Set theory: product of families of sets proof $\{A_i\}_{i \in I}$ and $\{B_i\}_{i \in I}$ are families of nonempty sets with the same index set, $I$. Prove that if $\prod_{i \in I}A_i \subset \prod_{i \in I}B_i$, then $A_i \subset B_i$ for all $i \in I$.
I am struggling with this proof question. I know the product is a set of functions, but how would I start the proof?
 A: As already noticed - and it is the important ingredient, really - the relevant definition is: $$\prod_{i\in I} A_i:=\left\{f:I\to \bigcup_{i\in I} A_i\,:\, \forall i\in I,\ f(i)\in A_i\right\}$$
Which gives you that $f\in\prod_{i\in I}A_i\subseteq \prod_{i\in I}B_i\implies f(i)\in B_i\wedge f(i)\in A_i$.
However, it might be worth noticing that, set-theoretically, there is another important ingredient: you'll need to use axiom of choice, because your exercise implies it in ZF.
In fact, suppose AC were false. Thus, there is a family of non-empty sets $\{A_i\}_{i\in I}$ such that $\prod_{i\in I}A_i=\emptyset$. Therefore, for any set $x$, it holds $\prod_{i\in I}A_i\subseteq \{x\}^I$. By your lemma, for all $i\in I$, $\emptyset\neq A_i\subseteq \{x\}$ and hence $A_i=\{x\}$. But this is absurd, because then $\prod_{i\in I} A_i=\{x\}^I=\{f\}$, where $f:I\to \{x\}$ is the constant function $\equiv x$.
Where do you need AC? Well, for a fixed $i$ and $a\in A_i$, you know that if some $f\in\prod_{i\in I} A_i$ has $f(i)=a$, then $a\in B_i$. What is there left to prove?
A: Let $a_i\in A_i$. Since $\prod_i A_i \twoheadrightarrow A_i$ is epi there is $f\in \prod_i A_i$ with $f(i) = a_i$. By assumption $f$ in $\prod_i B_i$. Hence $f(i) \in B_i$ and $f(i) = a_i$ implies $a_i\in B_i$.
A: A member of $\prod_{i \in I}A_i$ is a map $f$ that selects a member of $A_i$ for each $i \in I$.
If $\prod_{i \in I}A_i \subseteq \prod_{i \in I}B_i$, then $f \in \prod_{i \in I}B_i$.
All $a_i \in A_i$ may be chosen by $f$.
A: This is really just the definition of the Cartesian product on sets.
Note that we can also prove that if $A_i\subseteq B_i\;\;\forall i\in I$, then $\prod_\limits{i\in I} A_i \subseteq \prod_\limits{i\in I} B_i$, because $x_i\in A_i \implies x_i\in B_i$.
