(1). Prove Theorem 19.2.

Box topology. Suppose $\prod_{\alpha \in J} X_\alpha$ has the box topology. Let $\mathcal{C}$ be the collection of all sets of the form $\prod_{\alpha \in J} B_\alpha$ where $B_\alpha \in \mathcal{B}\alpha$ for each $\alpha$. Now suppose $U$ is open in $\prod X\alpha$ and $x \in U$. By definition of the box topology, there is some basis element $B = \prod U_\alpha$ where each $U_\alpha$ is open in $X_\alpha$ and $x \in B \subset U$. Then $x_\alpha \in U_\alpha$ for all $\alpha$. Then for each $\alpha$, there is some $B_\alpha \in \mathcal{B}\alpha$ such that $x\alpha \in B_\alpha \subset U_\alpha$. Then $x \in \prod B_\alpha \subset \prod U_\alpha \subset U$. And clearly we have $\prod B_\alpha \in \mathcal{C}$. So given an open set $U$ and $x \in U$, we found an element $C$ of $\mathcal{C}$ such that $x \in C \subset U$, so $\mathcal{C}$ is a basis for the box topology on $\prod X_\alpha$.

Product topology. Suppose $\prod_{\alpha \in J} X_\alpha$ has the product topology. Let $\mathcal{C}$ be the collection of all sets of the form $\prod_{\alpha \in J} B_\alpha$ where $B_\alpha \in \mathcal{B}\alpha$ for finitely many indices $\alpha$, and $B\alpha = X_\alpha$ otherwise. Suppose $U$ is open and $x \in U$. By definition of the product topology, there is some $B = \prod U_\alpha$ such that $x \in B \subset U$ where each $U_\alpha$ is open in $X_\alpha$ and $U_\alpha = X_\alpha$ for all but finitely many $\alpha$. Let $\alpha_1, \ldots, \alpha_n$ be the indices for which $U_\alpha \neq X_\alpha$. For each $U_{\alpha_i}$, we can take a basis element $B_{\alpha_i}$ for the topology of $X_{\alpha_i}$ so that $x_{\alpha_i} \in B_{\alpha_i} \subset U_{\alpha_i}$. Then we construct the set $\prod_{\alpha \in J} B_\alpha$ where for $\alpha \in \{ \alpha_1, \ldots, \alpha_n \}$ we have $B_\alpha = B_{\alpha_i}$ and for all other $\alpha \in J$, we have $B_\alpha = X_\alpha$. By construction, $x \in \prod B_\alpha \subset \prod U_\alpha \subset U$ and clearly $\prod B_\alpha \in \mathcal{C}$. So $\mathcal{C}$ is a basis for the product topology on $\prod X_\alpha$.

(6). Let $\textbf{x}1, \textbf{x}2, \ldots$ be a sequence of points of the product space $\prod X\alpha$. Show that this sequence converges to the point $\textbf{x}$ if and only if the sequence $\pi\alpha (\textbf{x}1), \pi\alpha(\textbf{x}2), \ldots$ converges to $\pi\alpha(\textbf{x})$ for each $\alpha$.

First direction. First, suppose $\textbf{x}1, \textbf{x}2, \ldots$ converges to the sequence $\textbf{x}$ in $\prod X\alpha$ under the product topology. Let $\beta$ be an arbitrary index. We show that the sequence $\pi{\beta} (\textbf{x}1), \pi{\beta} (\textbf{x}2), \ldots$ converges to the point $\pi{\beta} (\textbf{x})$ in $X_\beta$.

Suppose $U$ is a neighborhood of $\pi_{\beta} (\textbf{x})$ in $X_{\beta}$. Setting $V = \pi_{\beta}^{-1} (U)$, we have that $V$ is an open neighborhood of $\textbf{x}$ in the product topology on $\prod X_{\alpha}$ (since it's a subbasis element). Then by convergence of the sequence $\textbf{x}1, \textbf{x}2, \ldots$, there is some $N > 0$ such that for all $n \geq N$, we have $\textbf{x}n \in V$. Then applying $\pi{\beta}$, we have that for $n \geq N$, $\pi{\beta}(\textbf{x}n) \in U$. Thus the sequence $\pi\beta (\textbf{x}1), \pi\beta (\textbf{x}2), \ldots$ converges to $\pi\beta (\textbf{x})$ in $X\beta$.

Second direction. Now we prove the converse. Suppose that there is some $\textbf{x}$ in $\prod X_\alpha$ such that for every $\alpha$, the sequence $\pi_\alpha (\textbf{x}1), \pi\alpha(\textbf{x}2), \ldots$ converges to $\pi\alpha (\textbf{x})$ in $X_\alpha$. We show that the sequence $\textbf{x}_1, \textbf{x}_2, \ldots$ converges to $\textbf{x}$.

Suppose $U$ is a neighborhood of $\textbf{x}$ in $\prod X_\alpha$ under the product topology. Then there is a basis element $B$ such that $\textbf{x} \in B \subset U$. Note that by the definition of the product topology, $B$ can be expressed as:

$B = \prod U_\alpha$ where $U_\alpha$ is open in $X_\alpha$, and $U_\alpha = X_\alpha$ for all but finitely many indices $\alpha$.

Let $\beta_1, \ldots, \beta_m$ be the indices for which $U_\alpha \neq X_\alpha$. By convergence of each $\pi_\alpha$ sequence, for each $\beta_i$ there is some $N_i$ such that for $n \geq N_i$, we have $\pi_{\beta_i} (\textbf{x}n) \in U{\beta_i}$. Also note that for all $\alpha \notin \{\beta_1 , \ldots, \beta_m\}$, we trivially have that for all $n > 0$, $\pi_\alpha (\textbf{x}n) \in U\alpha$ since $U_\alpha = X_\alpha$. Then we set $N = \text{max}(\{N_1, \ldots, N_m \})$. Then for $n \geq N$, we have $\pi_\alpha (\textbf{x}n) \in U\alpha$ for all $\alpha$. This implies that $\textbf{x}n \in \prod U\alpha \subset U$.

Thus given a neighborhood $U$ of $\textbf{x}$, there is some $N$ such that for $n \geq N$, we have $\textbf{x}_n \in U$, so the sequence converges to $\textbf{x}$.

(7). Let $\mathbb{R}^\infty$ be the subset of $\mathbb{R}^\omega$ consisting of all sequences that are "eventually zero", that is, all sequences $(x_1, x_2, \ldots)$ such that $x_i \neq 0$ for only finitely many values of $i$. What is the closure of $\mathbb{R}^\infty$ in $\mathbb{R}^\omega$ under the box and product topologies?

Under the box topology, $\overline{\mathbb{R}^\infty} = \mathbb{R}^\infty$.

To prove this, we show that $\overline{\mathbb{R}^\infty} \subset \mathbb{R}^\infty$ since we already know that $\mathbb{R}^\infty \subset \overline{\mathbb{R}^\infty}$.

Suppose $x \notin \mathbb{R}^\infty$. We will show that $x \notin \overline{\mathbb{R}^\infty}$. Note $x$ is not eventually zero, i.e. $x_i \neq 0$ for infinitely many values of $i$.

Now note that for each $x_i \neq 0$, there is some neighborhood $U_i \subset \mathbb{R}$ of $x_i$ such that $U_i$ is disjoint from $\{ 0 \}$, e.g. if $x_i < 0$ set $U_i = (-\infty, \frac{x_i}{2})$ and otherwise set $U_i = (\frac{x_i}{2}, +\infty)$. Then define $V_i$ for each positive integer $i$ by

$V_i = U_i$ if $x_i \neq 0$ and $V_i = \mathbb{R}$ otherwise.

The product $\prod V_i$ is open in the box topology. But note there is no $y \in \prod V_i$ that is eventually zero, because infinitely many $V_i$ are disjoint from $\{ 0 \}$, thus infinitely many $y_i$ must be nonzero. So we have an open neighborhood of $x$ that is disjoint from $\mathbb{R}^\infty$, meaning $x \notin \overline{\mathbb{R}^\infty}$.

Under the product topology, $\overline{\mathbb{R}^\infty} = \mathbb{R}^\omega$.

For suppose $x \in \mathbb{R}^\omega$, and suppose $U$ is a neighborhood of $x$. There is a basis element $B$ such that $x \in B \subset U$ where $B = \prod U_i$, for $U_i$ open in $\mathbb{R}$, where $U_i = \mathbb{R}$ for all but finitely many values of $i$. Then let $y \in \mathbb{R}^\omega$ be the sequence where $y_i = x_i$ for all $i$ where $U_i \neq \mathbb{R}$, and let $y_i = 0$ otherwise. Then $y \in \prod U_i$, and $y$ is eventually zero. So every neighborhood of $x$ intersects with $\mathbb{R}^\infty$, meaning it's in the closure of that set.

(9). Show that the choice axiom is equivalent to the statement that for any indexed family $\{ A_\alpha \}_{\alpha \in J}$ of nonempty sets, with $J \neq 0$, the cartesian product

$\prod_{\alpha \in J} A_\alpha$

is nonempty.

Assume the choice axiom (as given on page 59). Suppose that $\{A_\alpha\}{\alpha \in J}$ is an indexed family of nonempty sets with $J \neq 0$. For each $\alpha \in J$, we define the set $B\alpha$ as the set of ordered pairs

$B_\alpha = \{ (x, \alpha) \mid x \in A_\alpha \}$

Then for each $\alpha$, since $A_\alpha$ is nonempty, $B_\alpha$ is nonempty. Also, for $\alpha_1 \neq \alpha_2$, clearly $B_{\alpha_1}$ and $B_{\alpha_2}$ are disjoint.

Then setting $\mathcal{A} = \{{B_\alpha} \}{\alpha \in J}$, we have that $\mathcal{A}$ is a collection of disjoint nonempty sets. Then there is a set $C$ such that $C \subset \bigcup{\alpha \in J} B_\alpha$, and for each $B_\alpha \in \mathcal{A}$, the set $C \cap B_\alpha$ contains a single element. Then we construct a function $f: J \rightarrow \bigcup_{\alpha \in J} A_\alpha$ where for each $\alpha$, given the single element $(u, \alpha) \in C \cap B_\alpha$, we set $f(\alpha) = u$. Then each $f(\alpha) \in A_\alpha$. But note that the cartesian product

$\prod_{\alpha \in J} A_\alpha$

is just the set of all maps $g$ from $J$ into $\bigcup_{\alpha \in J} A_\alpha$, where $g(\alpha) \in A_\alpha$. The function $f$ we have constructed is one such map, thus the cartesian product is nonempty.

Converse. Suppose that the statement given in the question is true. Let $\mathcal{A}$ be a collection of disjoint nonempty sets. Let $J$ be an index set for $\mathcal{A}$ (a trivial option is just to use the set $\mathcal{A}$ itself). If $\mathcal{A}$ is empty, the empty set qualifies as the desired set $C$, so we suppose $\mathcal{A}$ is nonempty. Then the collection $\mathcal{A}$ is really an indexed family $\{ A_\alpha \}{\alpha \in J}$ of nonempty sets with $J \neq 0$, so the cartesian product $\prod{\alpha \in J} A_\alpha$ is nonempty. Take $(x_\alpha){\alpha \in J}$ in this cartesian product. Just let $C = \{ x\alpha \mid \alpha \in J\}$ and $C$ has the desired properties. For $C$ is contained in the union of the elements of $\mathcal{A}$, and for each $A_\alpha \in \mathcal{A}$, since each of the $A_\alpha$ are disjoint, $C \cap A_\alpha$ contains a single element.