixing conditions of axiom A and of axiom A'. | |
| Jun 17, 2010 at 14:09 | comment | added | Gérard Lang | Again, thank you very much for the case FZ=GU. Concerning tra | |
| Jun 17, 2010 at 12:31 | comment | added | Joel David Hamkins | Gérard, I think that in ZF then the GU is equivalently formalized in the proper class version, since the universes continue to be $H_\kappa$ for inaccessible $\kappa$. The trick I use above for Tarski sets doesn't work at all for Grothendieck universes, since they must be transitive. (Indeed, it seems somewhat strange to me that Tarski didn't insist on transitivity.) Note also that as Solovay explains (citing Blass), there are various notions of what it means to be inaccessible, which are equivalent in ZFC but not in ZF, and this is part of the explanation for why TA implies AC. | |
| Jun 17, 2010 at 6:56 | vote | accept | Gérard Lang | ||
| Jun 17, 2010 at 6:56 | comment | added | Gérard Lang | then every sixteen axioms obtained as a conjunction of every possible choice of Bi-s are all equivalents. This seems a very interesting result,but the proof does not seem so easy. In particular, if we take B1=A1, B2=A'2, B3=A'3 and B4=A4, then Tarski asserts that the Union axiom is derivable from thus form of axiom B. | |
| Jun 17, 2010 at 6:52 | comment | added | Gérard Lang | A-1: Thank you very much for your very clear answer and proof concerning the TA case. Would the answer concerning the GU case be the same within ZF (so without AC ?). C-3: Thank you very much for the translation from german. So, Tarski is asserting that if we take for i=1,2,3,4 Bi to be either Ai or A'i, where Ai-s are the four conditions of his axiom a and a'i are the four conditions of his axiom A', thenevery sixteen axioms obtained as a | |
| Jun 16, 2010 at 16:49 | history | edited | Joel David Hamkins | CC BY-SA 2.5 |
Replaced TG with TA
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| Jun 16, 2010 at 16:31 | history | answered | Joel David Hamkins | CC BY-SA 2.5 |