Talk:Tarski–Grothendieck set theory

From Wikipedia, the free encyclopedia
Jump to: navigation, search
WikiProject Mathematics (Rated Start-class, Low-priority)
WikiProject Mathematics
This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of Mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.
Mathematics rating:
Start Class
Low Priority
 Field: Foundations, logic, and set theory


Questions[edit]

Is TG simply ZF minus Infinity, augmented by Tarski's axiom? Does that axiom also ensure the existence of infinite sets? What is known about the metamathematics of Tarski's axiom? Why is Grothendieck's name associated with TG? Has anyone written on TG outside of the Journal of Formalized Mathematics?132.181.160.42 03:38, 10 August 2006 (UTC)

This theory looks like very sloppy work to me, if this article correctly represents it. JRSpriggs 03:00, 21 August 2006 (UTC)
ad existence of infinite sets: yes, for any ordinal, Tarski's axiom gives you a limit ordinal containing it (http://mmlquery.mizar.org/mml/current/ordinal1.html#T51); the smallest containing the empty set is omega (http://mmlquery.mizar.org/mml/current/ordinal1.html#D12) JosefUrban 19:00, 8 June 2007 (UTC)
ad usage by Grothendieck: http://modular.fas.harvard.edu/sga/sga/4-1/4-1t_185.html;
and as for "sloppiness", I do not know how to measure this, but provided that this is used by two top-level mathematicians of 20. century, I'd be a bit cautious with such words (and if used at all, I'd certainly try to justify them) JosefUrban 19:09, 8 June 2007 (UTC)
I can't read Mizar, and I don't see how Tarski's axiom implies the axiom of infinity. It looks to me that Vω, which is a Grothendieck universe, is a model of these axioms, since all of its subsets are either hereditarily finite or have cardinality ω. — Charles Stewart (talk) 12:26, 24 June 2009 (UTC)
Tarski's axiom basically says that every set is a member of a set which is a Grothendieck universe. Vω is not a model of this axiom, because no nonempty element of Vω is a Grothendieck universe. More generally, every nonempty Grothendieck universe is infinite, which is why Tarski's axiom implies the axiom of infinity. — Emil J. 12:42, 24 June 2009 (UTC)
Ah, yes, of course, how silly of me. Thanks. So ZF+Tarski's axiom has the same theory as ZFC+the inaccessible cardinal axiom. — Charles Stewart (talk) 13:24, 24 June 2009 (UTC)

Implies axiom of choice?[edit]

"Tarski's axiom implies the Axiom of Choice"

why/how?does anyone have a proof of this?

16:25, 28 March 2007 (UTC)

yes, a verified one: http://mizar.uwb.edu.pl/JFM/Vol1/wellord2.html (or in full detail: http://mmlquery.mizar.org/mml/current/wellord2.html#T26, http://mmlquery.mizar.org/mml/current/wellord2.html#T28). JosefUrban 18:15, 8 June 2007 (UTC)

Unclear comment about definitions[edit]

"Any standard exposition of ZFC (e.g., Suppes 1960) reveals that given the four axioms above, all of the Definitional Axiom is redundant except for the existence of power sets and union sets, for which there are explicit ZFC axioms."

This isn't clear to me. How would a definitional axiom (e.g., the definition of a singleton) be "redundant"? I've deleted this passage from the article. — Preceding unsigned comment added by Jessealama (talkcontribs) 17:32, 7 June 2011 (UTC)

How do you get the empty set?[edit]

The present version of the article presents an axiom system which includes no obvious method for obtaining the empty set. If there is any method for getting it from the given axioms, please show it here. Otherwise, an empty set axiom should be added to the article. JRSpriggs (talk) 07:29, 12 March 2014 (UTC)

Similarly, it seems very likely that the set y given by Tarski's axiom is intended to be a transitive set, but there is no obvious way to prove that it is such from the given closure conditions. Why not say that it must contain the elements of its elements? JRSpriggs (talk) 08:29, 12 March 2014 (UTC)