Stuck on: 3.(b)

(1). (a) In $\mathbb{R}^n$, define

$$ d'(\textbf{x}, \textbf{y}) = \vert x_1 - y_1\vert + \cdots + \vert x_n - y_n \vert $$

Show that $d'$ is a metric that induces the usual topology of $\mathbb{R}^n$. Sketch the basis elements under $d'$ when $n = 2$.

Clearly, $d'(\textbf{x}, \textbf{y}) \geq 0$ for all $\textbf{x}, \textbf{y} \in \mathbb{R}^n$ and $d'(\textbf{x}, \textbf{y}) = 0$ iff $\textbf{x} = \textbf{y}$. Also, clearly $d'(\textbf{x}, \textbf{y}) = d'(\textbf{y}, \textbf{x})$ for all $\textbf{x}, \textbf{y}$.

Now we prove the triangle inequality. Note that

$$ \begin{array} {rl} d'(\textbf{x}, \textbf{y}) + d'(\textbf{y}, \textbf{z}) &= \vert x_1 - y_1\vert + \cdots + \vert x_n - y _n \vert + \vert y_1 - z_1 \vert + \cdots + \vert y_n - z_n \vert \\ &= \vert x_1 - y_1\vert + \vert y_1 - z_1 \vert + \cdots + \vert x_n - y _n \vert + \vert y_n - z_n \vert \\ & \geq \vert x_1 - z_1 \vert + \cdots + \vert x_n - z_n \vert = d'(\textbf{x}, \textbf{z}) \end{array} $$

Hence $d'(\textbf{x}, \textbf{y}) + d'(\textbf{y}, \textbf{z}) \geq d'(\textbf{x}, \textbf{z})$ as required.

(From here on we use non-boldface $x$ and $y$ to denote points of $\mathbb{R}^n$.)

Now comparing $d'$ to the square metric $\rho$, we have, by simple algebra:

$$ \rho(x, y) \leq d'(x, y) \leq n \rho(x, y) $$

So given $B_\rho (x, \epsilon)$ for some $\epsilon > 0$, note that $B_{d'}(x, \epsilon) \subset B_\rho (x, \epsilon)$. Because if $y \in B_{d'} (x, \epsilon)$ then $\rho(x, y) \leq d'(x, y) < \epsilon$ thus $y \in B_\rho (x, \epsilon)$. Thus the topology under $d'$ is finer than the topology under $\rho$.

Now given $B_{d'} (x, \epsilon)$ for some $\epsilon > 0$, set $\delta = \epsilon/n$. Then if $y \in B_\rho (x, \epsilon/n)$, we have

$$ \rho(x, y) < \epsilon / n \implies n\rho(x, y) < \epsilon \implies d'(x,y) < \epsilon $$

So $y \in B_{d'} (x, \delta)$ as required. So the topology under $\rho$ is finer than that of $d'$. Hence the topologies are equal.

The basis elements for $d'$ in $\mathbb{R}^2$ are diamonds:

(2). Show that $\mathbb{R} \times \mathbb{R}$ in the dictionary order topology is metrizable.

We show that $\mathbb{R}^2$ is metrizable in the dictionary order topology with the metric:

$$ d\left((x_1, x_2), (y_1, y_2)\right) = \begin{cases} \min \{1, \vert x_2 - y_2 \vert\} & \text{if } x_1 = y_1 \\ 1 & \text{otherwise} \end{cases} $$