Equivalence of Definitions of Cartesian Product of Indexed Family

$(1)$ 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 $2$.

$(2)$ 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 $1$.