10
$\begingroup$

It is known that ZFC needs infinitely many axioms, but NBG (Neuman-Bernays-Gödel set theory) is finitely axiomatizable (as first-order theories of course). But both theories agree completely on the set part of their universe (as far as I have read).

How could this be? How can describing even more objects (proper classes in NBG) while keeping the complexity of some part can reduce the effort to describe this structure? Is there some plausible and evident explanation of this observation? Maybe some philosophical insight from someone who knows the proofs of these statements?

Maybe, is it because the proper classes allow NBG to quantify over predicates in some sense? Something for which ZFC usually needs axiom schemas? If so, why isn't NBG absolutely favorable to ZFC as foundation of math? I mean we also prefer set theory to Peano arithmetic because the latter one allows us to quantify over subsets (in some sense) despite it is a first-order theory (I know we prefer ZFC over PA for tons of other reasons too).

Note:

I know of this and this question, but I ask specifically why the finite axiom system is not a convincing reason for NBG. However,the question on which to prefer, ZFC or NBG is secondary. Please concentrate on the finitely vs. infinitely many axioms part and how this can be.

$\endgroup$
12
  • 2
    $\begingroup$ @Dan Is it really two-sorted? I mean being a set is just a property a class can have. It is like saying ZFC is two-sorted because being finite is some property a set can have. $\endgroup$ – M. Winter May 17 '17 at 17:45
  • 7
    $\begingroup$ There is something to be said about a theory which proves the consistency of its finite fragments. It's a theory that allows you to "almost touch the meta-theory", in the sense that consistency proofs can be nicely internalized (e.g. use forcing over countable transitive models of "arbitrary large fragments of ZFC"). This fails when your theory is finitely axiomatizable. $\endgroup$ – Asaf Karagila♦ May 17 '17 at 18:00
  • 1
    $\begingroup$ The issue is that comprehension is highly restricted in NBG (you apply it to formulas without quantifiers over classes). This version of comprehension can be finitely axiomatized and suffices to give us the ZF version, but does not allow you to form many natural classes you'd like. MK solves this issue, at the cost of losing the finite axiomatizability. You may enjoy this answer and the subsequent comments. $\endgroup$ – Andrés E. Caicedo May 17 '17 at 19:25
  • 2
    $\begingroup$ I would separate out your two questions, because there are two distinct questions here, one purely mathematical and one philosophical. Answers to the mathematical question ('why isn't there a contradiction between the theorem that ZFC isn't finitely axiomatizable and the fact that NBG, a conservative extension of ZFC, is finitely axiomatizable?') will surely inform the philosophical question, so I think it's well worth focusing on just that. $\endgroup$ – Steven Stadnicki May 17 '17 at 19:47
  • 3
    $\begingroup$ You may be interested to know that this isn't a set theory issue, but rather a general logic issue. For example, the same thing happens with arithmetic: PA is not finitely axiomatizable, and in fact no consistent extension of PA in the language of PA is finitely axiomatizable, but the theory ACA$_0$ in the larger language of second-order arithmetic is finitely axiomatizable. (Incidentally, despite the name of the language, ACA$_0$ is indeed a first-order theory.) (continued) $\endgroup$ – Noah Schweber Apr 12 '18 at 19:26
-1
$\begingroup$

I will discuss this in two parts:

  1. Why is ZFC finitely axiomatizable in NBG.

  2. Why is class comprehension finitely axiomatizable.

The first part is simply because, by class comprehensions, every first order predicate is some class. The axioms of ZFC is restricted to first order predicates, and therefore definable in NBG.

As for part 2. NBG allows as to quantify over first order predicates. It is a second order language, applied to discuss first order objects. Just as separation and replacement are restricted to first order object, is is class comprehension, and it is therefore definable with a second order formula.

As for why NBG is not favored. In a sense it is. Many statements in Set Theory discuss proper classes. However the basis of Set Theory is sets, and ZFC is more intuitive in this aspect. And as many classes can be described in terms of sets or formulas, many results in NBG are derivable from ZFC plus the intuitive assumption "There is such thing as a class." The two theories are indeed equiconsistent.

$\endgroup$
1
  • 3
    $\begingroup$ General tip. If your text is three short paragraphs long, it doesn't need an abstract, and rarely an introduction longer than a single sentence. $\endgroup$ – Asaf Karagila♦ May 10 '19 at 19:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.