**Problem 1: Show there is a bijective correspondence of \(A \times B\) with \(B \times A\)**

Since \((a, b) \in A \times B \iff (b, a) \in B \times A\), we can map any element
\((a, b)\) to the element \((b, a)\) and vice versa, and so there is clearly a bijective
correspondence between the two sets.

**(a) Show that if \(B_i \subset A_i\) for all \(i\), then \(B \subset A\)****(b) Show the converse of (a) holds if \(B\) is nonempty.****(c) Show that if \(A\) is nonempty, each \(A_i\) is nonempty. Does the converse hold?****(d) What is the relation between the set \(A \cup B\) and the cartesian product \[\prod\limits_{i \in \mathbb{Z}_{+}} A_{i} \cup B_{i}.\] What is the relation between the set \(A \cap B\) and the cartesian product \[\prod\limits_{i \in \mathbb{Z}_{+}} A_{i} \cap B_{i}.\]**

(a) Remembering that
\[\prod\limits_{i \in \mathbb{Z}_{+}} S_{i} = S_1 \times S_2 \times \cdots\]
is defined as the set of all \(\omega\)-tuples \((\omega_1, \omega_2,
\cdots)\) of elements of \(S_1 \cup S_2 \cup \cdots\) such that \(\omega_1
\in S_1, \omega_2 \in S_2, \cdots\), it follows that for any given
\(\omega_{i} \in B_{i}\), \(\omega_{i}\) must also be in \(A_{i}\), and so any
\((\omega_1, \omega_2, \cdots) \in B\) must also be in \(A\).

(b) Suppose \(B \subset A\) and \(B \subset A \not\Longrightarrow B_{i}
\subset A_{i} \ \forall i\). That would allow for the existence of some
\(\omega\)-tuple \(\omega\) such that \(\omega_i \in B_1 \cup B_2 \cup \cdots\) and
\(\omega_{i} \not\in A_1 \cup A_2 \cup \cdots\). This is a clear contradiction, as
such a tuple would be an element of \(B\) and not an element of \(A\).

(c) Suppose \(A_{k} =
\varnothing\) for some \(k \in \mathbb{Z}_{+}\). Then, since \(A\) is
defined as the set of all \(\omega\)-tuples such that \(\omega_{i}
\in A_{i}\), and since there is no element \(\omega_{k}\) such
that \(\omega_{k} \in A_{k}\), there can be no such tuple. Thus,
Every \(A_{i}\) must be nonempty. The converse is also true. If
each \(A_{i}\) is nonempty, then there are elements \(\omega_{i} \in
A_{i}\) for each \(i\), and so the \(\omega\)-tuple \(\omega\) can
exist.

(d) The fourth part of the problem is asking us to describe the
relationship between \(A \cup B\) and \(\prod\limits_{i \in \mathbb{Z}_{+}}
A_{i} \cup B_{i}\), and \(A \cap B\) and \(\prod\limits_{i \in \mathbb{Z}_{+}}
A_{i} \cap B_{i}\).

In short, \(A \cup B \subset \prod\limits_{i \in \mathbb{Z}_{+}} A_{i} \cup B_{i}\): \[ \omega \in A \cup B \implies \omega \in A \textrm{ or } B \implies \omega_{i} \in A_{i} \textrm{ or } \omega_{i} \in B_{i} \implies \prod\limits_{i \in \mathbb{Z}_{+}} A_{i} \cup B_{i} .\] However, so long as \(A_1 \cup A_2 \cup \cdots \not\subset B_1 \cup B_2 \cup \cdots\) and \(B_1 \cup B_2 \cup \cdots \not\subset A_1 \cup A_2 \cup \cdots\), there can exist some element \(\omega \in \prod\limits_{i \in \mathbb{Z}_{+}} A_{i} \cup B_{i}\) such that some \(\omega_{i} \in A_1 \cup A_2 \cup \cdots \textrm{ and } \omega_{i} \not\in B_1 \cup B_2 \cup \cdots\) for some \(i\), and \(\omega_{k} \not\in A_1 \cup A_2 \cup \cdots \textrm{ and } \omega_{i} \in B_1 \cup B_2 \cup \cdots\) for some \(k\). This element would not be in either \(B\) or \(A\), proving the claim.\\ On the other hand, \(\prod\limits_{i \in \mathbb{Z}_{+}} A_{i} \cap B_{i} = A \cap B\): \[ \omega \in A \cap B \iff \omega \in A \textrm{ and } \omega \in B \iff \omega_i \in A_i \textrm{ and } \omega_{i} \in B_{i} \ \forall i \iff \omega \in \prod\limits_{i \in \mathbb{Z}_{+}} A_{i} \cap B_{i} \]

In short, \(A \cup B \subset \prod\limits_{i \in \mathbb{Z}_{+}} A_{i} \cup B_{i}\): \[ \omega \in A \cup B \implies \omega \in A \textrm{ or } B \implies \omega_{i} \in A_{i} \textrm{ or } \omega_{i} \in B_{i} \implies \prod\limits_{i \in \mathbb{Z}_{+}} A_{i} \cup B_{i} .\] However, so long as \(A_1 \cup A_2 \cup \cdots \not\subset B_1 \cup B_2 \cup \cdots\) and \(B_1 \cup B_2 \cup \cdots \not\subset A_1 \cup A_2 \cup \cdots\), there can exist some element \(\omega \in \prod\limits_{i \in \mathbb{Z}_{+}} A_{i} \cup B_{i}\) such that some \(\omega_{i} \in A_1 \cup A_2 \cup \cdots \textrm{ and } \omega_{i} \not\in B_1 \cup B_2 \cup \cdots\) for some \(i\), and \(\omega_{k} \not\in A_1 \cup A_2 \cup \cdots \textrm{ and } \omega_{i} \in B_1 \cup B_2 \cup \cdots\) for some \(k\). This element would not be in either \(B\) or \(A\), proving the claim.\\ On the other hand, \(\prod\limits_{i \in \mathbb{Z}_{+}} A_{i} \cap B_{i} = A \cap B\): \[ \omega \in A \cap B \iff \omega \in A \textrm{ and } \omega \in B \iff \omega_i \in A_i \textrm{ and } \omega_{i} \in B_{i} \ \forall i \iff \omega \in \prod\limits_{i \in \mathbb{Z}_{+}} A_{i} \cap B_{i} \]