7

Imagine we're developing all of mathematics from scratch. We settle on using a set-theoretic foundation.

Early on, we assert that an ordered pair $(x,y)$ can be abbreviated $xy$ whenever there is no ambiguity, and we define that a relation is a set of ordered pairs (alternatively, a subset of a Cartesian product). Furthermore, for any relation $f$, we define that $f\langle X \rangle = \{y \,|\,\exists x \in X : xy \in f\}$. We go on to prove the proposition that $f\langle X \cup Y \rangle = f\langle X \rangle \cup f\langle Y \rangle$.

A little later in the piece, we're looking at powersets and related ideas. We define that $\beta$ denotes the inclusion relation (a proper class). So $\beta^{-1}\langle \{Y\}\rangle$ denotes the powerset of $Y$. Symbolically, $\beta^{-1}\langle \{Y\}\rangle = \mathcal{P}(Y)$. And it follows easily that $\beta^{-1}\langle \{Y,Z\} \rangle = \mathcal{P}(Y) \cup \mathcal{P}(Z)$, by our proposition.

However, the above paragraph is bogus. The expression $\beta^{-1}\langle \{Y\}\rangle$ needn't denote the powerset of $Y$, because we haven't defined the meaning of this expression in the case where $\beta^{-1}$ is a proper class. Thus, we have to include "redundant" definition. To make things worse, our proposition was only proved for all relations, which doesn't include proper class relations. So now we need a "redundant" proposition (we copy-and-paste the proof!).

Clearly, this is very silly. Now the obvious solution is to have allowed relations to be proper classes in the first place. But this just pushes the problem up; now, it is relations between proper classes that our proposition has missed. Thus, we still have to deal with redundancy. My question is, how can we make sure our definitions/propositions don't need to be "doubled up".

goblin GONE
  • 67,744
  • That's an interesting choice to use $xy$ and $\langle X\rangle$. Where does it come from? – Asaf Karagila Jan 25 '13 at 19:38
  • $xy$ from graph theory. $\langle X \rangle$ was made up, to avoid ambiguity. It doesn't mean anything on its own, but $f(X)$ is different from $f\langle X \rangle$. As an aside, I'm not especially satisfied with this "angled bracket" notation. – goblin GONE Jan 26 '13 at 06:24
  • http://en.wikipedia.org/wiki/Type_theory – Kaveh Feb 04 '13 at 08:01

1 Answers1

3

You run into a problem because you want something that would work both for classes and sets, but in $\mathsf{ZFC}$ there is a huge difference between classes and sets. The latter is an object whereas the former might not be an actual object of the universe.

One way out is to talk about "collections" without specifying whether this is a class or a set, but this is not fully accurate because there is no well-defined notion of a collection. For example one can talk about the collection of all classes, but this collection is neither a set nor a class. It's a metatheoretical, or perhaps metamathematical notion.

Another way is to define things schematically in a similar way to the way we write the replacement schema. Whenever a formula defines something such and such then it is a relation. But this too amounts to arguments in the metatheory, because now we can't really say "Any well-ordered relation on $\omega$ has an initial segment isomorphic to $\langle\omega,\in\rangle$". Because this is a statement about all relations, but relations are no longer collections of ordered pairs -- they are definable collections. So this is a proof in the metatheory going over all the formulas.

(It is possible that there is a way to overcome this difficulty and write a formula which automatically generates relations over a particular set, without an appeal to a metatheoretical argument, but I don't see one right now.)

So we are again stuck. There is another way out, simply use a theory whose spectrum includes classes. Something like $\mathsf{NBG}$ which is a conservative extension of $\mathsf{ZFC}$ (it does not prove new theorems about sets), and allows classes. One can also consider to use the Tarski-Grothendieck set theory with universes, which is roughly the same as $\mathsf{ZFC}$+"there is a proper class of inaccessible cardinals". This allows us to treat classes as sets of another universe so the definitions go smoothly, but do note that the consistency strength of this theory is stronger than just using $\mathsf{ZFC}$.

Lastly, one can decide to use modern approaches to foundations like type theory based approaches, e.g. $\mathsf{SEAR}$, or other structural set theories such as $\mathsf{ETCS}$. Those might be prone to problems of their own, but I cannot really say what these problems might be due to a lack of knowledge on the topic.

From the meta-mathematical point I should add that we don't often write concrete proofs. We write schemata of proofs. Every time we say an ordered pair we don't really talk about $\{\{a\},\{a,b\}\}$. We actually say "fix $\varphi_p(x,y,z)$ which states that $z$ has the properties required from the ordered pair $\langle x,y\rangle$..." then every time you refer to ordered pairs you refer to this formula. But this is not really dependent in the actual formula.

Similarly for functions we can require different definitions, and the process goes the same. Of course we often take the union of functions, but we can redefine this operation just as well - and everything works as before, even if it is now somewhat cumbersome.

In this aspect there is no reason to expect a proper proof of "lack of redundancy", and when we talk about class relations we often treat the proofs schematically. We know the proof works because the relations have the wanted properties, and then we don't really care anymore.

As the final line implies, we have to verify that the relation we intend to use have the properties which fit into the schema we intend to use. For example well-foundedness or set-like properties.

Further reading:

  1. difference between class, set , family and collection
  2. Difference between a class and a set
  3. What can I do with proper classes?
Community
  • 1
Asaf Karagila
  • 393,674
  • Well lets assume Tarski-Grothendieck set theory. How does this save us from the problem of copy-and-paste? – goblin GONE Jan 30 '13 at 23:22
  • You move to a larger universe. Now the classes of the smaller universes are sets, and everything which is true about sets apply to them. In particular all the definitions of relations and so on. – Asaf Karagila Jan 30 '13 at 23:26
  • Give me an example that uses the proposition proved in order to conclude that $\beta^{-1}{Y,Z}=\mathcal{P}(Y)\cup\mathcal{P}(Z)$. – goblin GONE Jan 30 '13 at 23:42
  • We work in the universe $U$, then $\beta = {\langle X,Y\rangle\mid X\subseteq Y}$ is a set in some larger universe $U'$, therefore for $Y,Z\in U$ we have that in $U'$ it holds, $\beta^{-1}{Y,Z}=\mathcal P(Y)\cup\mathcal P(Z)$ which is a set in $U$, as wanted. – Asaf Karagila Jan 30 '13 at 23:44
  • By "universe" you mean a model for TG? – goblin GONE Jan 31 '13 at 00:21
  • Not just that, I was talking about TG+Universes. This is an additional axiom which states that there is always a larger set which is a model of TG. This is exactly what allows us to switch from classes to sets. As I said, this is similar to having a proper class of inaccessible cardinals, we can simply treat our universes as the sets of rank smaller than a fixed inaccessible, and the classes are sets in a larger universe - e.g. the model defined at the next inaccessible. – Asaf Karagila Jan 31 '13 at 00:23
  • But I want to stress this again, universes and inaccessible cardinals are more than what $\matsf{ZFC}$ can prove. It's an additional assumption, which may or may not be to your liking. Most people are just fine with that, I think. – Asaf Karagila Jan 31 '13 at 00:24
  • Can you point me to an introductory text/article on TG+Universes? I'm still not sure that the problem is solved.... EDIT: Yeah I'm fine with it, TG seems less arbitrary to me than ZFC. EDIT2: Why not add Universes to ZFC? – goblin GONE Jan 31 '13 at 00:28
  • Well, http://en.wikipedia.org/wiki/Tarski%E2%80%93Grothendieck_set_theory should be a reasonable start. As for the universes for ZFC, we often work with large cardinals (much stronger than "proper class of inaccessibles"), and we often make the assumption that there is some model of ZFC to work over when needed; but the most important reason is that it's not really needed. In the sense that we don't really care that we have to define something again, and at some point it just comes along naturally. Much like any other mathematical habit. – Asaf Karagila Jan 31 '13 at 00:39
  • You see set theorists know the limitations and know the ways to work with them without resorting to "more" consistency strength. Other mathematicians usually don't care about that, or don't even get to such problems (e.g. "all finite groups" is a class, but all finite groups whose underlying set is a set of integers is certainly a set), or that they use something like TG+Universes, or in rare cases they actually need a lot more, in which case they usually turn to set theorists and spark some interesting work (e.g. very large cardinals appear naturally in modern homotopy theory constructions). – Asaf Karagila Jan 31 '13 at 00:46
  • "The most important reason is that it's not really needed. In the sense that we don't really care that we have to define something again." We're going to have to agree to disagree on this point, I'm afraid. In any given mathematical work, a thing should be defined and/or proved at most once. – goblin GONE Jan 31 '13 at 00:47
  • Oh, I agree about that. And I don't think there is any set theorist which haven't sat down at least once to verify this "redundancy" is not a bad thing to work with. I believe that you are likely to find it in most set theory books which discuss classes. No one said anything about leaving something hanging in the air, just like you don't bother to prove that $2+2=4$ every time you write it, set theorists don't bother to prove this every time. – Asaf Karagila Jan 31 '13 at 00:52
  • Let me give you a problem from a something I'm currently working on, to make things more concrete. If $P$ is a poset (viewed as a structure) with underlying set $P^U$ and $Q \subseteq P^U$, then $Q$ is associated with a poset in a natural way, called the induced poset. In general, lets denote this poset $Q \bar{\cap} P$, although I doubt that will typeset nicely. Anyway, let $\mathbb{S}$ denote the poset of all sets ordered by inclusion. Then whenever $K$ is a collection of sets, we have that $K \bar{\cap} \mathbb{S}$ is the "obvious" poset associated with $K$.... – goblin GONE Jan 31 '13 at 04:41
  • ... so for example $\mathcal{P}(X)\bar\cap\mathbb{S}$ is the natural poset associated with the powerset of of $X$. Now the question is, how do I justify reusing results proved under the above definitions, if at some point I want to talk about, say, the collection of all sets that are universes? There's no reason why this collection shouldn't form a poset, as far as I can see, and thus my results for posets should hold. – goblin GONE Jan 31 '13 at 04:49
  • @user18921: Well, a class is really just a definable collection. You can justify this in the meta theory instead of an internal justification (i.e. fix a model and prove this from a much larger universe). Or you can sometimes prove that $P$ is set-like, that means that for every $x$, $P^{-1}(x)$ is a set in the universe. In this case you can use the fact that there is a set which contains all the relevant information (e.g. every nonempty class of ordinals has a minimal element, we simply reflect it down to some set of ordinals). – Asaf Karagila Jan 31 '13 at 09:09
  • Asaf sorry to be a pain, but I'm still not understanding. Suppose my definition 1 is "a relation is a set of ordered pairs." Then my definition 2 is "for any relation $R$, write $Ry$ in order to mean ${x ,|, (x,y) \in R}$." Then I emphasize that this notation is valid for non-alphabetical symbols. So if $\leq$ is the usual non-strict ordering on $\mathbb{R}$, then $(\leq x)$ is the set of all lower bounds of ${x}$. Now, given all this, am I allowed to write $(\subseteq X)$ for the powerset of $X$? Do I need to give this special mention? What should I say? – goblin GONE Feb 02 '13 at 19:42
  • @user18921: No special mention is really needed. The class $Rx$ is definable from the class $R$, which is itself a definable collection. So the whole thing is valid on a syntactic level. If you later want to insist on $Rx$ being a set, for example, then you need to show that the axioms of your set theory prove that it is impossible for this not to be true, e.g. if there are class-many subsets below $X$ then it does not have a power set, which is a contradiction to the power set axiom. – Asaf Karagila Feb 02 '13 at 20:00
  • Okay. It is what it is. – goblin GONE Feb 02 '13 at 20:01