# Do the constant functions from a proper class to a set form a set?

In axomiatic set theory one can define functions $$f : C \to D$$ between any two classes $$C, D$$ (even if one or both of them are proper classes).

Let $$C$$ be a proper class and $$D$$ be a set. Consider all constant functions $$f : C \to D$$. Is it legit to say that these functions form a set?

There is a $$1$$-$$1$$-correspondence between these function and the elements $$d \in D$$ which suggests that these functions form a set. On the other hand, these functions can formally can be regarded as product classes $$C \times \{d\}$$ with $$d \in D$$, thus they cannot be elements of a set.

Is there any variant of the axioms of set theory which allows to form a set of the above constant functions?

Moreover, I think it is okay to form the class $$\mathcal C(C,D)$$ of constant functions. Then we have a bijection between this class and the set $$D$$. It seems weird to have a bijection between a proper class and a set.

• There is a one-to-one correspondence of the constant functions $D\to C$ with the elements of $C$, not with the elements of $D$. Is that a typo? Commented Dec 3, 2023 at 16:40
• (My guess is OP intended to say "Consider all constant functions $f: C \to D$", which would fix both problems in addition to the fact that $f$ is supposed to be "to a set" and fits with the usual "alphabetical order" convention) Commented Dec 3, 2023 at 17:15
• @spaceisdarkgreen There was a typo which I corrected. Izaak van Dongen's guess was correct. Commented Dec 3, 2023 at 23:28
• Ther are only set-many such functions, but the domain of each is a proper class, thus so is each such function — too big to be a member of anything. Commented Dec 7, 2023 at 22:42

The main issue here is that the "collection" of all constant functions $$C\to D$$ (when $$C$$ is a proper class and $$D$$ is a set) is not even a class. Let me describe how two common theories (fail to) deal with this "collection".

In ZFC, where every object is a set, proper classes and class functions are not actually first-class objects. Instead, they are meta-theoretic objects. If I write down a property $$P$$, I can conceive of the class of all objects satisfying $$P$$, but this class doesn't exist as an object itself (unless I can prove that there is a set whose elements are exactly the objects satisfying $$P$$). Now given two classes (defined by properties $$P$$ and $$Q$$, I can describe a way of assignment a unique object satisfying $$Q$$ to each object satisfying $$P$$, but this assignment (called a class function) does not exist as an object itself (unless $$\{x\mid P(x)\}$$ and $$\{x\mid Q(x)\}$$ are sets). Thus "the collection of (constant) class functions $$C\to D$$" is not even a class, since its elements are not first-class objects. At the meta-theoretic level, we can write down a particular class function and prove it's constant. But we don't have a way of quantifying over classes in ZFC, so "all constant class functions" doesn't even make sense. And given a particular class function, it may not even be clear whether it is constant - for example, I could write down a class function whose behavior depends on whether the Continuum Hypothesis is true, in such a way that we cannot prove whether or not it is a constant function.

In NBG, both sets and proper classes exist as first-class objects, so if $$X$$ and $$Y$$ are classes, I can treat a class function $$f\colon X\to Y$$ as a first-class object. But if either $$X$$ or $$Y$$ is a proper class, $$f$$ will be a proper class, and hence $$f$$ cannot be an element of a class. So again, there is no object (set or class) whose elements are class functions $$X\to Y$$.

In both theories, the question "is the collection of all constant class functions $$C\to D$$ a set?" does not even make sense, because there is no such collection.

On the other hand, assuming $$D$$ is a set, we do have a way of talking about "the constant class function with value $$d$$" for each $$d\in D$$.

For example, in ZFC, we can use a parametric definition: the constant function $$F_d$$ is defined by $$y = F_d(x) \iff y=d$$. We can prove that for all $$d\in D$$, this definition gives a class function $$C\to D$$, and if $$d\neq d'$$, $$F_d\neq F_{d'}$$. Moreover, for each class function $$G\colon C\to D$$, we can prove (as a theorem schema) that if $$G$$ is constant, then there exists $$d\in D$$ such that $$G = F_d$$. Taken together, all this allows us to think of the "collection" of constant class functions as being in a kind of meta-theoretic bijection with the elements of $$D$$. We can use the elements of $$D$$ as definable "codes" for constant class functions, and the collection of codes is a set ($$D$$ itself). This will be sufficient in practice to treat the "collection" of constant class functions $$C\to D$$ as if it were a set.

• Thank you for this profound answer. I understand that in the standard axiomatic frameworks the "collection" of constant class functions $C→D$ is not a class, let alone a set. Neverheless there is a "stale taste" that a solution requires mental acrobatics to cover something which seems so obvious. But "seeming obvious" is of course not a reasonable concept. Commented Dec 3, 2023 at 23:50
• @KritikerderElche There is still one "out", but it's not widely used these days. If you invent a theory with three kinds of object - sets, arbitrary classes of sets, and arbitrary collections of classes - your desired collection exists as the final kind only. You're welcome, of course, to have infinitely many layers like this. And that's the point of stratification.
– J.G.
Commented Dec 3, 2023 at 23:55
• @J.G. Thank you, interesting point of view! Commented Dec 4, 2023 at 9:02
• Another common approach, along the lines of @J.G.'s comment, and which is widely used these days: Assume we have a whole heirarchy of universes. Agree to work in some universe $U$, so that every time we would say "set", we say "$U$-small set", i.e., "element of $U$". Now among the subsets of $U$, some are $U$-small, but others are not - the non-$U$-small subsets are the analogue of proper classes. But these are still sets, and in fact they are $U'$-small for some larger universe $U'$. Commented Dec 4, 2023 at 14:27
• This trick of allowing ourselves to move to larger universes is a nice bookkeeping device that avoids having to introduce new names for objects at each level ("set", "class", "hyperclass", "conglomerate", whatever...) @KritikerderElche See also here and here Commented Dec 4, 2023 at 14:31