Show that if S and R are sets then $S^R$ is also a set.

Question

So I have to show that if S and R are sets then $S^R$ is also a set.

Answer 1

I assume that $S^R$ is the set of functions $R \rightarrow S$, a "function" is defined as a binary relation, and we work in ZFC.

Assuming that we know that $R\times S$ exists, it's powerset $P(R\times S)$ also exists by the powerset axiom. Then, using axiom schema of separation on the set $P(R\times S)$ and a predicate "$f$ is a function from $R$ to $S$", we get exactly the set of functions $R \rightarrow S$.

The predicate "$f$ is a function from $R$ to $S$" can be defined as follows:

$$\varphi(f) = \forall x \in R (\exists y\in S :(x,y)\in f)\wedge(\forall y_1, y_2 \in S: (x,y_1) \in f \wedge (x,y_2) \in f \Rightarrow y_1 = y_2)$$

That is, every $x\in R$ has an image $y$ in $S$ and only one $y$ is the image of $x$.