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Can you unpack 2 for me? How can there only be finitely many axioms if NBG is 'equivalent' in proving power to ZFC?
by cliffy 9y ago
Can you unpack 2 for me? How can there only be finitely many axioms if NBG is 'equivalent' in proving power to ZFC?
- VyseofArcadia 9y agoUnfortunately I'm not a set theorist, so I haven't done any deep dives into the NBG/ZFC equivalence. My area of expertise (Lie theory) doesn't really require detailed knowledge along those lines. I'm only really aware of it because I'm a category theory enthusiast. That said, my intuition is that NBG is finitely axiomatizeable with equivalent strength precisely because of #1: it has a bigger ontology. From what I remember of grad school set theory, working with proper classes in ZFC amounts to clunky manipulations of formulas in the language of set theory, because there is no such thing as a class in ZFC. A proper class is a well-formed formula that fails to describe a set. It can be described in the language, but cannot be constructed by the axioms. Axiom schemas give one axiom for all formulas in the language of ZFC. Meanwhile, classes are an actual object in the language of NBG, so an axiom of NBG can just say "for all classes x..." There's a wikipedia article[1] that actually goes into a good bit of detail in the case of the axiom schema of specification. [1] https://en.wikipedia.org/wiki/Axiom_schema_of_specification https://en.wikipedia.org/wiki/Axiom_schema_of_specification