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Forgive me if this question is quite naïve; I've studied axiomatic set theory in the context of ZF, but my knowledge of NF(U) goes little beyond its axioms, what it means for a formula to be stratified, and stuff that I've read on websites here and there.

What makes NF appeal to me (more so than ZF) is that it uses fewer axioms and it resolves Russell's paradox in a way which better matches my intuition of how I think a 'set' should behave; certainly it's more intuitive than Foundation+Separation+Replacement+... in ZF. The downside of NF is that it proves $\neg$AC, which is a shame. But by adding urelements, which I can almost force myself to accept, we get NFU, which is:

  • Consistent;
  • Consistent with the Axiom of Choice;
  • Consistent with the Axiom of Infinity;
  • More intuitive than ZF;

And according to this page, NFU can "safely be extended as far as you think ZFC can be extended".

Now ZF(C) has its advantages, constructing new sets from old ones and whatnot, but it still hasn't been proved to be consistent. Wikipedia [citation needed] says: "A common argument against the use of NFU as a foundation for mathematics is that our reasons for relying on it have to do with our intuition that ZFC is correct." $-$ is that all there is to it?

My question is:

Why is ZF(C) the paradigm under which 'mathematics' is done, rather than NFU(+Inf)(+AC), which we know to be consistent?

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Very related: math.stackexchange.com/questions/152758/… –  Asaf Karagila Sep 9 '12 at 15:35
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@CliveNewstead: Gödel's incompleteness results means we can't hope for a proof of the consistency of ZF from within ZF or any of its subtheories. We have reasons to believe it is consistent, not least that no one has found an inconsistency in a hundred years. That's not a proof of its consistency. And surely the same must hold of NFU unless it's a supremely weak theory. –  Benedict Eastaugh Sep 9 '12 at 15:44
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Part of the problem with NFU is the underlying view that mathematics is a branch of logic. Quine (and others such as Russell) got it exactly backwards. –  André Nicolas Sep 9 '12 at 16:40
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Why is ZF less intuitive than NFU? –  Trevor Wilson Sep 9 '12 at 16:51
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@Clive: okay. Your reference asserts that the consistency of NFU can be proven in "elementary number theory (Z)," whatever that means. Maybe it means Peano arithmetic, in which case I think this is a misleading statement. If you're willing to question the consistency of ZFC, why aren't you also willing to question the consistency of PA? –  Qiaochu Yuan Sep 9 '12 at 18:57

3 Answers 3

up vote 22 down vote accepted

I've never studied NF or NFU in any great detail, but I have found some points rather subtle and potentially not worth the effort to work around.

  1. Stratification. NF cannot escape its roots in Russellian type theory. This means that it is difficult to directly compare sets of different "types". For example, consider the "maps" $X \to \mathscr{P}(X)$ used in Cantor's theorem. This is in some sense illegal in NF: you can only have maps between sets of the same type, and the type of $\mathscr{P}(X)$ is one higher than that of $X$. The standard NF circumlocution here is to talk about maps $\mathscr{P}_1 (X) \to \mathscr{P} (X)$ instead, where $\mathscr{P}_1 (X)$ denotes the set of singleton subsets of $X$. Cantor's proof carries through without a problem once this workaround is adopted... but one has to realise that this is a different theorem. In particular, if $V$ is the universal set, we can only prove that there is no surjection $\mathscr{P}_1 (V) \to \mathscr{P}(V)$ – a far cry from the indubitable fact that $\mathscr{P}(V) \subseteq V$! (In fact, if you have even one urelement, then $\mathscr{P}(V) \subsetneq V$.)

  2. Failure of cartesian-closedness. Consider the evaluation "map" $\textrm{ev} : Y^X \times X \to Y$. The graph of this "map" cannot be a set in NF for various reasons. [1, 2] Roughly speaking, if $X$ is a set of type $n$, then $Y$ must also be a set of type $n$, in which case $Y^X \subseteq \mathscr{P}(X \times Y)$ is a set of type $n + 1$; but then we are unable to write down a stratified definition of $Y^X \times X$. Thus, the category of NF sets is not cartesian closed!

  3. Local types. What I have said so far appears to contradict the (finite) axiomatisation given by Holmes [3], which says e.g. that you can always form the set $X \times Y$ or $Y^X$. But this is only an apparent contradiction. The fact of the matter is that sets in NF do not actually have a type. I like to think of NF as being "locally typed": sets only gain types when they interact with other sets. Thus, $Y^X \times X$ exists (in some sense), but we cannot make much use of it. Holmes writes,

    Another fact about stratification restrictions should be noted: a variable free in $\phi$ in a set definition $\{ x \mid \phi \}$ may appear with more than one type without preventing the set from existing, as long as this set definition is not itself embedded in a further set definition. The reason for this is that such a set definition can be made stratified by distinguishing all free variables: the resulting definition is supposed to work for all assignments of values to those free variables, including those in which some of the free variables (even ones of different type) are identified with one another. For example, the set $\{ x, \{ y \} \}$ has a stratified definition; the set $\{ x, \{ x \} \}$ has an unstratified definition, but the existence of $\{ x, \{ y \} \}$ for all values of $x$ and $y$ ensures the existence of $\{ x, \{ x \} \}$ for any $x$. But a term $\{ x, \{ x \} \}$ cannot appear in the definition of a further set in which $x$ is bound.

  4. Weaknesses. Holmes [4] points out that bare NFU has the same consistency strength as PA, and NFU + Infinity + AC has the same consistency strength as Mac Lane set theory (MAC) (i.e. Zermelo set theory but with only $\Delta_0$-separation). It is true that in many cases we only need $\Delta_0$-separation to form the sets we want – but when induction seeps into the picture we have to start worrying.

    Consider the vector space $V$ of all real polynomials. There is no doubt that we can form the dual space $V^* = \textrm{Hom}_\mathbb{R} (V, \mathbb{R})$ in MAC – but Mathias [5] says that MAC cannot prove "the $n$-th iterated dual space exists for all natural numbers $n$". This is because the cardinalities of the iterated dual space grow too fast – after all, $V_{\omega + \omega}$ is a model of MAC. This seems contrived, but it is enough for me to start worrying about whether important inductive constructions can be carried out in MAC. And if MAC is not strong enough to do this bit of mathematics, why should NFU + Infinity + Choice be? (The two theories are not biinterpretable, so it's not literally true that they prove the same theorems.)

Personally, I'd rather have unrestricted comprehension. But then I may as well ask for a pony...

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To be precise, you may as well ask for a pony that belongs to exactly those people who do not have a pony :) –  Trevor Wilson Sep 9 '12 at 18:49
    
Feferman [2011] proposes a system that extends both ZFC and NFU, so apparently you can have (a bit of) cake and eat it too! –  Zhen Lin Sep 10 '12 at 3:29
    
Can't point #2 be circumvented easily enough, by asking for ev : $(X \to Y) \times \mathscr{P}_1 (X) \to \mathscr{P}_1 (Y)$ instead? It's a different set from the one we were after originally, but it seems like you could use it the same way. –  Tanner Swett Sep 19 '12 at 5:38
    
Sure, you can define that map. But there's no internal bijection $X \to \mathscr{P}_1(X)$, so for category-theoretic purposes it doesn't satisfy the requirements. –  Zhen Lin Sep 19 '12 at 6:57
    
There are extensions of NFU that are quite strong and equivalent to ZF plus certain large cardinals, so weakness isn't really an objection. One of those extensions is an axiom scheme of strongly Cantorian separation; the strongly Cantorian sets already form a topos, so this takes most of the awkwardness out of the small (but still reasonably large) sets of NFU. Not being Cartesian closed is really its most damning feature. –  Malice Vidrine Feb 17 at 15:01

The problem with NF(U) is that it forces you to deal with the logic all the time. You can't really use anything and just make sure it's bounded by some set.

You need to keep verifying whether the defining formulas are stratified, and if you can or cannot use them to define new sets. Of course, you can use classes (in the sense of definable collections which are not sets) but this is not how we want a foundational set theory to work. We want to have the sets for mathematics.

I have to admit my own ignorance about how much of the mathematics it is possible to develop within the confines of the logic-pushing theories of NF and NFU, however there are two points to give here:

Mathematics, and indeed much of the human world, is based on "first come, first serves". It takes time and an incredible force to revolutionize the world. ZF was quite natural in many ways, and it was very usable for other people outside set theory. Of course you can always start working in NF(U) and hope it will catch on, write books and so on.

However as things stand ZF is the king of the hill, and the fact is that most mathematicians don't really care what is the set theoretical framework. They just want to do their mathematics. It happens that I am working on a project related to model-categories and models of ZFC nowadays. I see how little care we have to many foundational questions. Do we take things which are internally definable? Externally definable? Do we assume there is a model and do this and that?

No. We simply do the mathematics and solve the problems as they come along. ZF is very good and natural for that. Here is an example, I want to argue something about the collection of topologies of the real numbers which are Hausdorff. I don't need to argue that the defining formula of being a Hausdorff topology over $\mathbb R$ has some syntactical properties and show that this collection is a set. I simply state it and move along.

We also don't know about the relative consistency of the theories. It is possible that NFU is stronger or weaker than ZF, it could be that both are equiconsistent. Of course this is just a philosophical question how and where do you put your foundational "jump" and believe that your mathematics is consistent (and you have to assume something, anyway).

The proof of the consistency of NFU assumes the consistency of some other theory, and even the consistency of first-order logic. We don't know, we will never know whether or not our extremely-abstract-foundational theories are consistent. We will always have to assume that they are.

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Yes. I find it very strange that I work in model-categories. :-) –  Asaf Karagila Sep 9 '12 at 15:49
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NFU is known to be of weaker consistency strength than PA. (!) Holmes has a very readable survey of alternative set theories. Models of NFU can be constructed from models of Zermelo equipped with an automorphism that moves a rank. Forster conjectures that plain NF is also quite weak. –  Zhen Lin Sep 9 '12 at 16:51
    
I strongly agree with your fourth paragraph. To me all questions of this form read like some variant of "why use (whatever political or economical system I don't like) over (this other one which ought to be better)?" It's not purely, possibly not even primarily, a question of which system is better once in place; one also has to keep in mind the enormous practical issue of transition costs, and whether anyone feels like paying them. (Similarly why the United States doesn't use the metric system.) –  Qiaochu Yuan Sep 9 '12 at 18:46
    
Concerning "ZF is the king of the hill", some 25 years ago, when I learned a bit of set theory, the "official" story was that ZF was pretty much gone, and everybody used NBG, since that is (supposedly) more convenient to work with [and equivalent as far as the "set" part is concerned, iirc]. Do you happen to know whether that was a wild exaggeration reflecting only the authors' personal preference, or the pendulum has swung back to ZF, or people don't always distinguish between ZF and NBG in this context? –  Daniel Fischer Feb 17 at 15:35
    
@Daniel: Probably a wild exaggeration. Category theorists would tell you that working in NBG is easier because of the inherent mechanism for dealing with classes. However it quickly becomes completely unsatisfactory once you want to consider $2$-classes (classes whose elements are allowed to be proper classes). I asked a question about the set theoretical foundations for category theory in this aspect. The answer I got is that Tarski-Grothendieck+Universes (which is equivalent to $\sf ZFC$+Proper class of inaccessible cardinals) is probably the easiest to work with. –  Asaf Karagila Feb 17 at 15:42

Most mathematicians (whatever they officially claim) are not pure formalists: they are guided by various conceptions of the mathematical universes they purport to describe. That goes for set theorists too.

Zermelo set theory with choice was introduced (wasn't it?) as a compendium of the principles about set-building that mathematicians actually appealed to. Later, Replacement was thrown in the mix, giving us what we now call ZFC.

Now, at the outset it all looked a bit ad hoc (cooked up to give us a paradox-avoiding toolkit for the working mathematician). And the fact that it "worked" wasn't initially regarded as enough to quiet the search for alternatives.

However, already in Gödel's 1933 lecture the suggestion is emerging that ZFC (with Foundation) is a theory that actually fits a lovely and natural conception of sets as forming a well-founded cumulative hierarchy. It took (it seems in retrospect) a surprising time for this conception to become canonical: but nowadays most people imbibe it with their first set-theory course -- think of all those diagrams of the universe of the pure sets in a V-shaped stack of levels, with the von Neumann ordinals as the vertical spine. So, in short, ZFC now comes with a nice intuitively graspable story about the universe of sets that it purports to give us a (partial) description of. No wonder it can seem very attractive.

What, on the other hand, does the universe of sets NF purports to describe "look like"? NF, it seems, comes with no such motivating story -- or at least, not one with such intuitive and natural appeal. (Perhaps is no coincidence that NF was proposed by that arch nominalist Quine, who would put no value on our purported intuitive grasp of the structure of cumulative hierarchy.)

So there's a prima facie contrast: a 'natural' intuitive conception underpinning ZFC vs. (what can seem like) a paradox-blocking syntactic trick giving us NF. That's one rationale -- isn't it? -- for the continued predominance of standard set theory (or of a close variant like Scott-Potter, whose attraction is precisely that it more explicitly starts from the idea of a cumulative hierarchy). I am not here endorsing that rationale: I'm just suggesting that, rightly or wrongly, something like this plays a significant part in the story.

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I know Holmes does try to offer a rationale for NF-derivatives based on the property that stratified sentences are exactly the sentences preserved under a certain kind of auotomorphism, and many of these automorphisms are objects of the theory. So there's a sense in which stratified set theories are the ones whose universes are a very neat kind of symmetric. Admittedly, this is a less obvious and less intuitive picture than that of a cumulative hierarchy. –  Malice Vidrine Feb 17 at 15:18
    
Yes I know that story from Thomas Forster -- but to be honest I've never found it that intuitively helpful ... Still, I guess I should slightly edit what I said! –  Peter Smith Feb 17 at 16:48

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