Set of All Mappings of Cartesian Product
Theorem
Let $R$, $S$, and $T$ be sets.
Then:
- $R^{S \times T} \sim \paren {R^S}^T$
where $R^{S \times T}$ denotes the set of all mappings from $S \times T$ to $R$.
Proof
Define the mapping $F: \paren {R^S}^T \to R^{S \times T}$ as follows:
- $\map {\map F f} {x, y} = \map {\paren {\map f x} } y$ for all $x \in S , y \in T$.
Suppose $\map F {f_1} = \map F {f_2}$.
Then $\map {\paren {\map {f_1} x} } y = \map {\paren {\map {f_2} x} } y$ for all $x \in S , y \in T$ by the definition of $F$.
Therefore, $\map {f_1} x = \map {f_2} x$ and $f_1 = f_2$ by Equality of Mappings.
It follows that $F$ is an injection.
Take any $g \in R^{S \times T}$.
Define a function $f$ as $\map {\paren {\map f x} } y = \map g {x, y}$.
It follows that:
\(\ds \map {\map F f} {x, y}\) | \(=\) | \(\ds \map {\paren {\map f x} } y\) | Definition of $F$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map g {x, y}\) | Definition of $f$ |
Therefore, $F$ is a surjection.
Thus, $F$ is a bijection.
It follows that $ R^{S \times T} \sim \paren {R^S}^T$ by the definition of set equivalence.
$\blacksquare$
Sources
- 1971: Gaisi Takeuti and Wilson M. Zaring: Introduction to Axiomatic Set Theory: $\S 10.50$