J. Munkres. Topology (2013) Chapter I: Set Theory and Logic. 5: Cartesian Products

Definition: Family

Let \(A\) be a set. A family of elements of \(A\) is a function \(a\) with codomain \(A\). We call its domain \(I\) the index set, and we write \(a_i\) for the value of \(a\) at \(i \in I\). We write the family as \(\{a_i\}_{i \in I}\), and call it a family indexed by \(I\).

A sequence is a family indexed by \(\mathbb{Z}_+\).

Definition: Euclidean \(m\)-space

For \(m \in \mathbb{Z}_+\), we call \(\mathbb{R}^m\) Euclidean \(m\)-space.

Exercises

Exercise 1

Show that there is a bijective correspondence of \(A \times B\) with \(B \times A\).

Solution

The function mapping \((a,b)\) to \((b,a)\) is such a correspondence.

Exercise 2

(a)

Show that if \(n > 1\) there is a bijective correspondence of

\[ A_1 \times \cdots \times A_n \text{ with } (A_1 \times \cdots \times A_{n-1}) \times A_n \enspace . \]

Solution

The function mapping \((a_1, \ldots, a_n)\) to \(((a_1, \ldots, a_{n-1}), a_n)\) is such a correspondence.

(b)

Given the indexed family \( \{A_1, A_2, \ldots\} \), let \(B_i = A_{2i-1} \times A_{2i}\) for each positive integer \(i\). Show that there is a bijective correspondence of \(A_1 \times A_2 \times \cdots\) with \(B_1 \times B_2 \times \cdots\).

Solution

The function that maps \(\{a_n\}_{n \in \mathbb{Z}_+}\) to \(\{(a_{2_{i-1}},a_{2i})\}_{i \in \mathbb{Z}_+}\) is such a correspondence.

Exercise 3

Let \(A = A_1 \times A_2 \times \cdots\) and \(B = B_1 \times B_2 \times \cdots\).

(a)

Show that if \(B_i \subseteq A_i\) for all \(i\), then \(B \subseteq A\).

Solution

If \(\{b_i\}_{i \in \mathbb{Z}_+}\) is a family in \(B\) then we have \(b_i \in A_i\) for all \(i\), and so \(\{b_i\}_{i \in \mathbb{Z}_+} \in A\).

(b)

Show that the converse of (a) holds if \(B\) is nonempty.

Solution

  • Assume \(B\) is nonempty.
  • Pick \(\{b_i\}_{i \in \mathbb{Z}_+} \in B\).
  • Assume \(B \subseteq A\).
  • Let \(i \in \mathbb{Z}_+\).
  • Let \(x \in B_i\). Prove: \(x \in A_i\).
  • Let \(\{b'_j\}_{j \in I}\) be the family with \(b'_i = x\) and \(b'_j = b_j\) if \(j \neq i\).
  • \(\{b'_j\}_{j \in I} \in B\)
  • \(\{b'_j\}_{j \in I} \in A\)
  • \(b'_i = x \in A_i\)
  • \(\Box\)

(c)

Show that if \(A\) is nonempty, each \(A_i\) is nonempty. Does the converse hold?

Solution

If \(\{a_i\}_{i \in \mathbb{Z}_+} \in A\) then \(a_i \in A_i\) for all \(i\), and so \(A_i \neq \emptyset\). The converse requires the Axiom of Choice.

(d)

What is the relation between \(A \cup B\) 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

We have \(A \cup B \subseteq (A_1 \cup B_1) \times (A_2 \cup B_2) \times \cdots\). Equality does not hold in general: take \(A_i = \{0\}\) for all \(i\) and \(B_i = \{1\}\) for all \(i\). Then \((0,1,0,1,\ldots) \in (A_1 \cup B_1) \times (A_2 \cup B_2) \times \cdots\) but is not in \(A \cup B\).

We have \(A \cap B = (A_1 \cap B_1) \times (A_2 \cap B_2) \times \cdots\).

Exercise 4

Let \(m,n \in \mathbb{Z}_+\). Let \(X \neq \emptyset\).

(a)

If \(m \leq n\), find an injective map \(f : X^m \rightarrow X^n\).

Solution

Pick \(a \in X\). Define \(f\) by \(f(x_1, \ldots, x_m) = (x_1, \ldots, x_m, a, a, \ldots, a)\).

(b)

Find a bijective map \(g : X^m \times X^n \rightarrow X^{m+n}\).

Solution

Define \(g((x_1, \ldots, x_m), (y_1, \ldots, y_n)) = (x_1, \ldots, x_m, y_1, \ldots, y_n)\).

(c)

Find an injective map \(h : X^n \rightarrow X^\omega\).

Solution

Pick \(a \in X\). Define \(h\) by \(h(x_1, \ldots, x_n) = (x_1, \ldots, x_n, a, a, \ldots)\).

(d)

Find a bijective map \(k : X^n \times X^\omega \rightarrow X^\omega\).

Solution

Define \(k((x_1, \ldots, x_n),(y_1, y_2, \ldots)) = (x_1, \ldots, x_n, y_1, y_2, \ldots)\).

(e)

Find a bijective map \(l : X^\omega \times X^\omega \rightarrow X^\omega\).

Solution

Define \(l((x_1, x_2, \ldots), (y_1, y_2, \ldots)) = (x_1, y_1, x_2, y_2, \ldots)\).

(f)

If \(A \subseteq B\), find an injective map \(m : X^A \rightarrow X^B\).

Solution

Pick \(x_0 \in X\). Define \(m : X^A \rightarrow X^B\) by \(m(\{x_a\}_{a \in A}) = \{y_b\}_{b \in B}\), where \(y_b = x_b\) if \(b \in A\) and \(x_0\) otherwise.

Exercise 5

Which of the following subsets of \(\mathbb{R}^\omega\) can be expressed as the Cartesian product of subsets of \(\mathbb{R}\)?

(a)

\[ \{ \vec{x} \mid x_i \text{ is an integer for all } i \} \]

Solution

This set is \(\mathbb{Z}^\omega\).

(b)

\[ \{ \vec{x} \mid x_i \geq i \text{ for all } i \} \]

Solution

This set is \( [1, + \infty) \times [2, + \infty) \times \cdots \).

(c)

\[ \{ \vec{x} \mid x_i \text{ is an integer for all } i \geq 100 \} \]

Solution

This set is \( \mathbb{R}^{99} \times \mathbb{Z}^\omega\).

(d)

\[ \{ \vec{x} \mid x_2 = x_3 \} \]

Solution

This set cannot be expressed as a Cartesian product.

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