Talk:General set theory

message: The article said that there were only two existential quantifiers in the three axioms. This is incorrect. The property φ may contain some existential quantifiers. Furthermore, the axiom of extensionality
:\forall x \forall y [\forall z [z \in x \leftrightarrow z \in y] \rightarrow x = y]\,
contains an existential quantifier because it is logically equivalent to
:\forall x \forall y [x \neq y \rightarrow \exists z ([z \in x \land z \notin y] \lor [z \notin x \land z \in y])]\,.
JRSpriggs (talk) 02:08, 4 June 2009 (UTC)
replies: text: It's patently absurd to count quantifiers when neither theory can be finitely axiomatized. Both GST and ZF use and require infinitely many quantifiers, not two, three, seven, or what have you. — Emil J. 10:27, 4 June 2009 (UTC)

text: And even if they could be counted, what one would probably be interested in is the depth of quantifier alternation, not the mere number of quantifiers. I have a vague sense that counting existential quantifiers in this way is a Boolos-ism. I have seen other articles where editors, not aware that Boolos was writing in a abbreviated way, misinterpreted what he was saying. Maybe that is the case here. — Carl (CBM · talk) 11:44, 4 June 2009 (UTC)
replies: text: To EmilJ: Thank you for fixing many errors in this article. JRSpriggs (talk) 08:58, 5 June 2009 (UTC)
thread_title:  Third existential quantifier 

message: The article discusses the successor function S(x)=x\cup\{x\}. However, I didn't understand why "Adjunction implies that if x is a set, so is S(x)...". To use Adjunction, we need to assert that \{x\} is a set. In ZFC, Pairing and Separation assert that. In GST, we don't have Pairing. None of the 3 axioms does that. What did I miss? Tomlee2060 (talk) 15:46, 12 November 2010 (UTC)
replies: text: You apparently misunderstand the adjunction axiom. It says that for any sets x and y, x\cup\{y\} is also a set. Specialize with x = y to get the successor of x.—Emil J. 11:28, 13 November 2010 (UTC)
replies: text: Thanks. Indeed, I misread the Axiom. —Tomlee2060 (talk) 04:06, 15 November 2010 (UTC)
thread_title:  Problem with the successor function 

message: GST does not seem to include any induction- or foundation-like axiom (scheme), how does it interpret Peano Arithmetic? 2003:C5:8F16:A154:4970:6F4A:E249:2A4E (talk) 18:32, 9 September 2020 (UTC)
thread_title:  Peano arithmetic 

message: In the Peano Arithmetic section:
:The most remarkable fact about ST (and hence GST), is that these tiny fragments of set theory give rise to such rich metamathematics. While ST is a small fragment of the well-known canonical set theories ZFC and NBG, ST interprets Robinson arithmetic (Q), so that ST inherits the nontrivial metamathematics of Q. For example, ST is essentially undecidable because Q is, and every consistent theory whose theorems include the ST axioms is also essentially undecidable.[2] This includes GST and every axiomatic set theory worth thinking about, assuming these are consistent. In fact, the undecidability of ST implies the undecidability of first-order logic with a single binary predicate letter.
There is no mention of ST anywhere before this in the article. Maybe ST is Burgess's equivalent of Tarski's S', mentioned later in the article as "GST with Empty Set replacing the axiom schema of specification"? But if so, it needs to be mentioned earlier. Especially if so much of this article is going to be about the surprising power of ST, rather than that of GST. And of course there needs to be some justification, or at least mention, of the fact that anything that can be done with ST can still be done if you replace empty set with the axiom schema of separation, not just left for people to guess.
Because I'm only _guessing_ that this is what ST refers to, I don't want to edit the article in a possibly incorrect way; someone who actually knows what's intended needs to fix it. --157.131.154.133 (talk) 03:32, 22 December 2020 (UTC)
thread_title:  ST