(a) If $f$ mapping $\mathbb{Z}_{+}$ into $\cup_{i\in\mathbb{Z}_{+}}B_{i}$ is such that $\forall i$ : $f(i)\in B_{i}$ , then it maps $\mathbb{Z}_{+}$ into $\cup_{i\in\mathbb{Z}_{+}}A_{i}\supset\cup_{i\in\mathbb{Z}_{+}}B_{i}$ , and $\forall i$ : $f(i)\in B_{i}\subset A_{i}$ .

(b) If $B$ is non-empty then all $B_{i}$ are non-empty. If there is $i$ such that $B_{i}\subset A_{i}$ does not hold, then there is some $x$ in $B_{i}-A_{i}$ , and there is an element $f$ in $B$ such that $f(i)=x\notin A_{i}$ . Hence, $f\in B-A$ . Contradiction.

(c) If $A$ is non-empty, then there is $f\in A$ such that $\forall i\in\mathbb{Z}_{+}$ , $f(i)\in A_{i}$ , i.e. for $i\in\mathbb{Z}_{+}$ , $A_{i}$ is non-empty. The converse. The comment in parentheses means that one needs to be careful in making an argument for the converse statement. In this case we will need to implicitly rely on the axiom of choice to argue that we can select one element $x_{i}$ from each set $A_{i}$ so that $f(i)=x_{i}\in A$ . If we had an uncountable number of sets, the use of the axiom of choice would be crucial. In fact, the axiom of choice is equivalent to a more general version of the converse statement.

(d) Both imply the same set of functions $f:\mathbb{Z}_{+}\rightarrow\cup_{i\in\mathbb{Z}_{+}}A_{i}\cup\cup_{i\in\mathbb{Z}_{+}}B_{i}$ such that $f(i)\in A_{i}\cup B_{i}$ , but $A\cup B$ also requires that either for all $i\in\mathbb{Z}_{+}$ , $f(i)\in\cup_{i\in\mathbb{Z}_{+}}A_{i}$ , or for all $i\in\mathbb{Z}_{+}$ , $f(i)\in\cup_{i\in\mathbb{Z}_{+}}B_{i}$ . So $A\cup B\subset\prod_{i\in\mathbb{Z}_{+}}(A_{i}\cup B_{i})$ . But for the intersection we have the same set of functions $f:\mathbb{Z}_{+}\rightarrow\cup_{i\in\mathbb{Z}_{+}}A_{i}\cap\cup_{i\in\mathbb{Z}_{+}}B_{i}$ such that each $f(i)$ must be in $A_{i}\cap B_{i}$ .

Consider the following example. Let all $A_{i}$ be the set of even numbers, and all $B_{i}$ be the set of number divisible by $5$ . Then $A$ is the set of all sequences of even numbers, and $B$ is the set of all sequences of numbers divisible by $5$ . $A_{i}\cup B_{i}$ is the set of all numbers that are either even or divisible by $5$ , so $\prod_{i\in\mathbb{Z}_{+}}(A_{i}\cup B_{i})$ contains all sequences of numbers that are either even or divisible by $5$ , while $A\cup B$ is the set of all sequences such that either all numbers in the sequence are even, or all numbers in the sequence are divisible by $5$ . Clearly, the latter is a subset of the former, and, for example, $(5,2,2,\ldots)\in\prod_{i\in\mathbb{Z}_{+}}(A_{i}\cup B_{i})-(A\cup B)$ . On the other hand, $A_{i}\cap B_{i}$ consists of the numbers divisible by $10$ , and both $\prod_{i\in\mathbb{Z}_{+}}(A_{i}\cap B_{i})$ and $A\cap B$ contain the sequences of such numbers only.