Spivak, Ch. 15, Problem 25. Interpretation of proof that $|\sin{x}-\sin{y}|<|x-y|$ for all numbers $x\neq y$. 


*Prove that $|\sin{x}-\sin{y}|<|x-y|$ for all numbers $x\neq y$.


Let $x\neq y$. The Mean Value Theorem tells us that there is some $c\in (x,y)$ such that
$$\frac{\sin{x}-\sin{y}}{x-y}=\sin'(c)=\cos(c)$$
$$\left | \frac{\sin{x}-\sin{y}}{x-y} \right | = | \cos{c}| \leq 1$$
$$|\sin{x}-\sin{y}|\leq |x-y|$$
Equality holds if $|\cos{c}|=1$, which means $c=k\pi, k\in\mathbb{Z}$.
The claim in the problem statement is essentially that it never occurs that we have $x\neq y$ and a $c\in (x,y)$ from the MVT such that $|\cos{c}|=1$, correct?
To prove this we take two numbers $x<y$, and a third number $z\in (x,y)$ chosen such that there is no number of form $\pi k\in (x,z)$.
Then
$$|\sin{y}-\sin{x}|=|\sin{y}-\sin{z}+\sin{z}-\sin{x}|$$
$$=|(y-z)\cos{c_1}+(z-x)\cos{c_2}|\tag{1}$$
where $c_1\in(z,y)$ with $|\cos{c_1}|\leq 1$, and $c_2\in (x,z)$, so $c_2\neq k\pi \implies |\cos{c_2}|<1$.
Therefore
$$|\sin{y}-\sin{x}|<|(y-z)\cdot 1 +(z-x)\cdot 1| \tag{2}$$
$$= |y-x|$$
 A: Note that, for $x\ne y$
$$|\sin x- \sin y |=|2\cos\frac{x+y}2\sin\frac{x-y}2 |\le 
2|\sin\frac{x-y}2 |<|x-y|$$
A: Let $x<y$. Without loss of generality assume that
$\sin x\le \sin y$. The
$$
|\sin y- \sin x|=\sin y- \sin x=\int_x^y \cos t\,dt=\\
(y-x)-\int_x^y (1-\cos t)\,dt= \\
=
(y-x)-\frac{1}{2}\int_x^y \sin^2 (t/2)\,dt.
$$
It suffices to show that $\int_x^y \sin^2 (t/2)\,dt>0,$ whenever $y>x$.
But for the continuous function is nonnegative, $\sin^2 (t/2)\ge 0$, for all $t$, and not identically zero in any interval. Hence $\int_x^y \sin^2 (t/2)\,dt>0$, whenever $y>x$.
A: I think you are correct. Just clean up your last line by making it rigorous:
Therefore
$$\vert \sin{y}-\sin{x}\vert = \vert(y-z)\cos{c_1}+(z-x)\cos{c_2}\vert $$
$$ \leq \vert y-z \vert\ \vert \cos{c_1}\vert\ + \vert z-x \vert\ \vert \cos{c_2}\vert $$
$$ < \vert y-z \vert + \vert z-x \vert$$
$$ = (y-z) + (z-x)\quad \text{ because } x<z<y $$
$$ = y-x = \vert y - x\vert. $$
A: For $r>0$ and $S\subset \Bbb R,$ say that $S$ is $r$-discrete iff $|x-y|\ge r$ whenever $x,y$ are unequal members of S.
Suppose $r>0$ and $S$ is an $r$-discrete subset of $\Bbb R$. Exercise: If $f:\Bbb R\to \Bbb R$ is differentiable and if $f'(x)>0$ for all $x\in\Bbb R\setminus S$ then $f$ is strictly increasing. (Hint: If $x<y$ there are only finitely many members of $ (x,y)\cap S$).
(1). Let $f(x)=x-\sin x.$ Then $f'(x)=1-\cos x.$ Let $S=\{2n\pi: n\in\Bbb Z\}.$ Now $[x\ne y\land \frac {\sin x-\sin y}{x-y}=1]\implies [x\ne y\land f(x)=f(y)],$ which is impossible as $f$ is strictly increasing.
(2). Let $f(x)=x+\sin x.$ Then $f'(x)=1+\cos x.$ Let $S=\{(2n+1)\pi: n\in\Bbb Z\}.$ Now $[x\ne y\land \frac {\sin x -\sin y}{x-y}=-1]\implies [x\ne y\land f(x)=f(y)],$ which is impossible as $f$ is strictly increasing.
