Let \(A = A_{1} \times A_{2} \times \cdots\) and \(B = B_{1} \times B_{2} \times \cdots\) .
Show that if \(B_{i} \subset A_{i}\) for all \(i\), then \(B \subset A\) . (Strictly speaking, if we are given a function mapping the index set \(\mathbb{Z}_{+}\) into the union of the sets \(B_{i}\), we must change its range before it can be considered as a function mapping \(\mathbb{Z}_{+}\) into the union of the sets \(A_{i}\) . We shall ignore this technicality when dealing with cartesian products).
Show the converse of (a) holds if \(B\) is nonempty.
Show that if \(A\) is nonempty, each \(A_i\) is nonempty. Does the converse hold? (We will return to this question in the exercises of §19.)
What is the relation between the set \(A \cup B\) and the cartesian product of the sets \(A_i \cup B_i\) ? What is the relation between the set \(A \cap B\) and the cartesian product of the sets \(A_i \cap B_i\) ?
Solution
Write \[ A=\prod_{i\ge1}A_i, \qquad B=\prod_{i\ge1}B_i. \]
If \(B_i\subset A_i\) for every \(i\) and \(x\in B\), then \(x_i\in B_i\subset A_i\) for every \(i\). Hence \(x\in A\), so \[ B\subset A. \]
Suppose \(B\ne\varnothing\) and \(B\subset A\). Fix \(i\) and \(b\in B_i\). Choose \(x=(x_j)\in B\), and form \(y\) by replacing only the \(i\)-th coordinate of \(x\) by \(b\). Then \(y_j\in B_j\) for every \(j\), so \(y\in B\subset A\). Therefore \(b=y_i\in A_i\). Since \(b\in B_i\) was arbitrary, \[ B_i\subset A_i. \]
If \(A\ne\varnothing\), choose \(x\in A\). Then \(x_i\in A_i\) for every \(i\), so every \(A_i\) is nonempty.
The converse is a choice statement: from the mere nonemptiness of all \(A_i\), producing one element of \(\prod_iA_i\) requires choosing one element from each \(A_i\). For a countable family this is the countable axiom of choice. Thus the converse holds under the axiom of choice; it is not provable in ZF set theory without some form of choice.
- If \(x\in A\cup B\), then either \(x_i\in A_i\) for all \(i\) or \(x_i\in B_i\) for all \(i\). In either case \(x_i\in A_i\cup B_i\) for every \(i\), hence \[ A\cup B\subset\prod_i(A_i\cup B_i). \] The inclusion can be strict because the product on the right permits different coordinates to come from different sides.
For intersections, \[ \begin{aligned} x\in A\cap B &\iff x_i\in A_i\text{ and }x_i\in B_i\text{ for every }i\\ &\iff x_i\in A_i\cap B_i\text{ for every }i, \end{aligned} \] so \[ \boxed{A\cap B=\prod_i(A_i\cap B_i).} \]