Euclidean Metric on $\mathbb{R}$ - Section 1.2 Micheal Searcoid

Question

Self studying the book Metric Spaces by Searcoid. Not able to understand the following sentence :

A metric on a set $X$ is defined on $X\times X$, not on $X$. Nonetheless, some metrics, such as the Euclidean metric on $\mathbb{R}$, are determined by a function that is defined on $X$ itself.

What does it mean that the Euclidean metric is not defined on $X\times X$ but on $X\text{ ?}$ Don't we require $2$ points in $\mathbb R$ to define the metric?

On an earlier page he indicated that the Euclidean space is defined on $\mathbb{R}\times \mathbb{R}$ :

The most familiar example is that determined by the absolute-value function on $\mathbb{R}$; it is the function $(a,b)\mapsto |a-b|$ defined on $\mathbb{R}\times \mathbb{R}$. This metric is called the Euclidean metric on $ \mathbb{R}$.

Answer 1

Consider a function $\varphi\colon X\to\mathbb{R}$ and define $$ d_\varphi(x,y)=|\varphi(x)-\varphi(y)| $$ Such a function $d_\varphi$ is a metric on $X$ so long as $\varphi$ is injective. Indeed $d_\varphi(x,y)\ge0$ is obvious; if $d_\varphi(x,y)=0$, then $\varphi(x)=\varphi(y)$ and, by injectivity we conclude $x=y$.

The property $d_\varphi(y,x)=d_\varphi(x,y)$ is obvious. The triangular inequality follows from $$ d_\varphi(x,z)=|\varphi(x)-\varphi(y)+\varphi(y)-\varphi(z)| \le|\varphi(x)-\varphi(y)|+|\varphi(y)-\varphi(z)|= d_\varphi(x,y)+d_\varphi(y,z) $$

Not all metrics can be defined this way. For instance the discrete metric can't.

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However, the author is possibly referring to normed spaces: if $V$ is a normed space, we can deduce a metric by setting $d(x,y)=\|x-y\|$.

The structure of normed space can also be weakened: what's actually needed is a group (usually written additively) $G$ together with a function $\nu\colon G\to\mathbb{R}$ satisfying

1. $\nu(x)\ge0$, for every $x\in G$;
2. $\nu(x)=0$ if and only if $x=0$;
3. $\nu(x+y)\le\nu(x)+\nu(y)$.

Then $(x,y)\mapsto\nu(x-y)$ defines a metric on $G$.