Munkres - Cartesian Products

Adar Kahiri
June 2020

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.

Problem 3: Let \(A = A_1 \times A_2 \times \cdots\) and \(B = B_1 \times B_2 \times \cdots\)
(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, \(\prod\limits_{i \in \mathbb{Z}_{+}} A_{i} \cap B_{i} = A \cap B\): \[ \omega \in A \cap B \iff \omega \in A \textrm{ and } \omega \in B \iff \omega_i \in A_i \textrm{ and } \omega_{i} \in B_{i} \ \forall i \iff \omega \in \prod\limits_{i \in \mathbb{Z}_{+}} A_{i} \cap B_{i} \]