Exercise 1.1.2 (Real number line is metric space)

We verify the properties from Definition 1.1.2 of metric spaces. To do so, we will make use of Definition 5.4.5 of absolute value and the related Proposition 4.3.3, both from Analysis I.

1.

For any $x\in R$ we have $$d(x,x)=|x-x|=\left|0\right|=0$$

by the definition of absolute value.

2.

(Positivity) Let $x,y\in R$ such that $x\ne y$. By trichotomy of order, we either have $x-y>0$ or $x-y<0$. In the former case, $|x-y|=x-y>0$ by hypothesis. In the latter case, $|x-y|=-(x-y)$ and the latter is positive by Proposition 5.4.4.

3.

(Symmetry) Let $x,y\in R$. By Proposition 5.3.11 (Laws of algebra), we have $$d(x,y)=|x-y|=\{\begin{array}{cc}x-y,\hfill & x-y>0\hfill \\ 0,\hfill & x-y=0\hfill \\ -(x-y),\hfill & x-y<0\hfill \end{array}=\{\begin{array}{cc}-(y-x),\hfill & y-x<0\hfill \\ 0,\hfill & y-x=0\hfill \\ y-x,\hfill & y-x>0\hfill \end{array}=|y-x|=d(y,x).$$

4.

(Triangle inequality) Let $x,y,z\in R$. Triangle inequality for the metric $d$ can be easily deduced from the triangle inequality for the absolute value: $$d(x,z)=|x-z|=\left|(x-y)+(y-z)\right|\le \left|(x-y)\right|+\left|(y-z)\right|=d(x,y)+d(y,z).$$

(Triangle inequality for absolute value can be easily verified case-by-case:

$$|a+b|=\{\begin{array}{cc}a+b,\hfill & a+b\ge 0\hfill \\ -a+(-b),\hfill & a+b<0\hfill \end{array}\le \{\begin{array}{cc}\left|a\right|+\left|b\right|,\hfill & a+b\ge 0\hfill \\ \left|a\right|+\left|b\right|,\hfill & a+b<0\hfill \end{array}=\left|a\right|+\left|b\right|$$

since $-x,x\le \left|x\right|$ for any $x\in R$.)