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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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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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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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    It's my pleasure! :)2014-12-03