Indexed Cartesian Space is Set of all Mappings

Theorem

Let $I$ be an indexing set.

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

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

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

Proof

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

By definition of the Cartesian product of $\family {S_i}_{i \mathop \in I}$:

$(1): \quad \ds \prod_{i \mathop \in I} S_i := \set {f: \paren {f: I \to \bigcup_{i \mathop \in I} S_i} \land \paren {\forall i \in I: \paren {\map f i \in S_i} } }$

where $f$ denotes a mapping.

We have that that:

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

From Set Union is Idempotent:

$\ds \bigcup_{i \mathop \in I} S = S$

Hence when $S_i = S$ for all $i \in I$, $(1)$ can be written:

$\ds \prod_{i \mathop \in I} S = \set {f: \paren {f: I \to S} \land \paren {\forall i \in I: \paren {\map f i \in S_i} } }$

which is exactly the definition of $S^I$ towards which we are aiming.

$\blacksquare$

Sources

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