Need to prove this lemma in order to classify isometres in $\mathbb{R}$

Question

I am trying to use this lemma to prove that if an $f \in Isom_o(\mathbb{R})$, then $f(x)=x$ or $f(x)=-x$.. . The Lemma I want to use says: Suppose $f:\mathbb{R} \rightarrow \mathbb{R}$ is an isometry with $$f(0)=0$$ and $$f(1)=1$$ then $f=$id. I do not understand the proof of the lemma. I understand a similar lemma for the same situation in the plane, but this does not seem to click.

Proof

$$d(x,0)= |x-0|=|x|=d(f(x),f(0))$$ and $$d(x,1)=|x-1|=d(f(x),f(1))$$

I must be missing something, but how does this prove the lemma?

Answer 1

Every real number is uniquely determined by its distances from $0$ and $1$. Suppose that $d(x,0)=a$ and $d(x,1)=b$; then you can check that

• if $b<1$, then $x=a$;
• if $b\ge 1$ and $a<b$, then $x=-a$;
• if $b\ge 1$ and $a>b$, then $x=a$; and
• these are the only possibilities.

Thus, if $f$ is an isometry such that $f(0)=0$ and $f(1)=1$, and $d(x,0)=a$ and $d(x,1)=b$, then

$$d(f(x),0)=d(f(x),f(0))=d(x,0)=a$$

and

$$d(f(x),1)=d(f(x),f(1))=d(x,1)=b\;,$$

so $f(x)=x$, since $f(x)$ has the same distances from $0$ and $1$ as $x$ has.