Prove that every isometry from $\mathbb{R}$ to itself is either $j_a$ or $i \circ j_a$.
Here, $j_a$ is defined as $x\mapsto x+a$, and $i$ is defined by $x\mapsto -x$. Also, we're assuming the usual metric $d(x,y)=|x-y|$.
What I've tried: Let $f$ be an isometry from $\mathbb{R}$ to itself. I was able to prove that if $f$ is crescent, then $f=j_a$, for some $a\in \mathbb{R}$. Likewise, if $d$ is decrescent then $f=i\circ j_a$, for some $a\in \mathbb{R}$. Therefore, we're left to show that $f$ must be everywhere monotic. If we can prove this then we're done. I've tried dividing into some cases ($x$ positive, $x$ negative, etc.) so I could get rid of that annoying absolute value, but this quickly proved to be a tiresome and confusing approach, so I lost myself in the details halfway and couldn't finish it.
Question: I'm looking for a proof that if $f$ is an isometry $f$ from $\mathbb{R}$ to itself then it must be one of the functions above. It would be even better if I could get a proof that $f$ must be everywhere monotic, so I can just finish the solution myself. Thanks in advance.
Let $f:\mathbb{R} \to \mathbb{R}$ be an isometry. By composing with some suitable $j_a$, assume without loss of generality that $f(0) = 0$. Then for any fixed $x\in \mathbb{R}$, we have $f(x) = \pm x$ (with the sign depending on $x$). Since $f(x)/x : \mathbb{R}^{\not = 0} \to \{\pm 1\}$ is continuous away from $0$, the sign must be constant on each set $\mathbb{R}^{> 0}, \mathbb{R}^{< 0}$. But an isometry is clearly injective, so the sign must be constant everywhere.