# Adar Kahiri

## Cartesian Product

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. $$\tag*{\blacksquare}$$

Problem 3: Let $$A = A_1 \times A_2 \times \cdots$$ and $$B = B_1 \times B_2 \times \cdots$$

• (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 $$\prod_{i \in \mathbb{Z}_+} A_i \cup B_i$$? What is the relation between the set $$A \cap B$$ and the cartesian product $$\prod_{i \in \mathbb{Z}_{+}} A_{i} \cap B_{i}$$.

(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, $$\prod\limits_{i \in \mathbb{Z}_{+}} A_{i} \cap B_{i} = A \cap B$$: $$\omega \in A \cap B \iff \omega \in A \textrm{ and } \omega \in B \iff \omega_i \in A_i \textrm{ and } \omega_{i} \in B_{i} \ \forall i \iff \omega \in \prod\limits_{i \in \mathbb{Z}_{+}} A_{i} \cap B_{i}$$

$\tag*{\blacksquare}$