Cartesian Product of Subsets/Family of Nonempty Subsets

Theorem
Let $\family {S_i}_{i \mathop \in I}$ be a family of sets where $I$ is an arbitrary index set.

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

Let $\family {T_i}_{i \mathop \in I}$ be another family of sets.

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

Let $T_i \neq \O$ for all $i \in I$.

Then:
 * $T \subseteq S \iff \forall i \in I : T_i \subseteq S_i$.

Proof
From Leigh.Samphier/Sandbox/Cartesian Product of Subsets/Family of Subsets:
 * $\paren{\forall i \in I : T_i \subseteq S_i} \implies T \subseteq S$

So it remains to show that:
 * $T \subseteq S \implies \forall i \in I : T_i \subseteq S_i$.

Let $T \subseteq S$.

Let $x_j \in T_j$ for some $j \in I$.

Let $\map x j = x_j$

Suppose $k \in I: k \ne j$.

As $T_k \ne \O$ it is possible to use the axiom of choice to choose $\map x k \in T_k$.

Then:
 * $x \in T \subseteq S$

By definition of the Cartesian product $S$:
 * $\forall i \in I: \map x i \in S_i$

In particular:
 * $\map x j = x_j \in S_j$

Since $x_j$ was an arbitrary element of $T_j$ then $T_j \subseteq S_j$.

Since $j$ was an arbitrary element of $I$ then:
 * $\forall i \in I : T_i \subseteq S_i$

The result follows.