Let $C_1$ and $C_2$ be any two Cantor sets with different ratio $α$ ($0<α<1$, where $1/3$ is the standard ratio).Show that there exist a function $F:[0,1]\to [0,1]$ with $F$ being
continuous and bijective
monotonically increasing
$F$ maps $C_1$ surjectively onto $C_2$
From now on, I have constructed a continuous map on $C_1$, and I want to extend it to the whole space. Here is some idea. Since $C_1$ is closed set in $[0,1]$ we can extend continuously to a function on $[0,1]$, so this map satisfies condition continuous, surjectively onto $C_2$. My idea is that we can adjust that any non-bijective part to a line segment connecting the two points in the cantor set $C_1$. However, the interior part of this line segment must not have any cantor set's point, but we know there are limit points in the cantor set. Therefore, it is impossible to connecting a line segment (or eventually become a curve (infinite line segments), not a line segment. Can someone help me extend the map to $[0,1]$ with explicitly writing, or any other proof is very appreciated. Thanks in advance.
The Cantor set $C_1$ is closed and contains the endpoints $0,1$. It is well-known that a nonempty open subset of the real line is expressible as $\bigsqcup_{n\in F}(a_n,b_n)$ where $-\infty\le a_n<b_n\le\infty$ for each $n\in F$, all the intervals are disjoint and $F$ is some countable (possibly finite) set which corresponds to the connected components of the open set. This is true of $\Bbb R\setminus C_1$; thinking about connected components, it is clear that $(-\infty,0)$ and $(1,\infty)$ are two of these intervals; let's throw away those. It's easy to find $[0,1]\setminus C_1=\bigsqcup_{n\in\Bbb N}(a_n,b_n)$, then. You could find such a decomposition via the explicit construction of the Cantor set but I'd like to emphasise this is a general phenomenon.
You have a monotone homeomorphism $F:C_1\to C_2$ and would like to extend it to $[0,1]$ i.e. define $F$ suitably on each of the $(a_n,b_n)$. We really can use line segments as you suggested! It's easy to define such a thing $(a_n,b_n)\to[0,1]$, just take $(a_n,b_n)\in x\mapsto\frac{x-a_n}{b_n-a_n}\cdot(F(b_n)-F(a_n))+F(a_n)\in[0,1]$ and since all the intervals $(a_n,b_n)$ are disjoint and, together with $C_1$, cover $[0,1]$, we obtain a new function $F':[0,1]\to[0,1]$. It is obvious that $F'$ is a monotone increasing map. Working with a similar decomposition of $[0,1]\setminus C_2$ into intervals and using the assumption that $F(C_1)=C_2$ it is not too hard to see $F'$ is surjective, hence bijective. The last thing to do is determine the continuity of $F'$.
For any $n\in\Bbb N$ and a point $y\in(a_n,b_n)$, $F'$ is clearly continuous at $y$. The slightly tricky thing is at the (fractal-like) set of endpoints which turns out to just be $C_1$ itself. Although we know $F$ is continuous on $C_1$ it doesn't immediately follow $F'$ is continuous on $C_1$ too just from the fact $F'$ extends $F$. However, this is true in this case; take $x\in C_1$, $\epsilon>0$, $\delta>0$ such that $|x-y|<\delta$ and $x,y\in C_1$ implies $|F(x)-F(y)|<\epsilon/3$. Then if $y\in[0,1]$ has $|x-y|<\delta$, we either have $|F'(x)-F'(y)|=|F(x)-F(y)|<\epsilon/3<\epsilon$ in the case $y\in C_1$ or we have $y\in(a_n,b_n)$ for some $n$ and we have: $$\begin{align}|F'(y)-F(x)|&=F'(y)-F(x)\\&=(F'(y)-F(a_n))+(F(a_n)-F(x))\\&\le(F(b_n)-F(x))+(F(x)-F(a_n))+(F(a_n)-F(x))\\&<\epsilon/3+\epsilon/3+\epsilon/3\\&=\epsilon\end{align}$$In the case that $x<a_n<y$ or a very similar approach in the case $y<b_n<x$ (to ensure $|a_n-x|$ and $|b_n-x|$ are also $<\delta$).