14

In logic (especially in model theory and set theory), one defines a model $\mathcal A$ for a given signature $S$ to be given by a underlying set $A$ and some additional structure (by which I mean: for every constant symbol in $S$ an element of $A$, for every $n$-ary function symbol in $S$ a function $A^n\to A$, and so on). In set theory, one often considers models satisfying some set theory axioms (often one takes ZFC); these models have the form $(V, \in)$, where $V$ is called a universe and $\in$ a binary relation on $V$.

I find it philosophically unsatisfying that set-theoretic universes should be sets, since within a set-theoretic universe satisfying ZFC, this universe thinks that it is a proper class, if you know what I mean.

Why does one not allow models/universes to be proper classes?

Why are the set-theoretic universes sets?

Caleb Stanford
  • 45,895
Treppe
  • 143
  • 1
    This answer may be relevant to your question: http://math.stackexchange.com/a/1368646/630 – Carl Mummert Aug 14 '16 at 16:05
  • I wonder whether this question is a duplicate of http://math.stackexchange.com/q/56726/630 . I am not voting at the moment, to see if others agree. – Carl Mummert Aug 14 '16 at 16:07
  • 2
    @Carl I don't think it's a duplicate of that question. – Stefan Mesken Aug 14 '16 at 16:11
  • 1
    @CarlMummert Definitely strongly related. However the other question partly comes across as not understanding how a set theory can have a model at all, and is more specific to set theory. This question is more about why we insist models be sets more generally. – Caleb Stanford Aug 14 '16 at 16:19

1 Answers1

11

One of the reasons is that the $\sf ZFC$ axioms are not equipped for handling classes properly. For example $\sf ZF$ proves that all the axioms of $\sf ZFC$ hold in the inner model $L$, which is a proper class. But we cannot define a truth definition for that class, as that would violate Tarski's theorem. And the proof that all the axioms hold in $L$ is not a single proof, but rather a schema of proofs.

Sure, we can handle some class models relatively okay (e.g. the surreal numbers, which have a relatively simple theory), but arbitrarily speaking? We want our foundational theory to have access to the truth definition of a structure. Otherwise, what sort of foundation does the theory provide us?

If you notice, almost all the statements about class models of set theory are meta-theoretic statements.

Yes, we could use class-theories like $\sf KM$ (Kelley-Morse) and $\sf NBG$ (von Neumann-Goedel-Bernays), and this might allow us to extend the reach of truth definitions for various class models; but certainly not for the entire universe, as that would contradict Tarski's theorem about the undefinability of the truth. So the universe of set theory will, essentially, always be an internal class.

Asaf Karagila
  • 393,674
  • 1
    Thanks. But I already do not understand your first sentence: aren't the axioms of Morse-Kelley set theory enough to handle classes? – Treppe Aug 14 '16 at 16:14
  • A little bit. And indeed in MK we can define the truth predicate for the class of sets with $\in$. This costs us, though, as it means there is a collection of natural numbers which is a proper class. – Asaf Karagila Aug 14 '16 at 16:16
  • @AsafKaragila Could you please elaborate on your last remark ? – Rene Schipperus Aug 14 '16 at 16:30
  • @Rene: You can read Joel Hamkins' blog post on the topic right here: http://jdh.hamkins.org/km-implies-conzfc/ – Asaf Karagila Aug 14 '16 at 16:33
  • @Asaf: Thanks, I think I slowly get your point of ZFC being our foundational theory which has "set" as the notion of collection of objects. So if we want to speak about a "collection of objects", we shall use "set" by default. But if we want to go one step further and talk about our universe of "collections"/"sets" itself, we have to use another term ("class"). – Treppe Aug 14 '16 at 16:38
  • @Treppe: Yes, exactly. – Asaf Karagila Aug 14 '16 at 16:39
  • @AsafKaragila Well I gave the Hamkins blog a superficial once over (less than what it deserves), it seems he defines a set $S$ using second order class theory and then shows that $S=\mathbb{N}$. – Rene Schipperus Aug 14 '16 at 16:50
  • 1
    @Rene: I think you were looking at the wrong place. The general idea is that now the Levy-Montague reflection theorem is not a meta-theorem, but rather a theorem, when applied to statements involving only sets. So in particular, you get a uniform definition for a truth predicate. This is the same trick as proving that in full second-order arithmetic we can define a first-order truth predicate. – Asaf Karagila Aug 14 '16 at 17:06
  • 1
    "A collection of natural numbers which is a proper class"? Doesn't MK, like NBG, have a rule that the intersection between a set and a class is a set? (That's what I got from Mendelson's brief description of it. anyway). – hmakholm left over Monica Aug 14 '16 at 17:13
  • 2
    @Henning: I think you're right (at least if you include the "Limitation of size" which appears in the Wikipedia entries). I misunderstood some things. My point, however, that the truth definition is accessible as a class, not as a set. In the sense that every parameter-free definition for that set has to appeal to proper classes. – Asaf Karagila Aug 14 '16 at 17:31
  • 2
    @AsafKaragila: It's not even limitation of size -- $\text{set}\cap\text{class}=\text{set}$ is simply how the Axiom of Separation looks in NBG. I think the real point here is that what MK can define truth for is just formulas where all quantifiers are restricted to sets (that is, the language of ZFC), whereas MK still can't define truth for its own full language. So seen from MK, a class model of ZFC is still a restricted notion the same way considering only set models within ZFC is a restricted notion. – hmakholm left over Monica Aug 14 '16 at 17:46
  • @Henning: That depends on axiomatization. And yes, you are correct that Kelley-Morse still doesn't provide a truth definition for its own full model. – Asaf Karagila Aug 14 '16 at 17:49