Exercise 20.2

Proof. In what follows let

be the standard bounded metric on $ℝ$, noting that this is a metric by Theorem 20.1. Now define the function $d:{ℝ}^2\times {ℝ}^2\to ℝ$ by

We claim that this is a metric on ${ℝ}^2$ that induces the dictionary order topology.

First we show that $d$ is a metric on ${ℝ}^2$. Clearly $d(x,y)\ge 0$ since both $1\ge 0$ and $\overline{d}(x_2,y_2)\ge 0$ since $\overline{d}$ is a metric. Moreover if $x=y$ then $x_1=y_1$ and $x_2=y_2$ so that $d(x,y)=\overline{d}(x_2,y_2)=0$. Conversely if $d(x,y)=0$ then clearly $d(x,y)\ne 1$ so that it must be that $x_1=y_1$ and $d(x,y)=\overline{d}(x_2,y_2)=0$ so that $x_2=y_2$ since $\overline{d}$ is a metric. From this it follows that $x=y$ since $x_1=y_1$ and $x_2=y_2$, which shows property (1) of a metric.

It is also obvious that $d(x,y)=d(y,x)$ since if $x_1\ne y_1$ then $d(x,y)=1=d(y,x)$. If $x_1=x_2$ then $d(x,y)=\overline{d}(x_2,y_2)=\overline{d}(y_2,x_2)=d(y,x)$ since $\overline{d}$ is a metric. This shows property (2) of a metric. Lastly, consider $x$, $y$, and $z$ in ${ℝ}^2$.

Case: $x_1\ne z_1$. Then $d(x,z)=1$ and it must be that either $y_1\ne x_1$ or $y_1\ne z_1$ since otherwise we would have that $x_1=y_1=z_1$. Thus either $d(x,y)=1$ or $d(y,z)=1$ and hence

since both $d(x,y)\ge 0$ and $d(y,z)\ge 0$.

Case: $x_1=z_1$. Then $d(x,z)=\overline{d}(x_2,z_2)$. If $y_1=x_1$ then $x_1=y_1=z_1$ so that

since $\overline{d}$ is a metric. If $y_1\ne x_1$ then $y_1\ne x_1=z_1$, and hence $d(x,y)=d(y,z)=1$ so that

since $\overline{d}$ is the bounded metric so that it is always at most 1.

Thus in all cases we have shown property (3) of a metric.

In what follows let $\prec$ denote the dictionary order on ${ℝ}^2$. To show that $d$ induces the dictionary order topology, first consider any point $x\in {ℝ}^2$ and any basis element $B$ of the dictionary order topology that contains $x$. Then of course $B=(a,b)$, where $a\prec x\prec b$ since the dictionary order has no largest or smallest elements in ${ℝ}^2$. Now define

and

and let $\delta =\min <!--\; FUNCTION\; APPLICATION\; -->\left\{1,{\delta }_a,{\delta }_b\right\}$. Clearly the set $B_d(x,\delta )$ is a basis element of the topology induced by $d$, and we claim that $x\in B_d(x,\delta )\subset B$.

That $x\in B_d(x,\delta )$ is obvious. So now consider any $y\in B_d(x,\delta )$ so that $d(x,y)<\delta \le 1$. Hence it cannot be that $x_1\ne y_1$ by definition, since $d(x,y)=1$ in that case, and so $x_1=y_1$. If $x_2=a_2$ then it has to be that $a_1<x_1$ since otherwise it would not be the case that $a\prec x$. Thus we have $a_1<x_1=y_1$ so that $a\prec y$.

On the other hand if $x_2\ne a_2$ then it must be that $a_1\le y_1$ since otherwise we would have $x_1=y_1<a_1$ so that $x\prec a$. If $a_1<y_1=x_1$ then of course $a\prec y$ so assume that $a_1=y_1=x_1$. The it must be that $a_2<x_2$ since $a\prec x$, and so $\left|x_2-a_2\right|=x_2-a_2$. Then, since $x_1=y_1$, we have that $\overline{d}(x_2,y_2)=d(x,y)<\delta \le 1$ so it must be that $d(x,y)=\overline{d}(x_2,y_2)=\left|x_2-y_2\right|$. Also ${\delta }_a=\left|x_2-a_2\right|=x_2-a_2$ since $x_2\ne a_2$. Hence we have $\left|x_2-y_2\right|=d(x,y)<\delta \le {\delta }_a=x_2-a_2$, from which it readily follows that $a_2<y_2$ so that again $a\prec y$.

Therefore in all cases $a\prec y$. Analogous arguments show that $y\prec b$ so that $y\in (a,b)=B$, which shows that $B_d(x,\delta )\subset B$ as desired since $y$ was arbitrary. This shows that the topology induced by $d$ is finer than the dictionary order topology by Lemma 13.3.

Now again suppose that $x\in {ℝ}^2$, and that ${𝜖}^{\prime }>0$ and $x^{\prime }\in {ℝ}^2$ such that $B_d(x^{\prime },{𝜖}^{\prime })$ is an arbitrary basis element of the metric topology induced by $d$ that contains $x$. It was shown after the definition of a metric topology in the text that there is another ball $B_d(x,𝜖)$ centered at $x$ such that $B_d(x,𝜖)\subset B_d(x^{\prime },{𝜖}^{\prime })$. Let $\delta =\min <!--\; FUNCTION\; APPLICATION\; -->\left\{1,𝜖\right\}$ and define $a=x_1\times (x_2-\delta )$ and $b=x_1\times (x_2+\delta )$. Set $B=(a,b)$, which is clearly a basis element of the dictionary order topology. So consider any $y\in B$ so that $a\prec y\prec b$. Clearly it must be that $a_1=y_1=b_1=x_1$ since otherwise we would have that $y\prec a$ or $b\prec y$. From this it follows that $a_2=x_2-\delta <y_2<x_2+\delta =b_2$ so that $-\delta <y_2-x_2<\delta$ and hence $\left|x_2-y_2\right|<\delta$. Moreover, since $y_1=x_1$ and $\delta \le 1$, it follows that $d(x,y)=\overline{d}(x_2,y_2)=\left|x_2-y_2\right|<\delta \le 𝜖$. This shows that $y\in B_d(x,𝜖)$, which shows that $B\subset B_d(x,𝜖)\subset B_d(x^{\prime },{𝜖}^{\prime })$ since $y$ was arbitrary. This proves that the dictionary order topology is finer than the topology induced by $d$ again by Lemma 13.3.

Since each is finer than the other the topologies must be the same, which shows that the dictionary order topology is metrizable as desired. □