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Finite axiomatization of second-order NBG

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First-order NBG set theory is finitely axiomatizable. The proof of this basically shows that the axiom schemas, in the presence of the other axioms, can be reduced to a finite set of cases, and are hence equivalent to a finite set of axioms.

Is this equivalence also valid in second-order NBG? Or does the reduction to a finite number of cases only work for definable predicates (e.g. the first order axiom schema)?

set-theory axioms foundations

share|cite|improve this question

asked Mar 16 at 23:18

Mike BattagliaMike Battaglia

1,5181127

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  What do you mean by "second-order NBG", exactly?
  $endgroup$
  – Eric Wofsey
  Mar 16 at 23:37
  

  
  
  
  


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  NBG with the axiom schemas replaced with a second order axiom. Is this term ambiguous?
  $endgroup$
  – Mike Battaglia
  Mar 16 at 23:42
  

  
  
  
  


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  Well, that is at least somewhat ambiguous (it's not obvious what second-order axiom you're using, given that an essential feature of NBG is that its comprehension schema is restricted to formulas without class quantifiers--is "second-order NBG" the same thing as "second-order MK"?). But in any case, if second-order NBG does not have any axiom schemas, it's unclear what you mean by your question.
  $endgroup$
  – Eric Wofsey
  Mar 16 at 23:46
  
  

  
  
  
  


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  I guess there is probably only one reasonable choice for the second-order axiom, and no hope of distinguishing second-order NBG from second-order MK. I still don't know what your question is--are you asking whether second-order class comprehension is equivalent to the finitely many first-order axioms used in NBG? (In that case the answer is obviously not, since the latter are only equivalent to first-order class comprehension.)
  $endgroup$
  – Eric Wofsey
  Mar 16 at 23:58
  

  
  
  
  


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  Both NBG and MK have a class comprehension schema; the difference is that MK allows formulas with class quantifiers in the schema. That distinction is lost if you replace the schema with a second-order axiom saying "every collection of sets is a class". (Also, NBG usually isn't stated with a replacement schema, since you can just use a class quantifier, and the necessary generality follows from the comprehension schema. This doesn't really make a difference for anything, though.)
  $endgroup$
  – Eric Wofsey
  Mar 17 at 0:13
  
  

  
  
  
  

 | 
show 2 more comments

0

$begingroup$

First-order NBG set theory is finitely axiomatizable. The proof of this basically shows that the axiom schemas, in the presence of the other axioms, can be reduced to a finite set of cases, and are hence equivalent to a finite set of axioms.

Is this equivalence also valid in second-order NBG? Or does the reduction to a finite number of cases only work for definable predicates (e.g. the first order axiom schema)?

set-theory axioms foundations

share|cite|improve this question

asked Mar 16 at 23:18

Mike BattagliaMike Battaglia

1,5181127

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  What do you mean by "second-order NBG", exactly?
  $endgroup$
  – Eric Wofsey
  Mar 16 at 23:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  NBG with the axiom schemas replaced with a second order axiom. Is this term ambiguous?
  $endgroup$
  – Mike Battaglia
  Mar 16 at 23:42
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Well, that is at least somewhat ambiguous (it's not obvious what second-order axiom you're using, given that an essential feature of NBG is that its comprehension schema is restricted to formulas without class quantifiers--is "second-order NBG" the same thing as "second-order MK"?). But in any case, if second-order NBG does not have any axiom schemas, it's unclear what you mean by your question.
  $endgroup$
  – Eric Wofsey
  Mar 16 at 23:46
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  I guess there is probably only one reasonable choice for the second-order axiom, and no hope of distinguishing second-order NBG from second-order MK. I still don't know what your question is--are you asking whether second-order class comprehension is equivalent to the finitely many first-order axioms used in NBG? (In that case the answer is obviously not, since the latter are only equivalent to first-order class comprehension.)
  $endgroup$
  – Eric Wofsey
  Mar 16 at 23:58
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Both NBG and MK have a class comprehension schema; the difference is that MK allows formulas with class quantifiers in the schema. That distinction is lost if you replace the schema with a second-order axiom saying "every collection of sets is a class". (Also, NBG usually isn't stated with a replacement schema, since you can just use a class quantifier, and the necessary generality follows from the comprehension schema. This doesn't really make a difference for anything, though.)
  $endgroup$
  – Eric Wofsey
  Mar 17 at 0:13
  
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

First-order NBG set theory is finitely axiomatizable. The proof of this basically shows that the axiom schemas, in the presence of the other axioms, can be reduced to a finite set of cases, and are hence equivalent to a finite set of axioms.

Is this equivalence also valid in second-order NBG? Or does the reduction to a finite number of cases only work for definable predicates (e.g. the first order axiom schema)?

set-theory axioms foundations

share|cite|improve this question

asked Mar 16 at 23:18

Mike BattagliaMike Battaglia

1,5181127

$endgroup$

share|cite|improve this question

asked Mar 16 at 23:18

Mike BattagliaMike Battaglia

1,5181127

share|cite|improve this question

asked Mar 16 at 23:18

Mike BattagliaMike Battaglia

1,5181127

asked Mar 16 at 23:18

Mike BattagliaMike Battaglia

1,5181127

asked Mar 16 at 23:18

Mike BattagliaMike Battaglia

1,5181127