Prove that $[0,\pi/2]$ is $d$-complete, where $d(x,y)=\left|\sin(x-y)\right|$.

Question

Let $d$ be the metric on $[0,\pi/2]$ defined as: $$d(x,y)=\left|\sin(x-y)\right|.$$ Prove that $[0,\pi/2]$ is $d$-complete.

Attempt. Let $(x_n)$ be a $d$-Cauchy sequence on $[0,\pi/2]$, meaning that for all $\varepsilon>0$ there is some $n_0$ such that for all $n,~m\geq n_0$ we have: $$d(x_n,x_m)= \left| \sin(x_n-x_m)\right|<\varepsilon.$$ At this point I expect to find a $|\cdot|-$Cauchy sequence defined on a $|\cdot|$-closed subset of the complete reals $(\mathbb{R},|\cdot|)$, in order to get a $|\cdot|$-limit $x_0$ and therefore (somehow) a $d-$limit. But $\left|\sin(x_n-x_m)\right|\leq |x_n-x_m|$, so $(x_n)$ may not be the sequence I am looking for.

Thanks in advance for the help.

Answer 1

Note that there is $C > 0$ such that $|x| \le C \cdot |\sin(x)|$ for all $x\in [-\frac{\pi}{2}, \frac{\pi}{2}]$. To do so it is enough to show that $\frac{|x|}{|\sin(x)|}$ is bounded on $x\in [-\frac{\pi}{2}, 0) \cap (0, \frac{\pi}{2}]$. Indeed, if you define $\frac{|x|}{|\sin(x)|} = 1$ for $x = 0$ you obtain a function which is continuous on $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

That is all we need. If $(x_n)$ is $d$-Cauchy, then it's also Cauchy in the ordinary sense (we apply above inequality to $x = x_n - x_m$ which is from $[-\frac{\pi}{2}, \frac{\pi}{2}]$). Hence there is a limit of $(x_n)$ which will also be a $d$-limit of $(x_n)$.

Answer 2

Compactness of $I=[0,\pi/2]$ implies that if $(x_n)$ is a Cauchy sequence in $I$, then $x_n\to x\in I.$

Continuity of the sine function now gives

$d(x_n,x)=|\sin (x-x_n)|=$

$|\sin x \cos x_n-\cos x \sin x_n|\to |\sin x\cos x-\cos x \sin x|=0$

as $n\to \infty, $ and therefore, $x_n$ converges to $x$ in the metric $d$.