Show that $\rho(x,y) = |\sin(x)-\sin(y)|$ is not a metric on $\mathbb{R}$?

Question

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 $\mathbb{R}$

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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!!

Answer 1

$\sin(0)=\sin(2\pi) \implies \rho (0,2\pi)=0$. Not a metric. (You didn't prove things that were necessary in (1))

Answer 2

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.

Answer 3

As to what conditions we must impose on the function, see this question

Condition on function $f:\mathbb{R}\rightarrow \mathbb{R}$ so that $(a,b)\mapsto | f(a) - f(b)|$ generates a metric on $\mathbb{R}$

Answer 4

$\rho(x,y)=0 $ should always imply $x=y$ but in this case it is not true because $\rho(0,\pi)=0 $.