# Equivalence of Definitions of Cartesian Product of Indexed Family

## Theorem

The following definitions of the concept of Cartesian Product of Family are equivalent:

### Definition 1

Let $I$ be an indexing set.

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

The Cartesian product of $\family {S_i}_{i \mathop \in I}$ is the set of all families $\family {s_i}_{i \mathop \in I}$ with $s_i \in S_i$ for each $i \in I$.

This can be denoted $\ds \prod_{i \mathop \in I} S_i$ or, if $I$ is understood, $\ds \prod_i S_i$.

### Definition 2

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

The Cartesian product of $\family {S_i}_{i \mathop \in I}$ is the set:

$\ds \prod_{i \mathop \in I} S_i := \set {f \in \paren {\bigcup_{i \mathop \in I} S_i}^I : \forall i \in I: \paren {\map f i \in S_i} }$

where $\ds \paren {\bigcup_{i \mathop \in I} S_i}^I$ denotes the set of all mappings from $I$ to $\ds \bigcup_{i \mathop \in I} S_i$.

## Proof

### $(1)$ implies $(2)$

Let $\ds \prod_{i \mathop \in I} S_i$ be a Cartesian product of $\family {S_i}_{i \mathop \in I}$ by definition $1$.

Then by definition:

$\ds \prod_{i \mathop \in I} S_i$ is the set of all families $\family {s_i}_{i \mathop \in I}$ with $s_i \in S_i$ for each $i \in I$

By definition, a indexed family of sets is a mapping from an indexing set $I$ to a set of sets $\SS$ such that:

$\forall i \in I: S_i \subseteq \SS$

Let $f: I \to \SS$ be an indexed family of sets.

For each $i \in I$, $\map f i$ is an element of $S_i$.

That is:

$\map f i \in S_i$

Hence:

$\Img f = f \sqbrk I \subseteq \ds \bigcup_{i \mathop \in I} S_i$

Thus:

$\ds \prod_{i \mathop \in I} S_i = \set {f \in \paren {\bigcup_{i \mathop \in I} S_i}^I : \forall i \in I: \paren {\map f i \in S_i} }$

Hence $\ds \prod_{i \mathop \in I} S_i$ is a Cartesian product of $\family {S_i}_{i \mathop \in I}$ by definition $2$.

$\Box$

### $(2)$ implies $(1)$

Let $\ds \prod_{i \mathop \in I} S_i$ be a Cartesian product of $\family {S_i}_{i \mathop \in I}$ by definition $2$.

Then by definition:

$\ds \prod_{i \mathop \in I} S_i := \set {f \in \paren {\bigcup_{i \mathop \in I} S_i}^I : \forall i \in I: \paren {\map f i \in S_i} }$

where $\ds \paren {\bigcup_{i \mathop \in I} S_i}^I$ denotes the set of all mappings from $I$ to $\ds \bigcup_{i \mathop \in I} S_i$.

Let $f \in \ds \prod_{i \mathop \in I} S_i$ be arbitrary.

Then:

$\forall i \in I: \map f i = s_i$ for some $s_i \in S_i$

That is, $\ds \prod_{i \mathop \in I} S_i$ is the set of all mapping from an indexing set $I$ to a set of sets $\ds \bigcup_{i \mathop \in I} S_i$.

Thus by definition of indexed family of sets:

$\ds \prod_{i \mathop \in I} S_i$ is the set of all families $\family {s_i}_{i \mathop \in I}$ with $s_i \in S_i$ for each $i \in I$

Thus $\ds \prod_{i \mathop \in I} S_i$ is a Cartesian product of $\family {S_i}_{i \mathop \in I}$ by definition $1$.

$\blacksquare$