§9 #3.

First, we show that $A$ is infinite. Suppose $A$ is finite. Then there is some $n \in \mathbb{Z}+$ and a bijection $g: \{ 1, \ldots, n \} \rightarrow A$. We know that there is an injective function $f{n+1}: \{ 1, \ldots, n+1\} \rightarrow A$. Then we can define the injective function

$f': \{1, \ldots, n+1 \} \rightarrow \{1, \ldots, n \} = g^{-1} \circ f_{n+1}$

which is a contradiction.

We can define an injective function $F: \mathbb{Z}_+ \rightarrow A$ without using the choice axiom. We define $F$ as follows.

$F(1) = f_1(1)$

Given $F(1), \ldots, F(n)$, define $F(n+1)$ as $f_{n+1} (i)$, where $i$ is the smallest element of $\{1, \ldots, n +1 \}$ such that $f_{n+1}(i) \notin \{F(1), F(2), \ldots F(n) \}$.

Note that such an element $i$ exists because the set $\{F(1), F(2), \ldots F(n) \} \subset A$ has cardinality $n$ and the image set $f_{n+1} (\{1, \ldots, n+1\})$ has cardinality $n+1$ (since all $f_i$ are injective).

Thus $F$ exists and $F$ is injective by construction.

§10 #5. Show that the well-ordering theorem implies the choice axiom.

Let $\mathcal{A}$ be a collection of disjoint nonempty sets. By the well-ordering theorem, for every $A \in \mathcal{A}$, there is an order relation on $A$ that is well-ordering. Then every $A$ (when considered as a nonempty subset of itself) has a smallest element. Then there exists a set $C$ consisting of the smallest element of each $A$, such that $C$ is a subset of $\bigcup_{A \in \mathcal{A}} A$ and for each $A$, the set $C \cap A$ contains one element.

§16 #3. Just to check

$A$ open in $\mathbb{R}$, open in $Y$

$B$ not open in $\mathbb{R}$, open in $Y$. since $B = U \cap Y$ for $U = (-2, -\frac{1}{2}) \cup (\frac{1}{2}, 2)$

$C$ not open in $\mathbb{R}$, not open in $Y$

$D$ not open in $\mathbb{R}$, not open in $Y$

$E$ open in $\mathbb{R}$ (since it doesn't contain $0$), open in $Y$ (since $E \cap Y = E$)

§16 #4. Prove that $\pi_1 : X \times Y \rightarrow X$ is an open map.

Suppose $W \subset X \times Y$ is open. Note that the basis for the product topology of $X \times Y$ is the set of all $U \times V$ where $U$ is open in $X$ and $V$ is open in $Y$ (definition). Then since an open set in a topology is just a union of basis elements for that topology (lemma 13.1), we have

$W = \bigcup_{\alpha \in J} \Big( U_\alpha \times V_\alpha \Big)$ for $U_\alpha$ open in $X$ and $V_\alpha$ open in $Y$.

Then $\pi_1 (W) = \bigcup_{\alpha \in J} U_\alpha$. And since each $U_\alpha$ is open in $X$, the union is open in $X$. Thus $\pi_1(W)$ is open in $X$, so $\pi_1$ is an open map.

§16 #8. Describe the topology $L$ inherits as a subspace of $\mathbb{R}_\mathcal{l} \times \mathbb{R}$ and as a subspace of $\mathbb{R}_l \times \mathbb{R}_l$.

A little unsure about this, but is the topology on $L$ as a subspace of $\mathbb{R}_\mathcal{l} \times \mathbb{R}$ just the topology derived from the basis of open or half-open intervals? And for $\mathbb{R}_l \times \mathbb{R}_l$ from the basis of half-open, open, or closed intervals? Idk, having trouble reasoning about this. Also feel a little woozy about defining intervals on an arbitrary line.

My thinking is that $\mathbb{R}_l \times \mathbb{R}$ has the basis: interiors of rectangles plus the left side of the rectangle (no corners); and $\mathbb{R}_l \times \mathbb{R}_l$ has the basis: interiors of rectangles plus the left and bottom side of the rectangle, including the bottom left corner.

§16 #9. Show that the dictionary order topology on the set $\mathbb{R} \times \mathbb{R}$ is the same as the product topology $\mathbb{R}_d \times \mathbb{R}$, where $\mathbb{R}_d$ denotes $\mathbb{R}$ in the discrete topology.

We use lemma 13.3 to show that the topologies are equal.

A basis element for $\mathbb{R} \times \mathbb{R}$ is an interval of the form $B = \big( \langle a, b \rangle, \langle c, d \rangle \big)$ where $\langle a, b \rangle < \langle c, d \rangle$. Because we're taking the dictionary order, we have that either $a < c$, or $a = c$ and $b < d$. So the set $B$ looks like either

$B = \{ a \} \times (b, d)$ or

$B = \Big( \{a \} \times (b, +\infty ) \Big) \cup \Big( (a, c) \times \mathbb{R} \Big) \cup \Big( \{c\} \times (-\infty, d) \Big)$

Also, a basis element for $\mathbb{R}_d \times \mathbb{R}$ is of the form

$B = \{ x \} \times (a, b)$

for $x \in \mathbb{R}$ and $a < b \in \mathbb{R}$.

Then, applying lemma 13.3, we have that given $\langle x, y\rangle$ and a basis element $B$ of $\mathbb{R}^2$ where $\langle x, y \rangle \in B$, we can find some $c < y < d$ such that $\langle x, y \rangle \in \{ x \} \times (c, d) = B' \subset B$. So $x$ is contained in some basis element $B'$ of $\mathbb{R}_d \times \mathbb{R}$ such that $B' \subset B$. Thus the product topology is finer than the order topology.

Conversely, given $\langle x, y\rangle$ and a basis element $B$ of $\mathbb{R}_d \times \mathbb{R}$ where $\langle x, y \rangle \in B$, note that $B$ itself is a basis element of the order topology on $\mathbb{R}^2$. Thus the order topology is finer than the product topology.

Thus the two topologies are the same.

Also, I think that this topology is strictly finer than the standard topology on $\mathbb{R}^2$, by a similar argument.