Can any isometry between two non-empty subsets of $\mathbb R$ be extended to an isometry of $\mathbb R$ onto itself?

Question

Let $A , B$ be non-empty subsets of $\mathbb R$ and $f:A \to B$ be a function such that $|f(a)-f(b)|=|a-b| , \forall a,b \in A$ , then is it true that there exists a function $g :\mathbb R \to \mathbb R$ such that $|g(x)-g(y)|=|x-y|,\forall x,y \in \mathbb R$ and $g_{|A}=f$ ? If the answer is yes , then can we make $g$ to be surjective also ?

Answer 1

Hint: Show that once you define $f$ on two points, it is uniquely determined on all other points. As an example, if $f(0)=1$ and $f(1)=2$, can you prove that $f(x)=x+1$ for all $x\in A$?