# Indexed Cartesian Space is Set of all Mappings

## Theorem

Let $I$ be an indexing set.

Let $\displaystyle \prod_{i \mathop \in I} S$ denote the cartesian space of $S$ indexed by $I$.

Then $\displaystyle \prod_{i \mathop \in I} S$ is the set of all mappings from $I$ to $S$, and hence the notation:

- $S^I := \displaystyle \prod_{i \mathop \in I} S$

## Proof

Recall the definition of the cartesian space of $S$ indexed by $I$:

Let $I$ be an indexing set.

Let $\family {S_i}_{i \mathop \in I}$ be an family of sets indexed by $I$.

Let $\displaystyle \prod_{i \mathop \in I} S_i$ be the Cartesian product of $\family {S_i}_{i \mathop \in I}$.

Let $S$ be a set such that:

- $\forall i \in I: S_i = S$

### Definition 1

The **Cartesian space of $S$ indexed by $I$** is the set of all families $\family {s_i}_{i \mathop \in I}$ with $s_i \in S$ for each $i \in I$:

- $S_I := \displaystyle \prod_I S = \set {\family {s_i}_{i \mathop \in I}: s_i \in S}$

### Definition 2

The **Cartesian space of $S$ indexed by $I$** is defined and denoted as:

- $\displaystyle S^I := \set {f: \paren {f: I \to S} \land \paren {\forall i \in I: \paren {\map f i \in S} } }$

## Sources

- 1975: Bert Mendelson:
*Introduction to Topology*(3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 10$: Arbitrary Products: Exercise $1$