Can set operations be set-theoretically defined as functions? ( Is there any set that could be the domain of these functions?)

Question

Usually, a binary operation on a set A is defined as a function from the cartesian product of A with itself ( " A cross A") ( or, from a subset of this cartesian product) to A.

If binary set operations ( for ex. union, intersection, etc.) were to be defined as functions, they would take as input ordered pairs of sets, and these ordered pairs should " come from" the cartesian product of the set of all sets with itself.

But, in order this cartesian product to exist, the set of all sets should first exist.

So, can set operations be defined as functions? And, if it is not the case, what is their official status?

Answer 1

You are right :

we cannot use the cartesian product of the set of all sets with itself in order to define the basic set operations.

Using Axiom schema of Separation we prove that:

$\exists ! C \ \forall x \ (x \in C \leftrightarrow x \in A \land x \in B)$.

Thus, we may "enlarge" the basic set language introducing a new symbol $\cap$:

$A \cap B = y \leftrightarrow \forall x \ (x \in y \leftrightarrow x \in A \land x \in B)$.

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Similarly, but using in addition the Union Axiom, we prove that:

$\exists ! C \ \forall x \ (x \in C \leftrightarrow x \in A \lor x \in B)$.

This justify the introduction of the new symbol $\cup$:

$A \cup B = y \leftrightarrow \forall x \ (x \in y \leftrightarrow x \in A \lor x \in B)$.

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Thus, what we have is a "recipe" that allows us to manufacture, for every pair of existing sets (i.e. sets whose existence has been already proved by the theory) a new set : their union (respectively: intersection).

Answer 2

You are absolutely right about the set of all sets, which doesn't exist since it leads to contradictions. Such "sets" which are too large to actually be a set but still very nice to consider them as some sort of collection of objects are called proper classes. There are several axiom systems for set theory that make the notion of class explicit, e.g. von Neumann–Bernays–Gödel axioms.

Ordinary functions $X \rightarrow Y$ are certain subsets of the cartesian product $X \times Y$. We can use the same idea to define class-functions: if $Set$ is the class of all sets, a class-function is a subset of $Set \times Set$. This way, $\cup,\cap,\times$ etc. (viewed as operations on sets) become class-functions.

Answer 3

According to my restricted knowledge of sets and working in ZFC:

It might be that $\Phi(x,y,z)$ is a formula with $3$ free variables $x,y,z$ such that for every ordered pair $\langle x,y\rangle$ there is exactly one $z$ with the property that $\Phi(x,y,z)$ is true.

For instance take for $\Phi(x,y,z)$ the formalization of the statement that $z=x\cup y$.

In this case $\Phi$ determines a proper class of ordered pairs $\langle\langle x,y\rangle,z\rangle$, but not a function.

This because a function is by definition a set of ordered pairs.

However, it can be recognized as an operation on sets.