Show that $\rho(x,y) = |\sin(x)-\sin(y)|$ is not a metric on $\mathbb{R}$
and in what condition must be imposed on a function $f:\mathbb{R}\to\mathbb{R}$ in order for $\rho(x,y)=|f(x)-f(y)|$ to be a metric on $\mathbb{R}$
My work:
(1) positive definiteness
if $x=y$, $\sin(x)=\sin(y)$
$\rho(x,y)=\rho(x,x)=|\sin(x)-\sin(y)|=0$
(2) symmetry
sine function is bounded [-1,1], let $\sin(x)=a , \sin(y)=b$
$|\sin(x)-\sin(y)|=|a-b|$,
$|\sin(y)-\sin(x)|=|b-a|$
Both $b$ and $a$ are real numbers, so $|a-b|=|b-a|$
(3) Triangle inequality
let $\sin(z) = c$
$\rho(x,y) = |\sin(x)-\sin(y)|=|a-b|$,
$\rho(y,z) = |\sin(y)-\sin(z)|=|b-c|$
Thus, $\rho(x,z) = |\sin(x)-\sin(z)|=|a-c|$
How come it is not a metric space???
and how can I continue to proof $|f(x)-f(y)|$
thanks!!
Best Answer
For $\rho$ to be a metric, we require that $\rho(x,y)=0$ if and only if $x=y$. You've showed one of these, but the periodic nature of the sine function makes the other impossible.
We call $\rho$ a pseudometric--which means it's basically a metric, but sometimes we'll have $\rho(x,y)=0$ when $x\neq y$. The absolute value properties mean that for any set $X$ and function $f:X\to\Bbb R$, we will always have that $$\rho_f(x,y):=|f(x)-f(y)|$$ is a pseudometric on $X$. $\rho_f$ is a metric on $X$ if and only if $f$ is a one-to-one function.