7
$\begingroup$

There are several set theories, e.g. ZFC and NF, which often have different axioms or are even outright contradictory. And yet most of other branches of mathematics, e.g. abstract algebra or topology, even though they use the concept of set, can work in any of these theories. How is this possible?

Why isn't abstract algebra based on ZFC be different than that based on NF, even though they both have different set axioms?

$\endgroup$
  • 9
    $\begingroup$ Selection bias: prospective foundational theories that can't express the generally accepted kinds of reasoning in a wide majority of fields simply won't get enough traction for you to know about them and wonder why they can do what they do. $\endgroup$ – Henning Makholm Apr 29 '16 at 14:29
  • 4
    $\begingroup$ If a theory wants to be called "foundational" it needs to justify this by being able to interpret faithfully the "everyday mathematics", at least to a certain degree. $\endgroup$ – Asaf Karagila Apr 29 '16 at 14:47
  • 4
    $\begingroup$ The above having been said, NF is not really used as a foundational theory. $\endgroup$ – Andrés E. Caicedo Apr 29 '16 at 15:46
  • 3
    $\begingroup$ NF is hard to work with, since the limitation of comprehension involves a lot more hands-on logic, at least compared to the separation/replacement/collection schemata in ZFC. Not only that, we don't even know what is the consistency strength of NF in relation to other theories. It is conjectured to be more or less like PA, which wouldn't make it a particularly attractive foundational system for most mathematicians, even more since it refutes the axiom of choice, and we also don't know how much choice can survive with NF before inconsistencies are added. $\endgroup$ – Asaf Karagila Apr 29 '16 at 19:16
  • 2
    $\begingroup$ @user1952009: Tychonoff's theorem, Hahn-Banach theorem, compactness theorem in logic, taking inverse of arbitrary surjections, Hamel bases, Zorn's lemma, continuity implies sequential continuity, uniqueness of algebraic closures (even for $\Bbb Q$), and the list goes on and on and on and on and on and on and on and on and on and on and on and on and on... and on... $\endgroup$ – Asaf Karagila Apr 29 '16 at 23:09
4
$\begingroup$

It is not strictly true that mathematics turns out the same in both theories. For instance, since NF refutes Choice, and it's unclear if something like the Ultrafilter Lemma is consistent, the Completeness Theorem for first order logic is not known to be provable in NF or any of its extensions. There are categories $\mathcal{C},\mathcal{D}$ that are "small" in the traditional sense, but for which the functor category $\mathcal{D}^\mathcal{C}$ is not an exponential. Not every group can be the quotient of a free group. And moving to NFU, which can support Choice and easily be extended to match ZFC in consistency strength, doesn't fix all of these examples.

But note that there's a theory called KF which is axiomatized by extensionality, pairing, union, power set, and a scheme of stratified $\Delta_0$ separation. KF+"There is a Dedekind infinite set" (we'll call this "KFI") is pretty weak, but strong enough to talk about a pretty surprising amount of arithmetic and analysis. Now observe that KFI+"There is a universal set" is NF; and KFI+Regularity+Full Separation is, more or less, Zermelo. The fact that both theories are extensions of a common theory which is adequate to a good deal of ordinary mathematics should make it clear why they agree so well on the basics, even when the extensions are incompatible.

$\endgroup$
  • $\begingroup$ I was waiting for you to write something here! $\endgroup$ – Asaf Karagila Apr 29 '16 at 22:02
  • $\begingroup$ @AsafKaragila: I get predictable when NF comes up :P $\endgroup$ – Malice Vidrine Apr 29 '16 at 22:14
  • 1
    $\begingroup$ We all have our vices, and, hrmm, very little choice when they come up. :-P $\endgroup$ – Asaf Karagila Apr 29 '16 at 22:18

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.