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On the Wikipedia article for codomain, in the third paragraph, it roughly says:

When the domain of a function is a proper class X, in which case there is formally no such thing as a triple (X, Y, F). (?) With such a definition functions do not have a codomain.

As a proper class is a class that cannot be a member of some class, i.e. cannot be a set, I was wondering why a function with its domain being a proper class does not have a codomain?

Thanks and regards!

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up vote 4 down vote accepted

My guess is: proper classes cannot belong to sets, and triples are sets, so the triple in question does not make sense.

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Thanks! I was wondering why triples are sets? Is a triple a tuple? Here it says a tuple is not a set – Tim Feb 14 '11 at 23:17
For the fine details of what things are... you should probably consult something other than Wikipedia. In the most usual presentation of set theory, everything is a set, so there "$X$ is not a set" is always false! One can consider forms of set theory which have tupling as a primitive operation... and it is up to the particular for of the theory you consider whether classes can or cannot be components of tuples. – Mariano Suárez-Alvarez Feb 14 '11 at 23:19
My humble opinion is, unless you are doing something which really requires you to think about this, you simply ignore Wikipedia and its tangential remark about the codomain of functions defined on proper classes: Wikipedia has a way of mixing absurdly unhelpful information with its generally useful information, specially for people who are starting on various subjects. – Mariano Suárez-Alvarez Feb 14 '11 at 23:21
@Tim: The remark in Wikipedia is simply to indicate that in tuples the order matters, not just the elements being listed. Formally, the set $\{2,3,3\}$ is the same as the set $\{3,2,2\}$ and they are the same as $\{2,3\}$, while the three tuples $(2,3,3),(3,2,2),(2,3)$ are all different. What they meant to say is: The tuple $(a,b)$ is not just the set $\{a,b\}$. Actually, there is a way using sets of making sense of tuples. The point is that all we care about is that $(a,b)=(c,d)$ iff $a=c$ and $b=d$, so we can "identify" the order in which $a,b$ are listed. (Cont.) – Andrés E. Caicedo Feb 15 '11 at 7:30
The usual set theoretic way of defining tuples is by setting $(a,b)=\{\{a\},\{a,b\}\}$. One can then show that indeed $(a,b)=(c,d)$ iff $a=c$ and $b=d$. Triples are then defined by, say $(a,b,c)=((a,b),c)$, and similarly for other tuples. Of course, it really does not matter whether ordered tuples are defined as above, and there are several alternatives. For some history and other options, see "Reconsidering ordered pairs", by Dana Scott and Dominic McCarty, The bulletin of symbolic logic, vol 14 (2008), 379-397, available at – Andrés E. Caicedo Feb 15 '11 at 7:37

Actually, if you can define proper classes as such, you are probably in a theory with classes like NBG, so you have the "class comprehension" axiom schema which says that any formula that does not quantify over classes actually defines a class.

Let $F$ be a function defined on a proper class $X$, and let $\phi$ the formula (with free variable $y$ and quantified variable $p$) $$ \exists p:p\in F \wedge \pi_2(p)=y $$ where $\pi_2$ is the projection on the second component: $$ \pi_2(p) = w \equiv \exists t\in p:(w\in t)\wedge (\forall (r,s \in p)\; r\neq s \implies w\notin r \vee w\notin s) $$ or as finite set operations: $$ ⋃\left( ⋃p \setminus ⋂p \right) $$ This formula defines a new class, that is in fact the codomain of $F$.

In general, classes cannot be put into sets, but you only need sets when there is an arbitrary number of classes, for instance you can define the product of two classes as a formula $$ X\times Y = \{z: \pi_1(z)\in X \wedge \pi_2(z)\in Y\} $$ where $\pi_1$ and $\pi_2$ are the usual projections.

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Here you have the difference between the definition of a function in set theory, which is just a collection of ordered pairs with a certain property, and the definition in category theory - to which the OP is appealing - as an ordered triplet. Even in set theories with classes, proper classes are not elements of other classes, so the ordered triplet whose elements are proper classes does not exist. – Asaf Karagila Dec 3 '14 at 3:27
No, you can consider a class of triplets and so on; but this is not the same as a triplet whose elements are classes. Sure, you can just say that you take $X\times Y\times f$, and then you can reconstruct the classes using projections, but that, I believe, misses the point of using that system of definitions and it strikes closer at home with the set theoretic definition of a function as a collection of ordered pairs satisfying some particular property. – Asaf Karagila Dec 3 '14 at 18:26
I think that you're missing out a particularly important chunk of the idea here. In the universe of set theory, things exists if they are sets. Since the concept of an ordered pair is internalized, it becomes clear that in order to be in an ordered pair (or triplet, or whatever) you need to be a set to begin with. Even when you allow proper classes, those are not objects which are elements of other objects. If you want to redefine ordered pairs so classes could be inside ordered pairs, then you need to essentially turn this from a particular notion into a schema, [...] – Asaf Karagila Dec 3 '14 at 19:51
[...] since a particular notion means a collection whose elements are ordered pairs. But elements are sets, even in set theories like Morse-Kelley or NBG. So we turn the idea of an ordered pair into a schema in the meta-theory. Which is fine, and doable, of course, but turns out clunky and requires us to change all our proofs to proof schemata rather than proofs. This hinges too much on the meta-theory, then. It's not a bad solution, but when the meta-theory gets overly involved in the set theory world, we usually raise an eyebrow or two, or three. And this is sufficiently non-essential [...] – Asaf Karagila Dec 3 '14 at 19:53
[...] at least in my opinion, that we can just say "You know what, proper classes just can't be inside ordered pairs". It even fits better philosophically, because if a class can be inside an ordered pair, why can't it be inside another class? If it can, you really just getting into type theory, or universes whichever you prefer, but that's not "plain set theory" anymore, not even "plain set theory with classes" anymore. It's something else, and that's fine, but it's not set theory anymore. – Asaf Karagila Dec 3 '14 at 19:55

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