Two non-empty, compact, perfect sets in real line, with empty interior, can be mapped into each other by an order preserving isomorphism of real line Let $A,B$ be non-empty, dense in itself (i.e. every point is a limit point), compact subsets of $\mathbb R$ with empty interior. Then how to show that there exists order preserving isomorphism  $f: \mathbb R \to \mathbb R$ such that $f(A)=B$ ?
I can see that both $A,B$ are perfect sets (https://en.wikipedia.org/wiki/Perfect_set) , so has cardinality of continuum. Also, it is enough to take one of $A$ or $B$ to be the Cantor set. I can show that $A, B$ are order homeomorphic with the cantor set, but I don't know if such a homeomorphism can be extended to whole real line.  
I am unable to derive anything else. 
Please help . Thanks in advance 
 A: Theorem. (Cantor). Let $X$  be countably infinite and linearly ordered by $<_X$ with no $<_X$- largest nor $<_X$-smallest member, and such that $<_X$ is order-dense . (That is, if $a<_Xa'$ then  there exists $a''$ with $a<_Xa''<_X a').$ Then $(X,<_X)$ is order-isomorphic to $(\Bbb Q,<)$ where $<$ is the usual order on $\Bbb Q$.
Corollary. Any two such linear orders are order-isomorphic to each other.
We have $\Bbb R$ \ $A= (-\infty, \min A)\cup (\max A,\infty)\cup (\cup F_A)$ where $F_A$ is a countably infinite family of pair-wise disjoint non-empty bounded open intervals whose closures are pair-wise disjoint. 
We have $\Bbb R$ \ $B=(-\infty,\min B)\cup (\max B,\infty)\cup (\cup F_B)$ where $F_B$ is a countably infinite family of pair-wise disjoint non-empty bounded open intervals whose closures are pair-wise disjoint.   
For $U,V\in F_A$ let $U<_A V \iff  \sup U<\inf V.$ 
For $U',V'\in F_B$ let $U'<_B V'\iff \sup U'<\inf V'.$ 
By the corollary  above,  there is an order-isomorphism  $\psi:F_A\to F_B.$ That is, $\psi$ is a bijection and  $U<_AV\iff \psi (U)<_B \psi (V).$
For $U\in F_A$ let $f$ map $\overline U$ linearly onto $\overline {\psi(U)}$ with $f (\inf U)=\inf (\psi (U)).$ Let $f$ map $(-\infty,\min  A]$ linearly onto $(-\infty,\min B].$ Let $f$ map $[\max A,\infty)$ linearly onto $[\max B,\infty).$
For $x\in A,$ if $x$ is not an end-point of any member of $F_A,$ and if $\min A \ne x \ne \max A$ then $x=\sup \{\sup U: U\in F_A\land \sup U<x\}$. Let $f(x)=\sup \{\sup \psi(U): U\in A\land \sup U<x\}. $
Remark: The corollary can be proven directly by elementary means.
